MHTCET 16 May 2024 Shift 2 Paper

Concept:For a function to be continuous at a point, the value of the function at that point must equal the limit of the function as $x$ approaches that point.Explanation:Since $f(x)$ is continuous on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$, it is continuous at $x = \frac{\pi}{4}$.Therefore, the value at that point equals the limit:$$k = f\left(\frac{\pi}{4}\right) = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}$$Direct substitution gives $\frac{0}{0}$, so we apply L'Hospital's rule.Differentiate the numerator: derivative of $\sqrt{2}\cos x - 1$ is $-\sqrt{2}\sin x$.Differentiate the denominator: derivative of $\cot x - 1$ is $-\csc^2 x$.Thus,$$k = \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2}\sin x}{-\csc^2 x} = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\sin x}{\csc^2 x}$$Substitute $x = \frac{\pi}{4}$:$$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \csc \frac{\pi}{4} = \sqrt{2}$$$$k = \frac{\sqrt{2} \times \frac{1}{\sqrt{2}}}{(\sqrt{2})^2} = \frac{1}{2}$$Answer:$k = \frac{1}{2}$, which matches option A.