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

Answer 1

Answer: 0.5

Since,$f(x)$ is continuous, then
$\underset{x\to \frac{\pi }{4}}{\lim }f(x)=f(\frac{\pi }{4})$
$\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}cosx-1}{cotx-1}=k$
Now by L- Hospital’s rule
$\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}sinx}{cose{c}^{2}x}=k$
$\Rightarrow \frac{\sqrt{2}(\frac{1}{\sqrt{2}})}{(\sqrt{2}{)}^2}=k$
$\Rightarrow k=\frac{1}{2}$