UPSC NDA I 2015 Mathematics Paper

Concept:A function is continuous at a point if the limit as $x$ approaches that point equals the function's value at that point.Explanation:Given $f(x)$ is continuous at $x = \frac{\pi}{2}$, so $\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 3$.The limit for $x \neq \frac{\pi}{2}$ is $\lim_{x \to \frac{\pi}{2}} \frac{\alpha \cos x}{\pi - 2x}$, which gives $\frac{0}{0}$ form.Apply L'Hospital's Rule: differentiate numerator and denominator separately.Derivative of numerator: $\frac{d}{dx}(\alpha \cos x) = -\alpha \sin x$.Derivative of denominator: $\frac{d}{dx}(\pi - 2x) = -2$.Thus the limit becomes $\lim_{x \to \frac{\pi}{2}} \frac{-\alpha \sin x}{-2} = \frac{\alpha}{2} \cdot \sin\left(\frac{\pi}{2}\right) = \frac{\alpha}{2} \cdot 1 = \frac{\alpha}{2}$.Set this limit equal to $3$ for continuity: $\frac{\alpha}{2} = 3 \implies \alpha = 6$.Answer:$\alpha = 6$, which corresponds to option A.