Question

$f\left(x\right)={\tan }^{-1}(\sin x+\cos x)$, $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$, is increasing in

• A. $\left(\frac{-\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$
• B. $\left(0,\frac{\mathrm{\pi }}{2}\right)$
• C. $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{4}\right)$
• D. $\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right)$

Answer 1

The correct option is C

$\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{4}\right)$

Explanation for the correct option:

Find the increasing domain of given function:

$f\left(x\right)={\tan }^{-1}(\sin x+\cos x)$

$f\text{'}\left(x\right)=\frac{(\cos x-\sin x)}{1+{\left(\sin x+\cos x\right)}^2}$ $\left[\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{dx}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\right]$

$\because$$f\left(x\right)$ is increasing if $f\text{'}\left(x\right)\ge 0$

$\Rightarrow$$\frac{(\cos x-\sin x)}{1+{\left(\sin x+\cos x\right)}^2}\ge 0$

$\Rightarrow$ $\cos x-\sin x\ge 0$

$\Rightarrow$ $\cos x\ge \sin x$

$\therefore f\left(x\right)$ is increasing in $\left(\frac{\mathbf{-}\mathbf{\pi }}{\mathbf{2}}\mathbf{,}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{4}}\right)$

Hence, Option ‘C’ is Correct.