1. If log x 2 y 2 = a, then dy / dx = x 2 + y 2 1] xy 2] y / x. 3] x / y 4] none of these

1. If log x 2 y 2 = a, then dy / dx = x 2 + y 2 1] xy 2] y / x 3] x / y 4] none of these

1. If log x 2 y 2 = a, then x 2 + y 2 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) + f 1 (-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) = f 1 (x) Substitute x = e f 1 (-e)(-1) = f 1 (e) f 1 (e) + f 1 (-e) =0 Answer 1] 0

3.If f (x) = e x g (x), g(0) = 2, g 1 (0) =1, then f 1 (0) is equal to 1]1 2] 3 3] 2 4] 0

3.Solution :Given that f (x) = e x g (x), g(0) = 2, g 1 (0) =1, Since f (x) = e x g (x), Diff. w.r.t x f 1 (x) = e x g 1 (x) + g(x) e x Substitute, x = 0 f 1 (0) = e 0 g 1 (0) + g(0) e 0 = 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 g 1 (a) =2,g(a) =b and fog = i (identity function).then f 1 (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 g 1 (a) =2,g(a) =b and fog = i (identity function).then f 1 (b) is equal to Since : fog = I (fog ) x = x f [g(x)] = x Diff.w.r.t x f 1 [ g (x)]. g 1 (x) = 1 Substitute : x = a f 1 [g(a)]. g 1 (a) = 1 f 1 (b). 2 = 1 f 1 (b) = 1/2 Answer 1] 1/ 2

6.If y = Tan -1 4x + Tan -1 2 + 3x 1 + 5x 2 3 2x then dy/dxis equal to 1] 1 2] 5 1 + x 2 1 + 25x 2 3] 1 4] -5 1 + 25x 2

6.Solution : y =Tan -1 4x + Tan -1 2 + 3x 1 + 5x 2 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+25x 2 Answer 2] 5 1+25x 2

(sin x).. 7. If y = (sinx) (sin x) then dy/dx= 1] y 2 2] y 2 sin x sinx(1-logy) 1 logy 3] y 2 cot x 4] y 2 tanx 1 -logy 1 logy

7. Solution: Given that y = (sinx) (sin x).. (sin x) Y = [f(x)] y then dy = y 2 f 1 (x) dx f(x) (1 log y ) Y = ( sin x ) y dy = y 2 cosx y 2 cot x dx (1 log y ) sin x 1 log y Answer : 3] y cot x 1-logy

8. If x y = e x 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 : x y = e x y y log x = (x y) log e e 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 f 1 (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 f 1 ( x ) = 2 1 1 1 + ( logx) 2 x

f 1 ( x ) = 2 1 1 1 + ( logx) 2 x Substitute : x = e f 1 (e) 2 1 2 1 1 + ( log e e) 2 e 2e e Answer : 1] 1 e

11. If 2x 2 + 4xy + 3y 2 = 0, then d 2 y / dx 2 = 1] 0 2] ½ 3] 1 4] ¾ (2x+y) 2

11. Solution :Given that 2x 2 + 4xy + 3y 2 = 0, ax 2 + 2hxy + by 2 = 0 then d 2 y = 0 dx 2 Answer : 1] 0

12. ltf (x + y) = f(x). f(y) for all x and y. suppose f(5) = 2, f 1 (0) =3, then f 1 (5)= 1] 3 2] 2 3] 6 4]-1

12. Solution:Given that f (x + y) = f(x). f(y) f(5) = 2, f 1 (0) =3, then f 1 (5)= consider f ( x + 5 ) = f (x) f (5) Diff. w.r.t x f 1 ( x + 5) = f 1 ( x) f (5) Sub x = 0 f 1 ( 0 + 5 ) = f 1 (0) f (5) f 1 (5) = 3 (2) = 6 Answer : 3] 6

13. The differential co-efficient of f(sinx) w.r.tx where f (x) = logxis 1] tan x 2] cotx 3] (cosx) 4] 1/x

13. Solution: The differential co-efficient of f (sinx) w.r.tx where f (x) = logxis Diff. f (sinx) w.r.t x f 1 ( sinx) cosx since f (x) = log x 1 cosx = cot x f 1 (x) = 1 sinx x Answer: 2] cotx

14. If y = 1 + x + x 2 +.. to dy/ dx = [ x < ] 1] x/y 2] x 2 / y 2 3] -y 2 4] y 2

14. Solution : Given that : y = 1 + x + x 2 +.. 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 = y 2 Answer : 4] y 2

15.If f (x) = ( 1 + x ) ( 1 + x 2 ) ( 1 + x 3 ) (1+x n ) then f 1 (o) = 1] 0 2] 1 3] -1 4] 2

15. Solution: Given that : f (x) = ( 1 + x ) ( 1 + x 2 ) ( 1 + x 3 ) (1+x n ) log f (x) = log ( 1+x) + log (1+x 2 ) + log (1+x n ) Diff. w.r.t x f 1 (x) = 1 2x n x n-1 f(x) 1 + x 1 + x 2 1 + x n f 1 (0) = 1 2(0) n(0) f(0) 1 + 0 1 + 0 1 + 0 f 1 (0) = 1 Answer : 2] 1

16. If y = Tan -1 log (e/x 2 ) + Tan 4 +2logx log (ex 2 ) 1-8 logx then d 2 y/dx 2 = 1] 2 1] 2 2] 1 3] 0 4] -1

16.y = Tan -1 log (e/x 2 ) +Tan -1 4 +2logx log (ex 2 ) 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 d 2 y = 0 Answer : 3] 0 dx 2 Answer : 3] 0

17.If y = (Tanx-x)+ (Tanx-x)+ (Tanx-x)+.to terms then dy/dx = 1] sec 2 x 2] sec 2 x 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 f 1 (x) dx 2y 1 dy sec 2 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 (x 3 ) = x 5 ( x 0 ) then the value of f 1 ( 27) is 1]15 2] 45 3] 0 4] 10

18. Solution: Given that f (x 3 ) = x 5 ( x 0 ) Diff. w.r.t x f 1 (x 3 ) 3x 2 = 5x 4 f 1 ( 3 3 ) 3. 3 2 = 5.3 4 f 1 (27 ) = 5 (3) = 15 Answer: 1] 15

19. If log sinx y = cosx then dy/ dx= 1] [ cosx cotx sinxlog sinx] 2] y [ cosxcotx+ sin x log sinx] 3] [cosx cotx sinxlog sinx] 4] y [cosxcotx sinxlog sinx]

19. Solution: log sinx y = cosx then = Y = (sinx) cosx y = [f (x)] g(x) dy= [ f(x)] g(x) g (x) f 1 (x) + g 1 (x) log f (x) dx f(x) dy= ( sinx) cosx cosx cosx -sinx log sin x dx sinx = y [cosx cotx sinxlog sinx] Answer: 4] y [cosx cotx sinxlog sinx]

20. If y = sinx. Sin2x. sin3x.. Sinnx then dy/dx= n 1] y k cot kx 2] y k cot kx k=1 k=1 n 3] y k TAnkx 4] y k Tan kx k=1 k=1 n n

20. Solution : If y = sinx. Sin2x sin3x.. Sinnx log y = log sinx+ log sin 2x +. + log sinnx y 1 cosx 2 cos2x 3 cos3x +..+ n cosnx y sinx sin2x sin3x sinnx Y 1 = y [ cotx+ 2cot 2x +. + k cot k x] n Y 1 = 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 + 2 3! terms then d 2 y /dx 2.. to 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! e x = 1 + x + x 2 + x 3 +. to 2! 3! y = e log x y = x dy/dx = 1 d 2 y dx 2 = 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 sec 2 x Tan 2 x y = secx + Tan x dy = secx Tan x + sec 2 x dx = secx [Tan x + secx] Answer: 3] secx [ Tan x + secx]

23. If g is the inverse function of f and f 1 (x) = 1 then g 1 (x) is equal to 1 + x n, 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 + x n, and also f = g -1 (fog) = I ( identity function) (fog) x = x f ( g (x)] = x diff. w.r.t x f 1 [g(x)] g 1 (x) = 1 1 g 1 (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 f 1 (3) = 2, then f 1 (-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 f 1 (-x) (-1) = - f 1 (x) f 1 (-x) = f 1 (x) since f 1 (3) = 2 f 1 (-3) = f 1 (3) f 1 (-3) = 2 Answer: 3] 2

25.A differentiable function f(x) is such that f (1) = 7 and f 1 (1) = 1/7. If f -1 exists and f -1 = g, then 1] g 1 (7) = 1/7 2] g 1 (7) = 7 3] g 1 (1) = 1/7 4] g 1 (1) = 8

25.Solution : Given that : f -1 = g, f(1) = 7, and f 1 (1) = 1/7 go f = I ( gof) x = x g [ f (x)] = x g 1 [(f(x)]. f 1 (x) =1 g 1 [f(1)]. f 1 (1) = 1 g 1 (7). 1/7 = 1 g 1 (7) = 7 Answer 2] g (7) = 7

-1 2 2 26. If x = e Tan [y x x] then dy/dx= 1] 2x [1 + Tan x (log x )] + x sec 2 ( logx) 2] 2 x [ 1 + Tan ( log x) ] + sec 2 ( log x ) 3] 2 x [ 1 + Tan ( logx) ] + x 2 sec 2 ( logx) 4] 2x [ 1 + tan ( logx) ] + sec 2 ( logx)

-1 2 2 26. Solution : Given that x = e Tan [y x x ] Tan (logx) = y x 2 x 2 x 2 Tan (logx) = y x 2 Y = x 2 + x 2 Tan ( logx) dy/dx=2x + x 2 sec 2 (logx) + Tan ( logx) ( 2x) x Answer :3] 2 x [ 1 + Tan ( logx) ] + x 2 sec 2 ( logx)

. 27. If x = a [θ-sinθ], y = a[1-cosθ].then dy/dx = 1] cotθ/2 2] Tan θ/2 3] ½ coesec 2 θ/2 4] -½ cosec 2 θ/2

. 27. Solution : Given that If x = a [θ-sinθ], y = a[1-cosθ]. dy/dx = a [1-cosθ] a sinθ = 2sin 2 θ/2 2 sin θ/2 cosθ/2 =Tan θ/2 Answer 2] Tan θ/2

. 28. If log y = m tan -1 x, then 1] (1 +x 2 ) y 2 + (2x + m ) y 1 = 0 2] (1 +x 2 ) y 2 + (2x - m ) y 1 = 0 3] (1 +x 2 ) y 2 - (2x + m ) y 1 = 0 4] (1 +x 2 ) y 2 - (2x - m ) y 1 = 0

. 28. Solution : Given that log y = m tan -1 x, y 1 = m y 1+x 2 (1+x 2 ) y 1 = my (1 + x 2 ) y 2 + y 1 2x = my 1 ( 1 + x 2 ) y 2 + 2 xy 1 my 1 = 0 ( 1 + x 2 ) y 2 + ( 2 x m ) y 1 = 0 Answer : 2] ( 1 + x) y + ( 2x m) y = 0

. 29. If y2 2x 2 = y, then dy/dxat (1, -1 ) is 1] -4/3 2] 4/3 3] ¾ 4] -3/4

. 29. Solution: Given that y 2-2x 2 - y = 0 dy/dx = - (diff. w.r.t x keeping y as constant) = - (-4x) (diff. w.r.t y keeping x as constant) (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) = e x 1] e x 2] log x 3] 1/x 4] 1

. 30. Solution :diff f (logx) w.r.t x = f 1 ( log x ) 1 f (x) = e x x f 1 (x) = e x = x 1 f 1 ( logx) = e logx = 1 x =x Answer: 4] 1

. 31. If f (x) = sin [π / 2 [x]-x 5 ], 1 <x<2, where[x] denotes the greatest integer.less than or equal to x, then f 1 (5π/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]-x 5 ], 1 < x < 2, [x]=1 f (x) = sin [π/2 (1) x 5 ], f 1 (x) = cos [π/2 x 5 ]. [-5x 4 ] f 1 (5π/2) = cos [π/2 5π/2 ) 5 ].(-5)(5π/2 ) 4 = - 5 (π/2) 4/5 Answer 2] - 5 (π/2) 4/5

. 32. If y = cos -1 2cosx -3sinx then dy = 1] 2 2] -2 3] -1 4] 1 13 dx

. 32. Solution : y = cos -1 2cosx -3sinx 13 y = cos -1 2 cosx - 3 sinx 13 13 cosα = 2/ 13 sinα3/ 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 = cos 2 3x/2 sin 2 3x/2, then d 2 y/dx 2 is 1] 9y 2] -3 1 y 2 3] 3 1 y 2 4] -9y

. 33. Solution : Given that y = cos 2 3x/2 sin 2 3x/2 y= cos 3x dy/dx = - 3 sin 3x d 2 y/dx 2 = -9 cos 3x = -9y Answer: 4]-9y

. 34. If y = log 5 ( log 5 x ) then dy/dx = 1] 1 x log 5 x 2] 1 x log 5 x.log 5 x 3] 1 x log 5 x.(log5) 2 4] none of these

. 34. Solution : Given that: y = log 5 ( log 5 x ) dy 1 dx (log5) (log 5 x) x (log5) 1 = x (log 5 x) (log5) 2 Answer:3] 1 x (log 5 x) (log5) 2

. 35. If f (x ) =1+nx+ n (n-1) x 2 + n ( n-1) (n-2) x 3 2 6 +.. + x n then f (1) = 1] n ( n-1 ) 2 n-1 2] (n-1)2 n-1 3] n (n-1)2 n-2 4] n (n-1)2 n

. 35. Solution : Given that: f (x ) = 1 + nx+ n (n-1) x+ n ( n-1) (n-2) x 2 +.+ x n f (x) = ( 1+x) n f 1 (x) = x ( 1 + x ) n-1 f 11 (x) = n ( n-1) ( 1+x) n-2 f 11 (1) = n ( n-1) ( 1+1) n-2 = n ( n-1) 2 n-2 2 6 (by binomial theorem) Answer : 3]n ( n-1) 2 n-2

36. If y = sec -1 x+1 + sin -1 x - 1, then dy = 1] x - 1 x + 1 ` 2] x + 1 3] 0 4] 1 x - 1 x-1 x+1 dx

. 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 y = π / 2 dy/dx = 0 x +1 x +1 Answer :3] 0

. 37.If y = tan -1 1+sinx+ 1 sinx, 0 <x< π/2 Then dy/dx = 1] ½ 2] -½ 3] x/2 4] x/2 1+sinx 1 sinx

. 37. Solution : Given that: y = tan -1 1+sinx+ 1 sinx 1+sinx 1 sinx 0 <x< π/2 0 <x/2< π/4 y = tan -1 Cos x/2 > sin x/2 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. 38. y = sin -1 (3x 4x 3 ), then dy/dx at x = 1/3, is 1] -9 2 4 2] 9 2 3] 9 2 4 4] 9/8

. 37. 38. Solution : Given that: y = sin -1 (3x 4x 3 ) y = 3 sin -1 x dy/dx = 3 1 - x dy/dx = 3 9 9 2 1-1/9 8 4 Answer :3] 9 2 4

. 37. 39. If y = tan -1 1 + tan -1 1 + tan -1 1 1 + x+x 2 x 2 +3x+3 x 2 +5x+7. to n terms then y 1 (0) = 1] -1/n 2 +1 2] n 2 /(n 2 +1) 1] -1/n 2 +1 2] n 2 /(n 2 +1) 3] n 2 /n 2 +1 4] n / n+1

. 37. 39. If y = tan -1 1 + tan -1 1 + tan -1 1 1 + x+x 2 x 2 +3x+3 x 2 +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)-tan 1 x + 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 y 1 = 1/1+(x+n) 2 1/1+x 2 y 1 (0)=1/(1+n 2 )-1 Answer: 2] n 2 /(n 2 +1)