Chapter-25

DIFFERENTIATION

PRACTICE SHEET
1. If u = sin1 (x  y ), x = 3t, y = t3, then what is the dy
derivative of u with respect to ? 8. If y= sin1 x + sin1, 1  x 2 what is equal to?
1
dx

(a) 3(1  t 2 ) (b) 3(1  t 2 ) 1 1
2
(a) cos1 x  cos1 1  x 2 (b) 
1 cos x cos 1  x 2

(c) 5(1  t 2 ) 2 (d) 5 (1  t2) 
(c) (d) 0
2 2
2. If y = x + ex, then what is d x equal to?
dy 2 9. If f (x) = loge [loge x], then what is f’ (e) equal to?
(a) e1 (b) e
ex (c) 1 (d) 0
(a) ex (b) 
(1  e x )3
dy
10. For the curve x  y  1, what is the value of at
ex ex dx
(c)  (d) 
(1  e x ) (1  e x ) 2 1 1
 , ?
4 4
1 1 dy
3. If, x  y  t  , x 2  y2  t 2  What is equal to? 1
t t 2
dx (a) (b) 1
2
1 1 (c) 1
(a) (b)  (d) 2
x x
1
1 1 11. If y  ,then what is dy equal to
(c) (d)  log10 x dx
x2 x2 (a) x (b) x loge 10
What is the derivative of cos 1  2 cos x  3sin x  ?
(log x 10) 2 (log10e )
4. (c)  (d) x log10 e
 13  x

(a)
1
(b)  1 12. If xy = ex  y, then dy /dx is equal to which one of the
1 x2 1 x2 following ?
(c) 0 (d) 1 ( x  y) y
(a) (b)
5. What is the solution of y’ = 1 + x + y2 + xy2, y(0) = 0? (1  log x) 2
(1  log x)
 x2  (x  y) (log x)
(a) y  tan 2   x (b) y = tan2 (x2 +x) (c) (d)
 2  (1  log x) (1  log x) 2
 x2 
(c) y = tan (x2 + x) (d) y  tan   x 13. If f  x   cos x,g  x   log x,and y   gof  x  , then what is
 2 
the value of
dy at x = 0 ?
dy dx
6. If x 1  y  y 1  x  0, what is equal to?
dx (a) 0 (b) 1
(c) 1 (d) 2
1 1
(a)  (b) 
1 x (1  x) 2 14. What is the derivative of log x 5 respect to log 5 x ?

(a)   log 5 x   log5 x 

2 2
1 x (b)
(c) (d)
(1  x) 2 1 x (c)   log x 5  log x 5
2 2
(d)
dy
If f  x   sin x , then what is f   x  equal to?
2 2
7. If x = sin t  t cost t and y = t sin t + cos t, then what is 15.
dx
    
(a) 4x sin x 2 cos x 2    
(b) 2sin x 2 cos x 2
at point t  ?
2 (c) 4sin  x  sin x
2 2
(d) 2x cos  x 
2 2


(a) 0 (b) 16. If f  x   tan x  e 2x  7x 3 , then what is the value of
2
 f  0 ?
(c)  (d) 1
2 (a) 2 (b) 1

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 (c) 0 (d) 3 dy
22. If x  y  2, then what is at y  1 equal to ?
17. If 3  3  3
x y xy
, then what is dy equal to? dx
dx (a) 5 (b) 4

3x  y  3x 3 xy
3 y
 1
(c) 2 (d) 1

If x  k    sin   and y  k 1  cos   , then what is

(a) (b)
3y 1 3 x 23.
3x  3 y 3  3y x
the derivative of y with respect to x at    / 2 ?
(c) x (d)
3  3y 1  3x  y (a) 1 (b) 0
(c) 1 (d) 2
18. If y  sin  msin 1 x  , then what is the value of
d2 y
If x  t , y  t , what is
2 3
24. equal to ?
d y / dx at x  0?
2 2
dx 2
(a) m (b) m2 3
(c) m2 + 2 (d) 0 (a) 1 (b)
2t
dy d2 y d2 x 3 3
19. If y  f  x  , p  and q  2 , then what is equal (c) (d)
dx dx dy 2 4t 2
to ?
If y  sin 1 
q q 4x  dy
(a)  (b)  25. 2 
then what is equal to?
p2 p3  1  4x  dx
1 1
1 q (a) (b) 
(c) (d) 1  4x 2 1  4x 2
q p2
4 4x
(c) (d)
If f  x   e  and g  x   log cos x, then what is
sin log cos x 
20. 1  4 x2 1  4 x2
the derivative of f  x  g  x  ? x
26. What is the derivative of x a 2  x 2  a 2 sin 1   ?
(a) f  x  cos g  x  (b) f  x  sin g  x   a

(c) g  x  cos f  x  (d) g  x  sin f  x   (a) a 2  x 2 (b) 2 a 2  x 2

(c) x2  a2 (d) 2 x  a
2 2
2
21. What is the derivative of sin x with respect to cos2 x?
(a) tan2 x (b) cot2 x
(c) 1 (d) 1

ANSWER KEYS
1. b 2. b 3. c 4. d 5. d 6. b 7. a 8. d 9. a 10. c
11. c 12. d 13. a 14. a 15. a 16. b 17. b 18. d 19. b 20. a
21. c 22. d 23. a 24. c 25. c 26. b

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 Solutions
Sol. 1. (b) 3 1 1
u = sin 1 (x  y) and x = 3t, y = 4t3 Where  = tan–1  2  =  0
  2
So, u = sin1(3t  4t3) 1 x 1  x2
Let t = sin    = sin1 t, =cos–1 (cos(x–)) = x –  Sol. 9. (a)
So, u = sin1, (3 sin  4 sin3) Hence, f' (x) = 1 ( is constant) The given function is : f (x) = loge [loge x]
= sin1 (sin 3) = 3 Sol. 5. (d) Differentiating w.r.t. x, we get
Hence, u = 3 sin1 t Given differential equation is 1 1 1 1 1
f '(x)  .  f '(e)  .   e1
du 1 dy loge x x loge e e e
3  3(1  t 2 ) 1/2 =1 + x + y2 + xy2
dx Sol. 10. (c)
dt 1 t2
dy Given function: x  y  1
Sol. 2. (b)  =(1+x) (1+y2)
Given that y = x + ex dx is an implicit function
Differentiating w.r.t.x dy Differentiating both sides w.r.t.x, we get
 =(1+x)dx 1 1 dy
dy
 1 e x 1  y2  0
dx 2 x 2 y dx
x2

dx

1 tan–1 y = +x+c dy y
dy 1  e x x  
Given that when x = 0, y(0)= 0. dx x
Differentiating w.r.t.
1e x 
Hence, c=0 dy 1 1
d2x Value of at x  , y 

dx  x2 
 
. dx 4 4
dy 2 x 2 dy y=tan   x
1 e  2   dy  1/ 4
      1
 dx  1 , 1 
x
e 1 ex 1/ 4
1  e . 1  e    1  e 
=– Sol. 6. (b)
x x x 3 Given equation 4 4

x 1 y  y 1 x  0 Sol. 11. (c)

Sol. 3. (c)
1
Given that x + y = t – 1/t and Can be written as: Differentiating the given function, y 
x 1  y  y 1  x log10 x
x2 + y2 = t2 + 1
t2 dy 1 1
Squaring both sides, we get We get,  . log10 e
(x+y)2 = x2 + y2 + 2xy x2 (1+y)=y2 (1+x) dx log10 x 2 x
 2  x2 + x2 y = y2 + y2 x
  t  1    t 2  1   2xy logx 102.log10 e
 t  2  x2 – y2 = y2 x – x2y 
dy

 t 
(x–y) (x+y)= – xy (x–y) dx x
–2 = 2xy  xy = – 1 x+y = – xy y(1+x)= – x Sol. 12. (d)
(x–y)2 = (x+y)2 – 4xy xy = ex–y
2 2
x
 1 1  1 y= which is in explicit from Taking log both sides, we get
=  t    4  1  t 2   2 4  t   1 x
 t t2  t y, log x = x–y
Differentiating w.r.t.x, we get
x–y = t + 1/t dy 1  x 1  x 1 1 y=
x
x = t, y = –1/t
xy = – 1
dx

1  x 2


1  x2  1  log x

dy
Sol. 7. (a) 1  log x   x 0  1 
x y0 As given

dy
  x
dx x = sin t  t cos t and y = t sin t + cost dx 1  log x  2
dy y 1 1 On differentiating w.r.t.x, we get

dx
 
x t2 x2

dx =
1  log x   1

log x

Sol. 4. (d) dt
=cos t – {cos t + t (–sin t)} 1  log x 2 1  log x 2
The given function dx Sol. 13. (a)
 =cot t – cos t + t sin t = 1 sin t y = gof(x) = g {f(x)}log(cos x)
 2cos x  3sin x 
f (x)  cos 1  
dt
 13  dy  dy  1
  sin x    tan x
Can be written as and = t cos t + sin t – sin t = t cos t dx cos x
dt
 2 3  Hence, =cot t   dy  =  tan 0 = 0
cos–1  cos x.  sin x.   
 dx at x  0
 
 
 dy 
 13 13   cot  0
 dt  t   2 Sol. 14. (a)
2 3
let  cos  and =sin 2 Let u = logx 5, and v = log5 x
13 13 Sol. 8. (d) u = loge 5 ,and v  loge x
3 Given function is : loge x loge 5
13 3 y=sin–1 x + sin–1 1 x 2 On differentiating w. r. t. x., we get
tan =  du log e 5 dv 1
2 2 on differentiating, w.r.t.x, we get  , and 
x  log e x 
2
dx dx x log e 5
13 dy

1

1
.
1
 2x 
dx
So, (i), cos–1 (cosx cos + sinx sin) 1  x2 1  1  x2 2 1  x2

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 du
2
du/dx log e 5  log e 5  d2 y dy y
   x log e     0   1
dv dv/dx x  log e x 2  log e x  dx 2 dx x
=  (logx 5)2 =  (log5 x)2 Sol. 19. (b) Sol. 23. (a)
x = k ( + sin) and y = k (1 + cos)
Sol. 15. (a) We have, dy  p
f  x   sin 2 x 2 , dx  dy = k(1 + cos) and dy =  k sin
  cos  x  d2 x d d
f   x  = 4x sin x
2 2 dx 1 1 dp
  2  2  
dy p dy p dy
 dy 2sin cos
Sol. 16. (b) k sin  2 2   tan 
2  
f  x   tan x  e 2x  7x 3 , 
d x 1 dp
 2
1 dp dx
 2
q
 3 dx k 1  cos   2 cos 2  2
dy 2 p dy p dx dy p 2
f '(x) = sec2x - 2e-2x - 21x2
Sol. 20. (a)   dy  
f '(0) = 1 - 2 = -1      tan  1
Sol. 17. (b) f(x) = esin(log cos x)  dx  4
2
3x + 3y = 3x+y f(x) =esin(log cos x).cos(log cos x). 1 (sin x) Sol. 24. (c)
On differentiating w. r. t. x., we get cos x
x  t 2 , y  t3 ,
3x log3 + 3y log3 dy =3(x+y) log3 1  dy  =  esin(log cos x). cos(log cos x). tan x
  dx dy dy 3
dx  dx  and g(x) = log cos x  2t,  3t 2 ,  t,
3x + 3y dy =3x+y + 3(x+y) dy g(x) = 1 (sin x) =  tan x dt dt dx 2
dx dx cos x d 2 y 3 dt 3
 
Hence, f '  x   e .cos  log cos x  .tan x
sin  log cos x 
 dy (3x+y + 3y) = 33+y  3x dx 2 2 dx 4t
dx g ' x   tan x Sol. 25. (c)
 dy  3 3  1  3
x y

xy
x y 1    =esin(log cosx).cos(log cos x)  4x 
y  sin 1  2 
dx 3 1  3
y x
1  3x    =f(x).cos[g(x)]  1  4x 
Sol. 21. (c) put 2x = tan t
Sol. 18. (d) Let u= sin2 x, and v = cos2 x
y  sin  msin 1 x  , y = Sin-1(Sin 2t) = 2t = 2tan-1 (2x)
 du =2 sin x cos x dy 4
put m = 2 dx  ,
1 dx 1  4x 2
y  Sin(2Sin x) dv
and   2sin x cos x Sol. 26. (b)
y  Sin(Sin 1 2x 1  x 2 )  2 x 1  x 2 dx dy 1 a2 1
 x. .  2x   a 2  x 2  .
dy 2x(2x)  du  du/dx  2sin x cos x  1 dx 2 a2  x2 x2 a
 2 1 x2  1 2
dx 2 1 x2 dv dv/dx 2sin x cos x a
2(1  x 2 )  2 x 2 2  4x 2 Sol. 22. (d)
dy
  x 2 a 2 .a 1
dx 1 x2 1 x2 x  y  2,   a2  x2  .
at y = 1 , x = 1 a x
2 2
a x
2 2 a
2x
1  x 2 (8x)   2  4x 2 
= a  x  a2  x2  2 a2  x2
2 On differentiating w. r. t. x., we get 2 2
d y
 2 1 x2 1 1 dy
dx 2 (1  x 2 )   0, a2  x2
2 x 2 y dx
put x = 0

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 NDA PYQ
If f  x   2 , then what is the double derivative of f ''(x)?
x

 
1. dy
10. If y  ln e mx  e  mx , then what is the value of at
(b) x  x  1 2
x 2 x 2 dx
(a) 2 (ln 2)
x2 x 2
x  0?
(c) 2 (ln 2) (d) 2 (ln10 2) (a) 1 (b) 0
[NDA (I) - 2011] (c) 1 (d) 2
[NDA (II) - 2012]
2.    
If y  1  x1/ 4 1  x1/ 2 1  x1/ 4 , then what is
dy
dx
equal 11. Consider the following statements :
dy
to ? I. If y  ln  sec x  tan x  , then  sec x
(a) 1 (b) 1 dx
(c) 0 (d) 2x dy
[NDA (II) – 2011] II. If y  ln  cosec x  cot x  , then  cosec x
dx
dy 
3. If y=ln tan x, then what is the value of at x  ? Which of the above statements is/are correct ?
dx 4 (a) Only I (b) Only II
(a) 0 (b) –1 (c) Both I and II (d) Neither I nor II
(c) 1/2 (d) 1 [NDA (II) - 2012]
[NDA-2011(2)] dy
If 2x  3y  7, then what is
3 2
dy 12. equal to (where,
4. If y = cos t and x = sin t, then what is the value of ? dx
dx y  0 )?
(a) xy (b) x/y
(c) y/x (d) x/y x2 x
(a) (b)
[NDA (I) - 2012] 2y 2y
x 1 dy x2
5. If y  , then what is the value of ? (c) (d) None of these
x 1 dx y
2 2 [NDA (I) - 2013]
(a) (b)
x 1  x  1
2 2 2
13. The derivation of sec x w.r.t tan x is
2 2 (a) 1 (b) 2
(c) (d)
 x  1
2
 x  1 (c) 2sec x tan x (d) 2sec x tan x
2

[NDA (I) - 2013]

[NDA (I) - 2012]
14. What is the differentiation on log x x ?
6. What is the rate of change of x 2  16 w.r.t. x2 at x = 3 ? [NDA (I) - 2013]
1 1 (a) 0 (b) 1
(a) (b) (c) 1/x (d) x
5 10
dy
1 1 15. If y  x , then what is the value of
x
at x  1?
(c) (d) dx
20 15
(a) 0 (b) 1
[NDA (I) - 2012]
(c) 1 (d) 2
7. If f(x)=x2 – 6x + 8 and there exists a point c in the interval [NDA (I) - 2013]
[2, 4] such that f’(c)=0, then what is the value of c?
d2 y b
(a) 2.5 (b) 2.8 16. If y  sin  ax  b  , then what is 2
at x   , where
(c) 3 (d) 3.5 dx a
[NDA-2012(1)] a and b are constants and a  0?
If f  x   2 , then what is the derivative of f '  x  ?
sin x
8. (a) 0 (b) 1
(a) 2
sin x
ln 2 (b) sin x 2 sin x 1 (c) sin  a  b  (d) sin  a  b 

(c) cos x 2 sin x 1

(d)None of these [NDA (I) - 2013]
17. What is the derivation of x3 w.r.t. x2 ?
[NDA (II) - 2012]
3x
dy x (a) 3x2 (b)
If x  y  1, such that   , then what should be
m m
9. 2
dx y
3
the value of m? (c) x (d)
(a) 0 (b) 1 2
(c) 2 (d) 1 [NDA (II) - 2013]
[NDA (II) - 2012]

SANDEEP SINGH BRAR Ph:- +91 9700900034 - 390 -

18. What is the derivation of sin  sin x  ? [NDA-2014(2)]
27. f'(a), f'(b), f'(c) are in
(a) cos  cos x  (b) cos  sin x  (a) A.P
(c) cos  sin x  cos x (d) cos  cos x  cos x
(b) G.P
(c) H.P
[NDA (II) - 2013] (d) Arithmetic co-geometric progression
19. What is the derivation of x  1 at x = 2? [NDA-2014(2)]
28. f”(a), f”(b), f”(c) are:
(a) 1 (b) 0 (a) in A.P. only
(c) 1 (d) does not exist (b) in G.P. only
[NDA (II) - 2013] (c) in both A.P. and G.P.
20. If f(x) = 2x2 + 3x - 5 , then what is f'(0) + 3f '(-1) = (d) neither in A.P. nor in G.P.
(a) 1 (b) 0 [NDA-2014(2)]
(c) 1 (d) 2 Direction (for next two) : Given that,
[NDA (II) - 2013]
d  1 x2  x4 
1  cos x    Ax  B
21. What is the derivative of dx  1  x  x 2 
1  cos x
29. What is the value of A ?
1 2x (a) 1 (b) 1
(a) sec
2 2 (c) 2 (d) 4
1 2 x
[NDA (I) - 2015]
(b)  cosec 30. What is the value of B ?
2 2 (a) 1 (b) 1
2 x (c) 2 (d) 4
(c)  cosec
2 [NDA (I) - 2015]
(d) None of the above  
What is the derivative of tan 1  1  x  1  with respect to
2

[NDA (I) - 2014] 31.  

 x 
dz
22. If z = fo f(x) where f(x) = x2, then what is equal to? 1
dx tan x ?
(a) x3 (b) 2x3 1
(c) 4x3 (d) 4x2 (a) 0 (b)
2
[NDA-2014(1)]
(c) 1 (d) x
Direction (for next two): Consider the curve
[NDA (I) - 2015]
x  a  cos    sin   and y  a  sin    cos   . Direction (for next three)
dy Consider the parametric equation
23. What is equal to? a(1  t 2 ) and 2at
dx x y
1 t 2
1 t2
(a) tan  (b) cot 
32. What does the equation represent?
(c) sin 2 (d) cos 2 (a) It represents a circle of diameter a
[NDA (II) - 2014] (b) It represents a circle of radius a
2
d y (c) It represents a parabola
24. What is equal to ? (d) None of the above
dx 2
[NDA (I) - 2015]
(a) sec2θ (b) -cosec2θ dy
33. What is equal to?
sec3  dx
(c) (d) None of these
a (a) y/x (b) - y/x
[NDA (II) - 2014] (c) x/y (d) - x/y
[NDA (I) - 2015]
dy
25. If y  x ln x  xe , then what is the value of
x
at x = 2

dx 34. What is d y2 equal to

dx
1?
(a) 1+ e (b) 1 e a2 a2
(a) (b)
(c) 1 + 2 e (d) None of these y2 x2
[NDA (II) - 2014] 2 2
(c)  a (d)  a
For the next three (03) items that follow: x2 y2
Let f(x)=ax2+bx+c such that f(1)=f(–1) and a, b, c are in [NDA (I) - 2015]
Arithmetic progression.  1  sin x  1  sin x  
26. What is the value of b? 35. If y  cot 1   , where 0 < x < ,
(a) –1  1  sin x  1  sin x  2
(b) 0 dy
then is equal to
(c) 1 dx
(d) cannot be determined due to insufficient data

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 1 (a) p3 (b) 3p3
(a) (b) 2 (c) 6p3 (d) 6p3
2
[NDA (I) - 2016]
(c) sin x  cos x (d) sin x  cos x 44. What is the value of p for which f ”(0) = 0?
[NDA (II) - 2015] 1
d 2s (a)  or 0 (b) 1 or 0
36. If s  t  1, , then
2
is equal to 6
dt 2
1
1 1 (c) or 0 (d) 1 or 1
(a) (b) 2 6
s s [NDA (I) - 2016]
1 1 For the next four (4) items that follow:
(c) 3 (d) 4 Let f:RR be a function such that f(x) = x3 + x2 f’(1) + xf
s s
”(2) + f "'(3) for xR
[NDA (II) - 2015]
45. What is f(1) equal to:
37. The derivative of ln  x  sin x  with respect to (a) 2 (b) 1
 x  cos x  is (c) 0 (d) 4
[NDA (I) - 2016]
(a) 1  cos x (b) 1  cos x 46. What is f ’(1) is equal to ?
 x  sin x 1  sin x   x  sin x 1  sin x  (a) 5 (b) 1
(c) 1 (d) 0
(c) 1  cos x (d) 1  cos x
[NDA (I) - 2016]
 x  sin x 1  cos x   x  sin x 1  cos x  47. What is f '''(10) equal to?
[NDA (II) - 2015] (a) 1 (b) 5
dy y (c) 6 (d) 0
38. If xayb=(x–y)a+b, then the value of  is equal to:
dx x [NDA (I) - 2016]
(a) a/b (b) b/a 48. Consider the following:
(c) 1 (d) 0 1.f(2) = f(1)  f(0)
[NDA (II)-2015] 2.f ”(2)  2f’ (1) = 12
For the next two (2) items that follow: Which of the above is / are correct?
Consider the function (a) Only 1 (b) Only 2
f(x) = |x2  5x + 6| (c) Both 1 and 2 (d) Neither 1 nor 2
39. What is f  (4) equal to? [NDA (I) - 2016]
(a) 4 (b) 3 49. If y = log10 x + logx 10 + logx x + log10 10 then what is
(c) 3 (d) 2  dy 
[NDA (I) - 2016]   x = 10 equal to ?
 dx 
40. What is f  (2.5) equal to? (a) 10 (b) 2
(a) 3 (b) 2 (c) 1 (d) 0
(c) 0 (d) 2 [NDA (I) - 2016]
[NDA (I) - 2016] Direction (for next three): Consider the following for the
Direction (for next two): next two items that follow:
Consider the equation x + |y| = 2y Let f(x) = [|x|  |x  1|]2
41. Which of the following statements are not correct? 50. What is f ’(x) equal to when x > 1?
1.y as a function of x is not defined for all real x.
(a) 0 (b) 2x 1
2.y as a function of x is not continuous at x = 0
(c) 4x  2 (d) 8x  4
3. y as a function of x is differentiable for all x.
Select the correct answer using the code given below: [NDA (II) - 2016]
51. What is f ’(x) equal to when 0 < x < 1 ?
(a) 1 and 2 (b) 2 and 3
(c) 1 and 3 (d) 1,2 and 3 (a) 0 (b) 2x 1
[NDA (I) - 2016] (c) 4x  2 (d) 8x  4
42. What is the derivative of y as a function of x with respect x [NDA (II) - 2016]
for x < 0? 52. Which of the following equation is/are correct?
(a) 2 (b) 1 1. f(2) = f(5)
2. f ”(2) + f ” (0.5) + f ” (3) = 4
1 1
(c) (d) Select the correct answer using the codes given below:
2 3 (a) Only 1 (b) Only 2
[NDA (I) - 2016] (c) Only 1 and 2 (d) Neither 1 nor 2
For the next two (2) items that follow: [NDA (II) - 2016]
Consider the function 53. Let f(x+y) = f(x) f(y) for all x and y. Then what is f’(5) equal
x3 sin x cos x to [where f’ (x)] is the derivative of f(x)]?
f(x) = 6 1 0 where p is a constant. (a) f(5)f’(0) (b) f(5)f’(0)
(c) f(5) f(0) (d) f(5) +f’ (0)
p p2 p3
[NDA (I) - 2017]
43. What is the value of f ’(0)?

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 d 2x  2 x  0
54. What is equal to? 61. Consider the function f  x    x n x  what is f(0)
dy 2
 0 x  0
1 1 equal to ?
 d 2 y   dy  3  d 2 y   dy  2
(a) –     (b)  2    (a) 0 (b) 1
 dx 2   dx   dx   dx 
    (c) 1 (d) It does not exist
1 [NDA (I) - 2018]
 d 2 y   dy  3  d2y 
(c) –    (d)  2 
2

 dx 2   dx   dx 
62. If y = ex then, what is dy at x = π equal to
    dx
2
(b) 2π e
2
[NDA-2017(1)] (a) (1 + π) e
55. What is the derivative of log10 (5x2+3) with respect to x?
(c) 2 e (d) e
2 2

x log10 e 2x log10 e
(a) (b) [NDA (I) - 2018]
5x 2  3 5x 2  3
63. What is the derivative of the function f(x) =
10x log10 e 10x loge 10
(c) (d) 
5x 2  3 5x 2  3 e tan x  n  sec x   e nx at x  ?
4
[NDA-2017(1)]
(a) e/2 (b) e
 cos x  dy
 cos x 
cos x  (c) 2e (d) 4e
56. If y = , then is equal to:
dx [NDA (I) - 2018]
 y tan x
2
y 2 tan x  5  2 tan x  dy
If y  tan 1 
 2  5 tan x  ,
(a) (b) 64. then what is equal to ?
1  yIn  cos x  1  yIn  cos x    dx
y 2 tan x y 2 sin x 1
(c) (d) (a)  (b) 1
1  yIn  sin x  1  yIn  sin x  2 x
[NDA (II) - 2017] (c) 1 (d) 1
57. A function is defined in (0, ∞)by : 2 x
 1 x2 for 0  x  1 [NDA (II) - 2018]
f(x) = 
 Inx for 1  x  2 65. If u  e sin bx ax
and ve ax,
cos bx, then what is
In 2  1  0.5x for 2  x  
 u
du
v
dv
equal to ?
Which one of the following is correct in respect of the dx dx
derivative of the function, i.e., f’ (x)?
(a) f’(x) = 2x for 0 < x  1
(a) a e
2ax

(b) a  b
2 2
e ax

(b) f’(x) = 2x for 0 < x  1 (c) ab e

2ax
(d)  a  b  eax
(c) f’(x) = 2x for 0 < x < 1 [NDA (II) - 2018]
(d) f’(x) = 0 for 0 < x < ∞
dy
y  sin x , then what is the value of
x
[NDA (II) - 2017] 66. If at
1  x  1  1  x  1  dy dx
58. If y = sec    sin   ,then is equal to:

 x 1   x 1 dx x ?
(a) 0 (b) 1 6
x 1 x 1  

 6 ln 2    

(c) (d) 2 6
3 2 6 6 ln 2  3
x 1 x 1 (a) (b)
[NDA (II) - 2017] 6 6

 


 6 ln 2  
59. Consider the following statement; 
2 6
3 2 6 ln 2  3
1.Derivative of f(x) may not exist at some point.
(c) (d) 6
2.Derivative of f(x) may exist finitely at some point. 6 6
3.Derivative of f(x) may be infinite (geometrically) at some point. [NDA (II) - 2018]
Which of the following statements is / are correct?
 
(a) 1 and 2 (b) 2 and 3 67. What is d 1  sin 2x equal to, where x ?
(c) 1 and 3 (d) 1, 2, and 3 dx 4 2
[NDA (II) - 2017] (a) cos x  sin x (b)   cos x  sin x 
 2x  dy
60. If y = cos1
 2 
, then is equal to: (c)   cos x  sin x  (d)None of the above
1 x  dx
[NDA (II) - 2018]
2 2
(a)  for all |x| < 1 (b)  for all |x| > 1 68. Let f  x  y   f  x  f  y  and where f  x   1  xg  x    x  ,
1 x 2
1 x2
lim g  x   a and lim   x   b. Then what is f '(x) equal
2 x 0 x 0
(c) for all |x| < 1 (d) None of these
1 x2 to?
[NDA (II) - 2017] (a) 1  abf  x  (b) 1  ab
(c) ab (d) abf(x)

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 [NDA (II) - 2018] (c) 2e (d) −1
69. Derivative of sec2(tan-1x) w.r.t. x is [NDA (I) - 2021]
(a) 2x (b) x2 + 1 ex
(c) x + 1 (d) x2 80. What is the derivative of e with respect to ex?
x
[NDA (I) - 2019] e
70. If f(x) = sin(cosx) than f'(x) = (a) e (b) ex
(a) cos(cosx) (b) sin(-sinx) ex
(c) e ex (d) eex
(c) (sinx)cos(cosx) (d) (-sinx)cos(cosx)
[NDA (II) - 2021]
[NDA (I) - 2019]
81. If y=(1+x) (1+ x2) (1 + x4) (1 + x8) (1 + x16) then what is
(sin x )2
71. What is the derivative of 2 with respect to sinx? dy at x = 0 equal to?
2 2
(a) sin x.2(sin x ) .ln4 (b) 2sin x.2(sin x ) .ln4 dx
(c) ln(sin x)2(sin x )
2
(d) 2sin x cos x2(sin x )
2
(a) 0 (b) 1
(c) 2 (d) 4
[NDA (II) - 2019] [NDA (II) - 2021]
If y = cos x. cos4x.cos8x, then what is dy at x   equal
Direction (for next two): 1
Consider the equation xy = ex−y 82.
y dx 4
dy
72. What is at x = 1 equal to to?
dx (a) –1 (b) 0
(a) 0 (b) 1 (c) 1 (d) 3
(c) 2 (d) 4 [NDA (II) - 2021]
[NDA (II) - 2019] 83. Let f(x) be a polynomial function such that fof(x) = x4.
2
73. What is d y2 at x = 1 equal to ? What is f '(1)=
dx (a) 0 (b) 1
(a) 0 (b) 1 (c) 2 (d) 4
(c) 2 (d) 4 [NDA (II) - 2021]
𝑑𝑦
[NDA (II) - 2019] 84. If xyyx = 1, then what is at (1, 1) equal to?
𝑑𝑥
74. What is the derivative of tan-1xwith respect to cot-1x ? (a) –1 (b) 0
(a) −1 (b) 1 (c) 1 (d) 4
1 x [NDA (I) - 2022]
(c) (d)
1 x2 1 x2 85.
𝑑𝑦
If y = (xx)x, then what is the value of at x = 1?
[NDA 2020] 𝑑𝑥
1
dy equal to ? (a) (b) 1
2
75. If xmyn = am+n , then what is (c) 2 (d) 4
dx [NDA (I) - 2022]
(a) my (b)  my 86. Let y = [x + 1], –4 < x < – 3 where [.] is the greatest integer
nx nx function. What is the derivative of y with respect to x at x =
(c) mx (d)  mx – 3.5?
ny (a) –4 (b) –3.5
ny
(c) –3 (d) 0
[NDA 2020] [NDA (I) - 2022]
76. If eθφ = c + 4θφ, where c is an arbitrary constant and φ is a 𝑚
function of θ, then what is φdθ equal to ? 87. If the derivative of the function f(x)= +2nx + 1 vanishes
𝑥
(a) θ d φ (b) – θ d φ at x = 2, then what is the value of m + 8n?
(c) 4 θ d φ (d) −4 θ d φ (a) –2
[NDA 2020] (b) 0
77. What is the derivative of ex w.r.t. xe ? (c) 2
(d)cannot be determined due to insufficient data
(a)
xe x (b)
ex [NDA (I) - 2022]
ex e xe 88. Let f(x) be a function such that f'(x) = g(x) and f"(x) = –
ex f(x). Let h(x) = {f(x)}2 + {g(x)}2. Then consider the
(c)
xe x (d) following statements:
xe ex e 1. h'(3) = 0 2. h(1) = h(2)
[NDA (I) - 2021] Which of the statements given above is/are correct?
78. What is the derivative of sin(lnx) + cos(lnx) with respect to (a) 1 only (b) 2 only
x at x =e? (c) both 1 and 2 (d) neither 1 nor 2
(a) cos1  sin1 (b) sin1  cos1 [NDA 2022 (II)]
e e  x 2  x 1 dy
89. If y = ln 
2
 , then what is at x = 0 equal to?
cos1  sin1  x 2  x 1 dx
(c) (d) 0  
e (a) –2 (b) 0
[NDA (I) - 2021] (c) 1 (d) 2
79. If x = etcost and y = etsint, then what is the value of dx/dy [NDA 2022 (II)]
at t = 0 equal to ?
(a) 0 (b) 1

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 (a) –2 (b) –1
d  1  x 4  x 8 
 = ax + bx3, then which one of the
90. If (c) 0 (d) 2
dx  1  x 2  x 4 
 [NDA-2023 (2)]
following is correct?  2x  3 
98. If f’(x) = cos (ln x) and y = f   , then what is dy/dx
(a) a = b (b) a = 2b  x 
(c) a + b = 0 (d) 2a = b equal to?
[NDA 2022 (II)]
  2x  3     2x  3  
dy (a)cos  n   (b) –3/x2 sin  n  
If y = x x  16  8 ln x  x 2  16 , then what is
2
91. equal   x    x 
2 dx   2x  3     2x  3  
to? (c) 3/x2 cos  n   (d) –3/x2 cos  n  
  x    x 
(a) x x 2  16 (b) x- x 2  16 [NDA-2023 (2)]
(c) x  16
2
(d) 4 x  16 2 Consider the following for the next two (02) items that
follow:
[NDA 2022 (II)]
1
92. What is the derivative of sin2x with respect of cos2x? Let f(x) and g(x) be two functions such that g(x) = x 
(a) –1 (b) 1 x
(c) sin2x (d) cos2x 1 .
[NDA – 2023 (1)] and fog(x)  x 3 
x3
93. What is the derivative of cosec (x°)?
99. What is g[f(x)  3x] equal to?
(a) –cosec(x°) cot(x°) (b) –  cos ecx cotx
1 1
180 (a) x3  (b) x3 
 x3 x3
(c)  cos ecx cotx (d) – cos ecx  cotx 
180 180 1
(c) x 2  2 (d) x 2  1
[NDA – 2023 (1)] x x2
Consider the following for the next three (03) items that [NDA-2024 (1)]
follow: 100. What is f(x) equal to?
Let f(x) = |lnx|, x  1.
(a) 
2 (b) 2x  2
94. What is the derivative of f(x) at x = 0.5? 3
(a) –2 (b) –1 x x3
(c) 1 (d) 2 (c) 6x + 3 (d) 6x
[NDA – 2023 (1)] [NDA-2024 (1)]
95. What is the derivative of f(x) at x = 2? Direction: Consider the following for the two items given
(a) –1/2 (b) –1 below
(c) 1/2 (d) 2 let f: (–1,1) → R be a differentiable function with f(0) = –1
[NDA – 2023 (1)] and f’(0) = 1. Let h(x) = f(2f(x)+2) and g(x) = (h(x))2
96. What is the derivative of fof (x), where 1 < x < 2? 101. What is h’(0) equal to?
1 (a) –2 (b) –1
(a) (b) 1
(c) 0 (d) 2
ln x x ln x [NDA – 2024 (2)]
(c) – 1 (d) – 1 102. What is g’(0) equal to?
ln x x ln x (a) –4 (b) –2
[NDA – 2023 (1)] (c) 0 (d) 4
97. If f(x) = |ln|x|| where 0 < x < 1, then what is f’(0.5) equal [NDA – 2024 (2)]
to?

ANSWER KEY

1. a 2. b 3. d 4. d 5. b 6. b 7. c 8. d 9. c 10. b
11. c 12. c 13. a 14. a 15. b 16. a 17. b 18. c 19. c 20. b
21. b 22 c 23. a 24. c 25. c 26. b 27. a 28. c 29. c 30. a
31. b 32. b 33. d 34. d 35. a 36. c 37. a 38. d 39. c 40. b
41. d 42. d 43. d 44. a 45. d 46. a 47. c 48. c 49. d 50. a
51. d 52. a 53. a 54. c 55. c 56. a 57. c 58. a 59. d 60. a
61. a 62. b 63. c 64. a 65. a 66. a 67. a 68. d 69. a 70. d
71. a 72. a 73. b 74. a 75. b 76. b 77. a 78. a 79 b 80. a
81. b 82. a 83. c 84. a 85. b 86. d 87. d 88. c 89. b 90. d
91. c 92. a 93. b 94. a 95. c 96. d 97. a 98. c 99. a 100. d
101. d 102. a

SANDEEP SINGH BRAR Ph:- +91 9700900034 - 395 -

 Solutions
Sol.1. (a) dy dy mx m 1 Sol.14. (a)
f(x) = 2x mx m 1  mym 1 0  Let u = logx x = 1
dx dx mym 1
f(x) = 2x loge 2 m 1
 xm  y   du =0, and v = log x
f (x) = 2x log2 2. 2loge 2 = 2x (loge 2)2 = x   m   dx
Sol.2. (b) y m 1 y  x 
y = (1 + x1/4) (1 + x1/2)(1x1/4)  dv  1
Given,  x    x   y 
m
=(1 + x1/2)(1  x1/2) =1  x dx x
 m  
 dy  1
y  y  x   du  du/dx  0
m2 dv dv/dx
 x 
dx
Sol.3. (d)   1 Sol.15. (b)
y Given curve, y = xx
y  ln tan x
Which is true when m= 2. Taking log on both sides,
1 1
y . sec2 x Sol.10. (b) log y = x log x
tan x 2 tan x y = In(emx + emx) on differentiating both sides w. r. t. x., we get
put x = π/4 On differentiating it w. r. t. x., we get 1 dy 1
y=1 .  x. + log x
.  emx  e mx 
dy 1 d
Sol.4. (d)  y dx x
Given that, y = cot t and x = sin t dx emx  e mx dx
=  dy =(1+logx).y
Then, dy   sin t,and dx  cos t  m e  e
dx
mx  mx
 
 mx 
me mx  me  mx    mx
dt dt 1
 mx
 dy  x x (1+log x)
dy dy/dt  sin t x e mx
e  e  e  dx
Now,   
dx dx/dt cost y  1 1    dy  =(1)1 (1+log1) = 1(1+0) = 1
  dy   m  
Sol.5. (b)   0  dx at x 1
 dx at x  0  11 
We have, y = x  1 Sol.11. (c) Sol.16. (a)
x 1 Statement I.
Given curve,
Now, differentiating w. r. t. x., we get y = sin(ax + b), where a and b are constants, and
Given, y = In (sec x + tan x)
d 1 a ≠ 0.
dy 
x  1  x  1   x  1  x  1 On differentiating it w. r. t. x., we get
Now, differentiating both sides w. r. t. x., we get
 dx dx dy 1
 x  1   sec x  tan x  dy =cos(ax + b).a
dx  sec x  tan x 
2
dx
dx
=  x  1 .1   x  1 .1  x  1  x  1  2 = 1 (sec x. tan x + sec2 x) Again, differentiating both sides w. r. t. x., we
 x  1  x  1  x  1  sec x  tan x 
2 2 2
get
Sol.6. (b) 1 d 2 y =  sin(ax + b) a.a. = a2 sin (ax + b)
= sec x (tan x + sec x) = sec x
Let u = x 2  16, and v = x2  sec x  tan x  dx 2
=  a2 y.
Now, du  1
 2x
Statement II.
dx 2 x 2  16 Given, y = log (cosec x  cot x) [ y = sin (ax + b)]
dy 1 d (cosec x  cot x)   d2 y    b  
= x dv   2  a sin a.    b 
2

and  2x dx  cosec x  cot x  dx  dx at x   b   a  

x  16 2 dx a

Now, rate of charge of u w. r. t. x., is = 1 (cosec x. cot x + cosec2 x) =  a2 .sin(b + b) = a2. sin0 =  a2. 0=0
dx du/dx x 1  cosec x  cot x  Sol.17. (b)
   Let f(x) = x3 and g(x) = x2
dv dv/dx x 2  16 2x =cosec x.  cosec x  cot x  =cosec x
 cosec x  Now, df  x   3x 2 and dg  x   2x
du 1
 So, statements I and II both are true.
dx dx
dv 2 x 2  16
Sol.12. (c) Derivative of x3 w. r. t. x2
du
at  x  3 
1

1

1

1 Given curve is 2x3  3y2 = 7, y ≠ 0. df  x   df  x    dx 
 1 3x
     3x 
2

dv 2 9  16 2 25 2   10 Now, differentiating both sides w. r. t. x, we get dg  x       
 dx   dg x  2x 2
d  x  16 1 at x = 3
2
2.3.x2  3.2.y dy  0 Sol.18. (c)

dx  2
10 dx Let f(x) = sin(sin x)
dy dy x 2 On differentiating w. r. t. t. x, we get
Sol.7. (c) x  y
2
0  
f(x) = x2 – 6x + 8 dx dx y f(x) = cos (sin x). d (sin x)
f '(x) = 2x – 6 Sol.13. (a) dx
f '(c) = 2c – 6 = 0 Let u = sec2 x and v = tan2 x = cos x. cos (sin x)
c = 3 that is lie between [2, 4] Sol.19. (c)
Now, du =2 sec x. sec x. tan x = 2 sec2 x. tan x Let f(x) = |x 1|
Sol.8. (d) dx Redefined the function f(x),
f  x   2sin x ,
and dv =2tan x. sec2x f(x) = 1  x, x  1  f '  x   1, x  1
f'(x)  2sin x (ln 2) cos x    
dx  x  1, x  1  1, x  1

 du  d  sec x   du/dx  2sec x.tan x =1

Sol.9. (c) 2 2 f(2) = 1
Given, xm + ym = 1
On differentiating both sides w. r. t. x., we get dv d  tan 2 x  dv/dx 2 tan x.sec2 x Sol.20. (b)
f(x) = 2x2 + 3x - 5 ,

SANDEEP SINGH BRAR Ph:- +91 9700900034 - 396 -

f '(x) = 4x +3 so answer is AP and GP both d2 y a2
f '(0) + 3f '(-1) = 3 + 3(-1) = 0 2
 2
dx y
Sol.21. (b) Solutions (for next two)
Sol.35. (a)
x
We have, d  1  x  x  =Ax + B …(i)
2 4
2cos 2
1  cos x 2  cot x  2  y = cot1  1  sin x  1  sin x 
y  dx  1  x  x   
1  cos x 2sin 2 x 2  1  sin x  1  sin x 
2 Let us first divide x4+x2 +1 by x2 + x + 1 1
or y = cot
Thus, x  x  1 = x2  x + 1
4 2
dy 1 x 
  
2
  cosec2 x2  x 1 1  sin x  1  sin x
 
  
dx 2 2 
From equation (i), we have
 1  sin x  1  sin x 1  sin x  1  sin x 
Sol.22. (c) 
d (x2  x + 1) = Ax + B
f(x) = x2  
or y = cot1 1  sin x  1  sin x  2 1  sin x 
2
z = fof(x) = x4 dx
z '(x) = 4x3 2x  1 = Ax + B…(ii)  1  sin x  1  sin x 
Sol.23. (a) Sol.29. (c)
We have, x = a (cos + sin) On comparing, the coefficient of A, we get A = 2 or y = cot1  2 1  cos x  
and y = a (sin  cos) Sol.30. (a) 2sin x
On comparing the coefficient of constant terms  2 x 
 dx  a (sin +  cos + sin) in equation (ii), we get B = 1. =cot1  2 cos
d 2 
Sol.31. (b)  
 2sin x cos x 
 dx =a cos Let u = tan1  1  x  1 
2
 2 2
d  
 x   x
and dy  a (cos + sin  cos) 1
or y = cot1 cot 2 
d
and v = tan x  
To find : du
 dy
 a sin  dv on differentiating y = x
d 2
Consider, u = tan1  1  x  1 
2
dy dy 1
 dy   
a sin 
 d   x  dx 2
 tan 
dx dx a cos  Sol.36. (c)
On putting x = tan, we get
d
u = tan1  1  tan 2   1  s = t 2  1  s2  t 2  1
Sol.24. (c)  
 tan   2s ds  2t
We have, dy = tan  
 sec   1 
dt
dx = tan 1 1  1  cos  
   tan   ds t t
 d 2
y d  tan    sin    
 sec 2  dt s t2 1
dx 2   
=tan1  tan    
dx
=tan1  2sin 2
2  t  2t
 
 d y sec2   1    1 t2 
2
dx    2 2
   a cos    
 2sin cos  d 2s 2 t2 1
dx 2  a cos    d   2 
 
2
dt 2 t2 1
2

 d y  sec 
2 3
u = 1 tan1x
dx 2
a 2 = t2 1 t2 1 1
Sol.25. (c)  du  1  1  …(i)  
Given, y = x log x + xex dx
 
2  1 x2 
t 2  1.  t 2  1 s.s2 s3
On differentiating both sides w. r. t. x, we get
Now consider v = tan1 x Sol.37. (a)
dy 1 y1 = log (x + sin x)
 x.  log x  xe x  e x  dv  1 …(ii)
dx x y2 = (x + cos x)
dx 1  x 2
dy1 1 1  cos x 
 dy =1 + logx + xex + ex We have, du  du  dx [by using equation (i)  1  cos x  
dx dx  x  sin x   x  sin x 
dv dx dv
  dy  = 1 + log1 + 1.e1 + 1 and (ii)] …(i)
 
=1 1  dy2 = (1-sin x) …(ii)
 . 1  x 
 dx  x 1 2 1

=1 + 2e (log1=0) 2  1 x2  2 dx
Sol.32. (b) Dividing equation (i) by (ii), we get
dy1
Sol.26. (b) a(1  t 2 ) 2at
f(x) = ax2 + bx + c x y = dy1
 dx
1 t2 1 t2 dy 2 dy 2
f(1) = f(1)
after squaring and addition of both equations dx
a+b+c=ab+c
1  cos x
2b = 0 x2 + y2 = a2
b=0 this equation represent a circle. =  x  sin x  1  cos x 

Sol.27. (a) Sol.33. (d) 1  sin x   x  sin x 1  sin x 
f(x) = ax2 + bx + c x2 + y2 = a2 1
f '(x) = 2ax + b diff. both side w,r,t x Sol.38. (d)
f '(a) = 2aa + b, f '(b) = 2ab + b, f '(c) = 2ac + b 2x + 2yy ' = 0 xa yb = (x – y) a+b
given that a, b, c are in AP dy x taking log both side
than after multiply with 2a  a logx + b logy = (a + b) log(x – y)
dx y
2aa, 2ab, 2ac will be in AP differentiate w.r.t. x
now add b in each term Sol.34. (d) a b dy 1  dy 
d2 y y  xy ' y  x( x / y)   (a  b). .1  
2aa + b, 2ab + b, 2ac + b will be in AP   x y dx x y  dx 
Sol.28. (c) dx 2 y2 y2 by solving it
f '' (x) = 2a
f '' (a) = f '' (b) = f '' (c) = 2a

SANDEEP SINGH BRAR Ph:- +91 9700900034 - 397 -

 dy y Sol.45. (d) So, statement II incorrect.
 Putting this value in equation (v) Sol.53. (a)
dx x
dy y f(2) = 12 + 2 (5) f(x+y) = f(x).f(y)
 0 f(2) = 2 f(0) = f(0+0) = f(0)  f(0)
dx x
Sol.39. (c) and f  (3) = 6 or f(0) = f(0). f(0) + f(0).f(0){by product rule}
f(x) = |x2  5x + 6| Now the f(x) is or f(0) = 2f(0).f(0)
f(x) = |2x  5| f(x) = x3 + x2 (5) + 2x + 6 or f(0) = 0
f(4) = |2.45| f(x) = x3  5x2 + 2x + 6 Now f(5) = f(0+5) = f(0). f(5)
Sol.46. (a) f(5) = f(0). f(5) + f(5). f(0)
f(4) = |85|
Put x = 1 in f(x) f(5) = 0 + f(5). f(0) {from (i)}
f(4) = |3| = 3
f(1) = 1  5 + 2 + 6 f(5) = f(5). f(0)
Sol.40. (b)
f(1) = 4 Sol.54. (c)
f  (x) = |2|
Put x = 2 2
f (2.5) = 2 let y  f  x  , p  dy and q  d y ,
f(2) = 8  20 + 4 + 6 dx dx 2
Sol.41. (d)
f(2) =  2
This function is continuous for all values of x dx 1 d2 x 1 dp
and not differentiable at origin.
Sol.47. (c)   2  2
From equation (iii) dy p dy p dy
Sol.42. (d)
For x < 0 f (10) = 6 d2 x 1 dp 1 dp dx q
Now, f(0) = 6   2  2  3
dy f 0  h   f 0 dy 2 p dy p dx dy p
 lim Sol.48. (c)
dx h 0 5 I. f(2) =  2  d 2 y   dy  3
1
 h   0 f(1)  f(0) = 4  6 =  2 =   
 dx 2   dx 
= lim 3  lim
1 f(2) = f(1)  f(0) (True)  
h 0 h h 0 3
II. f(2)  2 f(1) = 2  2 (5) Sol.55. (c)
 
dy 1 = 2 + 10
 2x e  x
2

xe  x
2

dx 3 = 12
f(x) =
Sol.49. (d)
Sol.43. (d) 2 1  e x 1  e x
2 2

y = log10 x + logx 10 + logx x + log10 10

3 which is defined xR, except x = 0
x sin x cos x y = log10 x + 1  1  1
f(x) = 6 1 0 log10 x f(x) is differentiable on (,0)  (0, )
Sol.56. (a)
p p2 p3 y = log10 x + (log10 x)1 + 2  cos x .... 
dy 1 log10 e  (log10 x)2. 1 y =  cos x  cos x 
Expanding corresponding to C3  log10 e  0
f(x) = cos x (6p2 + p) + 0 + p3 dx x x …(i)
(x3  6sinx) dy 1 log e  (log x)2. 1 log e or y = (cos x)y
On differentiating w. r. t. to x.
 10 10 10
Taking log on both the sides,
dx x x
f(x) =  sin x (6p2 + p)  p3 (3x2 + 6cosx) log y = y.log (cos x)
At x = 10, differentiating w. r. t. x., we get
…(i)
 dy  2 1
 log10 10  (log10 10) . log10 e   sin x 
1 1 dy dy
  log  cos x  .
Put x = 0    y
f(0) = 0  p3 = (0 + 6 cos0°)  dx x 10 10 x y dx  cos x  dx
f(0) =  p3 (6) = 1 .1  12 . 1 .1  1  1  0
or dy log  cos x   1   y tan x
f(0) =  6p3 10 10 10 10
dx  y 
Sol.44. (a) Sol.50. (a)
Again differentiation of equation (i) We have, f(x) = (|x|)  (|x1|)2 or dy  y 2 tan x  y 2 tan x

f  (x) =  cos x(6p2 + p)  p3 (6x  6sin x) When x > 1, f(x) = [x  (x  1)]2 = 1 dx y.log  cos x   1 1  y.log  cos x 
f  (x) =  cos x (6p2 + p)  6p3 (x  sinx) f(x) = 0
Sol.57. (c)
Pitting x = 0 Sol.51. (d)
f  (0) =  cos 0° (6p2 + p)  6p3 (0  sin 0)  1 x2 for 0  x  1
 f(x) = (|x|  |x1|)2 
f  (0) =  (6p2 + p)  6p3 (0) When 0 < x < 1 f (x)   Inx for 1  x  2
f  (0) =  (6p2 + p) f(x) = [x  (x  1)}]2 = (2x  1)2 In 2  1  0.5x for 2  x  0

Since, f  (0) = 0 f(x) = 2 (2x  1) (2) = 8x  4
2x for 0  x  1
(6p2 + p) = 0 Sol.52. (a) 
f '(x)  1/ x for 1  x  2
p(6p + 1) = 0  1, x0  0.5 for 2  x  0
p = 0, p = 1 |x|  |x1| = 2x 1 0  x  1 
 function is not differentiable at x = 1
6  1, 1 x
Solutions (for next three)  so f(x) = –2x for 0 < x < 1
Sol.58. (a)
f(x) = x3 + x2 f(1) + xf(2) + f (3) f(x) =  1, when 0  x or x  1
On differentiating w. r. t. ‘x’  y = sec1  x  1   sin 1  x  1 
 2x  1 , when  x 1    
2

f(x) = 3x2 + 2x f(1) + f(2) …(i)  x 1   x 1

f(x) = 6x + 2f(1) …(ii) f(x) =  0, when 0  x or x  1
=cos1  x  1   sin 1  x  1 
f (x) = 6     
4  2x  1 , when
…(iii)   x 1  x 1  x 1
Putting x = 1 in equation (ii) and x = 3 in
equation (iii) f (x) = 0, when 0  x or x  1 
=  sin 1 x  cos 1 x   

f(1) = 3 + 2 f(1) + f(2) 8, when  x 1 2  2
f(1) + f(2) =  3 …(iv) I. f(2) = 1 and f(5) = 1
 dy  0
Now, f(2) = 12 + 2f (1) …(v) So, f(2) = f(5) = 1
dx
From equations (iv) and (v) Statement I is correct
Sol.59. (d)
f(1) + 12 + 2 f(1) =  3 II. f  (20 = 0 All statements are correct.
3f(1) =  15 f (0.5) = 8 and f  (3) = 0 Sol.60. (a)
f(1) =  5 Then, f (2) + f  (0.5) + f  (3) = 0 + 8 + 0 = 8

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y = cos1  2x  f(x) = dX
 ex e1
 2 
f x  h  f x f x f h  f x
1 x  dx
lim  lim
h 0 h 0
Put x = tan, h h dY ex xe x
 e1  e
y = cos1  2x  f  x  f  h   1 f  x   h g  h    h  dX ex ex
 2 
= lim  lim
1 x  h 0 h h 0 h Sol.78. (a)
Put x = tan, =abf (x) y = sin(lnx) + cos(lnx)
dy 1 1
y = cos1  2 tan   = cos1 (sin2) Sol.69. (a)  cos(ln x).  sin(ln x).
  f(x) = sec2(tan-1x) dx x x
 1  tan  
2

f(x) = 1 + tan2(tan-1x) at x = e
=cos1 cos    2  f(x) = 1 + x2 dy 1 1
   cos(lne).  sin(lne).
2  f '(x) = 2x dx e e
=   2    Sol.70. (d)
dy cos1  sin1
 where 0    2    f(x) = sin(cosx) 
2  2  f'(x) =cos (cosx) (–sinx) dx e
Sol.61. (a) Sol.71. (a) Sol.79. (b)

f(0) = lim    
f h f 0 x = et cost
dY
Y  2(sin x )   2(sin x ) 2sin x cos x ln 2
2 2
y = et sint
h 0 h dX dx
2  et ( sin t )  et cos t
= lim h log h  0 X  sin x 
dX
 cos x dt
h 0 h dx dy
Sol.62. (b)  et (cos t )  et sin t
dY dt
 2sin x.2(sin x ) ln 2  sin x.2(sin x ) ln 4
2 2
2
y= ex dX dx et ( sin t )  et cos t
Sol.72. (a) 
y'  2xex
2
dy et (cost)  et sin t
xy = ex−y
2 dx  sint  cos t
at x = π, y' = 2π e
taking log both sides 
y.logx = x − y ...(i) dy cost  sin t
Sol.63. (c) at t = 0
put x = 1 then y = 1
f(x) = etanx + n(sec x)  x differentiate (i) w.r.t. x dx
1
f(x) = etanx.sec2 x + sec x tan x  1 1 dy dy ..(ii) dy
sec x y.  log x.  1 
x dx dx Sol.80. (a)
f(/4) = 2e + 1  1 = 2e put x = y = 1 in this equation Y = e e  dY  e e .e x
x x
Sol.64. (a)
dy dx
y = tan1  5  2 tan x  0
  dx
X = ex  dX  e x
 2  5 tan x  Sol.73. (b)
dx
 5  Diff. equation (ii) of above question x

= tan1  2  tan x  dy dY e x .e x x

  x log y y   ee
dx d 2 y 1 dy d2y dX ex
 5   log x. 2  .  2
 1   2  tan x  x2 dx x dx dx Sol.81. (b)
    If y = (1 + x) (1 + x2) (1 + x4) (1 + x8) (1 + x16)
dy
5
put x = y = 1 and  0 in above equation then dy at x = 0
= tan1 tan1 tan x  tan 1 5  x dx T.f.
2 2 d2y dx
1 Multiply both sides by (1 – x)
 dy   1 dx 2
(1 – x) y = (1 – x) (1 + x) (1 + x2) (1 + x4) (1 +
dx 2 x Sol.74. (a) x4) (1 + x16)
Sol.65. (a)   (1 – x)y = (x – x6) (1 + x16)
d   co t 1 x 
u du  v dv d (tan 1 x)  2   1 ( 1 – x)y = 1 – x32

dx dx d (co t 1 x) d (co t 1 x) dy  32 x 31  y
=eax sin bx [aeax sin bx + beax cos bx] Sol.75. (b) (1 – x) 
dx 1 x
+ eax cos bx [aeax cos bx  beax sin bx] xmyn = am+n
y at x = 0 is 1
=e2ax [a sin2 bx + b sin bx cos bx + differentiate w.r.t. x
mxm-1 yn + nyn-1.xm y' =0 dy  323  1 = 1
a cos2 bx  b sin bx cos bx] 
=ae2ax xm-1.yn-1.my + nyn-1.xm-1 x. y' = 0 dx 1
y’ = −my/nx Sol.82. (a)
Sol.66. (a) Sol.76. (b) 1 𝑑𝑦 
y = cos x. cos4x. cos8x then at x = taking
y = (sin x)x e  c  4
𝑦 𝑑𝑥 4
log both sides
x  x
  cos x   log  subx  .  1
dy
   sin   diff. w.r.t. 𝛉 log y = log [cosx . cos4x cos 8x]
dx  sin x   d   d  log y = log cos x + log cos 4x + log cos 8x
/ 2 e       4   
 dy   1   1  d   d  1 dy
 dx  x   6   2    6 3  log 2 
       d 
e 
 4 
 d
   0 y dx
2/6  6 log 2  3  
  = 1  sin x   1
sin 4x.4  1  sin 8x.8
 6  d cos x cos 4x cos 8x
   0
Sol.67. (a) d dy
 y tan x  4 tan 4 x  8 tan 8 x 
For   x   , 1  sin 2 x  sin x  cos x  d  d dx
4 2 Sol.77. (a) T.f. dy at x  
Result = cos x + sin x Y = ex dx 4
Sol.68. (d) X = xe
Here 1 dy   1  4 tan   8 tan 2 
dY
 ex y dx
dx

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1 dy =[-1-0-0] dy  x 2  x  1   d  x 2  x  1  for 1 < x < 2 , f(x) is positive but fof(x) is
 2 ln 2   2  negative
 x  x  1  dx  x  x  1 
y dx dx so fof(x) = – ln(lnx)
1 dy differentiation of fof(x) =  1
 1 let differentiation of last step is g(x), than put x =
y dx 0 x ln x
 2 ln1g (0)  0
Sol.83. (c) dy Sol.97. (a)
fof(x) = x4 then f(x) = x2 dx f(x) = |ln|x||
f ’(x) = 2x Sol.90. (d) at x = 0.5 , lnx is negative
f ’(1) = 2 d  1  x 4  x8  so f(x) = – lnx
 
Sol.84. (a) dx  1  x 2  x 4  f’(x) = – 1/x
xy.yx = 1 f’(0.5) = – 2
take log both sides  
 
d  1  x2  x4 1  x2  x4   Sol.98. (c)
y log x + nlog y = 0 dx  1  x2  x4 
  2x  3   3
diff. w.r.t. x both sides y f   f 2  
1 1
y. + y' log x + log y + x. . y' = 0 
d
 
1  x  x  2 x  4 x  ax  bx3
2 4 3  x   x
𝑥 𝑦 dx dy  3  3 
put x = y =1 a = 2 and b = 4  f '  2   2 
dx  x  x 
1 + 0 + 0 + y' = 0 Sol.91. (c)
y' = –1 given that f’(x) = cos (ln x)
x x 2  16
Sol.85. (b) y  8 ln x  x 2  16  3   3 
f '  2    cos ln 2   
x 
2
y= x x  x   x 
dy x 2  16 x.2 x 1  2x 
  8 1   dy 3   3  3   2x  3  
x2 dx 2 2.2 x 2  16 x  x 2  16  2 x 2  16   cos ln 2     2 cos ln  
y= x dx x 2   x  x   x 
log y = x2 log x dy x 2  16 x2  1 
1 𝑑𝑦 1    8 

Sol.99. (a)
= x2. + 2x. log x dx 2 2 x 2  16  x 2  16  let
𝑦 𝑑𝑥 𝑥
𝑑𝑦 1
= y (x + 2x log x) solve by taking LCM g ( x)  x   y
𝑑𝑥
𝑝𝑢𝑡 𝑥 = 𝑦 = 1 dy x 2  16 x
  x 2  16
dx x 2  16 1
dy
1 fog(x)  x  3
3

dx Sol.92. (a) x
Sol.86. (d) differentiation of sin2x w.r.t. cos2x than
y= [x + 1] 1  1
3
d (sin2 x) 2 sin x cos x  1
  1 f ( y)  x 3    x    3 x  
 
G.I.f. is constant function between
x= – 3 to x = –4 d cos2 x  2 sin x cos x x3  x  x
so its derivative is 0 Sol.93. (b) f ( y)  y 3  3 y
Sol.87. (d)    so f(x) = x3 + 3x
𝑚 cos cex  cos ec x
f(x)= +2nx + 1
𝑥  180  g[f(x)  3x] = g(x3)
f'(x)= m  2n  0 differentiate by chain rule
= x3  1
x2 d d    x3
cos cex  cos ec x
 m dx dx  180 
=  2n  0 Sol.100. (d)
4        f(x) = x3 + 3x
8n = m  cos ec x  cot x f '(x) = 3x2
180  180   180 
m + 8n = m + m =2m f ''(x) = 6x

It depends upon m. so can not find due to  cos ecxcotx Sol.101. (d)
sufficient data 180 h(x) = f(2f(x)+2)
Sol.88. (c) Sol.94. (a) differentiate w.r.t. x
h(x) = {f(x)}2 + {g(x)}2 f(x) = |lnx| h'(x) = f '(2f(x)+2).2f '(x)
h’(x) = 2f(x).f '(x) + 2g(x).g '(x) ……(i) at x = 0.5 , lnx is negative h'(0) = f '(2f(0)+2).2f '(0)
given that f '(x) = g(x) so f(x) = – lnx given that f(0) = – 1
differentiate both sides f’(x) = – 1/x h'(0) = f '(0).2f '(0) = 2
f "(x) = g'(x) = – f(x) f’(0.5) = – 2 Sol.102. (a)
by equation (i) Sol.95. (c) g(x) = (h(x))2
h'(x) = 2f(x).g(x) – 2g(x).f(x) = 0 f(x) = |lnx| g '(x) = 2h(x).h'(x)
so h'(3) = 0 at x = 2 , lnx is positive
g '(0) = 2h(0).h'(0)
if h'(x) = 0, than h(x) is a constant function so f(x) = lnx
f’(x) = 1/x by above question h'(0) = 2
so h(1) = h(2)
Sol.89. (b) f’(2) = 1/2 and h(0) = f(2f(0)+2) = f(0) = - 1
 x2  x  1  Sol.96. (d) so g '(0) = – 4
y  ln 2  2  f(x) = |lnx|
 x  x 1 fof(x) = |ln|lnx||

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