Question

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by

$f(x) = \begin{cases} \frac{\sqrt{2}cos x - 1}{cot x - 1}, & x \ne \frac{\pi}{4} \\[2ex] k, & x = \frac{\pi}{4} \end{cases}$

is continuous, then $k$ is equal to:

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{1}{\sqrt{2}}$
• D. $2$

Answer 1

Answer: (A)

$\therefore $ function should be continuous at x = $\frac{\pi}{4}$
$\therefore \, \, \lim_{x \to \frac{\pi}{4}} f(x) \, = f\bigg(\frac{\pi}{4}\bigg)$
$\Rightarrow \, \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}cos x - 1}{cot x - 1}= k$
$\Rightarrow \, \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2}sin x }{-cosec^2 x}= k$ (Using L'H $\hat{o}$ pital rule)
$ \lim_{x \to \frac{\pi}{4}} \, \sqrt{2}sin^3 x = k$
$\Rightarrow \, \, k = \sqrt{2} \bigg(\frac{1}{\sqrt{2}}\bigg)^3 \, = \, \frac{1}{2}$