Question

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

$f(x)=\{\begin{array}{ll}\frac{\sqrt{2}cosx-1}{cotx-1},& x\ne \frac{\pi }{4}\\ k,& x=\frac{\pi }{4}\end{array}$

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 = $\frac{\pi }{4}$
$\therefore {\lim }_{x\to \frac{\pi }{4}}f(x)=f\left(\frac{\pi }{4}\right)$
$\Rightarrow {\lim }_{x\to \frac{\pi }{4}}\frac{\sqrt{2}cosx-1}{cotx-1}=k$
$\Rightarrow {\lim }_{x\to \frac{\pi }{4}}\frac{-\sqrt{2}sinx}{-cose{c}^{2}x}=k$ (Using L'H $\hat{o}$ pital rule)
${\lim }_{x\to \frac{\pi }{4}}\sqrt{2}si{n}^{3}x=k$
$\Rightarrow k=\sqrt{2}(\frac{1}{\sqrt{2}}{)}^3=\frac{1}{2}$