1. If log x2 – y2 = a, then dy / dx =

x2 + y2

1] xy 2] y / x

3] x / y 4] none of these

1. If log x2 – y2 = a, then

x2 + y2Solution :

Take y /x = k � y = k x

� dy/dx = k� dy/dx = k

� dy/dx = y / x

Answer : 2] y / x

2. If f is an even function and f 1 exists,

then f 1 (e) + f1 (-e) =

1] 0 2] 1

3] -1 4] e

2. Solution : Given that f is an even

function then f (-x) = f (x)

Diff. w.r.t ‘x’

f 1(-x) (-1) = f1 (x)f (-x) (-1) = f (x)

Substitute x = e f 1(-e) (-1) = f1 (e)

� f 1 (e) + f1 (-e) =0

Answer 1] 0

3.If f (x) = ex g (x), g(0) = 2, g1 (0) =1,

then f1 (0) is equal to

1]1 2] 3 3] 2 4] 01]1 2] 3 3] 2 4] 0

3.Solution : Given that f (x) = ex g (x),

g(0) = 2, g1 (0) =1,

Since f (x) = ex g (x),

Diff. w.r.t ‘x’

f1 (x) = ex g1 (x) + g(x) exf1 (x) = ex g1 (x) + g(x) ex

Substitute , x = 0

f1 (0) = e0 g1 (0) + g(0) e0

= 1(1) + 2 (1)

= 1 + 2 = 3

Answer : 2] 3

4. The derivative of an even function is always

1] an odd function

2] an even function

3] does not exist 3] does not exist

4] none of these

4. Solution : The derivative of an even

function is always an odd function

Answer : 1] odd function

5. Let f and g be differentiable functions

satisfying g1(a) =2,g(a) =b and fog = i

(identity function).then f1(b) is equal to

1] ½

2] 2 2] 2

3] 2/3

4] - ½

5. Solution : Given that Let f and g be differentiable

functions satisfying g1(a) =2,g(a) =b and fog = i

(identity function).then f1(b) is equal to

Since : fog = I � (fog ) x = x

� f [g(x)] = x Diff.w.r.t ‘x’

� f1 [ g (x)]. g1 (x) = 1� f1 [ g (x)]. g1 (x) = 1

Substitute : x = a f1[g(a)]. g1 (a) = 1

���� f1(b). 2 = 1

� f1(b) = 1/2

Answer 1] 1/ 2

6.If y = Tan-1 4x + Tan-1 2 + 3x

1 + 5x2 3 – 2x

then dy/dx is equal to

1] 1 2] 5

1 + x2 1 + 25x21 + x2 1 + 25x2

3] 1 4] -5

1 + 25x2

6.Solution : y =Tan-1 4x + Tan-1 2 + 3x

1 + 5x2 3 – 2x

Since: Tan-1 x ± Tan-1 y = Tan x ± y

1 + xy

Y = Tan-1 5x + x + Tan-1 2/3 +xY = Tan-1 5x + x + Tan-1 2/3 +x

1 – 5x.x 1–(2/3)x

Y = Tan-1 5x – Tan-1 x + Tan-1 2/3 + Tan-1 x

Y = Tan-1 5x +Tan-1 2/3

diff.w.r.t ‘x’

dy/dx = 5

1+25x21+25x2

Answer 2] 5

1+25x2

(sin x) ……….. ∞

7. If y = (sinx) (sin x) then dy/dx =

1] y2 2] y2 sin x

sinx (1-logy) 1 – logysinx (1-logy) 1 – logy

3] y2 cot x 4] y2 tanx

1 - logy 1 – logy

(sin x) ……….. ∞

7. Solution: Given that y = (sinx) (sin x)

Y = [f(x) ]y then dy = y2 f1 (x)

dx f(x) (1 – log y )

�Y = ( sin x )y�Y = ( sin x )y

� dy = y2 cosx y2 cot x

dx (1 –log y ) sin x 1 – log y

Answer : 3] y cot x

1-logy

8. If xy = ex – y, then dy / dx is equal to

1] y 2] x

(1 + log x)2 (1 + logx)2

3] log x 4] none of these

( 1 + logx )2

8. Solution : Given that : xy = ex – y

y log x = (x – y) log ee

y log x + y = x

� y ( 1 + log x ) = x

� y = x� y = x

1 + log x

Diff. y w.r.t ‘x’

dy log x

dx (1 + logx)2

Answer : 3] log x

(1 + logx)2

9. If 3 sin (xy) + 4 cos ( xy) = 5, then dy / dx

1] - y / x

2] 3 sin (xy) + 4 cos (xy)

3 cos (xy) – 4 sin ( xy)3 cos (xy) – 4 sin ( xy)

3] 3cos (xy) + 4 sin (xy)

4 cos (xy) – 3 sin (xy)

4] none of these

9. Solution : Given that

3 sin (xy) + 4 cos ( xy) = 5,

Take xy = k

Diff. w.r.t ‘x’

x dy + y = 0 x dy + y = 0

dx

dy = - y

dx x

Answer : 1] –y/x

10. If f (x)=cos-1 1 – ( log x )2 ,then f1(e) is

1 + ( log x )2

1] 1 / e

2] 1 2] 1

3] 2 / e

4] 2e

10. (x)=cos-1 1 – ( log x )2

1 + ( log x )2

cos-1 1 – f 2 ( x ) 2 Tan-1 [f(x)]

1 + f 2 (x )1 + f (x )

f (x) = 2 Tan-1 ( log x)

Diff. w.r.t ‘x’

f1 ( x ) = 2 1 1

1 + ( logx)2 x

f1 ( x ) = 2 1 1

1 + ( logx)2 x

Substitute : x = e

f1 (e) 2 1 2 1

1 + ( log e)2 e 2e e1 + ( logee)2 e 2e e

Answer : 1] 1

e

11. If 2x2 + 4xy + 3y2 = 0, then d2y / dx2 =

1] 0 2] ½ 3] 1 4] ¾

(2x+y)2(2x+y)

11. Solution : Given that 2x2 + 4xy + 3y2 = 0,

ax2 + 2hxy + by2 = 0 then d2y = 0

dx2

Answer : 1] 0 Answer : 1] 0

12. lt f (x + y) = f(x). f(y) for all x and y. suppose

f(5) = 2, f1 (0) =3, then f1(5)=

1] 3 2] 2 3] 6 4]-1

12. Solution: Given that f (x + y) = f(x). f(y)

f(5) = 2, f1 (0) =3, then f1(5)=

consider f ( x + 5 ) = f (x) f (5)

Diff. w.r.t ‘x’

f ( x + 5) = f ( x) f (5)

Diff. w.r.t ‘x’

f1 ( x + 5) = f1 ( x) f (5)

Sub x = 0

f1 ( 0 + 5 ) = f1 (0) f (5)

f1 (5) = 3 (2)

= 6

Answer : 3] 6

13. The differential co-efficient of f(sinx) w.r.t x

where f (x) = logx is

1] tan x 2] cotx 3] (cosx) 4] 1/x

13. Solution: The differential co-efficient of f (sinx)

w.r.t x where f (x) = logx is

Diff. f (sinx) w.r.t ‘x’

�f1 ( sinx ) cos x since f (x) = log x

� 1 cosx = cot x f1 (x) = 1 � 1 cosx = cot x f1 (x) = 1

sinx x

Answer: 2] cotx

14. If y = 1 + x + x2 + ….. to ∞ dy / dx = [ |x| < |]

1] x/y 2] x2 / y2

3] - y2 4] y2

14. Solution : Given that : y = 1 + x + x2 + ….. to ∞

[ |x| < |] which represents a geometric series

S ∞ = a , where |r|<|

1 – r

y = 1

1 - x 1 - x

Diff. y .w.r.t ‘x’

dy /dy = - 1 (-1) = 1

(1-x)2 (1-x)2

= y2

Answer : 4] y2

15.If f (x) = ( 1 + x ) ( 1 + x2 ) ( 1 + x3 ) ……(1+xn )

then f1 (o) =

1] 0

2] 1 2] 1

3] -1

4] 2

15. Solution : Given that :

f (x) = ( 1 + x ) ( 1 + x2 ) ( 1 + x3 )……(1+xn )

log f (x) = log ( 1+x) + log (1+x2) +………log (1+xn)

Diff. w.r.t ‘x’

f1 (x) = 1 2x ……… n xn-1f (x) = 1 2x ……… n x

f(x) 1 + x 1 + x2 1 + xn

f1 (0) = 1 2(0) ……… n(0)

f(0) 1 + 0 1 + 0 1 + 0

f1 (0) = 1

Answer : 2] 1

16. If y = Tan-1 log (e/x2) + Tan 4 +2logx

log (ex2) 1-8 logx

then d2y/dx2 =

1] 2 1] 2

2] 1

3] 0

4] -1

16.y = Tan-1 log (e/x2) +Tan-1 4 +2logx

log (ex2) 1-8 logx

y = Tan-1 1 – 2 logx + Tan-1 4 + 2 log x

1 + 2 log x 1 – 4 (2 logx )

Y= Tan-1 1-Tan-1 (2logx ) + Tan-1 4 + Tan-1 (2logx )Y= Tan 1-Tan (2logx ) + Tan 4 + Tan (2logx )

diff. w.r.t ‘x’

dy/dx = 0

d2y = 0 Answer : 3] 0

dx2

Answer : 3] 0

17.If y = (Tanx-x)+ (Tanx-x)+ (Tanx-x)+…….to ∞

terms then dy/dx =

1] sec2x 2] sec2x

2y-1 1-2y

3] Tanx 4] –Tan x

2y-1 2y-1

17. y = (Tanx-x)+ (Tanx-x)+ (Tanx-x)+…….to ∞

y = f(x) + f(x) + ………to ∞

dy f1 (x)dy f (x)

dx 2y – 1

dy sec2 x -1 Tan x

dx 2y-1 2y-1

Answer : 3] Tan x

2y-1

18. Let ‘f’ be a function defined for all x Є.R

If ‘f’ is differentiable and f (x3) = x5 ( x ≠ 0 )

then the value of f1 ( 27) is

1]15 1]15

2] 45

3] 0

4] 10

18. Solution : Given that f (x3) = x5 ( x ≠ 0 )

Diff. w.r.t ‘x’

f1 (x3 ) 3x2 = 5x4

f1 ( 33) 3. 32 = 5 .34f1 ( 33) 3. 32 = 5 .34

f1 (27 ) = 5 (3)

= 15

Answer: 1] 15

19. If logsinx y = cosx then dy/ dx =

1] [ cosx cotx – sinx log sinx ]

2] y [ cosx cotx + sin x log sinx ]

3] [cosx cotx – sinx log sinx ]3] [cosx cotx – sinx log sinx ]

4] y [cosx cotx – sinx log sinx]

19. Solution : logsinx y = cosx then =

Y = (sinx)cosx

y = [f (x)]g(x)

dy = [ f(x)] g(x) g (x) f1 (x) + g1 (x) log f (x)

dx f(x)dx f(x)

dy = ( sinx)cosx cosx cosx - sinx log sin x

dx sinx

= y [cosx cotx – sinx log sinx]

Answer: 4] y [cosx cotx – sinx log sinx]

20. If y = sinx. Sin2x. sin3x ….. Sinnx then dy/dx =

n n

1] y ∑ k cot kx 2] y ∑ k cot kxk=1 k=1

n n

3] y ∑ k TAn kx 4] y ∑ k Tan kxk=1 k=1

20. Solution : If y = sinx. Sin2x sin3x ….. Sinnx

log y = log sinx + log sin 2x +………. + log sinnx

y1 cosx 2 cos2x 3 cos3x + ……..+ n cosnx

y sinx sin2x sin3x sinnx

Y1 = y [ cotx + 2cot 2x + …. + k cot k x]

nn

Y1 = y ∑ k cot kx

K=1

Answer : 1] n

y ∑ k cot k x

K=1

21. If y = 1+ logx + (logx)2 + ( logx)3

+ ………….. to ∞

2 3!

terms then d2y /dx2

1] 1 / x 1] 1 / x

2] 2 / x

3] 1

4] 0

21. Solution :

y = 1+ logx + (logx)2 + ( logx)3

+ ………….. to ∞

2 3!

ex = 1 + x + x2 + x3 + ………. to ∞

2! 3!

y = elog xy = elog x

y = x

dy/dx = 1

d2y

dx2 = 0

Answer: 4] 0

22. If y = sec x + Tanx and o < x < π. then dy

secx – Tanx dx

1] sec x [ secx – Tanx ] 1] sec x [ secx – Tanx ]

2] Tan x [ secx + tanx ]

3] secx [secx + Tanx]

4] Tan x [ secx – Tanx]

22. Solution : If y = sec x + Tanx and o < x < π.

secx – Tanx

y = (sec x + Tanx)2

sec2x – Tan2x

y = secx + Tan xy = secx + Tan x

dy = secx Tan x + sec2x

dx

= secx [Tan x + secx]

Answer: 3] secx [ Tan x + secx]

23. If g is the inverse function of f and

f1 (x) = 1 then g1 (x) is equal to

1 + xn,

1] 1+(g (x))n

2] 1 – g (x)2] 1 – g (x)

3] 1 + g (x)

4] 1 – (g(x))n

23. Solution : Given that: g is the inverse function

of f and f` (x) = 1

1 + xn,

and also f = g-1

� (fog) = I ( identity function)

�(fog) x = x�(fog) x = x

� f ( g (x) ] = x

diff. w.r.t ‘x’

� f1 [g (x)] g1 (x) = 1

1 g1 (x) = 1

1+ (g(x)n

� g (x) = 1+ [g (x)n Answer 1] 1 + [ g (x)]n

24. If f (x) is an odd differentiable function

defined on (- ∞, ∞) such that f1 (3) = 2, then

f1 (-3) equals :

1] 0 1] 0

2] 1

3] 2

4] 4

24. Solution : Given that f (x) is an odd

differentiable function :

�f (-x) = - f (x)

Diff w.r.t ‘x’Diff w.r.t ‘x’

�f1 (-x) (-1) = - f1 (x)

�f1 (-x) = f1 (x) since f1 (3) = 2

�f1 (-3) = f1 (3)

f1 (-3) = 2

Answer: 3] 2

25.A differentiable function f(x) is such that

f (1) = 7 and f1 (1) = 1/7. If f-1 exists and f-1 = g,

then

1] g1 (7) = 1/7

2] g1 (7) = 7

3] g1 (1) = 1/7 3] g1 (1) = 1/7

4] g1 (1) = 8

25.Solution : Given that : f-1 = g, f(1) = 7,

and f1 (1) = 1/7

go f = I

( gof) x = x

� g [ f (x)] = x

� g1 [(f(x)]. f1 (x) =1� g1 [(f(x)]. f1 (x) =1

� g1 [f(1)]. f1 (1) = 1

� g1 (7). 1/7 = 1

� g1 (7) = 7

Answer 2] g (7) = 7

-1 2 2

26. If x = eTan [y – x x ] then dy/dx =

1] 2x [1 + Tan x (log x )] + x sec2 ( logx)

2] 2 x [ 1 + Tan ( log x) ] + sec2 ( log x )

3] 2 x [ 1 + Tan ( logx) ] + x2 sec2 ( logx)3] 2 x [ 1 + Tan ( logx) ] + x2 sec2 ( logx)

4] 2x [ 1 + tan ( logx ) ] + sec2 ( logx)

-1 2 2

26. Solution : Given that x = eTan [y – x x ]

Tan (logx) = y – x2

x2

x2 Tan (logx) = y –x2

�Y = x2 + x2 Tan ( logx)�Y = x + x Tan ( logx)

�dy/dx =2x + x2 sec2 (logx) + Tan ( logx ) ( 2x)

x

Answer :3] 2 x [ 1 + Tan ( logx) ] + x2 sec2 ( logx)

.

27. If x = a [θ-sinθ], y = a[1-cosθ].Then dy/dx =

1] cot θ/2 2] Tan θ/2

3] ½ coesec2 θ/2 4] -½ cosec2 θ/2

. 27. Solution : Given that

If x = a [θ-sinθ], y = a[1-cosθ].

dy/dx = a [1-cosθ]

a sin θ

= 2sin2 θ/2= 2sin2 θ/2

2 sin θ/2 cos θ/2

=Tan θ/2

Answer 2] Tan θ/2

. 28. If log y = m tan-1 x, then

1] (1 +x2) y2 + (2x + m ) y1 = 0

2] (1 +x2) y2 + (2x - m ) y1 = 0

3] (1 +x2) y2 - (2x + m ) y1 = 0

4] (1 +x2) y2 - (2x - m ) y1 = 0 4] (1 +x2) y2 - (2x - m ) y1 = 0

. 28. Solution : Given that

log y = m tan-1 x,

y1 = m

y 1+x2

(1+x2) y1 = my(1+x2) y1 = my

(1 + x2) y2 + y1 2x = my1

( 1 + x2 ) y2 + 2 xy1 – my1 = 0

( 1 + x2 ) y2 + ( 2 x – m ) y1 = 0

Answer : 2] ( 1 + x) y + ( 2x – m) y = 0

. 29. If y2 – 2x2 = y, then dy/dx at (1, - 1 ) is

1] -4/3

2] 4/3

3] ¾

4] -3/4 4] -3/4

. 29. Solution: Given that y2 - 2x2 - y = 0

dy/dx = - (diff. w.r.t ‘x’ keeping y as constant)

(diff. w.r.t ‘y’ keeping x as constant)

= - (-4x)

(2y-1)(2y-1)

dy/dx = 4x

2y-1

dy 4(1) (4) -4

dx (1,-1) 2(-1) -1 -3 3

Answer : 1] -4/3

. 30. Derivative of f (logx) w.r.t ‘x’ where

f (x) = ex

1] ex

2] log x

3] 1/x3] 1/x

4] 1

. 30. Solution :diff f (logx) w.r.t ‘x’

= f1 ( log x ) 1 f (x) = ex

x f1 (x) = ex

= x 1 f1 ( logx) = elogx

x =xx =x

= 1

Answer: 4] 1

. 31. If f (x) = sin [π / 2 [x]-x5], 1 <x<2,

where[x] denotes the greatest integer.less than

or equal to x, then f1(5 π/2) =

1] 5 (π /2)4/5

2] -5 (π /2)4/52] -5 (π /2)4/5

3] 0

4] none of these

. 31. Solution : Given that

f (x) = sin [π/2 [x]-x5], 1 < x < 2, [x]=1

f (x) = sin [π/2 (1) –x5],

f1 (x) = cos [π/2 –x5]. [-5x4]

f1 (5 π/2) = cos [π/2– 5 π/2 )5].(-5)(5 π/2 )4f1 (5 π/2) = cos [π/2– 5 π/2 )5].(-5)(5 π/2 )4

= - 5 (π/2)4/5

Answer 2] - 5 (π/2)4/5

. 32. If y = cos-1 2cosx -3sinx then dy =

√13 dx

1] 2

2] -2

3] -1 3] -1

4] 1

. 32. Solution : y = cos-1 2cosx -3sinx

√13

y = cos-1 2 cosx - 3 sinx

√13 √13

cosα = 2/√13 sin α 3/√13 cosα = 2/√13 sin α 3/√13

y = cos-1 [cosα cosx – sinα sinx]

y = cos-1 [cos(α + x)]

y = α + x

dy/dx = 1

Answer: 4] 1

. 33. If y = cos2 3x/2 – sin2 3x/2, then d2 y/dx2 is

1] 9y

2] -3 √1 – y2

3] 3 √ 1 – y2

4] -9y4] -9y

. 33. Solution : Given that

y = cos2 3x/2 – sin2 3x/2

y= cos 3x

� dy/dx = - 3 sin 3x

� d2y/dx2 = -9 cos 3x = -9y� d2y/dx2 = -9 cos 3x = -9y

Answer: 4]-9y

. 34. If y = log5 ( log5 x ) then dy/dx =

1] 1

x log5 x

2] 1

x log5 x.log5xx log5 x.log5x

3] 1

x log5 x.(log5)2

4] none of these

. 34. Solution : Given that:

y = log5 ( log5 x )

dy 1

dx (log5) (log5x) x (log5)

11

= x (log5x) (log5)2

Answer:3] 1

x (log5x) (log5)2

. 35. If f (x ) =1+nx+ n (n-1) x2 + n ( n-1) (n-2) x3

2 6

+ ………….. + xn then f “ (1) =

1] n ( n-1 ) 2n-1

2] (n-1)2n-12] (n-1)2n-1

3] n (n-1)2 n-2

4] n (n-1)2n

. 35. Solution : Given that:

f (x ) = 1 + nx + n (n-1) x+ n ( n-1) (n-2) x2

2 6

+ ………….+ xn

f (x) = ( 1+x)n (by binomial theorem)f (x) = ( 1+x)n (by binomial theorem)

f1 (x) = x ( 1 + x )n-1

f11 (x) = n ( n-1) ( 1+x)n-2

f11 (1) = n ( n-1) ( 1+1)n-2

= n ( n-1) 2n-2

Answer : 3] n ( n-1) 2n-2

36. If y = sec-1 x+1 + sin-1 x - 1 , then dy =

x-1 x+1 dx

1] x - 1

x + 1

` 2] x + 1` 2] x + 1

x - 1

3] 0

4] 1

. 36. Solution : Given that:

y = sec-1 x+1 + sin-1 x - 1

x -1 x +1

y = cos-1 x - 1 + sin-1 x - 1

x +1 x +1x +1 x +1

y = π / 2

dy/dx = 0

Answer :3] 0

. 37.If y = tan-1 √1+sinx+√1– sinx , 0 <x< π/2

√1+sinx –√1– sinx

Then dy/dx =

1] ½

2] -½ 2] -½

3] x/2

4] –x/2

. 37.

. 37. Solution : Given that:

y = tan-1 √1+sinx+√1– sinx

√1+sinx –√1– sinx

0 <x< π/2 � 0 <x/2< π/4

� Cos x/2 > sin x/2� Cos x/2 > sin x/2

y = tan-1 cosx/2 + sin x/2+ cos x/2-sinx/2

cosx/2 + sin x/2- cos x/2+sinx/2

y = tan-1 [cotx/2]

y = π/2 –x/2 � dy/dx = - ½

Answer :2] - ½

. 37.

. 37.

. 37.

. 38. y = sin-1(3x – 4x3), then dy/dx at x = 1/3, is

1] -9 √ 2

4

2] 9 √ 2

3] 9 √ 2

4

4] 9/8

. 37.

. 37.

. 37.

. 38. Solution : Given that:

y = sin-1(3x – 4x3)

y = 3 sin-1 x

dy/dx = 3

√1 - x√1 - x

dy/dx = 3 9 9 √2

√ 1 - 1/9 √ 8 4

Answer :3] 9 √ 2

4

. 37.

. 37.

. 37.

.

39. If y = tan-1 1 + tan-1 1 + tan-1 1

1 + x+x2 x2+3x+3 x2+5x+7

…. to n terms then y1 (0) =

1] -1/n2 +1 2] –n2 /(n2+1) 1] -1/n2 +1 2] –n2 /(n2+1)

3] n2 /n2 +1 4] n / n+1

. 37.

. 37.

. . 39. If y = tan-1 1 + tan-1 1 + tan-1 1

1 + x+x2 x2+3x+3 x2+5x+7

+………..+ to n terms

y = tan-1 (x+1)-x tan-1 (x+2) –(x+1) …………1+ (x+1)x 1+ ( x+2) ( x + 1)

tan-1 (x+n)-(x+n-1)tan-1 (x+n)-(x+n-1)

1 + ( x + n) ( x + n-1)

Y= tan-1 (x+1)- tan1x + tan-1 (x+2) –tan-1 (x+1) ……….+

tan-1 (x+n) –tan-1 (x+n-1)

Y=tan-1 (x+n) – tan-1 x

y1 = 1/1+(x+n)2 – 1/1+x2

y1 (0)=1/(1+n2)- 1 Answer: 2]–n2 /(n2+1)