Let $f(x)=\tan ^{-1}(g(x))$, where g(x) is monotonically increasing for $0<x<\frac{\pi}{2}$. Then, f(x), is
Answer & explanation
Correct answer: option 1
We have,
$f(x)=\tan ^{-1}(g(x)) \Rightarrow f'(x)=\frac{1}{1+\{g(x)\}^2} \times \frac{d}{d x}(g(x))$
For f(x) to be increasing, we must have
f'(x) > 0
$\Rightarrow \frac{1}{1+\{g(x)\}^2} g'(x)>0$
$\Rightarrow g'(x)>0$
$\Rightarrow x \in(0, \pi / 2)$ [Given]