If the function $f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x} & : x \neq \frac{\pi}{2} \\ 3 & : x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$, then $k$ is equal to

For a function to be continuous at a point $x = a$, the limit of the function as $x$ approaches $a$ must equal the value of the function at that point:

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

From the problem statement, we know that $f\left(\frac{\pi}{2}\right) = 3$. Therefore, we need to find $k$ such that:

$\lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = 3$

Evaluating the Limit using L'Hôpital's Rule:

Since substituting $x = \frac{\pi}{2}$ into the expression results in an indeterminate form of $\frac{0}{0}$, we can apply L'Hôpital's Rule (differentiating the numerator and denominator with respect to $x$):

• The derivative of the numerator $k\cos x$ is $-k\sin x$.
• The derivative of the denominator $\pi - 2x$ is $-2$.

Applying the rule:

$\lim_{x \to \frac{\pi}{2}} \frac{-k\sin x}{-2} = 3$

$\frac{-k\sin(\frac{\pi}{2})}{-2} = 3$

Since $\sin(\frac{\pi}{2}) = 1$:

$\frac{-k(1)}{-2} = 3$

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

$k = 6$