Let f and g be differentiable functions satisfying g(a)=b,g1(a)=2 and fog=I (identity function). Then f1(b) is equal to
fog=I⇒fogx=x ∀x
⇒f1(g(x))g1(x)=1∀x⇒f1(g(a))=1g1(a)=12⇒f1(b)=12 (∵g(a)=b)