Question

If the function f defined on $(\frac{\pi }{6},\frac{\pi }{3})$ by $f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{2}\cos x-1}{\cot x-1},& x\ne \frac{\pi }{4}\\ k,& x=\frac{\pi }{4}\end{array}$ is continuous, then k is equal to?

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

Answer 1

The correct option is A $\frac{1}{2}$
$\therefore$ function should be continuous at $x=\frac{\pi }{4}$

$\therefore \underset{x\to \frac{x}{4}}{lim}f\left(x\right)=f\left(\frac{\pi }{4}\right)$

$\Rightarrow {lim}_{x\to \frac{x}{4}}\frac{\sqrt{2}\cos x-1}{\cot x-1}=k$

$\Rightarrow \underset{x\to \frac{\pi }{4}}{lim}\frac{-\sqrt{2}\sin x}{-cose{c}^{2}x}=k$ (Using L'$\hat{o}$ pital rule)

$\underset{x\to \frac{\pi }{4}}{lim}\sqrt{2}{\sin }^{3}x=k$

$\Rightarrow k=\sqrt{2}{\left(\frac{1}{\sqrt{2}}\right)}^3=\frac{1}{2}$.