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Q)

Find the value of k if f(x) is continuous at $x = \frac{\pi}{2}$, when $f(x) =\begin{cases}\frac{k \cos x}{\pi - 2x }\;;\; x \neq \frac{\pi}{2} \\ 3 ,\; \; \; \; \; \; \; \; \; x = \frac{\pi}{2} \end{cases}$

$(A)\; k = -6$
$(B)\; k = 6$
$(C)\; k = 5$
$(D)\; k = -5$

The function is continuous at $x = \frac{\pi}{2}$
$\therefore \lim_{x \to \frac{\pi}{2}} f(x) = f \frac{\pi}{2}$
$\implies $$\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x} = 3$$$
$\implies k \lim_{x \to \frac {\pi}{2}} \frac{\sin( \frac{\pi}{2}-x)}{2 (\frac{\pi}{2} - x)} = 3$
\begin{align*}\implies \frac {k}{2} \times -1 &= 3 \\ \therefore k & =6 \end {align*}