Question

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

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

Answer 1

The correct option is B

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

Find the interval in which the given function is increasing

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

$$\begin{gathered} f\left(x\right)={\tan }^{-1}\left(\sin \left(x\right)+\cos \left(x\right)\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{1}{{\left(1+\sin \left(x\right)\cos \left(x\right)\right)}^2}\left(\cos \left(x\right)-\sin \left(x\right)\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{\cos \left(x\right)-\sin \left(x\right)}{1+\sin \left(2x\right)} \\ \Rightarrow f\text{'}\left(x\right)=-2\sin \left(2x\right)\cos \left(2x\right) \\ \Rightarrow f\text{'}\left(x\right)=-\sin \left(4x\right) \end{gathered}$$

For increasing function $f\text{'}\left(x\right)>0$

For $-\frac{\pi }{2}<x<\frac{\mathrm{\pi }}{4},\cos \left(x\right)>\sin \left(x\right)$

$y=f\left(x\right)$ is increasing in $\left(-\frac{\pi }{2},\frac{\mathrm{\pi }}{4}\right)$

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