If \(f(x) = \begin{cases}\displaystyle\frac{\tan\left(\frac{\pi}{4}-x\right)}{\cot 2x}, & x \ne \frac{\pi}{4}\\ k, & x = \frac{\pi}{4}\end{cases}\), is continuous at \(x = \displaystyle\frac{\pi}{4}\), then the value of \(k\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\)
For continuity at \(x = \frac{\pi}{4}\), k must equal the limit of f(x) as \(x \rightarrow \frac{\pi}{4}\).
Let \(x = \frac{\pi}{4} - h\), so \(h \rightarrow 0\). Then \(2x = \frac{\pi}{2} - 2h\) and \(\cot 2x = \tan 2h\):
\( \displaystyle \frac{\tan\left(\frac{\pi}{4}-x\right)}{\cot 2x} = \frac{\tan h}{\tan 2h} \)
Using \(\tan 2h = \frac{2\tan h}{1-\tan^2 h}\):
\( \displaystyle \frac{\tan h}{\tan 2h} = \frac{1-\tan^2 h}{2} \)
\( \displaystyle \lim_{h \rightarrow 0}\frac{1-\tan^2 h}{2} = \frac{1}{2} \)
So \(k = \frac{1}{2}\).
# Maths 04-08-2022 Slot 1, part M2: paper Q26-Q50 (Section B1), new rows from the paper PDF