If \(f(x) = \begin{cases}\displaystyle\frac{1 - \tan x}{4x - \pi} & x \ne \displaystyle\frac{\pi}{4}\\ k & x = \displaystyle\frac{\pi}{4}\end{cases}\) 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) → \(\displaystyle\frac{-1}{2}\)
For continuity at \(x = \frac{\pi}{4}\), k must equal the limit of f(x) as \(x \to \frac{\pi}{4}\). Let \(x = \frac{\pi}{4} + h\), so that \(h \to 0\) and \(4x - \pi = 4h\). By the tangent addition formula:
\(\displaystyle \tan\left(\frac{\pi}{4} + h\right) = \frac{1 + \tan h}{1 - \tan h}\)
\(\displaystyle 1 - \tan x = \frac{(1 - \tan h) - (1 + \tan h)}{1 - \tan h} = \frac{-2\tan h}{1 - \tan h}\)
So
\(\displaystyle \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{h \to 0} \frac{-2\tan h}{4h(1 - \tan h)}\)
\(\displaystyle = -\frac{1}{2}\cdot\lim_{h \to 0}\frac{\tan h}{h}\cdot\lim_{h \to 0}\frac{1}{1 - \tan h}\)
\(\displaystyle = -\frac{1}{2}\cdot 1\cdot 1 = -\frac{1}{2}\)
Hence \(k = -\frac{1}{2}\).