If the function \(f(x) = \begin{cases}\displaystyle\frac{\sin x - \cos x}{x - \frac{\pi}{4}}, & x \ne \displaystyle\frac{\pi}{4}\\ k, & x = \displaystyle\frac{\pi}{4}\end{cases}\) is continuous at \(x = \displaystyle\frac{\pi}{4}\), then the value of k is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\sqrt{2}\)
f is continuous at \(\frac{\pi}{4}\) when k equals the limit of f there.
Writing the numerator as a single sine:
\( \displaystyle \sin x - \cos x = \sqrt{2}\left(\sin x\cos\frac{\pi}{4} - \cos x\sin\frac{\pi}{4}\right) \)
\( \displaystyle = \sqrt{2}\,\sin\left(x - \frac{\pi}{4}\right) \)
Let \(h = x - \frac{\pi}{4}\), so \(h \to 0\). Using \(\lim_{h \to 0}\frac{\sin h}{h} = 1\):
\( \displaystyle \lim_{h \to 0}\frac{\sqrt{2}\sin h}{h} = \sqrt{2} \)
So \(k = \sqrt{2}\).