UPSC NDA I 2015 Mathematics Paper

Concept:The limit is the derivative of $G(x)$ at $x=1$, evaluated using L'Hôpital's rule for the $0/0$ form.Explanation:We have $G(x)=\sqrt{25-x^2}$ and need $\displaystyle \lim_{x \to 1}\frac{G(x)-G(1)}{x-1}$.First, $G(1)=\sqrt{25-1}=\sqrt{24}=2\sqrt{6}$.Substituting $x=1$ gives $\frac{0}{0}$, so apply L'Hôpital's rule: differentiate numerator and denominator separately.Derivative of numerator: $\frac{d}{dx}\sqrt{25-x^2}= \frac{1}{2\sqrt{25-x^2}}\cdot(-2x)=\frac{-x}{\sqrt{25-x^2}}$.Derivative of denominator: $\frac{d}{dx}(x-1)=1$.The limit becomes $\displaystyle \lim_{x \to 1}\frac{-x}{\sqrt{25-x^2}}$.Evaluating at $x=1$: $\frac{-1}{\sqrt{25-1}}=\frac{-1}{\sqrt{24}}$.Simplify: $\sqrt{24}=2\sqrt{6}$, so the limit equals $-\frac{1}{2\sqrt{6}}$.Answer:$-\frac{1}{2\sqrt{6}}$, which is option A.