The solution set of `f'(x)>g'(x)` where `f(x)=(1/2)5^(2x+1)` and `g(x)=5^x+4xlog5` is

The solution set of f'(x)>g'(x) where f(x)=((1)/(2))5^(2x+1) and g(x)=5^(x)+4x log5

If f@g=g@f, where f(x)=2x+5 and g(x)=3x+h, then: h=

Let F(x) = f(x) +g(x) , G(x) = f(x) - g( x) and H(x) =(f( x))/( g(x)) where f(x) = 1- 2 sin ^(2) x and g(x) =cos (2x) AA f: R to [-1,1] and g, R to [-1,1] Now answer the following If the solution of F(x) -G (x) =0 are x_1,x_2, x_3 --------- x_n Where x in [0,5 pi] then

If f(x)=4x-5,g(x)=x^(2) and h(x)=(1)/(x) , then f(g(h(x))) is :