The correct option is B [0,π2]
Given f(x)=cosx,g(x)=2x2−1
g(f(x))=2(f(x))2−1=2cos2x−1=cos2x
g(f(x)) is invertible iff cos2x is bijective
i.e., g(f(x)) is one-one and onto.
If cos2x is one-one, then exhaustive set of values for 2x is [nπ,(n+1)π] where n∈Z
⇒x∈[nπ2,(n+1)π2]
For n=0, x∈[0,π2]