If $f(x)=\left\{\begin{array}{cc}\frac{\sin (\cos x)-\cos x}{(\pi-2 x)^3}, & x \neq \pi / 2 \\ \quad ~~~~k \quad ~~~~~~, & x=\pi / 2\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then k =

$-\frac{1}{48}$

It is given that f(x) is continuous at $x=\frac{\pi}{2}$

∴  $k=\lim\limits_{x \rightarrow \pi / 2} f(x)$

$\Rightarrow k=\lim\limits_{x \rightarrow \pi / 2} \frac{\sin (\cos x)-\cos x}{(\pi-2 x)^3}$

$\Rightarrow k=\lim\limits_{x \rightarrow \pi / 2} \frac{\sin (\cos x)-\cos x}{\cos ^3 x} \times \frac{\sin ^3(\pi / 2-x)}{8(\pi / 2-x)^3}$

$\Rightarrow k=-\frac{1}{6} \times \frac{1}{8}=-\frac{1}{48} \quad \left[∵ \lim\limits_{x \rightarrow 0} \frac{\sin x-x}{x^3}=-\frac{1}{6}\right]$