the value of k is:

The correct option is -2/π.

Step-by-step Explanation:

From the image provided, we are given a piecewise function defined as:
$$f\left(x\right)=\{\begin{array}{cc}kx+1& \text{if }x\le \pi \\ \cos x& \text{if }x>\pi \end{array}$$
We need to find the value of $k$ such that the function is continuous at $x=\pi$.

For a function to be continuous at a point $x=c$, the left-hand limit (LHL) as $x$ approaches $c$, the right-hand limit (RHL) as $x$ approaches $c$, and the value of the function at $x=c$ must all be equal:
$$\underset{x\to {\pi }^{-}}{lim}f\left(x\right)=\underset{x\to {\pi }^{+}}{lim}f\left(x\right)=f\left(\pi \right)$$

Step 1: Calculate the Left-Hand Limit (LHL) and $f\left(\pi \right)$
For the interval $x\le \pi$, the function is defined by $f\left(x\right)=kx+1$. Evaluating this at $x=\pi$ gives:
$$f\left(\pi \right)=\underset{x\to {\pi }^{-}}{lim}(kx+1)=k\pi +1$$

Step 2: Calculate the Right-Hand Limit (RHL)
For the interval $x>\pi$, the function is defined by $f\left(x\right)=\cos x$. Taking the limit as $x$ approaches $\pi$ from the right:
$$\underset{x\to {\pi }^{+}}{lim}\cos x=\cos \pi$$
Since the value of trigonometric cosine at $\pi$ radians is $-1$:
$$\text{RHL}=-1$$

Step 3: Solve for k
Equating the LHL and RHL for continuity:
$$k\pi +1=-1$$
Subtracting 1 from both sides:
$$k\pi =-2$$
Dividing by $\pi$:
$$k=-\frac{2}{\pi }$$

Thus, the value of $k$ is indeed $-2/\pi$.