Problem Statement
Find the value of $k$ so that the function $f$ is continuous at $x = \frac{\pi}{2}$:
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$
Verified Solution & Marking Scheme
State Condition for Continuity
For $f$ to be continuous at $x = \frac{\pi}{2}$:
$\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 3$
Evaluate Limit Using Substitution
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$, $h \to 0$:
$\lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + h\right)}{\pi - 2\left(\frac{\pi}{2} + h\right)} = \lim_{h \to 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2}(1) = \frac{k}{2}$
Equate Limit to Function Value
$\frac{k}{2} = 3 \implies k = 6$