If the function `f(x)={{:(kx+1", if "xlepi),(cosx", if "xgtpi):}` is continuous at `x=pi`, then the value of 'k' is :

Options

A. -1
B. `-2/pi`
C. -2
D. None of these

Answer

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} kx + 1 & \text{if } x \leq \pi \\ \cos x & \text{if } x > \pi \end{cases} \]

is continuous at \( x = \pi \), we need to ensure that the left-hand limit and the right-hand limit at \( x = \pi \) are equal to the function value at \( x = \pi \).

### Step 1: Evaluate \( f(\pi) \)

First, we find \( f(\pi) \):
\[ f(\pi) = k\pi + 1 \quad \text{(since } \pi \leq \pi\text{)} \]

### Step 2: Evaluate the left-hand limit as \( x \) approaches \( \pi \)

Next, we calculate the left-hand limit:
\[ \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) = k\pi + 1 \]

### Step 3: Evaluate the right-hand limit as \( x \) approaches \( \pi \)

Now, we calculate the right-hand limit:
\[ \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x = \cos(\pi) = -1 \]

### Step 4: Set the limits equal for continuity

For \( f(x) \) to be continuous at \( x = \pi \), we need:
\[ \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) \]
This gives us the equation:
\[ k\pi + 1 = -1 \]

### Step 5: Solve for \( k \)

Now, we solve for \( k \):
\[ k\pi + 1 = -1 \\ k\pi = -1 - 1 \\ k\pi = -2 \\ k = \frac{-2}{\pi} \]

Thus, the value of \( k \) is:
\[ k = \frac{-2}{\pi} \]

### Final Answer:
The value of \( k \) is \( \frac{-2}{\pi} \).

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