If f(x)= ^ -1 (g(x)), where g(x) is monotonically increasing for … - Vidyadip

MCQ

If $\text{f}(\text{x})=\tan^{-1}(\text{g}(\text{x})),$ where $g(x)$ is monotonically increasing for $0 < \text{x} < \frac{\pi}{2}.$ Then, $f(x) $is :

✓
Increasing on $\Big(0,\frac{\pi}{2}\Big)$
B
Decreasing on $\Big(0,\frac{\pi}{2}\Big)$
C
Increasing on $\Big(0,\frac{\pi}{4}\Big)$ and decreasing on $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$
D
None of these.

✓

Answer

Correct option: A.

Increasing on $\Big(0,\frac{\pi}{2}\Big)$

Given : $g(x)$ is increasing on $\Big(0,\frac{\pi}{2}\Big).$ Then,
$\text{x}_1 < \text{x}_2,\forall\ \text{x}_1<\text{x}_2\in\Big(0,\frac{\pi}{2}\Big)$
$\Rightarrow\text{g}(\text{x}_1)<\text{g}(\text{x}_2)$
Taking $\tan^{-1}$ on both sides, we get
$\Rightarrow\tan^{-1}(\text{g}(\text{x}_1))<\tan^{-1}(\text{g}(\text{x}_2))$
$\Rightarrow\text{f}(\text{x}_1) < \text{f}(\text{x}_2),\ \forall\ \text{x}_1,\text{x}_2\in\Big(0,\frac{\pi}{2}\Big)$
So, $f(x)$ is increasing on $\Big(0,\frac{\pi}{2}\Big).$

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