Solution:

The provided function is: $f(x)=\left\{\begin{array}{lll}k x+1, \text { if } x \leq \pi \\ \cos x, \text { if } x>\pi\end{array}\right.$

$\lim _{x \rightarrow n^{+}} f(x)=\lim _{h \rightarrow 0} f(\pi+h)=\lim _{h \rightarrow 0} \cos (\pi+h)=-\cos h=-\cos0 = -1$

and $\lim _{x \rightarrow n^{-}} f(x)=\lim _{h \rightarrow 0} f(\pi-x)=\lim _{h \rightarrow 0} \cos (\pi-h)=-\cos h=-\cos 0=-1$

Now again,

$\lim _{x \rightarrow \pi} f(x)=\lim _{h \rightarrow 0}(k \pi+1)$

Since the provided function is continuous at $x=\pi$, we get

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

The value of $\mathrm{k}$ is $-2 / \pi$.