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CBSE XII Mathematics 2012 Solution (SET 2)

Mathematics

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 four marks

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 exact

requirement of the question.

(iv) There is no overall choice. However, internal choice has been provided in 4 questions of four

marks each and 2 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.

Section B

Q11. If , ,a b c

are three vectors such that 5, 12,and 13,and 0,a b c a b c

Find the value of

. . .a b b c c a

.

Ans. , ,a b c

are three vectors such that 5, 12 and 13.a b c

22 2 2

0

0 0 0

0

0

0 a

a b c

a b c a b c

a a b c b a b c c a b c

a a a b a c b a b b b c c a c b c c

a a b c a a b b b c c a b c c x x x

22 2

2 2 2

nd

2 . 0

5 12 13 2 . 0

25 144 169 2 . 0

338 2 . 0

2 . 338

338. 169

2

Thus, the v

x y y x

a b c a b b c c a

a b b c c a

a b b c c a

a b b c c a

a b b c c a

a b b c c a

alue of . –169.a b b c c a is


Q12. Solve the following differential equation:

2 22 2 0dy

x xy ydx

Ans.

2 2

2 2

2

2

2 2 0

2 2

2... i

2

dyx xy y

dx

dyx xy y

dx

dy xy y

dx x

This is a homogeneous equation of degree 2.

On taking y = vx, we obtain

... iidy dv

v xdx dx

From equations (ii) and (i), we obtain 2

2

2 2 2

2

2

2

2

2 ( ) ( )( )

2

2

2

2

2

2

dv x vx vxv x y vx

dx x

dv vx v xv x

dx x

dv vv x v

dx

dv vx

dx

dv dx

v x

On integrating both sides, we obtain

2

1

2

1 1 log C

2

1 1 log C

2

1 log C

2

1 log C = 0

2

dv dx

v x

xv

xy

x

xx

y

xx

y

Thus, 1

log C = 02

xx

y is the required solution.


Q13. How many times must a man toss a fair coin, so that the probability of having at least one head

is more than 80%?

Ans. Let p denote the probability that the man gets head.

Thus, 1

2p and

1 11

2 2q .

Let us assume that the man tosses the coin n times.

Let X denote the number of times for which the man gets head in n-trails.

P X = C ( 0,1,2,..., )

1 1 1 1C and

2 2 2 2

1C

2

n r n r

r

r n r

n

r

n

n

r

r p q r n

p q

Now,

0 0

0

0

80P X 1 (Given)

100

8P X 1 P X = 2 ... P X

10

8P X = 0 P X = 1 P X = 2 ... P X = P X = 0

10

81 P X = 0 Sum of all probabilities is 1

10

8P X = 0 1

10

2P X = 0

10

1P X = 0

5

1 1 1

2 2 5

1

n

n

n

n

n

C

C

1

2 5

1 11

2 5

1 1

2 5

n

n

n


It is known that 1 1

2 5

n

is true only when n ≥ 3.

Hence, the man must toss the coin at least 3 times.

Q14. If cos cos ,y x

x y finddy

dx.

Ans. The given function is cos cosy x

x y

Taking logarithm on both the sides, we obtain

logcos logcosy x x y

Differentiating both sides, we obtain

log cos log cos log cos log cos

1 1log cos cos log cos 1 cos

cos cos

log cos sin log cos sincos cos

log cos tan log cos tan

log cos t

dy d d dx y x y x x y

dx dx dx dx

dy d dx y x y x y

dx x dx y dx

dy y x dyx x y y

dx x y dx

dy dyx y x y x y

dx dx

x x

an tan log cos

tan log cos

tan log cos

dyy y x y

dx

dy y x y

dx x y x

OR

If sin y = x sin(a + y), prove that 2sin

sin

a ydy

dx a

Ans.

Given, sin sin

Differentiating both sides with respect to , we have

sin sin

cos sin sin

cos sin 1 cos

cos cos sin ... 1

y x a y

x

d dy x a y

dx dx

dy d dy a y x x a y

dx dx dx

dy dyy a y x a y

dx dx

dyy x a y a y

dx


sin siny x a y

sin

sin

yx

a y

… (2)

From (1) and (2), we have

2

2

2

2

sincos cos sin

sin

sin cos sin cos sin

sin sin

sin sin

sin

sin

dy yy a y a y

dx a y

dya y y y a y a y

dx

dya y y a y

dx

dya a y

dx

a ydy

dx a

Q15. Let A = R – {3}and B = R – {1}. Consider the function f : A B defined by 2

3

xf x

x

.

Show that is one-one and onto and hence find f–1

.

Ans. Given, A = R – {3} and B = R – {1}.

: A Bf is a function defined by 2

3

xf x

x

.

Let 1 2, Ax x .

Now,

1 2

1 2

1 2

1 2 1 2

1 2 2 1 1 2 2 1

1 2

2 2

3 3

2 3 3 2

2 3 6 3 2 6

f x f x

x x

x x

x x x x

x x x x x x x x

x x

f is one-one.

Again, let y be any arbitrary element of B.

2

3

f x y

xy

x


2 3

2 3

2 3

1 2 3

2 3, which is a real number for all 1.

1

x y x

x xy y

x xy y

x y y

yx y

y

Also, 2 3

31

y

y

for any y, because if we take

2 33

1

y

y

2 3 3 3

2 3, which is not possible.

y y

x R – {3} for all y R –{1}.

Thus, for all y R – {1} there exists 2 3

31

yx

y

R such that

2 32

2 3 2 3 2 21

2 31 2 3 3 33

1

y

y y yyf x f y

yy y y

y

Every element y in B has a pre-image x in A which is given by 2 3

1

yx

y

.

f is onto.

Hence, f is one-one and onto.

To find f–1

:

Let f(x) = y where x R – {3} and y R – {1}.

1

2

3

2 3

1

2 3

1

xy

x

yx

y

yf y

y

1 : 1 3f R R is defined by 1 2 3

1

xf x

x

for all x R – {1}.

Q16. 1 cosProve that tan , , .

1 sin 4 2 2 2

x xx

x


Ans.

1

2 2

1 1

2 2

1 1

cosWe have to prove that tan , , .

1 sin 4 2 2 2

cos sincos 2 2tan tan

1 sinsin cos 2sin cos

2 2 2 2

cos sin cos sincos 2 2 2 2

tan tan1 sin

x xx

x

x xx

x x x xx

x x x x

x

x

2

1 1

1 1

1

sin cos2 2

cos sincos 2 2tan tan

1 sincos sin

2 2

cos sin2 2

coscos 2tan tan

1 sincos sin

2 2

cos2

costan

1 sin

x x

x xx

x xx

x x

xx

x xx

x

x

x

1

1 1

1 1

1

1 tan2tan

1 tan2

tan tancos 4 2tan tan

1 sin1 tan tan

4 2

costan tan tan

1 sin 4 2

costan

1 sin 4 2

x

x

xx

xx

x x

x

x x

x

Hence, the result is proved.


OR

1 1 18 3 36Prove that sin sin cos .

17 5 85

Ans.

1 1 1

1 1

2

2

2

8 3 36We have to prove that sin sin cos .

17 5 85

8 3Let sin and sin

17 5

8 3sin and sin

17 5

cos 1 sin

8cos 1

17

15cos

17

Similarly,

cos 1 sin

3cos 1

x y

x y

x x

x

x

y y

y

2

5

4cos

5y

We know that,

cos (x + y) = cos x cos y – sin x sin y

1

1 1 1

15 4 8 3cos

17 5 17 5

60 24cos

85

36cos

85

36cos

85

8 3 36sin sin cos

17 5 85

x y

x y

x y

x y

Hence, the result is proved.


Q17. Find the point on the curve 3 11 5y x x at which the equation of tangent is y = x – 11.

Ans. The equation of the given curve is 3 11 5y x x .

Let P 1, 1x y be the point on the given curve.

Since 1, 1x y lies on 3 11 5y x x ,

3

1 1 111 5y x x … (1)

3 11 5y x x

Differentiating both sides with respect to x, we have

1 1

2

2

1

,

3 11

3 11x y

dyx

dx

dyx

dx

Given, the line y = x – 11 is tangent to the given curve.

Slope of the tangent to the curve at 1 1,x y = Slope of the line y = x – 11

1 1,

2

1

2

1

2

1

1

1

3 11 1

3 12

4

2

x y

dy

dx

x

x

x

x

When x1 = 2, we have

3

1

1

3

2 11 2 5 9 Using 1

When 2, we have

2 11 2 5 19 Using 1

y

x

y

So, the points are (2, –9) and (–2, 19).

Now, (2, –9) lies on y = x – 11 if –9 = 2 –11 which is true.

And, (–2, 19) lies on y = x – 11 if 19 = –2 –11 which is not true.

Thus, the required point is (2, –9).

OR

Using differentials, find the approximated value of 49.5 .


Ans.

Let

For 49,

49.5

49.5 49

0.5

y f x x

x

x x

x

x

For x = 49, we have 49 7y x .

Let dx = x = 0.5

49

49

Now,

1

2

1

2 49

1

14

We have

10.5 (For 49)

14

1

28

1

28

49.5

149.5 7

28

49.5 7.036 (approximately)

x

x

y x

dy

dx x

dy

dx

dy

dx

dydy dx

dx

dy x

dy

y y dy

y y

Thus, the approximate value of 49.5 is 7.036.

Q18. Evaluate: sin sin 2 sin3x x x dx


Ans.

I sin sin 2 sin 3

1sin 2sin 2 sin 3

2

1sin 2sin 3 sin 2

2

1sin cos 3 2 cos 3 2

2

1sin cos sin cos5

2

1 2sin cos 1 2sin cos5

2 2 2 2

1 1sin 2 2cos5 sin

4 4

1 cos 2 1sin 5

4 2 4

x x x dx

x x x dx

x x x dx

x x x x x dx

x x x x dx

x x x xdx dx

x dx x x dx

xx

sin 5

cos 2 1sin 6 sin 4

8 4

cos 2 1 cos 6 cos 4C

8 4 6 4

cos 2 cos 6 cos 4= C

8 24 16

12cos 6 3cos 4 6cos 2 C

48

x x x dx

xx x dx

x x x

x x x

x x x

OR

Evaluate: 2

2

1 1dx

x x

Ans.

2

2 2

2I =

1 1

2 A B CLet

11 1 1

dxx x

x

xx x x

Multiplying both sides by (1 – x) (1 + x2), we get

2 = A (1 + x2) + (Bx + C) (1– x)

2 = A (1 + x2) + B (x - x

2) + C (1– x)

2 = x2

(A – B) + x (B – C) + (A + C) … (1)

Comparing the coefficients of x2 , x and constant term on both sides of (1), we get

0 = A – B A = B … (2)


0 = B – C B = C … (3)

A + C = 2 A + A = 2 2A = 2 A = 1

So, B = A = 1 and C = B = 1

A = 1, B = 1, C = 1

22

2

2

2

2 2 2

2 2 2

2 1

2

2

2 1 1

1 11 1

2So, I =

1 1

1 1

1 1

1 1

1 1

log 1

1 1 1

1 2log 1

2 1 1

1log 1 log 1 tan C

2

Hence,

2 1log 1 log 1

21 1

x

x xx x

dxx x

xdx

x x

xdx dx

x x

x x dxdx

x x

x dxx dx

x x

x x x

dx x xx x

1tan Cx

Q19. Using properties of determinants, prove the following:

3 3 3

1 1 1

( ) ( ) ( )a b c a b b c c a a b c

a b c

Ans.

Let

3 3 3

1 1 1

a b c

a b c

.

Applying C1 C1 – C3 and C2 C2 – C3, we have:


3 3 3 3 3

2 2 2 2 3

2 2 2 2 3

0 0 1

0 0 1

0 0 1

1 1

a c b c c

a c b c c

a c b c c

a c a ac c b c b bc c c

c a b c c

a ac c b bc c c

Applying C1 C1 + C2, we have:

2 2 2 2 3

2 2 3

2 2 3

0 0 1

0 1

0 0 1

0 1

0 0 1

0 1

1

c a b c c

b a bc ac b bc c c

b c c a a b c

a b c b bc c c

a b b c c a a b c c

b bc c c

Expanding along C1, we have:

0 11

1a b b c c a a b c

c

a b b c c a a b c

Hence, the given result is proved.

Q20. If y = 3 cos (log x) + 4 sin(log x), show that

2

2

20

d y dyx x y

dx dx

Ans. y = 3 cos (log x) + 4 sin (log x) ... (1)


2

2

2

2

2

2

22

2

3sin log 4cos log

3sin log 4cos log

3sin log 4cos log

3cos log 4sin log

3cos log 4sin log

3cos log 4sin log

3cos lo

x xdy

dx x x

x xdy

dx x

dyx x x

dx

x xd y dyx

dx dx x x

x xd y dyx

dx dx x x

x xd y dyx

dx dx x

d y dyx x

dx dx

22

2

22

2

g 4sin log

3cos log 4sin log 0

0 3cos log 4sin log

x x

d y dyx x x x

dx dx

d y dyx x y y x x

dx dx

Q21. Find the equation of the line passing through the point (– 1, 3, – 2) and perpendicular to the

lines

2 1 1

and1 2 3 3 2 5

x y z x y z

Ans. Let the direction ratios of the required line be <a, b, c>.

The equations of the given lines are:

... 11 2 3

2 1 1and ... 2

3 2 5

x y z

x y z

Direction ratios of line (1) and (2) are <1, 2, 3> and <–3, 2, 5> respectively.

Since the required line is perpendicular to line (1) and line (2),

1 2 1 2 1 21 2 3 0 0

2 3 0 ... 3

and 3 2 5 0

3 2 5 0 ... 4

a b c a a b b c c

a b c

a b c

a b c

Solving (3) and (4), we have


2 3 0

3 2 5 0

10 6 9 5 2 6

4 , 14 , 8

a b c

a b c

a b ck

a k b k c k

The direction ratios of the required line are < 4k , –14k , 8k > or < 2, –7, 4>.

Equation of line passing through (–1, 3, –2) and having direction ratio’s < 2, –7, 4 > is

1 1 11 3 2

2 7 4

x x y y z zx y z

a b c

Thus, the equation of the required line is1 3 2

2 7 4

x y z

.

Q22. Find the particular solution of the following differential equation:

1 2 1; 0ydyx e y

dx

when x = 0

Ans.

1 2 1

2 1 1

Integrating both thesides,

2 1 1

2 1

y

y

y

y

y

dyx e

dx

dy dx

e x

dy dx

e x

e dy dx

e x

Let 2 – ey = t

– ey dy = dt

ey dy = –dt

1

dt dx

t x

– log t = log (x + 1) + c

– log (2 – ey) = log (x + 1) + c ........(1)

When x = 0, y = 0

Putting y = x = 0 in equation (1), we get c = 0

log (x + 1) + log (2 – ey) = 0

(x + 1) (2 – ey) = 1

So, (x + 1) (2 – ey) = 1 is the particular solution of 1 2 1ydy

x edx

.

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