If $f(x)$, defined by $f(x)=\left\{\begin{array}{lll}k x+1 & \text { if } & x \leq \pi \\ \cos x & \text { if } & x>\pi\end{array}\right.$ is continuous at $x=\pi$, then the value of $k$ is :

Left-hand limit as $x$ approaches $\pi$:

$\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-}(kx + 1) = k\pi + 1$

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Right-hand limit as $x$ approaches $\pi$:

$\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x = \cos \pi = -1$

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For continuity at $x = \pi$, these limits must be equal:

$k\pi + 1 = -1$

$k\pi = -2$

$k = \dfrac{-2}{\pi}$

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Verification: When $x = \pi$

Left side: $f(\pi) = k\pi + 1 = \dfrac{-2}{\pi} \cdot \pi + 1 = -2 + 1 = -1$

Right side: $\cos(\pi) = -1$

Since both sides equal $-1$, the value of $k$ is $\dfrac{-2}{\pi}$.