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CBSE XII Mathematics 2012 Solution (SET 2)

Mathematics General Instructions: (i) (ii)

(iii) (iv)

(v)

All questions are compulsory. 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. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 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. Use of calculators is not permitted.

Section B           Q11. If a , b , c are three vectors such that a  5, b 12,and c 13,and a  b  c  0, Find the value of       a .b  b . c  c . a . Ans.

      a  5, b  12 and c  13. a, b , c are three vectors such that     a b c  0          a  b  c  a  b  c  0 0  0              a  a b c b  a b c c  a b c  0                    a a  a b  a c  b  a  b b  b c  c  a  c b  c c  0 2       2       2   2      a  a b  c  a  a b  b  b c  c  a  b c  c  0  x  x  x and x  y  y  x



















2 2 2       a  b  c  2 a b  b c  c .a  0 2 2 2        5  12   13  2 a b  b c  c .a  0       25  144  169  2 a b  b c  c .a  0       338  2 a b  b c  c .a  0       2 a b  b c  c .a  338





















      338  a b  b c  c .a   169 2      Thus, the value of a b  b c  c .a is –169.





CBSE XII Mathematics 2012 Solution (SET 2) Q12. Solve the following differential equation: dy 2 x2  2 xy  y 2  0 dx Ans.

dy  2 xy  y 2  0 dx dy  2 x2  2 xy  y 2 dx dy 2 xy  y 2   ...  i  dx 2x2 This is a homogeneous equation of degree 2. On taking y = vx, we obtain dy dv vx ...  ii  dx dx From equations (ii) and (i), we obtain dv 2 x(vx)  (vx) 2 vx  ( y  vx) dx 2x2 dv 2vx 2  v 2 x 2 vx  dx 2x2 dv v2 vx v dx 2 2 dv v x  dx 2 dv dx  2  v 2x On integrating both sides, we obtain dv 1 dx  v2  2  x 2 x2

1 1  log x  C v 2 1 1   log x  C y 2 x    x 1   log x  C y 2 

x 1  log x  C = 0 y 2 x 1 Thus,  log x  C = 0 is the required solution. y 2 

CBSE XII Mathematics 2012 Solution (SET 2) 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. 1 1 1 Thus, p  and q  1   . 2 2 2 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 = r   n Cr p r q n r

(r  0,1, 2,..., n) n r

r

1   n Cr   2 

1  2   

1 1   p  2 and q  2   

n

1   Cr   2  n

Now, P  X  1 

80 100

(Given)

 P  X  1  P  X = 2   ...  P  X  n  

8 10

 P  X = 0   P  X = 1  P  X = 2   ...  P  X = n    P  X = 0  

 Sum of all probabilities is 1

8 10 8  P  X = 0  1  10 2  P  X = 0  10 1  P  X = 0  5  1  P  X = 0 

0

n 0

1  1   n C0     2  2  n

1  1  C0    2  5 n

n

1  1  1   2  5 n

1  1    2  5



1 5

8 10

CBSE XII Mathematics 2012 Solution (SET 2) n

1 1 It is known that    is true only when n ≥ 3. 2 5 Hence, the man must toss the coin at least 3 times.

Q14. If  cos x    cos y  , find y

Ans.

x

dy . dx

The given function is  cos x    cos y  Taking logarithm on both the sides, we obtain y log cos x  x log cos y Differentiating both sides, we obtain dy d d d log cos x   y   log cos x   log cos y   x   x   log cos y  dx dx dx dx dy 1 d 1 d  log cos x  y    cos x   log cos y 1  x    cos y  dx cos x dx cos y dx dy y x dy  log cos x     sin x   log cos y    sin y   dx cos x cos y dx dy dy  log cos x  y tan x  log cos y  x tan y dx dx dy   log cos x  x tan y   y tan x  log cos y dx dy y tan x  log cos y   dx x tan y  log cos x OR y

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

x

2 dy sin  a  y   dx sin a

Given, sin y  x sin  a  y  Differentiating both sides with respect to x, we have





d d sin y  x sin  a  y   dx dx dy d d  cos y  sin  a  y    x   x  sin  a  y   dx dx dx dy dy  cos y  sin  a  y  1  x  cos  a  y  dx dx dy cos y  x cos  a  y    sin  a  y   ... 1 dx 

CBSE XII Mathematics 2012 Solution (SET 2) sin y  x sin  a  y 

sin y … (2) sin  a  y  From (1) and (2), we have  dy  sin y cos  a  y    sin  a  y  cos y  dx  sin  a  y   dy  sin  a  y  cos y  sin y cos  a  y    sin 2  a  y  dx dy  sin  a  y  y    sin 2  a  y  dx dy  sin a  sin 2  a  y  dx 2 dy sin  a  y    dx sin a x

 x2  Q15. Let A = R – {3}and B = R – {1}. Consider the function f : A  B defined by f  x    .  x 3  Show that is one-one and onto and hence find f–1. Ans.

Given, A = R – {3} and B = R – {1}. f : A  B is a function defined by f  x  

Let x1 , x2  A . Now, f  x1   f  x2  

x1  2 x2  2  x1  3 x2  3

  x1  2   x2  3   x1  3  x2  2   x1 x2  2 x2  3x1  6  x1 x2  3x2  2 x1  6  x1  x2  f is one-one. Again, let y be any arbitrary element of B. f  x  y 

x2 y x 3

x2 . x 3

CBSE XII Mathematics 2012 Solution (SET 2)   x  2   y  x  3  x  2  xy  3 y  x  xy  2  3 y  x 1  y   2  3 y 2  3y , which is a real number for all y  1. 1 y 2  3y 2  3y Also,  3 for any y, because if we take 3 1 y 1 y  2  3y  3  3y x

 2  3, which is not possible.  x R – {3} for all y R –{1}.

Thus, for all y R – {1} there exists x 

2  3y  R  3 such that 1 y

 2  3y  2   2  3y   2  3y  2  2 y  1 y  f  x  f     2  3y  y 2  3 y  3  3 y   1 y      1 y  3   

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

2  3y . 1 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}. x2 y  x 3 2  3y x 1 y 2  3y  f 1  y   1 y

 f 1 : R  1  R  3 is defined by f 1  x  

Q16.

2  3x for all x  R – {1}. 1 x

 cos x   x    Prove that tan 1    , x   , .  1  sin x  4 2  2 2

CBSE XII Mathematics 2012 Solution (SET 2) Ans.  cos x   x    We have to prove that tan 1     , x   , .  2 2 1  sin x  4 2 x x   cos 2  sin 2     cos x 1 2 2 tan 1     tan  1  sin x   sin 2 x  cos 2 x  2sin x cos x   2 2 2 2  x x  x x    cos  sin cos  sin    cos x  2 2  2 2  1    tan 1     tan  2 x 1  sin x   x   sin 2  cos 2      x x  cos  sin   cos x  1 2 2  tan 1     tan  1  sin x   cos x  sin x   2 2 x x   cos 2  sin 2    x   cos  cos x   1  2  tan 1   tan    x x 1  sin x   cos  sin  2 2  x   cos   2 x  1  tan     cos x 1 2  tan 1     tan  1  sin x    1  tan x   2  x   tan  tan  1  cos x  1  4 2  tan     tan   1  sin x    1  tan tan x   4 2  cos x    x  1   tan 1    tan  tan      4 2  1  sin x   cos x   x  tan 1    1  sin x  4 2 Hence, the result is proved.

CBSE XII Mathematics 2012 Solution (SET 2) OR 8 3   36  Prove that sin 1    sin 1    cos 1  . 17  5   85 

Ans.

8 3   36  We have to prove that sin 1    sin 1    cos 1  . 17  5   85  8 3  Let sin 1    x and sin 1    y 17  5  8 3  sin x  and sin y  17 5 cos x  1  sin 2 x 2

8  cos x  1    17  15  cos x  17 Similarly, cos y  1  sin 2 y 2

3   cos y  1    5  4  cos y  5 We know that, cos (x + y) = cos x cos y – sin x sin y 15 4 8 3  cos  x  y      17 5 17 5 60  24  cos  x  y   85 36  cos  x  y   85  36   x  y  cos 1    85  8 3   36   sin 1    sin 1    cos 1   17  5   85  Hence, the result is proved.

CBSE XII Mathematics 2012 Solution (SET 2) Q17. Find the point on the curve y  x3 11x  5 at which the equation of tangent is y = x – 11. Ans.

The equation of the given curve is y  x3  11x  5 . Let P  x1, y1  be the point on the given curve. Since  x1, y1  lies on y  x3  11x  5 ,  y1  x13  11x1  5

… (1)

y  x  11x  5 Differentiating both sides with respect to x, we have dy  3x 2  11 dx  dy     3x12  11  dx  x1 , y1  Given, the line y = x – 11 is tangent to the given curve. Slope of the tangent to the curve at  x1 , y1  = Slope of the line y = x – 11 3

 dy    1  dx  x1 , y1   3x12  11  1  3x12  12  x12  4  x1   2 When x1 = 2, we have

y1   2   11  2   5  9 3

Using 1 

When x1  2, we have y   2   11  2   5  19 3

Using 1 

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 .

CBSE XII Mathematics 2012 Solution (SET 2) Ans.

Let y  f  x   x For x  49, x  x  49.5  x  49.5  49  x  0.5 For x = 49, we have y  x  49  7 . Let dx = x = 0.5 Now, y x dy 1  dx 2 x 1  dy      dx x  49 2 49 

1  dy      dx x  49 14 We have dy 

dy dx dx

1  0.5 14 1  dy  28 1  y  28  dy 

(For x  49)

 y  dy 

 49.5  y  y  49.5  7 

1 28

 49.5  7.036 (approximately)

Thus, the approximate value of Q18. Evaluate:  sin x sin 2 x sin 3x dx

49.5 is 7.036.

CBSE XII Mathematics 2012 Solution (SET 2) Ans. I    sin x sin 2 x sin 3 x  dx 1 sin x  2sin 2 x sin 3 x  dx 2 1   sin x  2sin 3 x sin 2 x  dx 2 1   sin x cos  3 x  2 x   cos  3 x  2 x   dx 2 1   sin x cos x  sin x cos 5 x  dx 2 1 2sin x cos x 1 2sin x cos 5 x   dx   dx 2 2 2 2 1 1   sin 2 x dx   2 cos 5 x sin x dx 4 4 1   cos 2 x  1      sin  5 x  x   sin  5 x  x   dx 4 2  4  cos 2 x 1     sin 6 x  sin 4 x  dx 8 4  cos 2 x 1   cos 6 x cos 4 x      C 8 4 6 4   cos 2 x cos 6 x cos 4 x =   C 8 24 16 1   2 cos 6 x  3cos 4 x  6 cos 2 x   C 48 OR 

Evaluate:

2

 1 x  1 x  dx 2

Ans. I= Let

2

 1  x  1  x dx 2

2 A Bx  C   2 1  x  1  x  1  x  1  x 2 

Multiplying both sides by (1 – x) (1 + x2), we get 2 = A (1 + x2) + (Bx + C) (1– x)  2 = A (1 + x2) + B (x - x2) + 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)

CBSE XII Mathematics 2012 Solution (SET 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 2 1 x 1    2 1  x  1  x 1  x 1  x 2



So, I =



2

 1  x  1  x  dx 2

x 1   1    dx 2  1  x 1  x  1 x 1  dx   dx 1 x 1  x2 log 1  x x dx   dx   2 2 1 1 x 1  x2 1 2x dx   log 1  x   dx   2 2 2 1 x 1  x2 1   log 1  x  log 1  x 2  tan 1 x  C 2 Hence, 2

1

 1  x  1  x  dx   log 1  x  2 log 1  x 2

2

 tan 1 x  C

Q19. Using properties of determinants, prove the following: 1 1 1

a

b

3

3

a

b

c   a  b  (b  c) (c  a) (a  b  c) c3

Ans. 1

1

1

Let   a

b

c .

b3

c3

a3

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

CBSE XII Mathematics 2012 Solution (SET 2) 0   ac

0

1

bc

c

a 3  c 3 b3  c 3 c 3 0  ac

0

1

bc

c

 a  c   a 2  ac  c 2   b  c   b 2  bc  c 2  0   c  a  b  c  1



 a 2  ac  c 2



c3

0

1

1

c

b

2

 bc  c 2



c3

Applying C1  C1 + C2, we have: 0

0

1

   c  a  b  c  0

1

c

b

2



 a 2   bc  ac 

b

2

 bc  c 2

c3

0

0

1

  b  c  c  a  a  b  0

1

c

 a  b  c

b

2

 bc  c 2



c3

0

0

1

  a  b  b  c  c  a  a  b  c  0

1

c

b

1

Expanding along C1, we have:    a  b  b  c  c  a  a  b  c  1

0

1

1

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 dy 2 d y x  x  y 0 2 dx dx Ans.



y = 3 cos (log x) + 4 sin (log x)

... (1)

2

 bc  c 2



c3

CBSE XII Mathematics 2012 Solution (SET 2) dy 3sin  log x  4 cos  log x    dx x x dy 3sin  log x   4 cos  log x    dx x dy x  3sin  log x   4 cos  log x  dx d 2 y dy 3cos  log x  4sin  log x  x 2    dx dx x x 2 d y dy 3cos  log x  4sin  log x  x 2    dx dx x x 2 d y dy 3cos  log x   4sin  log x  x 2   dx dx x 2 d y dy  x2 2  x  3cos  log x   4sin  log x  dx dx d2y dy  x 2 2  x  3cos  log x   4sin  log x   0 dx dx 2 d y dy  y  3cos  log x   4sin  log x    x2 2  x  y  0 dx dx 

Q21. Find the equation of the line ing through the point (– 1, 3, – 2) and perpendicular to the lines x y z x  2 y 1 z  1   and   1 2 3 3 2 5 Ans.

Let the direction ratios of the required line be . The equations of the given lines are: x y z   ... 1 1 2 3 x  2 y 1 z 1 and   ...  2  3 2 5 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  a  2  b  3  c  0  a1a2  b1b2  c1c2  0

 a  2b  3c  0

...  3

and  3  a  2  b  5  c  0  3a  2b  5c  0 ...  4  Solving (3) and (4), we have

CBSE XII Mathematics 2012 Solution (SET 2) a  2b  3c  0

3a  2b  5c  0 a b c   k 10  6 9  5 2  6  a  4k , b  14k , c  8k The direction ratios of the required line are < 4k , –14k , 8k > or < 2, –7, 4>. Equation of line ing through (–1, 3, –2) and having direction ratio’s < 2, –7, 4 > is x 1 y  3 z  2  x  x1 y  y1 z  z1     a  b  c  2 7 4   x 1 y  3 z  2 Thus, the equation of the required line is .   2 7 4

Q22. Find the particular solution of the following differential equation: dy  x 1  2e y 1; y  0 when x = 0 dx Ans.

dy  2e  y  1 dx dy dx  y  2e  1  x  1

 x  1

Integrating both the sides, dx 2e  1 x 1 y e dy dx   y 2e x 1 Let 2 – ey = t  – ey dy = dt  ey dy = –dt dt dx   t x 1  – 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 

dy

y



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

dy  2e y  1 . dx