1 65/1/3 P.T.O.Series:OSR/1RollNo.Code No.
65/1/3CandidatesmustwritetheCodeonthetitlepageoftheanswer-book.
Please check that this question paper contains 15 printed pages.
Code number given on the right hand side of the question paper
should be written on the title page of the answer-book by the
candidate. Pleasecheck that this question paper contains 29
questions. Please write down the Serial Number of the questions
before attempting it. 15 minutes time has been allotted to read
this question paper. The question paper will be distributed at
10.15 a.m.From 10.15 a.m. to 10.30 a.m., the student will read the
question paper only and will not write any answer on theanswer
script during this period.Mathematics[Time allowed : 3 hours]
[Maximum marks : 100]General Instructions:(i) All questions are
compulsory.(ii) The question paper consists of 29 questions divided
into three sections A, B and C. Section -A comprises of 10
questions of one mark each, Section - B comprises of 12 questions
of fourmarks each and Section - C comprises of 7 questions of six
marks each.(iii) All questions in Section - A are to be answered in
one word, one sentence or as per the
exactrequirementofthequestion.(iv) There is no overall choice.
However, internal choice has been provided in 4 questions of
fourmarks each and 2 questions of six marks each. You have to
attempt only one of the alternativesin all such questions.(v) Use
of calculators is not permitted. You may ask for logarithmic
tables, if required.Studymate Solutions to CBSE Board Examination
2013-2014DISCLAIMER: All model answers in this Solution to Board
paper are written by Studymate Subject Matter Experts.This is not
intended to be the official model solution to the questionpaper
provided by CBSE.The purpose of this solution is to provide a
guidance to students.2 65/1/3 P.T.O.1. If 23 4 1 7 0,5 0 1 10 5
(((+ = ((( yx find (x y)Sol.3 4 1 7 025 0 1 10 5yx (((+ = ((( 6 1 8
7 010 0 2 1 10 5yx+ +((= ((+ + y = 8 x = 2 10 = x y2. Solve the
following matrix equation for x :, [x 1] 1 0.2 0O ( = ( Sol.| |1 22
21 012 0x O (= ( | | 2 0 x O =x 2 = 0 2 x =3.2 5 6 2,8 7 3=xx write
the value of x.Sol.2 5 6 28 7 3xx=2x2 40 = 18 + 146 x = 4. Write
the antiderivative of 13 .| |+ |\ .xxSol.13 I x dxx= +}32 23 23xx
c| | | + + | |\ . 2 2 x x x c + +( ) 2 1 I x x c = + +5. If 1 11sin
sin cos 1.5 | |+ = |\ .xthen find the value of x.Sol.1 11sin sin
cos 15x | |+ = |\ .1 1 11sin cos sin 15x + =1 11cos sin2 5x = 1
11cos cos5x =15x =3 65/1/3 P.T.O.6. Let * be a binary operation, on
the set of all non-zero real numbers, given by a * b = 5ab for
alla, b e R {0}. Find the value of x, given that 2 * (x * 5) =
10.Sol.*5aba b =2* (x * 5) = 1052* 105x | | = |\ .2 * x= 10210255xx
= =7. Find the projection of the vector 3 7 + + i j kon the vector
2 3 6 . + i j kSol. 3 7 , 2 3 6a i j k b i j k = + + = + Projection
of or | |a ba bb= = 58. Write the vector equation of the plane,
passing through the point (a, b, c) and parallel to theplane .( ) 2
+ + =r i j k .Sol. Equation of plane( ). 0 r a n = ( ) ( ) ( ). . r
i j k ai b j ck i j k + + = + + + + ( ). r i j k a b c + + = + +9.
Evaluate :/20(sin cos ) .t}xe x x dxSol. ( )20sin cosxI e x x dx=
}( ) cos and'( ) sin f x x f x x = =
20cosxI e x (= = 1.10. Write a unit vector in the direction of
the sum of the vectors 2 2 5 = + a i j kand 2 7 . = + b i j kSol. 2
2 52 7 a i j k b i j k = + = + Unit vector = 4 3 12 4 3 1213 13 13
| | 16 9 144a b i j ki j ka b+ + = = + + + + SECTION-BQuestion
numbers 11 to 22 carry 4 marks each.11. Prove that, for any three
vectors, , a b c[ , , ] 2[ , , ] + + + = a b b c c a a b cSol. a b
b c c a (+ + + ( ) ( ) ( )a b b c c a (+ + + 4 65/1/3 P.T.O. ( )a b
b c b a c c c a (+ + + + ( ) ( ) + a b c b c aa b c b c a ((+ 2 a b
c ( OR11. Vectors, a band c are such that0 a b c + + = and | | 3,|
| 5 = =a band | | 7. =cFind the angle between a and b.Sol.0 a b c +
+ = | | 3, | | 5&||=7 a b c = = a b c + = ( ) ( ) ( ) ( ). . a
b a b c c + + = 2| | a a a b a b b b c + + + = 2 2 2| | 2| || |cos
| | | | a a b b c + + = 9 + 2(3)(5) cosu + 25 = 493 =12. Solve the
following differential equation :222( 1) 21dyx xydx x + =Sol. (
)2221 21dyx xydx x + =( )2 222 211dy xydx xx+ =It is linear
differential equation of typedyPy Qdx + =where 221xPx=
( )2221Qx=221.xdxPdxxI F e e= =}}( )2log| 1| 21xe x= = Solution
is given by ( )( )( )2 22221 11y x x dx cx = +}( )211 log1xy x cx =
++13. Evaluate : 6 62 2sin cossin .cos+}x xdxx xSol.6 62 2sin
cossin cosx xdxx x+}5 65/1/3 P.T.O.( ) ( )3 32 22 2sin cossin cosx
xdxx x+}( ) ( )2 2 4 4 2 22 2sin cos sin cos sin cossin cosx x x x
x xdxx x+ + }( )22 2 2 22 2sin cos 3sin cossin cosx x x xdxx x+ }2
213sin cosdxx x| | |\ .}2 22 2sin cos3sin cosx xdx x cx x+ +}2 2sec
cosec 3 xdx xdx x c + +} } tanx cot x 3x + cOR13.2( 3) 3 18 + }x x
x dxSol. ( )23 3 18I x x x dx = + }Put x2 + 3x 18 = t(2x + 3)dx =
dtLet x 3 = A(2x + 3) + BComparecoefficients:We get 12A =and 92B=I
( )2 21 92 3 3 183 18 2 2x x x dx x x dx = + + + } }=221 9 3 93 182
2 2 4tdt x x dx| | + + |\ .} }=32 22 1 2 9 3 92 3 2 2 2tx dx (| | |
| ( + || (\ . \ . }= ( )( )32 2 221 9 729 33 18 2 3 3 18 log 3 183
8 16 2x x x x x x x x c| |+ + + + + + + + |\ .14. Find the
intervals in which the function f (x) = 3x4 4x3 12x2 + 5 is(a)
strictlyincreasing(b) strictlydecreasingSol.4 3 2( ) 3 4 12 5 f x x
x x = +3 2'( ) 12 12 24 f x x x x = ( )212 2 x x x = ( ) ( ) ( ) '
12 2 1 f x x x x = +Put'( ) 00, 1,2 f x x = = Intervalsare( ) ( ) (
) , 1 ,( 1,0), 0,2 , 2, 6 65/1/3 P.T.O.Intervals Sign of( ) f x
Nature of( ) f x(1, 0 )( ) , 1 (0, 2)( ) 2, ve + ve ve+ veStrictly
Dec. Inc Dec. IncStrictly StrictlyStrictlyf(x) is strictly
increasing in( ) ( ) 1,0&2, , andf(x) is strictly decreasing
in( ) ( ) , 1&0,2 ORFind the equations of the tangent and
normal to the curve x = a sin3u and y = a cos3u at 4tu =Sol. x =a
sin3uy =a cos3u23 sin cosdxad =23 cos sindyad = 223 cos sincot3 sin
cosdy adx a = = 41dydx =(= (Equation of tangent at 4=y a cos34 = 1
(x a sin34)2
2 2 2a ay x y x + = + =equation of normal at 4=12 2 1 2a ay x| |
= |\ .0 y x =15. Let A = {1, 2, 3,..:.., 9} and R be the relation
in A A defined by (a, b) R (c, d) if a + d = b + c for(a, b), (c,
d) in A A. Prove that R is an equivalence relation. Also obtain the
equivalence class[(2, 5)].Sol. First Let us prove that the relation
is(i) Reflexive (ii) Symmetric(iii) Transitive(i) Reflexive :(a, b)
R (a, b) a + b = b + a which is true (additioniscommutative)
Givenrelationisreflexive.(ii) Symmetric :7 65/1/3 P.T.O.If (a, b) R
(c, d) a + d = b + cthen (c, d) R (a, b) c +b = d + a b + c = a + d
( addition is commutative) Givenrelationissymmetric.(iii)Transitive
:(a, b) R (c, d) a + d = b + c ...(i)and (c, d) R (e, f) c + f = d
+ e ....(ii)Adding (i) & (ii) we get a + f =b + e (a, b) R (e,
f) Givenrelationistransitive Givenrelationisequivalancerelation.The
equivalence class for [(2, 5)] is [(2,5),(5,2)].16. Prove that 11
sin 1 sincot ; 0, .2 4 1 sin 1 sinx x xxx x| |+ + t | |= e | | |+ \
.\ .Sol.11 sin 1 sincot1 sin 1 sinx xx x| |+ + | |+ \ .1cos sin cos
sin2 2 2 2cotcos sin cos sin2 2 2 2x x x xx x x x (+ + ( ( (+ +
1cot cot2 2x x( = ( OR16. Prove that 1 1 11 5 2 12tan sec 2tan .5 7
8 4 | |t | | | |+ + = | || |\ . \ .\ .Sol.1 1 11 5 2 12tan sec
2tan15 7 8 | || | | |+ + | || |\ . \ .\ .1 11 15 25 82 tan sec17140
( | |+ (| | |+ (|| | (| \ . ( \ . 1 11 5 22 tan sec3 7 | | ( | |+ |
|( |\ . \ .1 13 5 2tan sec4 7 | |+ | |\ .( )1 1 13 1tan tan tan 14
7 4 | | | |+ = = ||\ . \ .8 65/1/3 P.T.O.17. If y = xx, prove that
22210.| | = |\ .d y dy yy dx x dxSol. y = xx log y = x log x( )11
logdyxy dx= +( ) 1 logdyy xdx= +....(i)( )21211 logd yy x yx dx= +
+2112.y d y yyx y dx= +( )221210d y yyy x dx =18. Assume that each
born child is equally likely to be a boy or a girl. If a family has
two children,what is the conditional probability that both are
girls ? Given that(i) the youngest is a girl.(ii) atleast one is a
girl.Sol. S = {GG, GB, BG, BB}(i) A = Both are GirlsB = Youngest is
a GirlP (A/B) = 1( ) 142( ) 24= =P A BP B(ii) A : Both are GirlsB :
atleast one GirlP (A/B) = 1( ) 143( ) 34= =P A BP B19. Using
properties of determinants, prove the following :22 2 2 2211 11++ =
+ + ++x xy xzxy y yz x y zxz yz zSol. Applying R1 xR1, R2 yR2,
R3 zR3 and taking x,y,z common from C1, C2 , C3 we get2 2 22 2
22 2 2111x x xxyzy y yxyzz z z+A = ++R1 R1+ R2 + R32 2 2 2 2 2 2 2
22 2 22 2 21 1 111x y z x y z x y zy y yz z z+ + + + + + + + += ++9
65/1/3 P.T.O.( )2 2 2 2 2 22 2 21 1 11 11x y z y y yz z z= + + +
++C3 C3 C1 and C2 C2 C1 ( )2 2 2 221 0 01 1 00 1x y z yzA = + + +=(
)2 2 21 x y z + + +20. Differentiate 211 1tan| |+ | |\ .xxwith
respect to 122sin ,1| | |+\ .xx when x = 0.Sol. Let y =211 1tan put
tan| |+ | = |\ .xxx1sec 1tantany | |= |\ .11 costansiny | |= |\
.1tan tan2y| |= |\ .11 1 tan2 2y y x = =21 12 1dydx x| |= |+ \
.again122sin ,puttan1xz xx| |= = |+ \ .( )1 1sin sin2 2 2tan z x =
= =
221dzdx x=+Now,/ 1/ 4dy dy dxdz dz dx= =21. Find the particular
solution of the differential equation (2log 1),sin cos+=+dy x xdx y
y y given that 2t= ywhen x = 1.Sol. (sin cos ) (2log 1) y y y dy x
x dx + = +} }2cos cos 2 log2 + = + +} }xy y y dy x xdx c2 2 2cos
sin cos 2log .2 2 2x x xy y y y x c| | + + = + + |\ .2sin log y y x
x c = +When; 12t= = y x2ct= 2sin log2y y x xt= +10 65/1/3 P.T.O.22.
Show that lines ( ) (3 ) = + + r i j k i jand (4 ) (2 3 ) = + +r i
k i kintersect. Alsofindtheirpoint of intersection.Sol.1 1 13 1 0
+= = = x y z.... (1)4 0 12 0 3 += = = x y z... (2)Point on 1st
line[3 + 1, + 1, 1] ... (A)point on 2nd line[2 + 4, 0, 3 1] ...
(B)For point of intersection, A = BWe get = 1 and = 0 point of
intersection (4, 0, 1)SECTION-CQuestion numbers 23 to 29 carry 6
marks each.23. A dealer in rural area wishes to purchase a number
of sewing machines. He has only ` 5,760to invest and has space for
at most 20 items for storage. An electronic sewing machine costhim
` 360 and a manually operated sewing machine ` 240. He can sell an
electronic sewingmachine at a profit of ` 22 and a manually
operated sewing machine at a profit of ` 18.
Assumingthathecansellalltheitemsthathecanbuy,howshouldheinvesthismoneyinordertomaximize
his profit ? Make it as a LPP and solve it graphically.Sol. Let the
dealer buy x-electronic swing machine and y- manually operated
sewing machines x, y > 0Also x + y s 20[storage constraints]360x
+ 240 y s 5760i.e. 3x + 2y s 48objective function is z = 22 x +
18yPoint z x y = 22+ 18A(16,0) 352B(8,12) 392 (maxima)C(0,20)
36048121620240481216202428xyA (16, 0)B (8, 12)C (0, 20)Thedealer
shouldby 8 electronicand12 manual sewing machines.He willmake a
maximumprofit of ` 392.24. Acardfrom apackof52
playingcardsislost.From
theremainingcardsofthepackthreecardsaredrawnatrandom(withoutreplacement)andarefoundtobeallspades.Findtheprobability
of the lost card being a spade.Sol. Let A : getting three spade
cardsE1 : lost card is a spade P (E1) = 131521CCE2 : lost card is a
diamond 1312 521P(E ) =CC11 65/1/3 P.T.O.E3 : lost card is a club
1313 521P(E ) =CCE4 : lost card is a heart 1314 521P(E ) =CC41( ) (
) . ( | )== i iiP A P E P A E12 13 13 133 3 1 152 51 52 511 3 1 33
= + C C C CC C C CRequired probability = P (E1 | A)= 1 1P(E ).P(A/E
)P(A)12 133 152 511 312 13 13 133 3 1 152 51 52 511 3 1 3C CC CC C
C C3C C C C= + = 1049OR24. From a lot of 15 bulbs which include 5
defectives, a sample of 4 bulbs is drawn one by one
withreplacement.Findtheprobabilitydistributionofnumberofdefectivebulbs.Hencefindthemean
of the distribution.Sol. Let success be defined as getting a
defective bulb. If x indicates success then x = {0, 1, 2, 3,
4}Probability of success, 5 1p15 3= = and 1 2q 13 3= =We use P (x)
= nCx px qn x, where n is the number of trials.x 0 1 2 3 416 2 24 8
1P(x)81 81 81 81 813Mean, = 51( )= i iix P x = 4325. Find the area
of the region in the first quadrant enclosed by the x-axis, the
line y = x and thecircle x2 + y2 = 32.Sol. Solving y = x and x2 +
y2 = 32we get x = 4Required area = Area of region OCABO= 4 4 220
032 + } }x dx x dx(4, 0)C(4 2, 0)Coyxy = x= 4 2 422 10 43232 sin2 2
2 4 2 (+ + ( x x xx2 21 1(4) (0) 4 116sin 1 16 16sin2 2 2 2 (( | |=
+ + (|(\ . 8 16 8 162 4 t t ( | |= + + |(\ . = 4t sq. units.26.
Findthedistancebetweenthepoint(7,2,4)andtheplanedeterminedbythepointsA
(2, 5, 3), B (2, 3, 5) and C (5, 3, 3).12 65/1/3 P.T.O.Sol.
Equation of plane 2 5 34 8 8 03 2 0 + =x y zor 2x + 3y + 4z =
7Distance from (7, 2, 4) 292929= =OR26.
Findthedistanceofthepoint(1,5,10)fromthepointofintersectionoftheline
2 2 (3 4 2 ) = + + + +r i j k i j kand the plane .( ) 5. + =r i j
kSol. ThelineisQ (1,5,10)P( 2) ( 1) ( 2)3 4 2 + = = = x y zany
point on it is P (2 + 3, 1 + 4, 2 + 2) P lieson x y + z = 5 2 + 3 +
1 4 + 2 + 2 = 5i.e. = 0 the point of intersection is : P (2, 1, 2)
required distance from point Q = 13 units27. Two schools P and Q
want to award their selected students on the values of Discipline,
PolitenessandPunctuality.TheschoolPwantstoaward`xeach,`yeachand`zeachforthethreerespective
values to its 3, 2 and 1 students with a total award money of `
1,000. School Q wantsto spend ` 1,500 to award its 4, 1 and 3
students on the respective values (by giving the sameaward money
forthethreevaluesasbefore).If thetotalamountofawards for
oneprizeoneachvalueis`600,usingmatrices,findtheawardmoneyforeachvalue.Apartfromtheabove
three values, suggest one more value for awards.Sol. From thegiven
situationthe setof related equationsis3x + 2y + z = 10003x + y + 3z
= 1500x + y + z = 600The corresponding system ofmatrix equations
isAX = B where 3 2 1 10004 1 3 , and 15001 1 1 600xA X y Bz (((
(((= = = ((( ((( 3 2 1| | 4 1 3 51 1 1= = A
12 1 51 11 2 5| | 53 1 5A adjAA ( (= = ( ( 1100200300 ( (= =( (
X A Bx = ` 100, y = ` 200, z = ` 30028. Evaluate : /24 40sin cossin
cost+}x x xdxx x13 65/1/3 P.T.O.Sol./24 40sin cossin cost=+}x x xIx
x... (i)/24 40cos sin2cos sintt | | |\ .=+}x x xI dxx x...
(ii)Adding (i) and (ii) we get/24 40sin cos22 sin costt=+}x xI dxx
xPut sin2x = t 2 sinx cosx dx = dt sinx cosx dx = 2dt12 20122 2 (1
)t= + }dtIt t
1208 2 2 1t= +}dtIt t 1021162t= +}dtIt t1211200112tan1 116 16
161 12 22 4| | |t t t= = = || || | + |\ .\ .}tdtt29. Of all the
closed right circular cylindrical cans of volume 128t cm3, find the
dimensions of thecanwhichhasminimumsurfacearea.Sol. Given, V = 128t
= tr2hMinimize : S = 2trh + 2tr2Put 2128= hrhr
2256( ) 2t= + t S r rr2256( ) 4t' = + t S r rr32 256( ) 4 t'' =
+ t S rrFor CPs : S' (r) = 0 22564 0t + t = rrr = 4 cmNow S" (4)
> 0 at r = 4 is point of minima1288cm4 4t= =t h For minimum
surface area the can should have a radius of 4 cm and a height of 8
cm.