Let f:0,π2→[0,1] be a differentiable function such that f(0)=0,fπ2=1 then

Consider g(x)=sin-1f(x)-xg(0)=0,gπ2=0 A) By Rolle's theorem there is atleast one value of α∈0,π2 such that g'(α)=0⇒f'(α)1-f(α)2-1=0⇒f'(α)=1-(f(α))2 It is not true for all values of a B) Consider g(x)=f(x)-2xπg(0)=0,gπ2=0g'(x)=0 at least for one value of α∈0,π2⇒f'(α)=2π for atleast one value of α∈0,π2 C) Consider g(x)=[f(x)]2-4x2π2g(0)=0,gπ2=0⇒g'(x)=0 for atleast one value of α∈0,π2f(α)f'(α)=1π( true ) D) Consider g(x)=f(x)-4x2π2,g(0)=0 and gπ2=0f'(α)=8απ2 atleast for one value of α∈0,π2