1. If log
x2 – y2 = a, then
x2 + y2
dy / dx =
1] xy
2] y / x
3] x / y
4] none of these
1. If
log
x2 – y2 = a, then
x2 + y2
Solution :
Take y /x = k y = k x
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)
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] 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) 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
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
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
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]
3] 1
1
1 + x2
2]
5
1 + 25x2
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 +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+25x2
Answer 2]
5
1+25x2
7. If y = (sinx)
1]
(sin x)
(sin x)
y2
sinx (1-logy)
3] y2 cot x
1 - logy
……….. ∞
then dy/dx =
2]
y2 sin x
1 – logy
4] y2 tanx
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
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
(1 + log x)2
2]
x
(1 + logx)2
3]
log x
( 1 + logx )2
4] none of these
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
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] 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
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
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 )
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 + ( logee)2 e 2e
1
e
Answer : 1] 1
e
11. If 2x2 + 4xy + 3y2 = 0, then d2y / dx2 =
1] 0
2] ½
3]
1
(2x+y)2
4] ¾
11. Solution : Given that 2x2 + 4xy + 3y2 = 0,
ax2 + 2hxy + by2 = 0 then d2y = 0
dx2
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’
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
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
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
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-1
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
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 )
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
2y-1
4] –Tan x
2y-1
17. y = (Tanx-x)+ (Tanx-x)+ (Tanx-x)+…….to ∞
y = f(x) + f(x) + ………to ∞
dy
f1 (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
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 .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 ]
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)
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
1] y ∑ k cot kx
n
2] y ∑ k cot kx
k=1
k=1
n
n
3] y ∑ k TAn kx
k=1
4] y ∑ k Tan kx
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]
n
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
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 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 ]
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 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)
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
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
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’
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
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(1)]. f1 (1) = 1
g1 (7). 1/7 = 1
g1 (7) = 7
Answer 2] g (7) = 7
-1
26. If x = eTan
2
2
[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)
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)
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
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
.
28. Solution : Given that
log y = m tan-1 x,
y1 = m
y
1+x2
(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
.
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)
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/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
=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)4/5
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 )4
= - 5 (π/2)4/5
Answer 2] - 5 (π/2)4/5
.
32. If y = cos-1 2cosx -3sinx
√13
1] 2
2] -2
3] -1
4] 1
then dy =
dx
.
32. Solution : y = cos-1 2cosx -3sinx
√13
y = cos-1 2 cosx - 3 sinx
√13
√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] -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
Answer: 4]-9y
.
34. If y = log5 ( log5 x ) then dy/dx =
1]
1
x log5 x
2]
1
x 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)
1
= 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-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)
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
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 +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
3] x/2
4] –x/2
.
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
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. y = sin-1(3x – 4x3), then dy/dx at x = 1/3, is
38.
1] -9 √ 2
4
2] 9 √ 2
3] 9 √ 2
4
4] 9/8
.
37. Solution : Given that:
38.
y = sin-1(3x – 4x3)
y = 3 sin-1 x
dy/dx = 3
√1 - x
dy/dx =
3
9
√ 1 - 1/9 √ 8
9 √2
4
Answer :3] 9 √ 2
4
.
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
3] n2 /n2 +1
2] –n2 /(n2+1)
4] n / n+1
.
-1
39.
If
y
=
tan
37.
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)
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)