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

6

The correct answer is Option (1) → 6

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

Continuity at $x=\frac{\pi}{2}$ requires

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

As $x\to\frac{\pi}{2}$, both numerator and denominator tend to $0$

Apply L’Hospital’s Rule

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

$=\frac{k}{2}$

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

$k=6$

The value of $k$ is $6$.