$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{kx + 1,}&{if}&{x \le \pi }\\{\cos ,}&{if}&{x > \pi }\end{array}} \right.$ at $x = \pi .$

$\mathop {\lim }\limits_{x \to {\pi ^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to \pi + h\atop\scriptstyle h \to 0} f(\pi + h)$

$= \mathop {\lim }\limits_{\scriptstyle x \to \pi + h\atop\scriptstyle h \to 0} \cos (\pi + h) = \mathop {\lim }\limits_{\scriptstyle x \to \pi + h\atop\scriptstyle h \to 0} - \cosh = - \cos (0) = - 1$

$\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to \pi - h\atop\scriptstyle h \to 0} k(\pi - h) + 1 = k\pi + 1$
and $f(\pi ) = k\pi + 1$

Since the given function is continuous at $x = \pi$,

therefore, $\mathop {\lim }\limits_{x \to {\pi ^ + }} f(x) = \mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = f(\pi )$

$\Rightarrow$ $k + 1 = - 1 \Rightarrow k\pi = - 1 - 1 \Rightarrow k\pi = - 2 \Rightarrow k = \cfrac{{ - 2}}{\pi }$