Question

The function $f\left(x\right)={\tan }^{-1}\left(\sin \left(x\right)+\cos \left(x\right)\right)$ is an increasing function on the interval

• A. $\left(0,\mathrm{\pi }\right)$
• B. $\left(0,\frac{\mathrm{\pi }}{2}\right)$
• C. $\left(0,\frac{\mathrm{\pi }}{4}\right)$
• D. $\left(0,\frac{3\mathrm{\pi }}{4}\right)$
• E. $\left(0,\frac{5\mathrm{\pi }}{4}\right)$

Answer 1

The correct option is C

$\left(0,\frac{\mathrm{\pi }}{4}\right)$

Find the interval in which the given function is increasing

We know that a function $f\left(x\right)$ is increasing then $f\text{'}\left(x\right)>0$

$$\begin{gathered} f\left(x\right)={\tan }^{-1}\left(\sin \left(x\right)+\cos \left(x\right)\right) \\ \Rightarrow y={\tan }^{-1}\left(\sqrt{2}\sin \left(x+\frac{\pi }{4}\right)\right)\left[\mathrm{Le}ty=f\left(x\right)\right] \\ \Rightarrow \tan \left(y\right)=\sqrt{2}\sin \left(x+\frac{\pi }{4}\right) \\ \Rightarrow {\sec }^2\left(y\right)\frac{dy}{dx}=-\sqrt{2}\cos \left(x+\frac{\pi }{4}\right) \end{gathered}$$

So,

$$\begin{gathered} =\frac{dy}{dx}>0 \\ =\cos \left(x+\frac{\pi }{4}\right)>0 \\ =x\in \left(0,\frac{\pi }{4}\right) \end{gathered}$$

Hence option $\left(\mathrm{C}\right)$ is correct.