If the function \(f(x) = \begin{cases}\displaystyle\frac{k\cos x}{\pi-2x} & : x \ne \displaystyle\frac{\pi}{2}\\ 3 & : x = \displaystyle\frac{\pi}{2}\end{cases}\) is continuous at \(x = \displaystyle\frac{\pi}{2}\), then \(k\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6
For continuity at \(x = \frac{\pi}{2}\), the limit there must equal \(f\left(\frac{\pi}{2}\right) = 3\). Let \(x = \frac{\pi}{2} - h\), so that \(h \to 0\) as \(x \to \frac{\pi}{2}\):
\( \displaystyle \cos x = \cos\left(\frac{\pi}{2} - h\right) = \sin h \)
\( \displaystyle \pi - 2x = \pi - 2\left(\frac{\pi}{2} - h\right) = 2h \)
\( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = \lim_{h \to 0} \frac{k\sin h}{2h} = \frac{k}{2} \)
using the standard limit \(\displaystyle\lim_{h \to 0}\frac{\sin h}{h} = 1\). Hence:
\( \displaystyle \frac{k}{2} = 3 \)
\(k = 6\)
NCERT (2026-27 edition):
Continuity
