The value of $k(k>0)$ such that the length of the longest interval in which the function $f\left(x\right)={\sin }^{-1}|\sin kx|+{\cos }^{-1}(\cos \kappa x)$ is constant $\frac{\pi }{4}$ is/are

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

$\text{ Let }g\left(x\right)={\sin }^{-1}|\sin x|+{\cos }^{-1}(\cos x)$

$g\left(x\right)=\left\{\begin{array}{l}2x,0\le x\le \frac{\pi }{2}\\ \pi ,\frac{\pi }{2}<x\le \frac{3\pi }{2}\\ 4\pi -2x,\frac{3\pi }{2}<x\le 2\pi \end{array}\right.$

$g\left(x\right)$ is perodic with period $2\pi$ and is constant in the continous interval $\left[2n\pi +\frac{\pi }{2},2n\pi +\frac{3\pi }{2}\right]$

$\text{ (Where }n\in I\text{ and }f\left(x\right)=g\left(kx\right)\text{ ) }$

So, $f\left(x\right)$ is constant in the interval

$\left[\frac{2n\pi }{k}+\frac{\pi }{2k},\frac{2n\pi }{k}+\frac{3\pi }{2k}\right]$

$\Rightarrow \frac{\pi }{4}=\frac{3\pi }{2k}-\frac{\pi }{2k}\Rightarrow \frac{\pi }{k}=\frac{\pi }{4}\Rightarrow k=4$