Let `f(x)= int dx/((x^2+1)(x^2+9))` and `f(0)=0` if `,f(sqrt3)=5/36 k` then `k` is
(A) `pi/6`
(B) `pi/3`
(C) `pi/2`
(D) `pi/4`

Let f(x)=int(dx)/((x^(2)+1)(x^(2)+9)) and f(0)=0 if f(sqrt(3)) =(5)/(36)k, then k is

If int_0^pi x f(sinx) dx=A int_0^(pi/2) f(sinx)dx , then A is (A) pi/2 (B) pi (C) 0 (D) 2pi

f(x)=sin+sqrt(3)cosx is maximum when x= pi/3 (b) pi/4 (c) pi/6 (d) 0

Let f(x) = int_(x)^(x+(pi)/(3))|sin theta|d theta(x in [0,pi])

Let f(x) =(kcosx)/(pi-2x) if x!=pi/2 and f(x)=3 if x=pi/2 then find the value of k if lim_(x->pi/2) f(x)=f(pi/2)

Let f(x)=tan^(-1)(1/2 tan2x)+tan^(-1)(cotx)+tan^(-1)(cot^3x) then (1) f((3 pi)/8)=pi (2) f(pi/8)=0 (3) f(pi/8) =pi (4) f((3pi)/8)=0

If I_(1)=int_(3pi)^(0) f(cos^(2)x)dx and I_(2)=int_(pi)^(0) f(cos^(2)x) then

If f(x)=cos^(-1)((2x^2+1)/(x^2+1)), then rang of f(x) is [0,pi] (2) (0,pi/4] (0,pi/3] (4) [0,pi/2)

For f(x)=sin^2x ,x in (0,pi) point of inflection is: pi/6 (b) (2pi)/4 (c) (3pi)/4 (d) (4pi)/3

If f'(x)=x|sinx|AAx""in(0,pi/2)andf(0)=(1)/(sqrt3), then f(x) will be