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Maximum Time: 90 Minutes [Each right answer carries 4 marks and wrong –1] Maximum Marks: 180
1. The set of values of a for which the function f(x) = (4a – 3) (x + ln 5) + 2(a – 7) cot 2x x sin
2 2
possess critical points is
(a) (1, ) (b) [1, ) (c) ( , 2) (d) ( , 4/3) (2, ) Ans. (d)
Solution:
= (4a 3)

4a 3 4a 3

2. Let f(x) =
2 2
2 2

then the value of f(x) at x = (2n + 1) π , n I (the set of
integers) is equal to
Ans. (d)
Solution: 2
2 2
2 2

cos3x 1 2cos 3x

Then,
tan3x 4 1

= (–1) (–3 – 6) = 9 3. If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to
(a) 2
then f(x) = x + tan x
y = f–1(y) + tan (f–1 (y))
y = g(y) + tan (g(y)) or x = g(x) + tan (g(x)) …(i)
Differentiating both sides, then we get
1 = g’(x) + sec2 g(x). g’(x)
2 2

= 2
1
2 (g(x) x)
4. If 2n 2n n n(1 x ) (1 y ) a(x y ), then 2n
2n
then, (cos θ cos ) a (sin θ sin )
θ θ

2

or 1 n 1 n 1sin x sin y 2cot a
Differentiating both sides, we have
n 1 n 1

5. If 10 10x sec θ cos θ, y sec θ cos θ and 2
2 2dy (x 4) k(y 4),
dx
(a) 1
Ans. (d)
Solution:
2 2 2x 4 (sec θ cos θ) 4 (sec θ cos θ) …..(i)
Similarly, 2 10 10 2y 4 (sec θ cos θ) …(ii)
Now, dx

and 9 9dy 10 sec θ sec θtan θ 10 cos θ( sin θ)
dθ
= 10 1010 tan θ(sec θ cos θ)
dxdx tan θ(sec θ cos θ)
dθ
2 2
dx (sec θ cos θ) (x 4)

dx
dy dx dx
(a) 0 (b) 1 (c) 2 (d) none of these
Ans. (a)
dx dx dx dx dx dy

dx dy dy dy dy

dx dy dx dx dy

dx dx dy

7. If y = (1 + x) (1 + x2) (1 + x4)….. (1 + n2x ), then
dy
(a) 0 (b) – 1 (c) 1 (d) none of these
Ans. (c)
Since, y = (1 + x) (1 + x2) (1 + x4) … n2(1 x )
(1 – x) y = (1 – x2) (1 + x2) (1 + x4) … n2(1 x )
= (1 – x4) (1 + x4) … n2(1 x )
…………………….....

2
dx (1 x)
dx 1

8. If y2 = P(x) is a polynomial of degree 3, then 2 2
3
2
equals
(a) P’’’(x) + P’ x (b) P’’(x). P’’’(x) (c) P(x). P’’’(x) (d) none of these
Ans. (c)
(2y) y'' y'(2y') P''(x)
2y3y’’ = y2 P’’(x) – 2(yy’)2
= 2
4 [from Eq. (i)]
3 21 2y y'' P x P'' x {P' x }
2
dx
dx dx
x
dy
dx
(a) 14 (b) 7/8 (c) 1 (d) none of these
Ans. (b)

Subtracting Eq. (iv) from (iii), we get
t and x4 + y4 = t2 +
2
1
(a) – 1 (b) 0 (c) 1 (d) none of these
Ans. (c)
2
t t
x2y2 = – 1
dx
dx
11. If variables x and y are related by the equation x = y
0 2
2
1
dx
1 x x cot
(a) – 1 (b) 1 (c) log 2 (d) – log 2
Ans. (a0
2 2x
1 (x )
1 (x )

13. The solution set of f’(x) > g’(x) where f’(x) = (1/2) 52x + 1 and g(x) = 5x + 4x loge 5 is
(a) (1, ) (b) (0, 1) (c) [0, ) (d) (0, )
Ans. (c)
2x 1 x e e
1 5 log 5 2 5 log 5 4 log 5
2
(5x – 1) (5. 5x + 4) > 0
x5 1
y , at θ 3 1 cos 2θ dx

Ans. (b)
2dy cosec θ

dx is equal to
dy

16. If f’(x) = sin x + sin 4x. cos x, then 2 π f ' 2x
2
2 is equal to
(a) – 1 (b) 0 (c) 2 2π (d) none of these
Ans. (c)
d 2x 2

= 2 2 2π π sin 2x sin (8x 2π). cos 2x 4x
2 2
2 2
1 x x x 3x 3

+…+ up to n terms. Then y’(0) is equal to
(a) 2
Ans. (b)
1 11 1

+

= n
1

1 (x r 1)(x r)

2 2

n 1 n

18. If f(x) = (1 + x)n, then the value of nf ''(0) f (0)
f(0) f '(0) .... 2! n!
is
(a) n (b) 2n (c) 2n–1 (d) none of these
Ans. (b)
n 1 n 2

= nC0 + nC1 + nC2 + … + nCn = 2n
19. Let f be a function such that f(x + y) = f(x) + f(y) for all x and y and f(x) = (2x2 + 3x) g(x) for all x where
g(x) is continuous and g(0) = 3. Then f’(x) is equal to
(a) 9 (b) 3 (c) 6 (d) none of these
Ans. (a)
h
h
h
= 3g(0)
= 3.3
= 9
3

for 2 < x < 3 and [x] denotes the greatest integer less than or equal to x, then
(f ' π/3) is equal to
(a) π/3 (b) π/3 (c) π (d) none of these
Ans. (b)
3
3

0 y f t sin {k x t }dt, then
2 2
(a) 0 (b) y (c) k. f(x) (d) k2 f(x)
Ans. (c)
= x

…(i)
x
0
dy Im e . f x e f t e dt ike
dx
= x
Now, 2
x ikx ikx ikt ikx
0 Im ik. e . f x . e f t . e dt ike
= k f(x) – k2y [from Eq. (i)]
2
2
2
dx
22. If 6 6 3 3(1 x ) (1 y ) a(x y ) and 6
6
then
(a) f(x, y) = y/x (b) f(x, y) = y2/x2 (c) f(x, y) =2y2/x2 (d) f(x, y) = x2/y2
Ans. (d)
(cosθ cos ) a(sin θ sin )
θ θ θ θ

1 1 3 1 3 1θ cot a sin x sin y 2cot a
2
(a) 32/9 (b) 64/3 (c) 64/9 (d) none of these
Ans. (a)
x …(i)
or x
2 2
Differentiating both sides w. r. t. x, then
x2 F’(x) + F(x). 2x = 4x2 – 2F’(x)
Put x = 4
18F’(4) + 0 = 64 [ F(4) 0 , from Eq. (i)]
32

dx is equal to
Ans. (b)
2 dy dy 2 y x 2y 1 2
dx dx
1
1
Ans. (b)
(x x)
2|x x|
dx 2| 1 1|
= – 1
2
π/2 0 (3 2 sin t) cos t dt = 0, then
π, π
Ans. (c)
2 dy (3 2 sin x) . 1 cos y 0
dx

27. If f(x) = x3 + x2 f’(1) + x f’’(2) + f’’(3) for all x R , then
(a) f(0) + f(2) = f(1) (b) f(0) + f(3) 0 (c) f(1) + f(3) = – f(2) (d) none of these
Ans. (a)
f(x) = x3 + x2 f’(1) + x f’’(2) + f’’’(3)
f’(x) = 3x2 + 2x f’(1) + f’’(2)
f’(1) = 3 + 2f’(1) + f’’(2)
f’(1) + f’’(2) = – 3 …(i)
and f’’(x) = 6x + 2f’(1)
f’’(2) = 12 + 2f’(1)
–2f’(1) + f’’(2) = 12 …(ii)
Solving Eqs. (i) and (ii) we get
f’(1) = – 5 and f’’(2) = 2
and f’’’(x) = 6
f’’’(3) = 6 …(iii)
Substituting the values of f’(1), f’’(2) and f’’’(3) from Eq. (i), (ii) and (iii) in f(x)
f(x) = x3 – 5x2 + 2x + 6
f(0) = 6, f(1) = 4, f(2) = – 2, f(3) = – 6
Hence, f(0) + f(2) = f(1)
and f(1) + f(3) = f(2)
dx is equal to
x e
Ans. (a)
2 2
dx log x x
= e e
= 2
e
(1 t ) (1 t )
4
6
( 1 t ) ( 1 t )
2t t
dy 2 1 t
1 t
dx t t {1 1 t }

, then
(a) f is not derivable for all x, with |x| 1 (b) f is derivable at x = 1
(c) f is not derivable at x = – 1 (d) f is not derivable for all x, with |x| 1 Ans. (c) Solution:
1
2
1 x2x 1
0
2
2

31. If f(x – y), f(x). f(y) and f(x + y) are in AP, for all x, y and f(0) 0, then
(a) f(2) = f(–2) (b) f(3) + f(–3) = 0 (c) f’(2) + f’(–2) 0 (d) f’(3) = f’(–3)
Ans. (a)
2f(x) f(y) = f(x – y) + f(x + y) …(i)
Replacing x and y and y by x, then
2f(y) f(x) = f(y – x) + f(y + x) …(ii)
From Eqs. (i) and (ii), we get
f(x – y) = f(y – x)
Put y = 0, then
Differentiating w. r. t. x, then
f’(x) = – f’(–x)
f’(x) + f’(–x) = 0 f’(2) + f’(–2) = 0
32. Consider the equation x |y| 2y. What is the derivative of y as a function of x with respect to x for x < 0?
(a) 2 (b) 1 (c) 1
2 (d)
1 y x
3
Derivative of y as a function of x w. r. t. x for x < 0 is 1
. 3
33. If n 1f (x)
nf (x) e for all n N and f0(x) = x, then n
d {f (x)}
dx
(c) fn(x). fn–1(x)…f2(x). f1(x) (d) n
i i 1
dx
= n n 1 2 1 0
d f x . f x ....f x . f x (f (x))
dx

2
x
x
(b) has a value 0 when x= 1, x = 4
9
(d) has a differential coefficient 27e2 – 8 when x = e
Ans. (b)
d f t dt f x . 3x f x . 2x
dx
= x ln x (9x – 4)
Let z = x ln x (9x – 4)
Then dz

= 2(9e – 4) + 9e = 27e – 8
35. If 1 is a twice repeated root of the equation ax3 + bx2 + bx + d = 0 then
(a) a = b = d (b) a + b 0 (c) b + d = 0 (d) a d
Ans. (c)
Then f(1) = 0 and f’(1) = 0
a + 2b + d = 0 ….(i)
and 3a + 2b + b = 0 …(ii)
From Eqs. (i) and (ii)
a = d = – b
36. Differential coefficient of sin–1 x w. r. t. sin–1 (3x – 4 x3) is
(a) 1 π π
(b) π π

(d) π π

Ans. (a)
Solution:
2
1 1 v sin (3x 4x ) 3 sin x, x
2 2
2
dv 3 2 2
1 1 , x 1

37. If F(x) = f(x)g(x) and f’(x) g’(x) = c, then
(a) f g
F''' f ''' g'''
Differentiating both sides w. r. t. x, we get
F’(x) = f’(x). g(x) + g’(x). f(x)
F’(x) = f’(x) g’(x)

Again differentiating both sides w. r. t. x, we get
F’’(x) = f’’(x) g(x)
+ g’’(x). f(x) + 2f’(x). g’(x)
F’’(x) = f’’(x). g(x) + g’’(x). f(x) + 2c …(ii)
Dividing both sides by F(x) = f(x). g(x)
{ f’(x). g’(x) = c}
Then F''(x) F''(x) g''(x) 2c
F(x) f(x) g(x) f(x) g(x)
F f g fg
Differentiating both sides w. r. t. x, we get
f’(x) g’’(x) + g’(x) f’’(x) = 0
From Eq. (ii)
F’’(x) = f’’(x) . g(x) + g’’(x). f(x) + 2c
Differentiating both sides w. r. t. x, we getr
F’’’(x) = f’’(x). g’(x) + f’’’(x). g(x) + g’’(x). f’(x) + f(x). g’’’(x) + 0
= f’’’(x). g(x). g’’’(x). f(x) + 0
Now, dividing both sides by F(x)= f(x) g(x)
Then, F'''(x) f '''(x) g'''(x)
F(x) f(x) g(x)
h
On the basis of above information, answer the following questions:
38. If u = f(x), v = g(x), then the value of D* (u. v) is
(a) (D*u)v + (D*v)u (b) u2 D*v + v2D*u (c) D*u + D*v (d) uvD (u + v)
Ans. (b)
= 2 2 2 2
h
h
g (x) h
= u2D*v + v2D*u
39. If u = f(x), v = g(x) then the value of u
D * v
v
(b)
2
v
(d)
2
g x h g x lim
h
f x h g x x h f x lim
h g x h g x

=
2 2h 0
{f (x h) f (x)} {g (x h) g (x)} g x f (x)
h hlim g (x h)g (x)

=

g x D* f x f x D* g x
g x
v
40. D*(tan x) is equal to
(a) sec2 x (b) 2 sec2 x (c) tan x sec2 x (d) 2 tan x sec2 x
Ans. (d)
h
h 0 h 0
tan x h tan x lim lim (tan (x h) tanx)
h
= sec2 x. 2 tan x
= 2 tan x sec2 x
41. The value of D* f(x) at the point on the curve y = f(x) such that tangent to it are parallel to x-axis, then
(a) f(x) (b) zero (c) 2f(x) (d) xf(x)
Ans. (b)
h
f x h f x lim lim (f(x h) f(x))
h
= 0
42. The value of D*c, where c is constant is
(a) non-zero constant (b) 2 constant (c) does not exist (d) zero
Ans. (d)
h
= 0
43. If x = b 1 2y dy cos by y , then
b dx …………
(a) 2(by y )/ y (b) 2(by y )/ y (c) 2 2(by y )/ y (d) 2(by y )/ y
Ans. (a)
b (given)
Putting 2y b cos θ , we get
1 2 2 2 4x b cos cos θ b cos θ b cos θ
= bθ b cos θ sin θ
2 2dx b b (cos θ sin θ)
dθ
Now,

2cos θ
y

1
2e
x dx = f(k) – f(1), then one of the possible value of k–10 is …………….
(a) 4 (b) 6 (c) 5 (d) 3
Ans. (b)
(given)
Now,

16 121 1
x z
= f(16) – f(1)
f(k) = f(16)
One possible value of k is 16.
45. If f’(x) = 23x 4 and y = f(x3), then at x = 2, dy
dx equals …………………….
Ans. (c)
3 2 6 2dy f '(x ).3x 3x 4 . 3x
dx

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