Question

Let $f\left(x\right)={\tan }^{-1}\left\{\varphi \left(x\right)\right\},$ where $\varphi \left(x\right)$ is m.i. for $0<x<\frac{\pi }{2}.$ Then $f\left(x\right)$ is

• A. increasing in $(0,\frac{\pi }{2})$
• B. decreasing in $(0,\frac{\pi }{2})$
• C. increasing in $(0,\frac{\pi }{4})$ and decreasing in $(\frac{\pi }{4},\frac{\pi }{2})$
• D. none of these

Answer 1

The correct option is A increasing in $(0,\frac{\pi }{2})$

$f\left(x\right)={\tan }^{-1}\varphi \left(x\right)$
for calculating increasing and decreasing function, have to calculate derivative of $f\left(x\right)$
$f^{\prime }\left(x\right)=\frac{1}{1+{\varphi }^2\left(x\right)}$
$f^{\prime }\left(x\right)$ is always positive so function will be increasing in interval $(0,\frac{\pi }{2})$