MATHEMATICS SOLUTION No. titlepageoftheanswerbook Candiatesmustwritecodeonthe Please check that this...

® CBSE XII EXAMINATION-2018

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CLASS-XII / (CBSE)

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MATHEMATICS & SOLUTION

Time allowed : 3 hours Maximum Marks : 100

General Instructions :

(i) All questions are compulsory.

(ii) The question paper consists of 29 questions divided into four sections A, B, and C and D. Section Acomprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each,Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of sixmarks each.

(iii) All questions in Section A are to be answered in one word, one sentence of as per the exact requiremt ofthe question.

(iv) There is no overall choice. However, internal choice has been provided in 3 question of four marks eachand 3 questions of six marks each. You have to attempt only one of the alternatives in all such questions.

(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.

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CLASS-XII / (CBSE)

PAGE # 2

SECTION AQuestion number 1 to 4 carry 1 mark each.

Q 1. Find the value of ).3(cot3tan 11

Ans 1. tan�1 )3( � cot�1 )3(

= )3cot(3

1

= )6

(3

= 26

526

53

Q 2. If the matrix

01b

102

3a0

A is skew symmetric, find the values of �a� and �b�.

Ans 2.

01b

102

3a0

A

for skew symmetric metrix of A.a = �2 , b = 3

Q 3. Find the magnitude of each of the two vectors banda , having the same magnitude such that the

angle between them is 60° and their scalar product is29

.

Ans 3. |b||a|

ba

= 60,29

29

60cos|b||a|

29

21

|a| 2

9|a| 2

3|a|

3|a|

Q 4. If a * b denotes the larger of �a� and �b� and if a ° b = (a * b) + 3, then write the value of (5) ° (10),where * and ° are binary operations.

Ans 4. a ° b = (a * b) + 3

(5) ° (10) = (5 * 10) + 3= 10 + 3 = 13

SECTION BQuestion numbers 5 to 12 carry 2 marks each.

Q 5. Prove that : 3 sin�1x = sin�1 (3x � 4x3),

21

,21

x

P.T.O

http://www.pspd.resonance.ac.in






CLASS-XII / (CBSE)

PAGE # 3

P.T.O

Ans 5. 3 sin�1x = sin�1(3x � 4x3)

21

,21

x

Let sin�1x = t ...........(1)

x = sint

R.H.S. = sin�1(3x � 4x3)

= sin�1 [3sint � 4sin3t]

= sin�1 (sin3t)

= 3t

= 3sin�1x

= L.H.S.

3sin�1x = sin�1(3x � 4x3)

Q 6. Given

74

32A , compute AA�1 and show that 2A�1 = 9I �A.

Ans 6.

74

32A |A| = 74

32

= 14 �12 = 2

A�1 = |A|Aadj

A�1 =

24

37

2

1

L.H.S. 2A�1 =

24

37

Now 9I � A =

74

32

10

019

=

74

32

90

09

9I � A =

24

37

L.H.S. = R.H.S

Q 7. Differentiate

xsinxcos1

tan 1 with respect to x.

Ans 7. Let y =

xsin

xcos1tan 1

y =

2x

cos2x

sin2

2x

cos2tan

2

1

y =

2x

cottan 1







CLASS-XII / (CBSE)

PAGE # 4

P.T.O

y =

2x

2tantan 1

y = 2x

2

So

2

x

2dx

d

dx

dy

= 21

Q 8. The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005x3 �

0.02x2 + 30x + 5000. Find the marginal cost when 3 units are produced, where by marginal cost wemean the instantaneous rate of change of total cost at any level of output.

Ans 8. Cost (MC) = dx

)x(dC}5000x30x020x0050{

dxd 23

= 0.015x2 � 0.04x + 30

at x = 3 dx

)x(dC 0.015(3)2 � 0.04 (3) + 30

= 0.015 × 9 � 0.12 + 30

= 0.135 � 0.12 + 30

= 30.015The required marginal cost is 30.015 Rs.~ 30.02 Rs.

Q 9. Evaluate : dxxcos

xsin2x2cos2

2

Ans 9. dxxcos

xsin2x2cos2

2

dxxcos

xsin2xsin212

22

Cxtandxxsec2

Q 10. Find the differential equation represention the family of curves y = a ebx+5, where a and b are arbitarayconstants.

Ans 10. y = aebx+5

beadxdy 5bx

bydxdy

by

dxdy

Again diff. w.r. to x

0y

)dx/dy(dx/ydy2

222







CLASS-XII / (CBSE)

PAGE # 5

P.T.O

0dxdy

dx

ydy

2

2

2

Q 11. If is the angle between two vectors k�3j�2i� and k�j�2i�3 , find sin .

Ans 11. cos = 149941

343

|b||a|

ba

cos = 75

1410

sin2 = 1 � cos2

= 1 � 495

= 49

2549

= 2924

sin = 2924

Q 12. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, giventhat the red die resulted in number less than 4.

Ans 12. Red dice & Black dice

die 1st Black

die 2nd Red

A = { (2, 6) (6, 2) (3, 5), (5, 3), (4, 4) }B = { (1,1) (2, 1) (3, 1), (4, 1) (5, 1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3)

(6,3) }A B = (6,2), (5,3)

91

182

)B(n)BA(n

BA

P

SECTION C

Question numbers 13 to 23 carry 4 marks each.

Q 13. Using properties of determinants, prove that

)zxyzxyxyx3(9

1z311

11y31

x3111

Ans 13. L.H.S. =

1z311

11y31

x3111

(C1 C1 � C2)

=1z31z3

11y3

x3110

(C2 C2 � C3)







CLASS-XII / (CBSE)

PAGE # 6

P.T.O

=1z3z3

10y3

x31x30

(R3 R3 � R2)

=0z3y3z3

10y3

x31x30

(Common 3 from C1 & C2)

= 90zyz

10y

x31x0

= 9 )0yz()x31()yx0()x(0

= 9 [xy + xz + yz + 3xyz]

Q 14. If (x2 + y2)2 = xy, finddxdy

.

OR

If x = a (2 � sin 2) and y = a (1 � cos 2), find dxdy

when .3

Ans 14. (x2 + y2)2 = xy

ydxdy

x)dxdy

y2x2()yx(2 22

ydxdy

x)dxdy

yx()yx(4 22

ydxdy

xdxdy

yxydxdy

yxx4 3223

xy4x4ydxdy

xyyxdxdy

4 2332

xy4x4yxy4yx4dxdy 2332

xy4yx4

xy4x4ydxdy

32

23

ORx = a (2 � sin2)y = a (1 � cos2)

ddx

= a [2 � 2 cos2]

ddy

= a(0 � (�2sin2)

ddy

= 2a sin2

)2cos1(a22sina2

d/dxd/dy

dxdy

2sin2

cossin2dxdy







CLASS-XII / (CBSE)

PAGE # 7

cot

sincos

dxdy

when .3

thendxdy

= 3

1

3cot

Q 15. If y = sin (sin x), prove that 2

2

dx

yd+ tan x

dxdy

+ y cos2 x = 0.

Ans 15. y = sin(sinx) ................(1)

xcos)xcos(sindxdy

............(2)

xcos))x(sinsin(xcos)xsin()xcos(sindx

yd2

2

))x(sin(sinxcos)xcos(sinxsindx

yd 22

2

by equation (1) & (2)

xcosyxcos

1dxdy

xsindx

yd 22

2

0xcosydxdy

xtandx

yd 22

2

Q 16. Find the equations of the tangent and the normal, to the curve 16x2 + 9y2 = 145 at the point (x1,y1),

where x1 = 2 and y1 > 0.OR

Find the intervals in which the function f(x) = 12x24x5x4x 23

4

is (a) strictly increasing, (b)

strictly decreasing.

Ans 16. 16x2 + 9y2 = 145 ..........(1)

at (x1, y1)

145y9x16 21

21

at x1 = 2

16 × 4 + 145y9 21

64145y9 21 = 81

21y = 9

y1 = ± 3 y1 > 0

y1 = 3

So, P (x1 y1) (2, 3)

diff. w. r. to x of eqn (1)

32x + 18y 0dxdy

x32dxdy

y18

y18x32

dxdy







CLASS-XII / (CBSE)

PAGE # 8

slope of tangent at point (x1, y1)

1

1

y9x16

dxdy

= 39

216

2732

dxdy

equation of tangent at point (x1 y1)

y � y1 = )xx(dxdy

1

y � 3 = )2x(2732

equation of Normal at point (x1 y1)

y � y1 = )xx(

dxdy

11

y � 3 = )2x(3227

OR

f(x) = 12x24x5x4x 23

4

24x10x34x4

dx)x(df 2

3

24x10x3xdx

)x(df 23

for Increasingf (x) > 0

x3 � 3x2 � 10x + 24 > 0

x3 � 2x2 � x2 + 2x � 12x + 24 > 0

x2 (x�2) �x (x�2) �12 (x�2) > 0

(x �2) (x2 � x � 12) > 0

(x �2) (x � 4) (x + 3) > 0

(�3, 2) (4, )for decreasingf(x) < 0(� , �3) (2, 4)

Q 17. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as tohold a given quantity of water. Show that the cost of material will be least when depth of the tank is halfof its width. If the cost is to be borne by nearby settled lower income families, for whom water will beprovided, what kind of value is hidden in this question?







CLASS-XII / (CBSE)

PAGE # 9

Ans 17. Let the l, b, h of the open tank be. x, x, y

Volume of tank V = x2y,

Surface area of open tank S = x2 + 4xy

S = x2 + 4x

2x

V

S = x2 + xV4

2x

V4x2

dxdS

3x2V40dxdS

0x

V82

dx

Sd32

2

for all x

4V = 2x3

4x2y = 2x3

When x = 2y, S is minimum the depth of tank is half of its width. Water capacity depend on the

investment amount is low. As the families are of low income base the values associated with thisquestion is to design the tank of minimum cost so that maximum number of families can afford it andavail water supply. Obviously at is responsibility of one and all to think & work for better ment of poor.

Q 18. Find : dx)xsin1()xsin1(

xcos22

Ans 18. dx)xsin1()xsin1(

xcos22

sin x = tcosx dx = dt

dx)t1()t1(

22

Now 22 t1

CBt)t1(

A

)t1()t1(

2

2 = A(1+ t2) + (Bt + C) (1 � t)

at t = 12 = 2A + 0A = 1at t = 02 = A + C2 = 1 + CC = 1at t = �1

2 = 2A + 2 (�B + C)

2 = 2 + 2(�B + 1)

0 = �2B + 2

2B = 2







CLASS-XII / (CBSE)

PAGE # 10

B = 1

= dt1t

1tt1

12

= dt1t

1

)1t(2

t2t1

122

= � log |1�t | + Cttan|1t|log21 12

= � log | 1�sinx | + Cxsintan|1xsin|log21 12

Q 19. Find the particular solution of the differential equation ex tan y dx + (2 � ex) sec2 y dy = 0, given that

4y

when x = 0.

OR

Find the particular solution of the differential equation ,xsinxtany2dxdy

given that y = 0 when

.3

x

Ans 19. ex tany dx + (2 � ex) sec2y dy = 0

ex tany dx = (ex�2) sec2y dy

dyytanysec

dx2e

e 2

x

x

on integrating both side

dyytanysec

dx2e

e 2

x

x

Clogytanlog2elog x

at x = 0, y =4

log | 1 � 2 | = log1 + log C

log C = 0

log |ex �2| = log |tan y|

0ytan2e

logx

e

0x

eytan2e

ex � 2 = tan y

ex = 2 + tan y

OR

xsinxtany2dxdy

QPydxdy

is a linear D.E.







CLASS-XII / (CBSE)

PAGE # 11

P = 2 tanx

I.f. = dxxtan2e

= xseclogxseclog2 2

ee

I.f. = sec2x

y. I.f = dxQ)f.I(

y.sec2x = dxxsin.xsec2

y.sec2x = dxxsec.xtan

y sec2x = secx + C

at x = 3

, y = 0

0 = 2 + C

C = �2

Particular Soln of D.E. y sec2x = secx � 2

Q 20. Let k�5j�4i�b,k�j�5i�4a

and .k�j�i�3c

Find a vector d

which is perpendicular to both c

and b

and ad = 21.

Ans 20. k�j�5i�4a

k�5j�4i�b

k�j�i�3c

)bc(d

bc

541

113

k�j�i�

= k�13)16(j�)1(i�

= k�13j�16i�

d

= )k�13j�16i�(

given 21ad

21)k�j�5i�4()k�13j�16i�(

(4 � 80 +13) = 21

(17 � 80) = 21

(� 63) = 21

= 31

)k�13j�16i�(31

d

Q 21. Find the shortest distance between the lines

)k�3j�2i�()j�i�4(r

and ).k�5j�4i�2()k�2j�i�(r







CLASS-XII / (CBSE)

PAGE # 12

Ans 21. )k�3j�2i�(j�i�4r

)k�5j�4i�2(k�2j�i�r

)j�i�4()k�2j�i�(aa 12

k�2i�3aa 12

S.D. = |bb|

)bb()aa(

21

2112

21 bb

= 542

321

k�j�i�

= )44(k�)65(j�)1210(i�

= k�0j�i�2

| 21 bb

| = 514

S.D. = 5

)j�i�2()k�2i�3(

= 5

6

5

006

S.D. = 5

56

Q 22. Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number oftails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a �head� or �tail� is obtained. If

she obtained exactly one �tail�, what is the probability that she threw 3, 4, 5 or 6 with the die ?

Ans 22. E1 = Getting 1 or 2 in single throw of a die.

E2 = 3, 4, 5 or 6 in single throw of a die.

A = getting exactly one tail.

P(E1) = 31

62 , P(E2) = 6

4=

32

P(A | E1) = 83

21

21

C21

13

P (A | E2) = 21

Req. Prof. P(E2 | A) = )E|A(P)E(P)E|A(P)E(P)E|A(P)E(P

2211

22

=

21

32

83

31

21

32

= 118







CLASS-XII / (CBSE)

PAGE # 13

Q 23. Two number are selected at random (without replacement) from the first five positive integers. Let Xdenote the larger of the two numbers obtained. Find the mean and variance of X.

Ans 23. We observe that X can take values 2, 3, 4, 5

202

41

51

41

51

)2X(P

204

42

51

41

52

)3X(P

206

43

51

41

53

)4X(P

208

44

51

41

54

)5X(P

iX pi Xipi i2i pX

2202

204

208

3204

2012

2036

4206

2024

2096

5208

2040

20200

iiXp 2iiXp =

20340

= 2080

Mean )X( = iiXp = 42080

Var(X) = 2iiXp � ( iiXp )2

= 17 � 42

= 17 �16

Var(X) = 1SECTION D

Question number 24 to 29 carry 6 marks each.

Q 24. Let A = {x Z : 0 x 12}. Show thatR = {(a, b) : a, b A, | a � b | is divisible by 4} is an equivalence relation.

Find the set of all elements related to 1. Also write the equivalence class [2].OR

Show that the function f : R R defined by f(x) = Rx,1x

x2

is neither one-one nor onto. Also, if

g : R R is defined as g(x) = 2x � 1, find fog(x).

Ans 24. A = { x z : 0 x 12 } = { 0, 1, 2, .....12}R = { (a, b) : |a � b| is multiple of 4 }

Reflexivity :- Let a element a , a AA| a � a | = 0 , which is multiple of 4

(a, a) R







CLASS-XII / (CBSE)

PAGE # 14

So R is reflexiveSymmetry :- Let (a, b) R , a, b ANow (a, b) R ,

| a � b | is multiple of 4

| a � b | = 4 , N | b � a | = 4

(b , a) RSo R is symmetric.

(iii) Transitivity Let (a, b) & (b, c) R a, b, c A |a � b| = 4 & | b � c | = 4 ,N a � b = ± 4 ------(1) b � c = ± 4--------(2)

Eqn (1) and (2)

a � c = ± 4 ± 4

a � c is multiple of 4

|a � c| = 4k

(a, b) and (b, c) R (a, c) R So R is transitive

Hence R is equivalance relationLet x be an element of A Such that (x,1) R.then

|x � 1| = 4k

|x �1| = 0, 4, 8, 12

x �1 = 0, 4, 8, 12

x = 1, 5, 9Now set of all elements Related to 1 is { 1, 5, 9}equivalence class [2] = { 2, 6, 10 }

OR

f(x) = 1x

x2

for one � one

f(x) = f(y)

1y

y

1x

x22

xy2 + x = yx2 + y

xy2 � yx2 + x � y = 0

xy (y �x) + x � y = 0

0)xy1()yx(

x = y & xy = 1

x = y1

(x y)

It is many-one not one-one for onto or into

Let f(x) = y







CLASS-XII / (CBSE)

PAGE # 15

y1x

x2

x = x2y + y

x2y � x + y = 0

y2

y411x

2

x R, y Rbut at y = 0 x RSo It is into not onto

fog = f(g(x)) = 1)x(g

)x(g2

= 1)1x2(

1x22

= 1x41x4

1x22

= 2x4x4

1x22

Q 25. If

211

423

532

A find AA�1. Use it to solve the system of equations

2x �3y + 5z = 11

3x + 2y � 4z = �5

x + y � 2z = � 3.

ORUsing elementary row transformations, find the inverse of the matrix

542

752

321

A

Ans 25. | A | = 211

423

532

| A | = 2(�4 + 4) + 3 (�6 + 4) + 5(3 � 2)

| A | = 0 � 6 + 5 = �1 0 (unique Sol.)

Cofactor C = 13232

591

120

adj A = CT =

1351

2392

210







CLASS-XII / (CBSE)

PAGE # 16

A�1 = |A|)A(adj

AA�1 =

1351

2392

210

AX = B

X = A�1 B = B.|A|A.adj

X =

1

3

5

11

1351

2392

210

X =

392511

694522

650

X =

3

2

1

3

2

1

z

y

x

x = 1, y = 2, z = 3

ORA = IA

A

100

010

001

542

752

321

R2 R

2 � 2R

1 ; R

3 R

3 + 2R

1

A

102

012

001

100

110

321

R1 R

1 � 2R

2

A

102

012

025

100

110

101

R1 R

1 � R

3 ; R

2 R

2 � R

3

A

102

114

123

100

010

001

= BA







CLASS-XII / (CBSE)

PAGE # 17

B = A�1 =

102

114

123

Q 26. Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line y = xand the circle x2 + y2 = 32.

Ans 26. line y = x ................(1)

circle x2 + y2 = 32 ................ (2)

the area of the region OBMO

4

0

4

0xdxdxy

= unit8]x[21 4

02

Again the area of the region BMAB

dxx32dxy24

4

224

4

=

24

4

12

24

xsin)32(

21

x322x

=

2

1sin32

21

163224

)1(sin160)24(21 11

=

4

16422

16

= 8 � 8 � 4 = (4 � 8) unit

So required area OBAMO = 8 + (4 � 8) = 4. Ans.

Q 27. Evaluate : dxx2sin916xcosxsin4/

0

OR

Evaluate : ,dx)ex3x(3

1

x2 as the limit of the sum.

Ans 27. (i) dxx2sin916)xcosx(sin4/

0

= dx)}x2sin1(1{916

)xcosx(sin4/

0

= dx)xsinx(cos925

)xcosx(sin4/

02

Put cosx � sinx = t

(sinx + cosx) dx = � dt

Lower limit x = 0 t = (cos 0 � sin 0 = 1)

Upper limit x = 4

t = cos 4

� sin 4

= 0







CLASS-XII / (CBSE)

PAGE # 18

=

0

12t925

dt

=

1

02)t3(25

dt

=

1

022 )t3()5(

dt

= 31

t35t35

log)5(2

11

0

= 28

log301

1log3535

log301

(ii) 3

1

x2 dx)ex3x(

0h

b

alindx)x(f

h[f(a) + f(a+h) + f(a+2h) + .............+ f(a + (n�1)h)]

where h = n2

n13

hn

ab

Now I = 3

1

x3

1

3

1

2 dxedxx3dxx

h)1n(1(f.......)h21(f)h1(f)1(fhlim0h

I = }e)h21(3)h21{(}e)h1(3)h1{(}e)1(31{[hlim )h21(2)h1(212

0h

+..........+ {(1+(n�1)h}2 + 3 {1 + (n �1)h} + e{1+(n�1)h}]

= hlim0h [ 2222 }h)1n(1{.....)h21()h1(1{

+3 {1+(1 + h) + (1 + 2h) +.....+ (1+(n�1)h) } + }e......eee{ )h)1n(1(h21h11 ]

= hlim0h [{n + 2h (1 + 2 +......+ (n�1)} + h2 {12 + 22 + .....(n�1)2 }

+3 { n + h (1 + 2 + 3......+(n�1) } + e { 1 + eh + e2h +........+e(n�1)h }]

=

1e

1)e(e

2)1n(n

hn36

)1n2)(1n(n.h2

)1n(n.h2nhlim

h

nh2

0h

=

1e

1ee

6)1n2)(1n(nh

2)1n(n

.h5n4hlimh

nh2

0h

Put n2

h

= h

h1e

1elimne2

6)1n2()1n(n

n

42

)1n(nn2

5n4n

2limh

nh

0h20h







CLASS-XII / (CBSE)

PAGE # 19

Put nh = 2

= 1ee

n

1n21n68

21nn

n

208lim 2

22n

=

1ee

n

12

n1

134

n1

1108lim 22n

= 1ee234

108 2

= 1ee38

18 2

= 1ee362 2

Ans.

Q 28. Find the distance of the point (�1, �5, �10) from the point of intersection of the line

)k�2j�4i�3(k�2j�i�2r

and the plane .5)k�j�i�(r

Ans 28. k�2j�4i�3k�2j�i�2r

................ (1)

Cartesion form

22z

41y

32x

Gen. point at line P + 2, 4 � 1, 2 + 2

Given plane eq. 5k�j�i�.r

................ (2)

Let P is intersection point of the (1) & plane so P will satisfying the equation of plane (2)

So, 5k�j�i�k�22j�1423

5221423

+ 5 = 5 = 0Point P =(2, �1, 2)

distance between (�1, �5, �10) & (2, �1 ; 2) is

= 222 1025112 = unit13169144169

Q 29. A factory manufactures two types of screws A and B, each type requiring the use of two machines, anautomatic and a hand-operated. It takes 4 minurtes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws �A� while it takes 6 minutes on the automatic

and 3 minutes on the hand-operated machine to manufacture a packet of screws �B�. Each machine is

available for at most 4 hours on any day. The manufacturer can sell a packet of screws �A� at a profit of

70 paise and screws �B� at a profit of Rs. 1. Assuming that he can sell all the screws he manufactures,

how many packets of each type should the factory owner produce in a day in order to maximize hisprofit ? Formulate the above LPP and solve it graphically and find the maximum profit.

Ans 29. Let number of screw of type A = xnumber of screws of type B = y4x + 6y 240 2x + 3y 1206x + 3y 240 2x + 3y 800x 0, y 0







CLASS-XII / (CBSE)

PAGE # 20

Max. Z = 0.7 x + (1)yZ = 0.7 x + y

So maximum. profit = 41 Rs. occurs at x = 30, y = 20



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