MATHEMATICS

Time Limit : 3 Sitting Each of 70 Minutes duration approx.

 Question bank on Method of differentiation
There are 72 questions in this question bank.
Select the correct alternative : (Only one is correct)

1
Q.13 If g is the inverse of f & f  (x) = then g  (x) =
1 x 5

1 1
(A*) 1 + [g(x)]5 (B) (C)  (D) none
1  [g(x)]5 1  [g(x)]5

 n e2  d 2y
Q.25 If y = tan1  x 
+ tan1 3  2 n x then =
 nex 2  dx 2
  1  6 n x
(A) 2 (B) 1 (C*) 0 (D)  1

 3x  4  dy
Q.37 If y = f   & f  (x) = tan x2 then =
 5x  6  dx
2
 3x  4 1
(A) tan x3 (B*)  2 tan   .
 5x  6  (5x  6)2
 3 tan x 2  4 
(C) f   tan x2 (D) none
 5 tan x 2  6 

dy 1
Q.48 If y = sin1  x 1  x  x 1  x 2  & = + p, then p =
dx 2 x (1  x)

(A) 0 (B) sin1 x (C) sin1 x (D*) none of these

[Sol. 2
x = sin ; x = sin 
y = sin–1(sin cos + sin cos) = sin–1(sin( + ) =  +  = sin–1x + sin–1 x
1 1 3 1
Dy =  assuming x2 + x 1 i.e. 0 x < ]
2 2 x (1  x )
1 x 2
 2 2
Note sin–1  x 1  y  y 1  x 
 
–1
= sin x + sin y–1 if x2 + y2  1
= – (sin–1x + sin–1y) if x2 + y2  1 ]

 2x  1 dy
Q.59 If y = f  2
 & f  (x) = sin x then =
 x  1 dx

(A)
1  x  x2  2x  1
sin  2  (B*)

2 1  x  x2  sin  2x  1
2 2
2 2
1  x  2  x  1
1  x   x  1

1  x  x2  2x  1
(C) 2 sin   (D) none
1  x  2  x 2  1
 x10
Q.611 Let g is the inverse function of f & f  (x) = . If g(2) = a then g  (2) is equal to
1  x  2

5 1  a2 a 10 1  a 10
(A) 10 (B*) (C) (D)
2 a 10 1  a2 a2
[Sol. f [ g(x) ] = x  f  [ g (x)] . [g(x)] = 1  f  (a).g(2) = 1 [putting x = 2]
a 10 1  a2
given, f  (a) =  g  (2) = ]
1  a2 a 10
Alternative : g [f (x)] = x
g' [f (x)] . f ' (x) = 1
now g (2) = a  f (a) = 2
 g and f are inverse of each other
now f (x) = 2  g (2) = x = a
 g ' (2) . f ' (a) = 1
1 1 a 2
g ' (2) =  = 10 ]
f (a ) a
dy
Q.715 If sin (xy) + cos (xy) = 0 then =
dx
y y x x
(A) (B*)  (C)  (D) y
x x y

2x dy 
Q.820 If y = sin1 2 then dx  is :
1 x  x  2

2 2 2
(A) (B) (C*)  (D) none
5 5 5

 1  1
Q.922 The derivative of sec1  2
 w.r.t. 1  x 2 at x = is :
 2x  1 2
(A*) 4 (B) 1/4 (C) 1 (D) none

 d   3 d 2y 
Q.1023 If y2 = P(x), is a polynomial of degree 3, then 2    y . 2  equals :
 dx   dx 
(A) P  (x) + P  (x) (B) P  (x) . P  (x) (C*) P (x) . P  (x) (D) a constant
d
[Sol: 2 (y3 y2) = 2 (y3.y3 + 3 y2 y1y2). Now differentiate y2 = P(x) thrice)]
dx

Q.1124 Let f(x) be a quadratic expression which is positive for all real x . If
g(x) = f(x) + f (x) + f  (x), then for any real x, which one is correct .
(A) g(x) < 0 (B*) g(x) > 0 (C) g(x) = 0 (D) g(x)  0
 dy
Q.1227 If xp . yq = (x + y)p + q then is :
dx
(A) independent of p but dependent on q (B) dependent on p but independent of q
(C) dependent on both p & q (D*) independent of p & q both .
[Hint: Note that f (x, y) is a homogeneous function  dy/dx = y/x]

g (x) . cos x1 if x  0
Q.1328 Let f(x) =  where g(x) is an even function differentiable at x = 0, passing
0 if x  0
through the origin . Then f  (0) :
(A) is equal to 1 (B*) is equal to 0 (C) is equal to 2 (D) does not exist
[Hint : g (x) = g ( x)  g  (0) = 0 ]

1 1 1 dy np
Q.1430 If y = + + then at e m is equal to:
1  x n  m  x p m 1  xm  n  xp n 1  xm  p  xn p dx
(A) emnp (B) emn/p (C) enp/m (D*) none
[ Hint : multiply Numerator and Denominator by xm ]

log sin 2 x cos x

Q.1531 Lim
x0 x has the value equal to
log 2 x cos
sin
2
2
(A) 1 (B) 2 (C*) 4 (D) none of these
[Hint : Use base changing theorem and then use L' Hospital's rule]

Q.1632 If f is differentiable in (0, 6) & f  (4) = 5 then Limit

c h=
f ( 4)  f x 2
x2
2x
(A) 5 (B) 5/4 (C) 10 (D*) 20

Lim xm (ln x)n where m, n  N then :

Q.1733 Let l = x 0

(A*) l is independent of m and n (B) l is independent of m and depends on m

(C) l is independent of n and dependent on m (D) l is dependent on both m and n
n
Lim (ln x)
[Hint : l = x 0  m
x

Lim n! ( 1) n m
Using L' Hospital's rule l = x x =0]
0
mn

cos x x 1
f  (x)
Q.1834 Let f(x) = 2 sin x x 2 2x . Then Limit
x0 =
x
tan x x 1
(A) 2 (B*)  2 (C)  1 (D) 1
[Hint : C2  C2 – xC3  x2 (tan x – cos x)]  f' (x) = (tan x – cos x) 2x – x2 (sec2x + sin x) ]
 cos x sin x cos x
 
Q.1936 Let f(x) = cos 2x sin 2x 2 cos 2x then f    =
2
cos 3x sin 3x 3 cos 3x
(A) 0 (B) – 12 (C*) 4 (D) 12
[Hint : Differentiate column wise, where 1 = – 4; 2 = 0 and 3 = 8]
Q.2037 People living at Mars, instead of the usual definition of derivative D f(x), define a new kind of derivative,
D*f(x) by the formula
f 2 (x  h)  f 2 (x)
D*f(x) = Limit where f(x) means [f(x)]2. If f(x) = x lnx then
h 0 h
D * f ( x ) x  e has the value
(A) e (B) 2e (C*) 4e (D) none
[Hint: D*f(x) = 2f(x).f (x)
D*(x lnx) = 2x lnx (1 + lnx)]
f (x)  g (x)
Q.2138 If f(4) = g(4) = 2 ; f  (4) = 9 ; g  (4) = 6 then Limit
x4 is equal to :
x 2

3
(A*) 3 2 (B) (C) 0 (D) none
2
[Hint : Use L' Hospital's rule or first rationalise]
f (x  3h)  f (x  2h)
Q.2241 If f(x) is a differentiable function of x then Limit
h0 =
h
(A) f  (x) (B*) 5f  (x) (C) 0 (D) none
d 2x
Q.2345 If y = x + ex then is :
dy 2
ex ex 1
(A) ex (B*)  3 (C)  2 (D) 3
1 e  x
1 e  x
1 e 
x

d 2y
Q.2448 If x2y + y3 = 2 then the value of at the point (1, 1) is :
dx 2
3 3 5
(A)  (B*)  (C)  (D) none
4 8 12
g (x) . f (a )  g (a ) . f (x)
Q.2549 If f(a) = 2, f  (a) = 1, g(a) =  1, g  (a) = 2 then the value of Limit is:
xa xa
(A)  5 (B) 1/5 (C*) 5 (D) none

Q.2650 If f is twice differentiable such that f  (x)   f (x), f  (x)  g(x)

2 2
h  (x)   f (x)   g(x) and
h (0)  2 , h (1)  4
then the equation y = h(x) represents :
(A) a curve of degree 2 (B) a curve passing through the origin
(C*) a straight line with slope 2 (D) a straight line with y intercept equal to  2.
[Hint: h   (x) = 0 ]
Q.2756 The derivative of the function, f(x)=cos-1
R
S1 U + sin R
(2 cos x  3 sin x) V S1
1
U
(2 cos x  3 sin x) V
T13 W T13 W
3
w.r.t. 1  x 2 at x = is :
4
3 5 10
(A) (B) (C*) (D) 0
2 2 3
2 3 3
[Sol. Put cos  = ; sin  = ; tan  =
13 13 2
–1 –1
y = cos { cos(x + ) } + sin {cos(x – ) }

= cos–1{cos(x + ) + – cos–1{cos( – x)} (think !)
2

=x++ –+x
2

y = 2x + ; z = 1 x 2
2
dy
now compute ]
dz

Q.2857 Let f(x) be a polynomial in x . Then the second derivative of f(ex), is :

(A) f  (ex) . ex + f  (ex) (B) f  (ex) . e2x + f  (ex) . e2x
x
(C) f  (e ) e 2x (D*) f  (ex) . e2x + f  (ex) . ex
[Hint: y1 = f (ex.) ex ]

1 2x + 1
Q.2959 The solution set of f  (x) > g  (x), where f(x) = (5 ) & g(x) = 5x + 4x (ln 5) is :
2
(A) x > 1 (B) 0 < x < 1 (C) x  0 (D*) x > 0

x2  1 x2  1 dy
Q.3062 If y = sin1 2 + sec1
2 , x > 1 then is equal to :
x 1 x 1 dx

x x2
(A) (B) (C*) 0 (D) 1
4
x 1 x4  1

x2  1 x2  1 
[Hint : sec–1 2 = cos–1
2 hence given expression = ]
x 1 x 1 2

x x x x x x dy
Q.3165 If y = ......  then =
a  b a  b a b dx
a b a b
(A) (B) (C) (D*)
ab  2 ay ab  2 by ab  2 by ab  2 ay

x
[Hint: y = ]
x
a
by
Q.3266 Let f (x) be a polynomial function of second degree. If f (1) = f (–1) and a, b, c are in A.P., then f '(a),
f '(b) and f '(c) are in
(A) G.P. (B) H.P. (C) A.G.P. (D*) A.P.
[Sol. Let 2
f (x) = px + qx + r
f (1) = f (–1) gives p + q + r = p – q + r
hence q = 0
Hence f (x) = px2 + r
f ' (x) = 2px ....(1)
Given a, b, c are in A.P.
hence 2pa, 2pb, 2pc will also be in A.P.
or f ' (a), f ' (b), f ' (c) will also be in A.P.  (D) ]
y y1 y2
Q.3367 If y = sin mx then the value of y 3 y4 y 5 (where subscripts of y shows the order of derivatiive) is:
y6 y7 y8
(A) independent of x but dependent on m (B) dependent of x but independent of m
(C) dependent on both m & x (D*) independent of m & x .
[Hint : Differentiate column wise. D1 = 0 & D2 = 0.
F
Gn I
 mx J. lastly D is also zero  Ans is 0]
Note that Dn sin m x = mn. sin H2 K 3

y 
Q.3470 If x2 + y2 = R2 (R > 0) then k = where k in terms of R alone is equal to
2 3
1  y  
1 1 2 2
(A) – 2 (B*) – (C) (D) –
R R R R2
[Sol. 2x + 2yy' = 0
x
x + yy' = 0  y' = – ....(1)
y
1 + yy'' + (y')2 = 0
1  ( y' ) 2
y'' = –
y

y' ' 1  ( y' ) 2 1 1 1 1

now k = =– =– =– =– =– ]
1  (y' )  2 32
y 1  ( y' ) 2 y 1  ( y' ) 2
y 1
x2
2
y x 2 R
y2
Q.3571 If f & g are differentiable functions such that g  (a) = 2 & g(a) = b and if fog is an identity function then
f  (b) has the value equal to :
(A) 2/3 (B) 1 (C) 0 (D*) 1/2
[Hint : f(g(x)) = x ; f  (g(x)) . g  (x) = 1 ; f  (g(a)) . g  (a) = 1 ; f  (b) . 2 = 1
 f  (b) = 1/2 ]

x3
Q.3672 Given f(x) =  + x2 sin 1.5 a  x sin a . sin 2a  5 arc sin (a2  8a + 17) then :
3
(A) f(x) is not defined at x = sin 8 (B) f  (sin 8) > 0
(C) f  (x) is not defined at x = sin 8 (D*) f  (sin 8) < 0
 x3
[Sol. f (x) =  + x2 sin 6 – x sin4 . sin8 – 5 sin–1 ((a – 4)2 + 1)
3
f (x) = – x2 + 2x sin6 – sin4 sin8 (a = 4)
2
f (sin8) = – sin 8 + 2 sin6 sin8 – sin4 sin8
= sin8 [ – sin8 + 2 sin6 – sin4]
= – sin8 [sin8 + sin4 – 2sin6] = – sin8[2sin6 cos2 – 2sin6]
= 2 sin8 sin6 [ 1 – cos2] ]

Q.3776 A function f, defined for all positive real numbers, satisfies the equation f(x2) = x3 for every x > 0 . Then
the value of f  (4) =
(A) 12 (B*) 3 (C) 3/2 (D) cannot be determined
2 2
[Hint : 2x f  (x ) = 3 x ; 4 f  (x) = 12  f  (4) = 3 ]

Q.3877 Given : f(x) = 4x3  6x2 cos 2a + 3x sin 2a . sin 6a + n 2 a  a 2   then :

(A) f(x) is not defined at x = 1/2 (B) f  (1/2) < 0
(C) f  (x) is not defined at x = 1/2 (D*) f  (1/2) > 0
[Hint : 2a – a = – (a – 2a) = – ((a – 1) – 1) = 1 – (a – 1)2, hence f (x) can be defined only when a = 1.
2 2 2

Now f ' (x) = 12 x2 – 12 x cos 2 + 3 sin 2 sin 6

F
G1I
f' H2 JK= 3 – 6 cos 2 + 3 sin 2 sin 6 = 3 (1 + sin 2 sin 6) – 6 cos 2.
Note that cos 2 < 0 and 1 + sin 2 sin 6 > 0  D]

d 2y dy
Q.3978 If y = (A + Bx) emx + (m  1)2 ex then 2  2m + m2y is equal to :
dx dx
(A*) ex (B) emx (C) emx (D) e(1  m) x
[Hint : multiply given equation by emx & then differentiate twice ]

Q.4079 Suppose f (x) = eax + ebx, where a  b, and that f '' (x) – 2 f ' (x) – 15 f (x) = 0 for all x. Then the product
ab is equal to
(A) 25 (B) 9 (C*) – 15 (D) – 9
[Sol. 2 ax 2
(a – 2a – 15)e + (b – 2b – 15)e = 0 bx

 (a2 – 2a – 15) = 0 and b2 – 2b – 15 = 0

 (a – 5)(a + 3) = 0 and (b – 5)(b + 3) = 0
 a = 5 or – 3 and b = 5 or – 3
 a  b hence a = 5 and b = – 3
or a = – 3 and b = 5
 ab = – 15 Ans. ]
Q.4180 Let h (x) be differentiable for all x and let f (x) = (kx + ex) h(x) where k is some constant. If h (0) = 5,
h ' (0) = – 2 and f ' (0) = 18 then the value of k is equal to
(A) 5 (B) 4 (C*) 3 (D) 2.2
[Sol. f ' (x) = (kx + ex) h'(x) + h (x)(k + ex)
f ' (0) = h ' (0) + h (0)(k + 1)
18 = – 2 + 5(k + 1)  k=3 ]

Q.4283 Let ef(x) = ln x . If g(x) is the inverse function of f(x) then g  (x) equals to :
(A) ex (B) ex + x (C*) e ( x  ex ) (D) e(x + ln x)
 y
[Hint: Let f (x) = y  x = f –1(y) = g (y)  x  ee
dy
 = e e y ·e y = e ey  y
dx
x
x
hence g ' (x) = e e ]
dy
Q.4384 The equation y2exy = 9e–3·x2 defines y as a differentiable function of x. The value of for
dx
x = – 1 and y = 3 is
15 9
(A) – (B) – (C) 3 (D*) 15
2 5
 xy  dy  dy
[Sol. y2  e  x  y   + exy · 2y = 9e–3 · 2x
  dx  dx
put x = – 1 and y = 3
 3  dy  dy
9  e   1  3   + e–3 · 6 = – 9e–3 · 2
  dx  dx

 dy  dy
– 9   3 + 6 = – 18
 dx  dx
dy dy
3 = 45  = 15 Ans. ]
dx dx

Q.4485 Let f(x) = x x  

x
and g(x) = x
xx  then :
(A) f  (1) = 1 and g  (1) = 2 (B) f  (1) = 2 and g  (1) = 1
(C) f  (1) = 1 and g  (1) = 0 (D*) f  (1) = 1 and g  (1) = 1
Q.4588 The function f(x) = ex + x, being differentiable and one to one, has a differentiable inverse f–1(x). The
d –1
value of (f ) at the point f(l n2) is
dx
1 1 1
(A) (B*) (C) (D) none
n2 3 4
dx dx 1 dx  1 1
[Hint: y = ex + x ; diff. w.r.t y, 1 = (ex + 1) ; = x   = n 2 = ]
dy dy e 1 dy  x n 2 e  1 3

dy dy dx 1
[Alternate :  ex  1 ; 3  ]
dx dx x  n 2 dy 3

log sin|x| cos3 x 

Q.4689 If f (x) = for |x| < x0
x 3
log sin|3x| cos3  
 2
=4 for x = 0
  
then, the number of points of discontinuity of f in   ,  is
3 3
(A) 0 (B) 3 (C*) 2 (D) 4
 
[Hint: f(0) = 4; f is not defined at x = + , as base of the logarithm can't be one.
6
Now evaluate Lim
x0
f (x) using L' Hospital's rule to check the continuity at x = 0 ]

(a  x) a  x  (b  x) x  b dy
Q.4791 If y = then wherever it is defined is equal to :
a x  xb dx

x  (a  b) 2 x  (a  b) (a  b) 2 x  (a  b)
(A) (B*) (C)  (D)
(a  x) (x  b) 2 (a  x) (x  b) 2 (a  x) (x  b) 2 (a  x) (x  b)

[Hint: Read Nr =  ax   

3
xb 
3
and factorise]

d2 y dy
Q.4893 If y is a function of x then 2 + y = 0 . If x is a function of y then the equation becomes :
dx dx
3
d2 x dx d2 x  dx
(A) 2 +x =0 (B) +y   =0
dy dy d y2  dy
2 2
d2 x  dx d2 x  dx
(C*) y   =0 (D) x   =0
d y2  dy d y2  dy

d2y dy
[Sol. Given 2  y 0
dx dx

FI FI
dy 1 d2y d 1 G
G JJ d G1 J. dy 1d2x 1
now 
dx dx
 2 
dx dx dx G J Gdx JJ dx
dy G

dxF2
. 2.
IJdy dx
dy dy H K Hdy K dy
G
HK dy

d2 x
d2y d2 y
 (putting in (1) )
dx 2 dx F
G IJ 3

dy HK
d2x

d2y dy F
Gdy I d x
2 2

Fdx I
3
y
dx
=0  y
Hdx JK dy  0 2
C ]
G
Hdy JK
Q.4994 A function f (x) satisfies the condition, f (x) = f  (x) + f  (x) + f  (x) + ......  where f (x) is a
differentiable function indefinitely and dash denotes the order of derivative . If f (0) = 1, then f (x) is :
(A*) ex/2 (B) ex (C) e2x (D) e4x
[Sol. f (x) = f  (x) + f  (x) + f  (x) + ...... 
f ' ( x)  f ' ' ( x)  f ' ' ' ( x)  f ' ' ' ' ( x)  ........ 
  2 f ' (x) = f ' (x) + f '' (x) + f ''' (x) + ......
f '( x) 1
 2 f '( x)  f ( x)  
f ( x) 2
1
n f ( x)  x  c
2
if x = 0 ; f(0) = 1  c = 0
x
hence n f ( x) 
2
x
 f ( x)  e 2 ]

cos 6x  6 cos 4 x  15 cos 2 x  10 dy

Q.5099 If y = , then =
cos 5x  5 cos 3x  10 cos x dx
(A) 2 sinx + cosx (B*) –2sinx (C) cos2x (D) sin2x
[Sol. Nr = cos6x + (1+5) cos4x + (5+10) cos2x +10
= cos6x + cos4x + 5(cos4x + cos2x ) +10 (1 + cos2x)
= 2 cos5x cosx + 10 cos3x cosx + 20 cos2x
= 2cosx [cos5x + 5 cos 3x + 10 cosx ]
------Denominator------
Nr dy
 y  2 cos x    2 sin x  ( C) ]
Dr dx
3
d 2 x  dy  d2y
Q.51102 If   + 2 = K then the value of K is equal to
dy 2  dx  dx
(A) 1 (B) –1 (C) 2 (D*) 0
[Hint: K = 0]
 1
1 1
 
Q.52103 If f(x) = 2 sin 1  x  sin 2 x (1  x) where x   0 , 
2
then f ' (x) has the value equal to
2 2
(A) x (1  x) (B*) zero (C)  x (1 x) (D) 

[Hint : f(x) simplifies to   f ' (x) = 0 or directly differentiate f (x) to get zero]

1

x2 if x  0
 e
Q.53104 Let y = f(x) = 

 0 if x  0
Then which of the following can best represent the graph of y = f(x) ?

(A) (B) (C*) (D)

 2 2 3
Lim e 1 h 1h 1 h 2h
[Hint : f ' (0) = h0 = Lim 2 = 2 = Lim 2 = 0
h h0
e1 h 2
 e1 h . 3
h0
h 2 e1 h
h
1

Lim e
Hence f is differentiable at x = 0. Also x x2
1C


d2 y
Alternatively : check concavity by finding and eliminate D. ]
dx 2
1 1 1
 m n m n   m n   m n
     
n   m
Q.54105 Diffrential coefficient of  x
m n 
. x . x w.r.t. x is
    
     

(A) 1 (B*) 0 (C) – 1 (D) xmn

1
[Hint : exponent on x = (l  m) ( m  n)( n  l ) [ l2 – m2 + m2 – n2 + n2 – l2] = 0

 x0 = 1  y = 1  y' = 0 ]

Q.55107 Let f (x) be diffrentiable at x = h then Lim

bx  hgf ( x)  2 h f ( h)
is equal to
x h xh
(A*) f(h) + 2hf '(h) (B) 2 f(h) + hf '(h) (C) hf(h) + 2f '(h) (D) hf(h) – 2f '(h)
[Hint: Use L' Hospital's rule ]
d 3y
Q.56115 If y = at2 + 2bt + c and t = ax2 + 2bx + c, then equals
dx 3
(A) 24 a2 (at + b) (B) 24 a (ax + b)2 (C) 24 a (at + b)2 (D*) 24 a2 (ax + b)
d 3y
[Hint: for objective note that in y highest degree of x is 4 and therefore is a linear function of x.
dx 3
Which satisfies only in (D). ]

1 x x
Q.57116 Limit  a arc tan  b arc tan  has the value equal to
x  0 x x  a b 

ab (a 2  b 2 ) a 2  b2
(A) (B) 0 (C) (D*)
3 6a 2 b 2 3a 2 b2

 x x 
 a tan 1  b tan 1 
 a a 
[Sol. Using Lopital rule Limit  
x 0 x x
 
 

a1 1 b 1 1
. .  . .
 x  a 2 x  x  b 2 x
1  2  1  2   a2 b2 1  1 a 2  b2
 a   b  Lim  .
= Lim = x 0  2 2
x  3 = 3a 2 b2 ]
x 0 3
. x  (a  x ) x ( b  x )
2
 x
Q.58125 Let f (x) be defined for all x > 0 & be continuous. Let f(x) satisfy f    f ( x )  f ( y) for all x, y
 y
& f(e) = 1. Then :
1
 
(A) f(x) is bounded (B) f    0 as x  0
 x
(C) x.f(x)1 as x 0 (D*) f(x) = ln x

Q.59126 Suppose the function f (x) – f (2x) has the derivative 5 at x = 1 and derivative 7 at x = 2. The derivative
of the function f (x) – f (4x) at x = 1, has the value equal to
(A*) 19 (B) 9 (C) 17 (D) 14
[Sol. y = f (x) – f (2x)
y' = f ' (x) – 2 f ' (2x)
y'(1) = f ' (1) – 2 f ' (2) = 5 ....(1)
and y'(2) = f ' (2) – 2 f ' (4) = 7 ....(2)
now let y = f (x) – f (4x)
y' = f ' (x) – 4 f ' (4x)
y ' (1) = f ' (1) – 4 f ' (4) ....(3)
substituting the value of f ' (2) = 7 + 2 f ' (4) in (1)
f ' (1) – 2 [7 + 2 f ' (4)] = 5
f ' (1) – 4 f ' (4) = 19  (A) ]

x4  x2 1 dy
Q.60127 If y = 2 and = ax + b then the value of a + b is equal to
x  3x  1 dx
5 5 5 5
(A) cot (B*) cot (C) tan (D) tan
8 12 12 8
( x 2  1) 2  3x 2 ( x 2  1  3x )(x 2  1  3x )
[Sol. y= =
x 2  3x  1 x 2  1  3x
dy
= 2x – 3  a=2 & b=– 3
dx
 5
a+b=2– 3 = tan 12 = cot 12 Ans. ]

Q.61128 Suppose that h (x) = f (x)·g(x) and F(x) = f  g ( x )  , where f (2) = 3 ; g(2) = 5 ; g'(2) = 4 ;
f '(2) = –2 and f '(5) = 11, then
(A) F'(2) = 11 h'(2) (B*) F'(2) = 22h'(2) (C) F'(2) = 44 h'(2) (D) none

Q.62129 Let f (x) = x3 + 8x + 3

which one of the properties of the derivative enables you to conclude that f (x) has an inverse?
(A) f ' (x) is a polynomial of even degree. (B) f ' (x) is self inverse.
(C) domain of f ' (x) is the range of f ' (x). (D*) f ' (x) is always positive.
Q.63132 Which one of the following statements is NOT CORRECT ?
(A) The derivative of a diffrentiable periodic function is a periodic function with the same period.
(B*) If f (x) and g (x) both are defined on the entire number line and are aperiodic then the function
F(x) = f (x) . g (x) can not be periodic.
(C) Derivative of an even differentiable function is an odd function and derivative of an odd differentiable
function is an even function.
(D) Every function f (x) can be represented as the sum of an even and an odd function
[Hint : For (B): consider f (x) = x  x 2  1 and g (x) = x  x 2  1 then F(x) = 1 which is periodic
 False]
Select the correct alternatives : (More than one are correct)
dy
Q.64504 If y = tan x tan 2x tan 3x then has the value equal to :
dx
(A*) 3 sec2 3x tan x tan 2x + sec2 x tan 2x tan 3x + 2 sec2 2x tan 3x tan x
(B*) 2y (cosec 2x + 2 cosec 4x + 3 cosec 6x)
(C*) 3 sec2 3x  2 sec2 2x  sec2 x
(D) sec2 x + 2 sec2 2x + 3 sec2 3x
x  x dy
Q.65505 If y = e  e then equals
dx

e x  e x e x  e x 1 1
(A*) (B) (C*) y2  4 (D) y2  4
2 x 2x 2 x 2 x
2 dy
Q.66506 If y = xx then =
dx
2 2
(A ) 2 ln x . xx (B) (2 ln x + 1). xx
2  1 2 1
(C*) (2 ln x + 1). x x (D*) x x . ln ex2

dy
Q.67507 Let y = x  x  x  ......  then =
dx
1 x 1 y
(A*) (B) (C*) (D*)
2y  1 x  2y 1  4x 2x  y

dy 1 x dy y
[Hint : y2 = x + y  = 2 y  1 also y = y + 1  = 2x  y
dx dx
make a quadratic in y to get explicit function  C ]
dy
Q.68508 If 2x + 2y = 2x + y then has the value equal to :
dx

2y
(A*)  x (B*)
1
(C*) 1  2y (D*)

2x 1  2y 
2 1  2x 2 y
2 x
 1
Q.69509 The functions u = ex sin x ; v = ex cos x satisfy the equation :
du dv d2u
(A*) v u = u2 + v2 (B*) = 2v
dx dx dx2
d 2v
(C*) =  2u (D) none of these
dx 2
 x 2 x 1
Q.70511 Let f (x) = . x then :
x 11
(A * ) f  (10) = 1 (B*) f  (3/2) =  1
(C) domain of f (x) is x  1 (D) none
2

[Hint : f (x) =
 x 1  1 2 x 1
.x =
x1 1
.x = [
 x if x [1 , 2)
]
x if x  (2 , )
x1  1 x 11
Q.71512 Two functions f & g have first & second derivatives at x = 0 & satisfy the relations,
2
f(0) = , f  (0) = 2 g  (0) = 4g (0) , g  (0) = 5 f  (0) = 6 f(0) = 3 then :
g(0)

f (x) 15
(A*) if h(x) = then h  (0) = (B*) if k(x) = f(x) . g(x) sin x then k  (0) = 2
g(x) 4

g (x) 1
(C*) Limit
x0 = (D) none
f  (x) 2
n ( n x ) dy
Q.72514 If y = x ( n x ) , then is equal to :
dx
y y
(A)
x

n x n x  1  2 n x n  n x  (B*)
x
(ln x)ln (ln x) (2 ln (ln x) + 1)
y y n y
(C) ((ln x)2 + 2 ln (ln x)) (D*) (2 ln (ln x) + 1)
x n x x n x
n ( n x )
(n x )
[Sol. y= x
ny  (n x) n ( nx) . nx .....(1)
n ( ny)  n ( n x) . n( nx)  n( nx)
1 1 dy 2 n( nx) 1 1
.  . 
ny y dx n x x x nx
2 n( nx)  1

x nx
dy y ny
  . ( 2n ( nx)  1)  D
dx x nx
Substituting the value of y from (1)
dy y
 ( nx) n ( nx ) ( 2n ( nx)  1)  B ]
dx x