Question

Let $f\left(x\right)={\tan }^{-1}\left(g\left(x\right)\right),$ where g (x) is monotonically increasing for 0 < x < $\frac{\mathrm{\pi }}{2}.$ Then, f(x) is
(a) increasing on (0, π/2)
(b) decreasing on (0, π/2)
(c) increasing on (0, π/4) and decreasing on (π/4, π/2)
(d) none of these

Answer 1

(a) increasing on (0, $\mathrm{\pi }$/2)

$$\begin{gathered} \text{Given:}g\left(x\right)\text{is increasing on}\left(0,\frac{\mathrm{\pi }}{2}\right)\text{. Then,} \\ {x}_1<x_2,\forall x_1,x_2\in \left(0,\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow g\left(x_1\right)<g\left(x_2\right) \\ {\text{Taking tan}}^{-1}\text{on both the sides, we get} \\ {\tan }^{-1}\left(g\left(x_1\right)\right)<{\tan }^{-1}\left(g\left(x_2\right)\right) \\ \Rightarrow f\left(x_1\right)<f\left(x_2\right),\forall x_1,x_2\in \left(0,\frac{\mathrm{\pi }}{2}\right) \\ \text{So,}f\left(x\right)\text{is increasing on}\left(0,\frac{\mathrm{\pi }}{2}\right)\text{.} \end{gathered}$$