If the function $\mathrm{f}$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\frac{\sqrt{2} \cos \mathrm{x}-1}{\cot \mathrm{x}-1}, & \mathrm{x} \neq \frac{\pi}{4} \\ \mathrm{k}, & \mathrm{x}=\frac{\pi}{4}\end{array}\right.$
is continuous, then $\mathrm{k}$ is equal to:
Correct Option: , 2
Since, $f(x)$ is continuous, then
$\lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)$
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}=k$
Now by L- hospital's rule
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\operatorname{cosec}^{2} x}=k \Rightarrow \frac{\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)}{(\sqrt{2})^{2}}=k \Rightarrow k=\frac{1}{2}$
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