topper pre board-answersheet_cbse xii_math

7/28/2019 Topper Pre Board-Answersheet_CBSE XII_Math

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Class XII

Mathematics

Solutions

Section A

Ans.1 Let f -1(-4) = x.Then f(x) = -4 x2 = -4

x = 4 x = 2i, which is in C. f -1(-4) = {2i, -2i} (1 Mark)

Ans.2 1 13

sin sin cos x 15

Operating both sides by sin-1 we get,

1 1 1

1 1

3sin cos x sin 1

5 2

3x Since, sin x cos x (1 mark)

5 2

Ans.3 A =3 2

4 2

A = 3 24 2

3 24 2

= 1 2

4 4

A = KA 2I

1 2 3K 2K 2 0

4 4 4K 2K 0 2

1 2 3K 2 2K

4 4 4k 2K 2

Comparing the terms we get,K = 1 (1 Mark)

Ans.4n 1

adj A A

, where n is order of square matrix

Given A is a square matrix of order 3


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3 1 2adjA A A

Since, adjA 25

So A 5

(1 Mark)

Ans.5 A is a matrix of order 2 x 3 and B is a matrix of order 3 x 5.Hence, (AB) exists and it is of order 2 x 5.

(AB) is of order 5 x 2. (1 Mark)

Ans.6 2 sinxxsec xdx x tanx tanxdx x tanx dxcosx

x tanx log(cosx) c (1 Mark)

Ans.7 Let f(x) = sin7x f(-x) = sin7 (-x) = {sin(-x)}7 = {-sin x}7 = - sin7x = -f(x)So, f(x) is an odd function.

72

2

sin x dx = 0 (1 Mark)

Ans.8 Here, b 9 i j 6k

n 3i j 2k

The linex 2 y 1 z 3

9 6

is perpendicular to the plane 3x y 2z = 7 if

9 6

3 1 2

or = -3 (1 Mark)

Ans.9 5 3 ;a i j k

3 5b i j k

a b 6i 2j 8k

a b 3i j 4k 1 mark2

Ans.10 We know,2 2 2

a b a b 2 a.b

= (2)2 + (5)2 2(2) = 4 + 25 4 = 25

a b

= 5 (1 mark)SECTION B

Ans.11 Given, f: R Ris defined as f(x) = |x|g: R Ris defined as g(x) = [x]

fog5

2

= f

5g

2


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Now, g5

2

=

52.5 2

2

(1 Mark)

fog5

2

= f(2) = |2| = 2 (1 Mark)

gof 2 = g f 2 Now, f 2 = 2 = 2 (1 Mark)

gof 2 = g( 2 ) = 1 (1 Mark)OR

(a, b) R (c, d) a + d = b + c

For reflexive : (a, b) R (a, b) a + b = b + a, true in N

(1 mark)

For symmetric : (a, b) R (c, d) a + d = b + c c + b = d + a

(c, d) R (a, b), for (a, b), (c, d) N x N (1 mark)

For transitive: Let (a, b), (c, d), (e, f), N X N

(a, b) R (c, d) and (c, d) R (e, f)

(a, b) R (c, d) a + d = b + c (i)

(c, d) R (e, f)

c + f = d + e (ii)

Adding (i) and (ii), we get,

a + d + c + f = b + c + d + e a + f = b + e (a, b) R (e, f)

Hence, R is transitive. (1

12

marks)

Hence relation is an equivalence relation. (1/2 mark)

Ans.12 cot-1 1 sinx 1 sinx

1 sinx 1 sinx

= cot-1

2 2 2 2

2 2 2 2

x x x x x xsin cos sin2 sin cos sin2

2 2 2 2 2 2

x x x x x xsin cos sin2 sin cos sin2

2 2 2 2 2 2


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[Since, sin2A + cos2A = 1] (1 mark)

2 2 2 2

1

2 2 2 2

x x x x x x x xsin cos 2sin cos sin cos 2sin cos

2 2 2 2 2 2 2 2

cot x x x x x x x xsin cos 2sin cos sin cos 2sin cos

2 2 2 2 2 2 2 2

(1 mark)

[Since, sin2A = 2 sinA cosA]

= cot-1

2 2

2

x x x xcos sin cos sin

2 2 2 2

x x x xcos sin cos sin

2 2 2 2

(1 mark)

= cot-1

x2cos

2x

2sin2

= cot-1x

cot2

=x

2

(1 mark)

Hence, proved.

Ans.13


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2

2

2

1 2 33 2 2

2 3 2

2 2 3

1 2 3

2 2 2

2 2 2

2 2 2

1

a 1 ab ac

ab b 1 bc

ac bc c 1

Multiplying and dividing R ,R ,R by a,b,c respectivelya a a b a c

1ab b b b c

abcac bc c c

Taking a,b,ccommon from C ,C ,C respectively

a 1 a aabc

b b 1 b (1mark)abc

c c c 1

ApplyingR

1 2 3

2 2 2 2 2 2 2 2 2

2 2 2

2 2 2

2 2 2 2 2 2

2 2 2

2 2 1 3 3 1

2 2 2 2

2

2 2 2

R R R

a b c 1 a b c 1 a b c 1

b b 1 b

c c c 1

1 1 1

(a b c 1) b b 1 b (1mark)

c c c 1

Applying C C C and C C C

1 0 0

(a b c 1) b 1 0 (1mark)

c 0 1

Expanding along first row

(a b c 1) RHS (1mark)

Ans.14

2

3

2

sin a 1 x sinx, x 0

x

f x c , x 0x bx x

, x 0

bx

L.H.L.

x 0

sin a 1 x sinxlim

x


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x 0 x 0

sin a 1 x sinx 1lim a 1 lim mark

a 1 x x 2

a 1 1 a 2 1 mark

R.H.L2

3x 02

x bx xlim

bx

x 0

x 0

x 0

x 1 bx 1 1lim mark

2bx x

1 bx 1 1 bx 1lim

bx 1 bx 1

1 bx 1 1lim

bx 1 bx 1

1 1 mark2

Since f is continuous at x = 0,

L.H.L. = R.H.L. = f (0)

a + 2 = c =1

2

3 1a , c

2 2

1 mark

OR

tGiven: x=a cost+log tan ; y= a sint

2

tx a cos t log tan

2

dx 1 t 12a sin t .sec .

tdt 2 2tan

2

dx 1a sin t

t tdt2 sin cos

2 2

(1 Mark)

2dx 1 acos t

a sin tdt sin t sin t

(1 Mark)

y = a sin t


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dyacos t

dt (1 Mark)

dy

dy a cos t sin tdttan t

2dxdx cos tacos tdt sin t

(1 Mark)

Ans.15 We know that ex, sin x and cos x functions are continuous anddifferentiable everywhere.Product, sum and difference of two continuous functions is again acontinuous function.

So, f is also continuous in5

,4 4

.

Also, f(x) is differentiable in5

,4 4

(1 Mark)

Now, f 4

= 4e sin cos 0

4 4

f5

4

=5

4 5 5e sin cos 04 4

f4

= f5

4

(1 Mark)

Rolles theorem is applicable.f(x) = ex(sin x cos x) + ex( cos x + sin x) = 2ex sin x f(x) = 0 gives 2ex sin x = 0 sin x = 0

x = 0, (1 Mark)Now,

5,

4 4

The theorem is verified with x = (1 Mark)

Ans.16 Here,3 3

sin , cosx a t y b t (1)Differentiating (1) w.r.t. t,

2dx 3asin t.cos tdt

and

2dy 3bcos t.sintdt

(1 Mark)

2

2

dydy 3bcos t.sint bdt cott

dxdx a3asin t.cos tdt

(1 Mark)

Slope of the tangent at2

t

is


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2

dy bcot 0

dx a 2

(1 Mark)

Hence, equation of tangent is given by

3y b cos 0 or y 0

2

(1 Mark)

OR

f(x) =4sinx 2x xcosx 4sinx

x2 cosx 2 cosx

(1/2 mark)

f(x) =

2

2

cosx 2 cosx sin x4. 1

2 cosx

2 2

2

2

2

2

2cosx cos x sin x4. 1

2 cosx

2cosx 14 1

2 cosx

8cosx 4 4 cos x 4cosx

2 cosx

2

2 2

cosx 4 cosx4cosx cos x

2 cosx 2 cosx

(2 marks)

Now for increasing function f(x) > 0

f(x)

2

4 cosxcosx

2 cosx

2

4 cosx

2 cosx

is always +ve since |cos x|1 (1/2 mark)

f(x) > 0 if cos x is +ve

f(x) is increasing in3

0, ,22 2

(1 mark)

Ans.17 Let I =

cosxdx

1 sinx 2 sinx

Let sin x = t


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cos x dx = dt (1

2mark)

dtI

1 t 2 t

A B dt1 t 2 t

(1

2mark)

A(2 t) + B(1 t) = 1

Comparing the coefficients we get, A = 1 and B = -1 (1 mark)

1 1I dt dt

1 t 2 t

log 1 t log 2 t c (1 mark)

log 2 t log 1 t c

log 2 sinx log 1 sinx c

2 sinxlog +c (1 mark)

1 sinx

Ans.18 1 2tan y x dy 1 y dx

1

2

1

2 2

tan y x dx

dy1 y

dx tan y x

dy 1 y 1 y

1

2 2

dx x tan y

dy 1 y 1 y

Which is a linear differential equation with

1

2 2

1dy

12 tan y1 y

1 tan yP , Q 1 mark

1 y 1 y

I.F. e e 1 mark

Solution of differential equation is

x11 1tan y tan y2

tan ye e . dy

1 y

1mark

2

Let tan-1y = t 2

1dy dt

1 y


10/21

1tan y tx.e t.e dt

= t et - et + C 1 mark

1 1 1tan y 1 tan y tan y

x.e tan y.e e C

x = tan-1 y 1 + C1tan ye

1 mark2

Ans.19 The equation of the family of circles is(x a)2 + (y b)2 = 9 --- (1) ( Mark)

Differentiating both sides w.r.t. x, we get,

2(x a) + 2(y b)dy

dx= 0

(x a) = -(y b)dy

dx--- (2) (1 Mark)

Again, differentiating both sides w.r.t. x, we get,

1 + (y b)2

2

d y

dx+

2dy

dx

= 0

(y b) =

2

2

2

1dy

dx

d y

dx

--- (3) (1 Mark)

From (2), (x a) =

2

2

2

1dy

dxdy.dxd y

dx

( Mark)

Putting in (1), we get,2 2

2 2

2

2 2

2 2

dy dy1 1

dydx dx9

dxd y d y

dx dx

( Mark)

22

2

2

2

dy1 dydx

1 9dxd y

dx

3 22 2

2

dy d y1 9

dx dx

( Mark)


11/21

Ans.20 Let a and b

be two unit vectors

So, | a|

= 1 | b |

= 1

Also sum of these vectors is a unit vector, i.e., | a b | 1

2

2 2

| a b | 1

a b . a b 1

| a| |b | 2(a.b) 1

2| a.b | 1

(1 Marks)

Consider |a b|2

2

2 2

| a b | a b . a b

| a| |b | 2 a . b

1 1 1 3

(1 Marks)

| (a b) | 3

(1 Mark)

Ans.21 The given planes are

r. i 2j 3k 4 0

r. 2i j k 5 0

Converting in Cartesian form, we have: ( Mark)x + 2y + 3z 4 = 02x + y + z + 5 = 0Equation of plane passing through the intersection of these planes isgiven byx + 2y + 3z 4 + (2x + y + z + 5) = 0 (1 Mark) (1 + 2)x + (2 + )y + (3 + )z 4 + 5 = 0 (1) ( Mark)

This is perpendicular to the plane 5x + 3y 6z + 8 = 0 5 (1 + 2) + 3 (2 + ) 6 (3 + ) = 0 (1 Mark) -7 + 7 = 0 = 1 ( Mark)

Substituting in (1), the required equation of the plane is3x + 3y + 4z + 1 = 0 ( Mark)

Ans.22 Let E1 : The event that six comes on the die.

E2 : The event that six does not comes on the die.

A : The event that man report it is a six


12/21

1 2 11 1 5

P E and P E 1 P E 16 6 6

Probability that the man report that there is a six on the die given thesix comes on the die

1

AP

E

= Probability that man speaks truth

3

4 (1/2 Mark)

Probability that the man reports that there is a six on the die giventhat six does not comes on the die

2

AP

E

= Probability that man does not speak truth

31

4

1

4 (1/2 Mark)

By Bayes Theorem, we have:

1EPA

= Probability of getting a six given that man reports that there

is six on die

11

1 21 2

AP E P

E(1 Mark)

A AP E P P E P

E E

1 3

6 41 3 5 1

6 4 6 4

3(1 Mark)

8

Speaking truth pays in the long run. However, sometimes lie told for agood cause is not bad. (1 Mark)

OR


13/21

Let N = 729 be the number of families having six children.

Let probability of survival of a girl child be given by1

p3

and that of a boy

child is q =2

3. Let X be the number of girls in the family.

Then, X = 0, 1, 2, 3, 4, 5, 6

Probabilities of 0, 1, 2, .. , 6 girls in the family are given by the expansion

6

n 2 1q p

3 3

Where 6 r r

6r

2 1P r C

3 3

(1 Mark)

Probability of 2 girls and 4 boys in the family of 6 =

4 2

62

2 1 6 5 16 1 80P 2 C

3 3 2 1 81 9 243

(1 Mark)

Number of families having 2 girls and 4 boys in the backward state

= N P(2)

= 729 80

240243

(1 Mark)

Yes, a female child is neglected in the backward areas. Moral education tosociety to spread awareness among people and incentive for the female childlike free education, various schemes etc should be organized.

(1 Mark)

SECTION C

Ans.23 The given system of equations can be written as.

1

x1 1 1 4

12 1 3 0y

1 1 1 21

z

or AX = B


14/21

|A| =

1 1 1

2 1 3 1(1 3) 1(2 3) 1(2 1) 10 0

1 1 1

(1 mark)

A

-1

exists and the given system of equations has unique solution.A11 = 4 A12 = -5 A13 = 1

A21 = 2 A22 = 0 A23 = -2

A31 = 2 A32 = 5 A33 = 3

Adj A =

T4 5 1 4 2 2

2 0 2 5 0 5

2 5 3 1 2 3

(2 marks)

A-1 =1

AAdj A =

4 2 21

5 0 510

1 2 3

(1 mark)

X = A-1B =

4 2 2 41

5 0 5 010

1 2 3 2

(1 mark)

=

20 21

10 1

10 10 1

1

x= 2,

1

y= -1,

1

z= 1

x =1

2, y = -1, z = 1 (1 mark)

Ans.24 Let P(x, y) be the position of the helicopter and let the position of soldierbe at A(3, 2).

2 2

2

22 2

AP x 3 y 2

but y= x 2 is the eqn of the curve

AP x 3 x

(1 Mark)

Let AP2 = z = (x 3)2 + x4 (1 Mark)

3dz

2 x 3 4xdx


15/21

= 4x3 + 2x - 6 (1 Mark)For mxima or minima,

3dz 0 2x x 3 0dx

or (x 1)(2x2 + 2x + 3) = 0

x = 1 (1 Mark)2

22

2

2

d z12x 2

dx

d z0 at x 1

dx

z is Minimumat x 1

Now, when x = 1, y = x2 + 2 = 3 The required point is (1, 3) (1 Mark)

And distance AP = 2 2

1 3 3 2 5 (1 Mark)

OR

Let 2a cm be the length and b cm be the breadth of the rectangle. Then a cmis the radius of the semi-circle.

By hypothesis, perimeter

p 2a b b a

2b p 2 a

(1 Mark)

Also, A = Area of the window

A

1 2a 2a b

2

1 2a a p 2 a

2

1 2pa 4 a

2d

p 4 ada

(1 Mark)

(1 Mark)

For max or min,


16/21

A

A

d0 p 4 a 0

da

pa

4

2dAlso, 4 0

2da

(2 Marks)

The light will be maximum and for this, radius of the semicircle, isp

a .4

(1 Mark)

Ans.252 2

2 2

x y1

a b --- (1)

x y1

a b --- (2)

(1 Mark)

Equation (i) gives y1 =2

12

x

b

a

Equation (ii) gives y2 = 1

xb

a

(1 Mark)

Area of the smaller region

1 20

ay y dx (1 Mark)

2a ax xb 1 dx b 1 dx2 a0 0a

a ab x2 2a x dx b 1 dx

a a0 0

(1 Mark)


17/21

a a2 2 2 2

b x a x a x x1sin b x

a 2 2 a 2a00

2 2b a a1

sin 1 b aa 2 2a

ab ab 1. ab 2 sq. units.

2 2 2 4

(2 Marks)

Ans.261 x 1 x. 1 x 1 x

Let I dx dx dx1 x 1 x1 x. 1 x

(1 Mark)

= 1 2dx x

dx I I1 x 1 x

( Mark)

121

12 1 1

dxI 1 x dx1 x

2 1 x c 2 1 x c

(1 Mark)

2x

I dx1 x

Let x = sin2 dx = 2 sin cos d (1 Mark)

22

sin .2sin .cosI d 2 sin d

cos

(1 Mark)

2sin2

1 cos2 d sin cos c2

1

2sin x x 1 x c (1 Mark)

I = I1 I21

1

2 1 x sin x x 1 x C

1 x x 2 sin x C

( Mark)

OR

To evaluate: 2

2 x

0

x e dx

Here, f (x) = x + ex, a = 0, b = 2

So, nh = b a = 2 (1 mark)

Now, f (0) = 0 + e0 = 1

f (0+h) = f (h) = h + eh

f (0 + 2h) = f (2h) = 2h + e2h


18/21

f (0 + (n - 1)h) = f ((n 1)h) = (n 1)h + e(n-1)h (1 mark)

Now, 2

h 00

f x dx limh f 0 f 0 h f 0 2h f 0 n 1 h

2 n 1 h2 2 2 2 h 2hh o

nh

22 2 2hh o

nh

3hh o

nhhh o h o h o

limh h 2 h n 1 h 1 e e e

e 1limh h 1 2 n 1 1. 1 mark

e 1

h e 1n n 1 2n 1lim h 1 mark

6 e 1

nh nh h 2nh h hlim lim e 1 lim 1 m6 e 1

2

2 2

ark

2 2 0 4 0e 1 1

6

8 5e 1 e 1 mark

3 3

Ans.27 Given equation of line isx 1 3 y z 1

5 2 4

or

x 1 y 3 z 1

5 2 4

.

Vector equation of given line 2r a b (i 3j k) (5i 2 j 4k)

(1)

(1 Mark)Direction ratios of the line are (5, 2, 4)Vector equation of the line passing through (3, 0, 4) and parallel to the given line

1

r a b (3i 0j 4 k) (5i 2 j 4k) ... (2)

(1 Mark)

Now, distance between the lines (1) and (2) is given by 2 1

b a ad

b(1 Mark)

2 1

i j k

b a a 5 2 4 18i 23j 11k

2 3 3

(1 Mark)

2 1

b a ad

b


19/21

18i 23j 11k(1 Mark)

25 4 16

18i 23j 11k

45

324 529 121 974(1 Mark)

45 45

Ans.28 B1 : the bulb is manufactured by machine XB2 : the bulb is manufactured by machine YB3 : the bulb is manufactured by machine Z

P(B1) = 1000/(1000+2000+3000)= 1/6P (B2) =2000/(1000+2000+3000)= 1/3

P(B3) = 3000/(1000+2000+3000)= 1/21

1 mark2

P(E|B1) = Probability that the bulb drawn is defective given that it ismanufactured by machine A = 1% = 1/ 100Similarly, P(E| B2) = 1.5% = 15/ 1000 = 3/ 200

P(E| B3) =2%= 2/1001

1 mark2

1 11

1 1 2 2 3 3

P(B )P(E B )P(B |E)P(B )P(E B ) P(B )P(E B ) P(B )P(E B )

1 1

6 1001 1 1 3 1 2

6 100 3 200 2 100

1

61 1

16 2

1 1 3 marks1 3 6 10

Ans.29 Let x and y be number of packets of food P and Q respectively.

Linear programming problem is

Minimize Z = 6x + 3y (Vitamin A)

subject to


20/21

12x + 3y 240 or 4x + y 80

4x + 20y 460 or x + 5y 115

6x + 4y 300 or 3x + 2y 150

x 0, y 0 (2 marks)

Graphically the problem can be represented as

(2 marks)


21/21

Coordinates of corner points L, M, N are (2, 72), (15, 20) and (40, 15)

respectively.

We have:

z = 6x + 3y

L (2, 72)

M (15, 20)

N (40, 15)

228

150 Minimum

285

So, minimum value of z is at point (15, 20), i.e., when 15 packets of food P and 20packets of food Q are used. (2 marks)

topper pre board-answersheet_cbse xii_math

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