Let `f(x)=2^(2x-1)" and "g(x)=-2^(x)+2xlog2.` Then the set of points satisfying `f'(x)gtg'(x)`, is

Let f(x) = x^(2) and g(x) = sin x for all x in R. Then the set of all x satisfying (fogogog)(x) = (gogof)(x), where (fog)(x) = f(g(x)), is

Let f(x)=x+3ln(x-2)&g(x)=x+5ln(x-1) then the set of x satisfying the inequality f'(x)

Let f(x)=x+3ln(x-2)&g(x)=x+5ln(x-1), then the set of x satisfying the inequality f'(x) lt g'(x) is

Let f(x)=x^(2) and g(x)= sin x for all x in R . Then the set of all x satisfying (fogogof)(x)=(gogof)(x) , where (fog)(x)=f(g(x)) , is

Let f(x)=x^(2) and g(X)=sinx for all xepsilonR . Then the set of all x satisfying (f o g o g o f)(x)=(g o g o f)(x) , where (f o g)(x)=f(g(x)) is

Let f(x)= 2^(2x−1) and ϕ(x) = −2^x+2xlog2 . If f'(x)>ϕ'(x), then value of x is-

Let f(x)= 2^(2x−1) and ϕ(x) = −2^x+2xlog2 . If f'(x)>ϕ'(x), then value of x is-

if F(x)=x^2-2x+3 then the set of values of x satisfying f(x-1)=f(x+1)