SOLJPCT2180813 - 1 PAPER-A : (MATHEMATICS) 1. At the x! 5 6 " , ...................................... Sol. (4) x 3 sin 2 13 – $ + 3cos3x $13 f(x) % ] 13 , 13 [ – f ' ( ) * +," 6 5 = 2sin ' ( ) * +," 2 + 3cos ' ( ) * +," 2 = 2 at x = 6 5" , neither maximum nor minimum and not zero value occurs. 2. If the vectors ........................................... Sol. (1) c 1 1 1 b 1 1 1 a = 0 c 1 1 c – 1 1 – b 0 0 b – 1 1 – a = 0 -( a –1) (c(b –1 ) –(1 –c)) + ((1 –b)(1 –c)) = 0 -(a –1)(bc –1) + (1 –b)(1 –c) = 0 -(1 –b)(1 –c) = (1 –a)(bc –1) ...........(1) Required value a – 1 1 + b – 1 1 + c – 1 1 = ) c – 1 )( b – 1 )( a – 1 ( ] b – 1 c – 1 1 – bc )[ a – 1 ( # # = 1 3. Let R be a relation........................................ Sol. (3) If a is negative, then |a| $a is not true .R is not reflexive If a = 1, b = 2, then |a| $b but |b| $a is not true .R is not symmetric let a, b %R and b, c %R - |a| $b and |b| $c, then c /0 !c /|b| - c /b . c /|a| or |a| $c . (a, c) %R Hence R is transitive. 4. If f(x) = 0 1 0 2 3 4 ! 5 # 1 x 2 – x 5 1 x k 1 x 1 x 2 ,.................................. Sol. (2) ) h – 1 ( f lim 0 h6 = f(1) = ) h 1 ( f lim 0 h # 6 3 = k = 3 -k = 3 5. ( 2) 6. ( 3) HINTS & SOLUTIONS CUMULATIVE TEST-2 (CT-2) (JEE MAIN) T ARGET : JEE (MAIN+ADV ANCED) 2014 COURSE NAME : VIJETA (JP) 7. The number of solutions of.................. ... Sol. (4) L.H.S. %[–1, 1] but R.H.S. /2. Hence cannot be equal, no solution. 8. (1 ) 9. (2 ) 10. (3) 11. If z is a complex................................. Sol. (1) z = |z| 6 5 i e " = 7 8 9 : ; < " # " 6 5 sin i 6 5 cos 4 = 7 7 8 9 : : ; < # 2 i 2 3 – 4 = i 2 3 2 – # 12 . If A = {1, 3, 5, 7, 9, 11...................................... Sol. (3) (A = B) > B? = A A? = A = N 13. (1) 14. (2) 15 . The value of........... ............................ Sol. (3) –1 x 0 sin(2tan3x) lim sin 4x 6 × x 3 x 3 tan 2 × x 4 x 3 × x 4 sin x 4 1 – = 1 × 2 × 4 3 × 1 = 2 3 16 . The chord of contact....... ...................... Sol. (1) 1 2 1 2 2 2 2 2 a b c b a b – – ! - ! b ac !A M GM .. . . / 17. ( 1 ) 18. ( 1 ) 19. (4) 20. ( 2 ) 21 . The reciprocal of the ................................... Sol. (3) The shortest distance (SD) ! 2 1 4 2 5 3 2 3 4 3 2 5 2 3 4 3 2 5 – – – " " " i j k 78 1 ! DA TE : 18-08-2013 CODE : 0 & 3
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c) = 0- (1 – b)(1 – c) = (1 – a)(bc – 1) ...........(1)Required value
a –1
1+
b –1
1+
c –1
1=
)c –1)(b –1)(a –1(
]b –1c –11 –bc)[a –1( ##= 1
3. Let R be a relation........................................Sol. (3)
If a is negative, then |a| $ a is not true. R is not reflexiveIf a = 1, b = 2, then |a| $ b but |b| $ a is not true . R is notsymmetriclet a, b % R and b, c % R - |a| $ b and |b| $ c, then c / 0
! c / |b| - c / b . c / |a| or |a| $ c. (a, c) % R
Hence R is transitive.
4. If f(x) =01
02
3
4
!
5#
1x2 –x5
1xk
1x1x2
,..................................
Sol. (2)
)h –1(flim0h6 = f(1) = )h1(flim
0h#
63 = k = 3 - k = 3
5. (2 )
6. (3)
HINTS & SOLUTIONS
CUMULATIVE TEST-2 (CT-2)
(JEE MAIN)TARGET : JEE (MAIN+ADVANCED) 2014
COURSE NAME : VIJETA (JP)
7. The number of solutions of.....................
Sol. (4)
L.H.S. % [ –1, 1] but R.H.S. / 2. Hence cannot be equal, no
solution.
8. (1) 9. (2) 10. (3)
11. If z is a complex.................................Sol. (1)
z = |z| 6
5i
e
"
= 78
9:;
< "#
"
6
5sini
6
5cos4
=778
9
::;
<
# 2
i
2
3 –4 = i232 – #
12. If A = {1, 3, 5, 7, 9, 11......................................Sol. (3)
(A = B) > B? = AA? = A = N
13. (1)
14. (2)
15. The value of.......................................Sol. (3)
–1x 0
sin(2tan3x)lim
sin 4x6×
x3
x3tan2×
x4
x3×
x4sin
x41 –
= 1 × 2 ×4
3 × 1 =
2
3
16. The chord of contact.............................Sol. (1)
1
2
1
2
2 2 2 2a b c b a b – –!
- !b ac
!A M G M. . . ./
17. (1) 18. (1)
19. (4) 20. (2)
21. The reciprocal of the ...................................
34. Particle A ................. .......... ..Sol. (4)
1st separation will decrease then sep will increaseFor slope :- 1st there will be some V
app, then V
app will decrease
slowly and become zero, after Vsep
will increase gradually.
Hence slope will decrease till it become zero afterward it will
increase.
35. Three rings .............................Sol. (1)
Bz = B
y =
R2
0ED
Bnet
=2z
2y
2x BBB ##
= 3 R20ED
36. The distance .............................Sol. (2)
Potential in conductor is constantPotential in dielectric decreases at slower rate.
37. If x, y and .............................Sol. (2)
Resistance decreases - EG - xG
38. A reflecting .............................Sol. (2)
dx
dy = 2cos '
(
)*+
, L
x2 = –1
cosL
x2= –
2
1
x =3L ,
3L2
39. A particles with .............................Sol. (1)
U + k = Ewhen U = EK = 0
40. The equivalent .............................Sol. (1)
Circuit can be reduced to :
A B R R
2 R /3
2 R /3
2 R /32 R /3
2 R /3 2 R /3
R R
R R
R eq =H
35
22
41. A rod of length .............................Sol. (2)
Ldx e 1R e dx
A A A e
IJ JJ I, )! ! ! * '
+ (K K /0 0
0 0
..
L L
x L
' (
)*+
, IJ
!!.1e
e
L
AV
R
V I
0
00
42. The connection .............................
Sol. (1)
16µC 16µC
16µC 16µC
I + I+
+II+
24V
0V
12µC 24µC
12µC24µC
24V
0V
I + + I
I+ +I
Initially finally
. q through switch = 12 µC
43. Consider a long .............................Sol. (3)
q inner is equal to q outer in magnitude. E inner is greater than E outer always.. Pressure on A greater than Pressure on B . Force of interac-tion is equal and opposite. . Net force = 0.
44. The speed of .............................Sol. (2)
speed of boat in upstream = 15 km/hrDistance = 1.5 km
time t1 =
15
5.1 = 6 min
Dist. moved by float in this time
boat
1.5 km
floatvk
= 5 ×10
1 =
2
1 km
speed of boat w.r.t. float = 20 km/hr, t2 =
20km2 = 6 min.
45. The speed of .............................
Sol. (2)
Distance = v0 × 1 +
2
v0 × 1 +
4
v0 × 1 + ............
= v0
' (
)*+
, ### ..............
4
1
2
11
= 2v0
46. The focal length .............................
Sol. (2)Given f
0 = 2 cm, f
e = 5 cm
uo = -3, vo = 6 cm mo = -2for normal adjustment
ue = -5 cm, v
e = infinity m
e = 5
47. A bead is released .............................