Let f and g be differentiable functions satisfying g'(a) = 2, g(a) = b and fog = I(identity function). Then, f'(b) is equal to
Answer & explanation
Correct answer: option 3
We have,
fog = I
⇒ fog(x) = I(x) for all x
⇒ f(g(x)) = x for all x
⇒ $\frac{d}{d x}\{f(g(x))\}=1$ for all x
$\Rightarrow \frac{d}{d g(x)}\{f(g(x))\} . \frac{d g(x)}{d x}=1$ for all x
$\Rightarrow f'(g(x)) g'(x)=1$ for all x
$\Rightarrow f'(g(a)) g'(a)=1 \Rightarrow f'(b) \times 2=1 \Rightarrow f'(b)=\frac{1}{2}$