Class 12th CBSE Mathematics

Question Paper Solution 2012-13 Code No. 65/1

Section A Solution 1:

1 πtan 3

3

1 π 5πcot 3 π

6 6

1 1 π 5π 3π πtan 3 cot 3

3 6 6 2

Solution 2:

1 1 3tan 2 sin 2 cos

2

= 1 πtan 2 sin 2

6 1 3 π

cos2 6

= 1 πtan 2 sin

3

= 1 3tan 2

2

π 3sin

3 2

= 1 πtan 3

3 1 π

tan 33

Solution 3:

TA A For skew symmetric matrix

0 1 2

A 1 0 3

x 3 0

1

0 1 x

A 1 0 3

2 3 0

1A A on comparing, –x = – 2

x 2

Solution 4:

1 1A

1 1

2 1 1 1 1A .

1 1 1 1

= 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

= 2 2

2 2

= 2 1 1

1 1

A2 = 2A

K 2

Solution 5: y = mx,

Differentiating both sides w.r.t m

dym

dx

Solution 6:

2 3 5

6 0 4

1 5 7

32a 5 32A 22 22

32 32a . A 5 22 110

Solution 7:

Since it is external division

2 a b 1 3a 2b 2a 2b 3a 2b

2 1 1

= a 4b

Solution 8: Since a is a unit vector a = 1

S X a . X a 15

22X a 15

2X 1 15

2X 16 X 4

Solution 9:

From a pt (x, y, z), length of perpendicular to plane ax + by + cz + d = 0 is

1 1 1

2 2 2

ax by cz d

a b c

The reqd length = 2 2 2

2 0 3 0 6 0 21

2 2 6 =

213

7 units

Solution 10:

Let marginal revenge = m, Total revenue = R,

dRm

dx

dm

dx (3x2 + 36x + 5) = 6x + 36

At x = 5 m = 6(5) + 36 = 66

Marginal revenue at x = 5 is 66

The question indicates that the firm cares for its workers.

Section B

Solution 11:

For f(x) = x2 + 4 f : R 4, to be invertible, f(x) has to be one-one onto. For one-

one.

f(x1) = f(x2) for 1 2 1 2x x x x R .

2 21 2x 4 x 4

1 2x x

x1 = x2

which is

possible

x1 = – x2

which is not possible as

x1, x2 both have to be (+)

ve

Hence, only x1 = x2 holds true, f(x) is one-one

For Onto: f(x) = x2 + 4

df x 2x

dx which is (+ve) x R i.e. for the domain of f(x)

f(x) is an increasing function x R

Min f(x) at x = 0, Max. f(x) at x =

At x = 0 f(x) = 4

At x = f(x) =

The given function is onto

f(x) is invertible

Let f(x) = y x = f–1 (y)

2x 4 y 2x y 4 .

x y 4 .

1f y y 4 .

Solution 12:

1 1 3cos tan x sin cot

4

1 1

2

1 4cos cos sin tan

31 x

1

2

1 4sin sin

51 x

1

2

1 1

1tan x cos

1 x

cot x tan 1 / x

2

1 4

51 x

2 251 x

16

2 9x

16

3x

4

Solution 13:

Let

x x y x 2y

x 2y x x y

x y x 2y x

1 1 2 3C C C C

x x y x 2y x y x 2y

x 2y x x y x x y

x y x 2y x x 2y x

3 y x y x 2y

3 x y x x y

3 x y x 2y x

1 x y x 2y

3 x y 1 x x y

1 x y x

2 2 1R R R , 3 3 1R R R

1 x y x 2y

3 x y 0 x x y x y x 2y

0 x 2y x y x x 2y

1 x y x 2y

3 x y 0 y y

0 y 2y

Expanding with respect to C1

3 x y y 2y y y

= 3 (x + y) [2y2 + y2]

= 9y2 (x + y) Hence proved.

Solution 14:

x y xy e

Taking log to base ‘e’ on both sides.

x y xe elog y log e

x log y = (y – x) loge e be elog a b log a

x log y y x …(1) alog a 1

Differentiating both sides with respect to x

d

dx (x log y) =

d

dx (y – x)

d

dx . log y + x.

d

dx log y =

dy dx

dx dx

d dv duu.v u , v

dx dx dx

x dy dylog y 1

y dx dx e

d d 1x 1, log x

dx dx x

dy x1 log y 1

dx y …(3)

From (1), we have,

x log y = y – x

log y = y

x – 1.

1 + log y = y

x

x 1

y 1 log y ... (2)

Sub. x

y From (2) in (3),

1 + log y = dy

dx

11

1 log y

1 + log y = dy 1 log y 1

dx 1 log y

1 log y 1 log ydy

dx log y

21 log ydy

dx log y

Hence, Proved

Solution 15:

Let f(x) = x 1 x

1

x

2 .3sin

1 36

x x1

x

2. 2 .3f x sin

1 36

= x

1

2x

2. 6sin

1 6

Put x6 tan θ 1 xθ tan 6

1

2

2 tan θf x sin

1 tan θ

= 1sin sin 2θ 2

2 tanθsin 2θ

1 tan θ

f x 2θ = 1 x2 tan 6

1 xd df x 2 tan 6

dx dx 1 x

e2

d 1 dtan t x x . log a

dx t t dx

= xe2

x

26 log 6

1 6

the reqd. Ans. is x

e2

x

2.6 . log 6

1 6

Solution 16:

For continuity verify at x = 0,

x 0 x 0Lt f x Lt f x f 0

2 0 1f 0 1

0 1

x 0x 0

2 0 12x 1Lt f x Lt 1

x 1 0 1

x 0 x 0

1 kx 1 kx 1 kx 1 kxLt f x Lt

x 1 kx 1 kx

= x 0

1 kx 1 kxLt

x 1 kx 1 kx

= x 0

2k xLt

x 1 kx 1 kx

= 1

2k K1 1

x 0 x 0Lt f x Lt f x f 0

K = –1 = – 1

K 1

Solution 17:

I = cos 2x cos 2αdx

cos x cos α

=

2x 2α 2α 2x2 sin sin

2 2

x α α x2 sin sin

2 2

dx.

cos C cos D

C D D C2 sin sin

2 2

= sin x α sin α x dx

dxα x α x

sin sin2 2

x α x α α x α x2 sin cos 2.sin cos

2 2 2 2dx

α x α xsin sin

2 2

I = x α α x

4 cos cos dx2 2

A A

sin A 2 sin cos2 2

= α x α x

2 2 cos cos dx2 2

2 cos A cos B cos A B cos A B

= α x α x α x α x

2 cos cos dx2 2 2 2

= 2 cos α dx 2 cos x dx dx x, cos x dx sin x

= 2 x cos α 2 sinx c

Solution 18:

Let I = 5

1dx

x x 3

I = 4

5 5

xdx

x x 3

Let x5 + 3 = t , x5 = t – 3

5x4 dx = dt

x4 dx = 1

5 dt

1 dt

It 3 t 5

= 1 3

dt5 3 t 3 t

I = 1 1 1

dt15 t 3 t

1 1 1 1I dt dt

15 t 3 15 t

= 1 1

log t 3 log t C15 15

= 5 51 1log x log x 3 C

15 15

5

5

1 1 1dx log x log x 3 C

3 15x x 3.

Solution 19:

Let 2π

sinx0

1I dx

1 e … (1)

From Properties of Definite integral, we know that

a a

0 0f x dx f a x dx

2π

sin 2π xo

1I dx

1 e sin 2π x sin x

2π

sin x0

1I dx

1 e

2π sin x

sin x 10

1.eI dx

e ...(2)

Adding (1) & (2)

2π sin x

sin x sin x

0

e 1I I dx

1 e 1 e

2π sin x

sin x0

1 e2I dx

1 e

2π

02I dx

2π

02I x

b b

aadx x

2I 2π I n

2π

sin x0

1dx π

1 e

Solution 20:

a i j 7k

b 5i j λk

a b 5 1 1 1 1 j 7 λ k

a b 6i 2j 7 λ k

a b 1 5 i 1 1 j 7 λ k

a b 4i 7 λ k

a b is to a b so, a b a b 0

6i 2j 7 λ k . 4i 7 λ k 0

–24 + 0 + 49 – 2λ =0

25 = 2λ λ 5

Solution 21:

let the Normal to the required plane be a line parallel to another line with direction

ratios proportional to (l, m, n) Since, the given plane with normal of direction ratios

proportional to (1, –2, 4) is to the reqd plane, so

l 2m 4n 0

The reqd. plane passes through A (2, 1, –1) B (–1, 3, 4), so, the vector AB lies on the

plane, AB is to the normal required.

3l 2m 5n 0 … (1)

l 2m 4n 0 ..(2)

Solving (1), (2) simultaneously,

l = m = n

2 5 –3 2

–2 4 1 –2

l m nK

18 17 4 (say)

l = 18k, m = 17k, n = 4k, so, the reqd. normal has dr’s proportional to (18, 17, 4)

So, the vector equation of the plane is

μ . r a . n

μ . 18 i 17 j 4k 2 i j k . 18 i 17 j 4k

μ . 18 i 17 j 4k = 49.

Solution 22:

Let X be event of A reacting school in time, Y be the event of B reaching school in time.

P(X) = 3/7 P(Y) = 5/7

Since x, y are 4 dependent events, so 3 5 15

P X Y P X . P Y7 7 49

P(only one reaching in time) = PC only) + P(y only)

= P X P X Y P Y P X Y

= 3 15 5 15

7 49 7 49

= 8 30

7 49

= 56 30 26

49 49

Section C

Solution 23:

Clearly for max. area, the rectangle has to be symmetrical about both the minor and

major axes of ellipse.

2 2

2 2

x y1

a b

2 2 2

2 2

y a x

b a

2 2by a x

a

Area of the rectangle = 2x. 2y = 4xy =

2 2b4x. a x

a

2 2 44ba x x

a

2

2 2 4aa x x k

4b(say)

clearly, since a, 4b are constants, is max when k is maximum.

2

2

dk d k0, 0

dx dx

2 3dk2a x 4x 0

dx

x = 0 2a2 – 4x2 = 0

2 2x a /2

ax

2

22 2

2

d k2a 12x

dx

At x = 0, 2

2

2

d k2a 0

dx

At 2 2

2 2

2

a d k ax 2a 12 4a 0

dx 22

2 22 2

2

a d k ax 2a 12 4a 0

dx 22

Since x has be (+) ve, so that area 4xy is (+) ve

ax

2

2 2b by a x y

a 2

Area = 4xy = a b

4 2ab2 2

Solution 24:

x x 0y x

x x 0

for x < 0, y = x2, y = x

x = x2 x = 1, 0

for x < 0, y = x2, y = x

–x = x2, x = –1, 0

Since the area is exactly symmetrical about y axis, so Area reqd = OACO + OBDO,

= 2 (OACO)

Reqd Area = 2 A A

O O

line dx Parabola dx

= 1 1

2

0 0

2 x dx x dx

=

1 12 3

0 0

x x2

2 3

2 32x x

xdx , x dx2 3

Reqd Area = 1 1

22 3

= 1

26

1Area sq. units

3

Solution 25:

1 2tan y x dy 1 y dx

1 2 dxtan y x 1 y

dy

2 1dx1 y x tan y

dy

1

2 2

dx 1 1x tan y

dy 1 y y 1

On comparing with

dxRx S

dy

2

1R

1 y

1

2

tan yS

y 1

Integrated Factor = Rdy

e

x I. F S IF dy

I.F = 12

1dy

tan1 ye e y 1

2

1dy tan y

1 y

1 11

tan y tan

2

tan yx e . e dy

y 1

Put 1tan y t

2

1

1 y dy = dt

tt

I IIx. e t . e dt

Using Integration by parts

dv duu.vdx u vdx dx

dx dx

tt td.e dt

x.e t. e dt dtdt dt

t t tx.e t .e e dt

t t tx . e t e e

x t 1

1x tan y 1

tan x 1 y

Solution 26: The reqd. plane passes through (1, 2, 3) and the intersection of the plane

1 2μ . i j 2k 5 0, P , μ . 3i j k 60 P

So, the reqd. plane, P

O = P = P1 + 2λ P

O = P = μ . i j 2k 5 λ μ . 3i j k 6

P = μ i j 2k 3λi λj λk 5 6λ 0

P μ 1 3λ i λ 1 j λ 2 k 5 6λ 0

since it passes through (1, 2, 3)

μ 1 2j 3k

0 1 2j 3k . 1 3λ 1 λ 1 j λ 2 k 5 6λ

0 1 3λ 2λ 2 3λ 6 5 6λ

2λ 0 λ 0

The reqd plane is

0 μ 1 3 0 i 0 1 j 0 2 k 5 6 0

0 μ i j 2k 5

The reqd plane is μ i j 2k 5

Solution 27:

Let x be the event of A winning, Y be the even of B winning , Z be the event of getting a 6

on the die

P(z) = 1/6

1 5 5 1P x

6 6 6 6………………

=

21 25 25

1 ...6 36 36

= 1 1 36 6

P x256 11 6 11136

This can be explained by that x can happen i.e. A can win when he gets 6 on his first

attempt or when he not get 6 on his first attempt and B also not get 6 on first attempt

and then A gets 6 i.e. on its second attempt and so on and so forth similarly,

5 1 5 5 5 1P 4

6 6 6 6 6 6 + --------------

=

25 25 25

136 36 36

= 5 1 5 36 5

2536 36 11 11136

A (a winning) = 6/11

p (B winning) = 5/11

Since, Probability of A winnings greater than probability of B winning, the referee had

not been fair.

Solution 29:

x + y + z = 12

3z + 3y + 2x = 33

x + z = 2y

x + y + z = 12 … (1)

2x + 3y + 3z = 33 …(2)

x – 2y + z = 0 …(3)

1 1 1 x 12

A 2 3 3 X y B 33

1 2 1 z 0

1 1 1

A 2 3 3

1 2 1

2 2 1 3 3 1C C C , C C C

1 0 0

A 2 1 1 A 3 0

1 3 0

A is invertible

Now,

adj A =

T9 3 7

3 0 3

0 1 1

adj A =

9 3 0

1 0 1

7 3 1

1 1A

A adj A

1

9 3 01

A 1 0 13

7 3 1

AX = B

Pre multiply by 1A

1A . AX AB 1 1A A A.A 1

1X A B

x 9 3 0 121

y 1 0 1 333

z 7 3 1 0

x 9 12 3 331

y 1 12 0 03

z 7 12 3 33

x 9 31

y 12 43

z 15 5

x = 3

y = 4

z = 5

Checking

x + y + z = 12 2x + 3y + 3z = 33

3 + 4 + 5 = 12 2(3) + 3(4) + 3(5) = 33

x – 2y + z = 0

3 – 2(4) + 5 = 0

Hence x = 3, y = 4, z = 5

the other quality can be discipline.