Q77 / 150 · MHT CET 2024 (16 May Shift 2) · Mathematics

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

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

Solution

$\mathrm{f}(x)$ is continuous on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$. $\Rightarrow \mathrm{f}(x)$ is continuous at $x=\frac{\pi}{4}$. $\therefore \quad \mathrm{f}\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ Applying L'Hospital's rule on R.H.S., we get $\begin{aligned} f\left(\frac{\pi}{4}\right) & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}(-\sin x)}{-\operatorname{cosec}^2 x} \\ & =\frac{\sqrt{2} \sin \frac{\pi}{4}}{\operatorname{cosec}^2 \frac{\pi}{4}} \\ & =\frac{\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)}{(\sqrt{2})^2} \\ & =\frac{1}{2}\end{aligned}$

Hence, the correct answer is option: 'A'.

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