If `f"(x)>0AAx in R ,f'(3)=0,a n dg(x)=f("tan"hat2x-2"tan"x+4),0

If f''(x) gt forall in R, f(3)=0 and g(x) =f(tan^(2)x-2tanx+4y)0ltxlt(pi)/(2) ,then g(x) is increasing in

If f(0) = f'(0) = 0 and f''(x) = tan^(2)x then f(x) is

If f(x) is continuous at x=0 , where f(x)=(sin (a+x)-sin (a-x))/(tan (a+x)-tan (a-x)), x != 0 , then f(0)=

Let f (x) = int x ^(2) cos ^(2)x (2x + 6 tan x - 2x tan ^(2) x ) dx and f (x) passes through the point (pi, 0) If f : R-(2n +1) pi/2 to R then f (x) be a :

Let f (x) = int x ^(2) cos ^(2)x (2x + 6 tan x - 2x tan ^(2) x ) dx and f (x) passes through the point (pi, 0) The number of solution (s) of the equation f (x) =x ^(3) in [0, 2pi] be:

Given f(0)=0;f'(0)=0 and f''(x)=sec^(2)x+sec^(2)x tan^(2)x+ then f(x)=

If f:R rarr R is continous function satisying f(0) =1 and f(2x) -f(x) =x, AAx in R , then lim_(nrarroo) (f(x)-f((x)/(2^(n)))) is equal to

If f(x) is continuous at x=0 , where f(x)=(log(2+x)-log(2-x))/(tan x) , for x!=0 , then f(0)=