If $f(x)=\begin{cases}\frac{k~cos~x}{\pi-2x},&x
e\frac{\pi}{2}\ 3&,x=\frac{\pi}{2}\end{cases}$ is continuous at $x=\frac{\pi}{2}$ then k is:

Q: If $f(x)=\begin{cases}\frac{k~cos~x}{\pi-2x},&x\ne\frac{\pi}{2}\\ 3&,x=\frac{\pi}{2}\end{cases}$ is continuous at $x=\frac{\pi}{2}$ then k is:

If $f(x)=\begin{cases}\frac{k~cos~x}{\pi-2x},&x\ne\frac{\pi}{2}\\ 3&,x=\frac{\pi}{2}\end{cases}$ is continuous at $x=\frac{\pi}{2}$ then k is:

Question

Accepted Answer

To determine the value of $ k $ that makes the function $ f(x) $ continuous at $ x = \frac{\pi}{2} $, we need to ensure that the limit of $ f(x) $ as $ x $ approaches $\frac{\pi}{2}$ is equal to $ f\left(\frac{\pi}{2}\right) $.

The function $ f(x) $ is defined as:

$$
f(x) = 
\begin{cases} 
\frac{k \cos x}{\pi - 2x}, & x \ne \frac{\pi}{2} \
3, & x = \frac{\pi}{2} 
\end{cases}
$$

For continuity at $ x = \frac{\pi}{2} $, we need:

$$
\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 3
$$

First, we calculate the limit of $ f(x) $ as $ x $ approaches $\frac{\pi}{2}$:

$$
\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}
$$

As $ x \to \frac{\pi}{2} $, $\cos x \to 0$ and $\pi - 2x \to 0$, resulting in an indeterminate form $\frac{0}{0}$. We can apply L'Hôpital's Rule to resolve this:

Differentiate the numerator and the denominator:

- The derivative of $ k \cos x $ is $-k \sin x$.
- The derivative of $\pi - 2x$ is $-2$.

Applying L'Hôpital's Rule:

$$
\lim_{x \to \frac{\pi}{2}} \frac{-k \sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{k \sin x}{2}
$$

As $ x \to \frac{\pi}{2} $, $\sin x \to 1$. Therefore, the limit becomes:

$$
\frac{k \cdot 1}{2} = \frac{k}{2}
$$

For continuity, this limit must equal the value of the function at $ x = \frac{\pi}{2} $, which is 3:

$$
\frac{k}{2} = 3
$$

Solving for $ k $:

$$
k = 6
$$

Thus, the correct value of $ k $ is 6, which corresponds to Option 2.

**Explanation for Other Options:**

- **Option 1 (k = 3):** If $ k = 3 $, then $\frac{k}{2} = \frac{3}{2}$, which does not equal 3. Therefore, this option does not satisfy the continuity condition.
  
- **Option 3 (k = 9):** If $ k = 9 $, then $\frac{k}{2} = \frac{9}{2} = 4.5$, which does not equal 3. Therefore, this option does not satisfy the continuity condition.
  
- **Option 4 (k = 12):** If $ k = 12 $, then $\frac{k}{2} = \frac{12}{2} = 6$, which does not equal 3. Therefore, this option does not satisfy the continuity condition.

In conclusion, only Option 2 (k = 6) ensures that the function $ f(x) $ is continuous at $ x = \frac{\pi}{2} $.