Aieee 2012 Solution (12th may)
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Detailed solution given below
Q1
1)
2)
3)
4)
Ans 3
Q2
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2)
3)
4)
Ans 4
Q3
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Ans 2
Q4
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Ans 2
Q5
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Ans 4
Q6
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Ans 2
Q7
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Ans 4
Q8
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Ans 3
Q9
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Ans 3
Q10
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Ans 4
Q11
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Ans 1
Q12
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Ans 1
Q13
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Ans 4
Q14
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Ans 4
Q15
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Ans 3
Q16
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Ans 4
Q17
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Ans 1
Q18
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Ans 4
Q19
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Ans 1
Q20
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Ans 2
Q21
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Ans 3
Q22
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Ans 4
Q23
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Ans 4
Q24
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Ans 2
Q25
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Ans 3
Q26
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Ans 4
Q27
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Ans 4
Q28
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3)
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Ans 1
Q29
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4)
Ans 1
Q30
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4)
Ans 3
Q61
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3)
4)
Ans 4
Q62
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Ans 2
Q63
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4)
Ans 1
Q64
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3)
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Ans 4
Q65
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3)
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Ans 2
Q66
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2)
3)
4)
Ans 3
Q67
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2)
3)
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Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
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2)
3)
4)
Ans 2
Q70
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2)
3)
4)
Ans 1
Q71
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3)
4)
Ans 3
Q72
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3)
4)
Ans 2
Q73
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3)
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Ans 4
Q74
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3)
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Ans 3
Q75
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3)
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Ans 3
Q76
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3)
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Ans 2
Q77
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3)
4)
Ans 4
Q78
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2)
3)
4)
Ans 1
Q79
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2)
3)
4)
Ans 1
Q80
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2)
3)
4)
Ans 2
Q81
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2)
3)
4)
Ans 2
Q82
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3)
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Ans 4
Q83
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3)
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Ans 4
Q84
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3)
4)
Ans 2
Q85
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Ans 3
Q86
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Ans 2
Q87
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Ans 1
Q88
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4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q3
1)
2)
3)
4)
Ans 2
Q4
1)
2)
3)
4)
Ans 2
Q5
1)
2)
3)
4)
Ans 4
Q6
1)
2)
3)
4)
Ans 2
Q7
1)
2)
3)
4)
Ans 4
Q8
1)
2)
3)
4)
Ans 3
Q9
1)
2)
3)
4)
Ans 3
Q10
1)
2)
3)
4)
Ans 4
Q11
1)
2)
3)
4)
Ans 1
Q12
1)
2)
3)
4)
Ans 1
Q13
1)
2)
3)
4)
Ans 4
Q14
1)
2)
3)
4)
Ans 4
Q15
1)
2)
3)
4)
Ans 3
Q16
1)
2)
3)
4)
Ans 4
Q17
1)
2)
3)
4)
Ans 1
Q18
1)
2)
3)
4)
Ans 4
Q19
1)
2)
3)
4)
Ans 1
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q5
1)
2)
3)
4)
Ans 4
Q6
1)
2)
3)
4)
Ans 2
Q7
1)
2)
3)
4)
Ans 4
Q8
1)
2)
3)
4)
Ans 3
Q9
1)
2)
3)
4)
Ans 3
Q10
1)
2)
3)
4)
Ans 4
Q11
1)
2)
3)
4)
Ans 1
Q12
1)
2)
3)
4)
Ans 1
Q13
1)
2)
3)
4)
Ans 4
Q14
1)
2)
3)
4)
Ans 4
Q15
1)
2)
3)
4)
Ans 3
Q16
1)
2)
3)
4)
Ans 4
Q17
1)
2)
3)
4)
Ans 1
Q18
1)
2)
3)
4)
Ans 4
Q19
1)
2)
3)
4)
Ans 1
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q7
1)
2)
3)
4)
Ans 4
Q8
1)
2)
3)
4)
Ans 3
Q9
1)
2)
3)
4)
Ans 3
Q10
1)
2)
3)
4)
Ans 4
Q11
1)
2)
3)
4)
Ans 1
Q12
1)
2)
3)
4)
Ans 1
Q13
1)
2)
3)
4)
Ans 4
Q14
1)
2)
3)
4)
Ans 4
Q15
1)
2)
3)
4)
Ans 3
Q16
1)
2)
3)
4)
Ans 4
Q17
1)
2)
3)
4)
Ans 1
Q18
1)
2)
3)
4)
Ans 4
Q19
1)
2)
3)
4)
Ans 1
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q10
1)
2)
3)
4)
Ans 4
Q11
1)
2)
3)
4)
Ans 1
Q12
1)
2)
3)
4)
Ans 1
Q13
1)
2)
3)
4)
Ans 4
Q14
1)
2)
3)
4)
Ans 4
Q15
1)
2)
3)
4)
Ans 3
Q16
1)
2)
3)
4)
Ans 4
Q17
1)
2)
3)
4)
Ans 1
Q18
1)
2)
3)
4)
Ans 4
Q19
1)
2)
3)
4)
Ans 1
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q12
1)
2)
3)
4)
Ans 1
Q13
1)
2)
3)
4)
Ans 4
Q14
1)
2)
3)
4)
Ans 4
Q15
1)
2)
3)
4)
Ans 3
Q16
1)
2)
3)
4)
Ans 4
Q17
1)
2)
3)
4)
Ans 1
Q18
1)
2)
3)
4)
Ans 4
Q19
1)
2)
3)
4)
Ans 1
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q14
1)
2)
3)
4)
Ans 4
Q15
1)
2)
3)
4)
Ans 3
Q16
1)
2)
3)
4)
Ans 4
Q17
1)
2)
3)
4)
Ans 1
Q18
1)
2)
3)
4)
Ans 4
Q19
1)
2)
3)
4)
Ans 1
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q17
1)
2)
3)
4)
Ans 1
Q18
1)
2)
3)
4)
Ans 4
Q19
1)
2)
3)
4)
Ans 1
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q20
1)
2)
3)
4)
Ans 2
Q21
1)
2)
3)
4)
Ans 3
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q22
1)
2)
3)
4)
Ans 4
Q23
1)
2)
3)
4)
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q24
1)
2)
3)
4)
Ans 2
Q25
1)
2)
3)
4)
Ans 3
Q26
1)
2)
3)
4)
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q27
1)
2)
3)
4)
Ans 4
Q28
1)
2)
3)
4)
Ans 1
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q29
1)
2)
3)
4)
Ans 1
Q30
1)
2)
3)
4)
Ans 3
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q61
1)
2)
3)
4)
Ans 4
Q62
1)
2)
3)
4)
Ans 2
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q63
1)
2)
3)
4)
Ans 1
Q64
1)
2)
3)
4)
Ans 4
Q65
1)
2)
3)
4)
Ans 2
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q66
1)
2)
3)
4)
Ans 3
Q67
1)
2)
3)
4)
Ans 2
Q68
1)
2)
3)
4)
Ans 2
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q69
1)
2)
3)
4)
Ans 2
Q70
1)
2)
3)
4)
Ans 1
Q71
1)
2)
3)
4)
Ans 3
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q72
1)
2)
3)
4)
Ans 2
Q73
1)
2)
3)
4)
Ans 4
Q74
1)
2)
3)
4)
Ans 3
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q75
1)
2)
3)
4)
Ans 3
Q76
1)
2)
3)
4)
Ans 2
Q77
1)
2)
3)
4)
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q78
1)
2)
3)
4)
Ans 1
Q79
1)
2)
3)
4)
Ans 1
Q80
1)
2)
3)
4)
Ans 2
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q81
1)
2)
3)
4)
Ans 2
Q82
1)
2)
3)
4)
Ans 4
Q83
1)
2)
3)
4)
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q84
1)
2)
3)
4)
Ans 2
Q85
1)
2)
3)
4)
Ans 3
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
Q86
1)
2)
3)
4)
Ans 2
Q87
1)
2)
3)
4)
Ans 1
Q88
1)
2)
3)
4)
Ans 3
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
By
Saurav gupta
Electronics amp telecomm engg(2nd year)
Jadavpur university
Q89
1)
2)
3)
4)
Ans 2
Q90
1)
2)
3)
4)
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q1 Case 1
PV = Pᵢ x 2V
Pᵢ = P2
Case 2
PV⁵ sup3 = Pₐ (2V)⁵ sup3
Pₐ = P 2ˉ⁵ sup3
PₐPᵢ = 2ˉsup2 sup3
Ans 3
Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains
unchanged
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q3 dMdt =αv
now
F = 0
dPdt=0
Mdvdt + vdMdt = 0
Ma + v x αv = 0
a = -vsup2αM
Ans 2
Q4 Case 1
N = Nᵢ( frac12 ) ⁰⁵
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
100 = 1600 ( frac12 ) ⁸ ⁰⁵
T frac12 = 2 sec
Case 2
N = 1600 frac12 )⁶ sup2
= 200
Ans 2
Q5 R = 2radich1 (H-h1)
For same range
h1 = h2
Ans 4
Q6
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
RP = 1dθ = 122λD
dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2
= 223 x 10ˉ⁵
X(min) 80 = 223 x 10ˉ⁵
X(min) = 179 x 10ˉsup3
Hence minimum spacing should be 179 x 10ˉsup3 to be resolved
Ans 2
Q7 fact
Ans 4
Q8
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
I (max)I(min) = [(a+b)(a-b)]sup2
W1w2 = (ab)sup2
125 = (ab)sup2
ab = 15
I (max)I(min) = 94
Ans 3
Q9 The maximum energy of photoelectrons is
hυ-ф
Ans 3
Q10 L = constant
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Mωrsup2 = c
Now
T =mωsup2r
= (mrsup2ω)sup2mrsup3
= Lsup2mrsup3
= Arˉsup3
Where A = Lsup2m
Ans 4
Q11 r= (llrsquo - 1)R
given r =R
llrsquo ndash 1 = 1
llrsquo = 2
lrsquo = l2
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q12 V dvdx = -Gmxsup2
˳int V dv = -Gm intʳdxxsup2
Solving we get
Vsup2 = 2Gmr[1 ndash 1radic2]
Ans 1
Q13 Sinθ = 1μ
For total internal reflection
θ ge 45⁰
radic2r
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
sinθ ge 1radic2
1μ ge 1radic2
μ le radic2
μ le 1414
hence only green amp blue colours can be seen
Ans 4
Q14 V =2 rsup2(ρ-σ)g9ᶯ
V proprsup2
Ans 4
Q15 ωprop1radicc
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
ω1ω2=radicc2c1
now
c propϵ
ω1ω2=radicϵ2ϵ1
υ1υ2=radicϵ2ϵ1
10⁴(10⁴-50) = radicϵ21
ϵ2 =101
Ans 3
Q16 Repulsive force
Also no two electric field lines can intersect
Ans 4
Q17
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
E= cB
= 3x10⁸x482x10ˉsup1sup1
= 001446 NC
λ=cυ
= 360m
Ans 1
Q18
h d
case 1
d-h = vsup22g
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
h = d ndash vsup22g---(1)
case 2
usup2 = k x 2gd klt1
vsup2 = 2kgd ndash 2gh
h = kd ndash vsup22g---(2)
Ans 4
Q19
R prop lA
RrsquoR = (lrsquol)(AArsquo)
= frac12 x frac12
= frac14
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q20 Capacitanceofearth=4πϵ˳R
=4πx885x10ˉsup1sup2 x 64 x 10⁶
=711μF
Statement 2---false
Ans 2
Q21 Y = 1[1 + (x - t2)sup2]
Eqᶯ of wave
partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)
Solving we get
C= frac12
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q22 Force applied by you + frictional force = force
applied by horse
Statement 2---false
Ans 4
Q23 Forceunitlength=μ˳IIrsquo2πd
= 4 x 10ˉ⁵
Total force = 8 x 10ˉ⁵N
Ans 4
Q24 Case 1
V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Case 2
Vrsquo = radic2gh
5v˳4=radic2gh
Substituting this value in eqᶯ (1)
1 + ksup2Rsup2 = 2516
K =34 R
Ans 2
Q25 λ=4l(2n-1)
υ=330(2n-1)4l
l= 5(2n-1)16 m
put n=2
l = 9375cm
Ans 3
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q26 Temp will increase
Ans 4
Q27 Vprop1n
Ans 4
Q28 Y=(Arsquo+Brsquo)rsquo
= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law
=AB
Ans 1
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q29 KE(final) = KE(initial) + work done
frac12 x 2 x 5sup2 + area of F-x diagram
=25 + 2 ndash 2 + 45
= 295
Ans 1
Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2
Ans 3
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q61 a(b x c) = 0
=gt a | b x c
ie a is || to the plane of b and c
Ans 4
Q62 ax+2by+3b = 0
bx-2ay-3a=0
solving we get
x=0 y = -32
Ans 2
Q63
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
σsup2= Σxᵢsup2n ndash xsup2
x = [1 + 3 +5 +helliphellipupto n terms]n
= n
Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms
= n[4nsup2 - 1]3
Statement 2---false
σsup2= (nsup2-1)3
Ans 1
Q64 1 k 3
3 k -2 = 0
2 3 -4
K = 332
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 4
Q65 Point of intersection= (12) amp (44)
Area = int⁴ [2radicx ndash (2x+4)3] dx
=13
Ans 2
Q66 Let th eqᶯ of plane be
ax+by+cz=d
it passes through (07-7)
ax+b(y-7)+c(z+7)=0-----(1)
it also passes through (-13-2)
1
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+4b+5c = 0----(2)
now
dr of normal to the plane will be perpendicular to the line
-3a+2b+c = 0-----(3)
Solving (2) amp(3)
a1 = b1 = c1
substituting this value in eqᶯ (1)
x+y+z = 0
Ans 3
Q67 C1=(41)
C2= (-3-4)
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
R1= 4
R2= 5
|C1C2| = radic74
R1+R2 = 9
R1+R2 gt |C1C2|
Ans 2
Q68 Lim sin(πcossup2x)
X -gt 0 xsup2
C1 C2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Differentiating num amp den (00 form Lrsquo Hospital rule)
U have to differentiate 2 times
= π
Ans 2
Q69
F(x) = intdxsin⁶x
Put tanx= t
Secsup2x dx = dt
1 + tsup2 dx = dt
dx = dt1+tsup2
Sinx = tradictsup2+1
Substituting these values
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F = int (1+tsup2)sup2t⁶ dt
= -[1t + 23tsup3 + 15t⁵] + c
= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo
Ans 2
Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)
= -2 [(radic3-1)2radic2]
= (1-radic3)radic2
Ans 1
Q71 Ysup2 = 3(1- xsup216)
dxdy|sup2 sup3 sup2 = -4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
(y-32) = 4(x-2)
Y = 4x ndash 132
am = -132
a4 = -132
a =-26
ysup2= -104x
Ans 3
Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx
log[(2-x)(2+x)] is an odd funcᶯ
[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q73 Q(42)
P(-42)
Eqᶯ of PQ x = 4
Ysup2 = x
dydx = frac14
Slope of normal = -4
Ans 4
Q74
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
F(x) = xsup3 + 1
Frsquo(x) = 3xsup2
Frsquorsquo(x)= 6x
Now
Frsquo(x) = 0
3xsup2 = 0
X = 0
Frsquorsquo(0)=0
Not local extremum point
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Statement 1---false
Ans 3
Q75 dydx + 2x(xsup2-1)= x(xsup2-1)
IF = exp int 2x(xsup2-1) dx
= xsup2-1
Ans 3
Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
When n is even
S = n(n+1)sup22---given
Let n = 2m
S (even)= 2m(2m+1)sup22
S (odd) = 2m(2m+1)sup22 + (2m+1)th term
= 2m(2m+1)sup22 + (2m+1)sup2
=(2m+1)sup2[m+1]
Now we put 2m+1 = n
= nsup2(n+1)2
Ans 2
Q77 axsup2 + bx + c=0
1 is a root
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
a+b+c=0
Y = 4axsup2+3bx+2c
4axsup2 + 3bx + 2(-a-b) = 0
D = 9bsup2 + 32(a+b) ne0
Hence two roots
Ans 4
Q78 P amp R are parallel as direction cosines are same
Ans 1
Q79 White-2 black-1
P=23
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q80 AB = I
Ans 2
Q81 [(x-1)x]ᶯ(1-x)ᶯ
= (-1x)ᶯ[1-x]sup2ᶯ
= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]
sup2ᶯCn
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q82 Men and women has to sit alternately
67
Ans 4
Q83 ~(Pq)
= ~(~P v q)
= p Ʌ~q
Can also check it by truth table
Ans 4
Q84 L1||L4 L2||L3 L1||L2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
L1 | L3 L2 | L4
Ans 2
Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3
For x gt 0
F(x) = asinx + beˣ + cxsup3
Frsquo(x) = acosx + beˣ + 3cxsup2
For x lt 0
F(x) = -asinx +beˉˣ - cxsup3
Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2
Lim x0+ f(x) = a + b
Lim x0ˉ f(x) = -a-b
For f(x) to be differentiable a=0 b=0
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 3
Q86 AM1 = AM2
(tp + tq)2 = (tr + ts)2
a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d
p + q = r + s
Ans 2
Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
P(-126)
Q(6λ+23λ+3-4λ-4)
dr of PQ =(6λ+33λ+1-4λ-10)
since these 2 lines are |
(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0
λ = -1
Q = (-400)
|PQ| = 7
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Ans 1
Q88 Statement 1---true
Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1
Ans 3
Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2
= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)
= 2(|Z1|sup2+|Z2|sup2)
Ans 2
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION
BY
SAURAV GUPTA
ELECTRONICS amp TELECOMM ENGG (2ND YEAR)
JADAVPUR UNIVERSITY
Q90 drdt = 01 cmhr
A = πrsup2
dAdt = 2πr drdt
= 10π cmsup2hr
Ans 4
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