If \(f(x) =\begin{cases}\displaystyle\frac{\tan(\frac{\pi}{4}-x)}{\cot 2x}, & x\ne\frac{\pi}{4}\\ 2K+1, & x=\frac{\pi}{4}\end{cases}\) is continuous at \(x=\frac{\pi}{4}\), then the value of K is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle-\frac{1}{4}\)
For continuity at \(x = \frac{\pi}{4}\), the limit there must equal \(f\left(\frac{\pi}{4}\right) = 2K + 1\).
Let \(x = \frac{\pi}{4} - t\), so \(t \to 0\) as \(x \to \frac{\pi}{4}\). Then
\(\displaystyle \tan\left(\frac{\pi}{4} - x\right) = \tan t\)
\(\displaystyle \cot 2x = \cot\left(\frac{\pi}{2} - 2t\right) = \tan 2t\)
Using the standard limit \(\displaystyle\lim_{t\to 0}\frac{\tan t}{t} = 1\):
\(\displaystyle \lim_{t \to 0}\frac{\tan t}{\tan 2t} = \lim_{t \to 0}\frac{\tan t}{t}\cdot\frac{2t}{\tan 2t}\cdot\frac{1}{2} = \frac{1}{2}\)
\(\displaystyle 2K + 1 = \frac{1}{2}\)
\(\displaystyle K = -\frac{1}{4}\)