UPSC NDA I 2015 Mathematics Paper

Concept:To find where a function attains its minimum, we use the first derivative test: set $f'(x) = 0$ to find critical points, then use the second derivative test: if $f''(x) > 0$, the point is a local minimum.Explanation:Given $f(x)=\frac{x^{2}-1}{x^{2}+1}$, with $x \in \mathbb{R}$.Apply the quotient rule: $f'(x) = \frac{2x(x^{2}+1) - 2x(x^{2}-1)}{(x^{2}+1)^{2}} = \frac{4x}{(x^{2}+1)^{2}}$.Set $f'(x)=0 \implies 4x=0 \implies x=0$. This is the only critical point.Now compute the second derivative to classify the critical point.Differentiate $f'(x)$: $f''(x) = \frac{4\cdot (x^{2}+1)^{2} - 4x \cdot 2(x^{2}+1)\cdot 2x}{(x^{2}+1)^{4}}$Simplify: $f''(x) = \frac{4(x^{2}+1)^{2} - 16x^{2}(x^{2}+1)}{(x^{2}+1)^{4}} = \frac{4(x^{2}+1) - 16x^{2}}{(x^{2}+1)^{3}} = \frac{4 - 12x^{2}}{(x^{2}+1)^{3}}$.Evaluate at $x=0$: $f''(0) = \frac{4 - 0}{1^{3}} = 4 > 0$.Since $f''(0) > 0$, the function has a local minimum at $x=0$.Thus, $f(x)$ attains its minimum value at $x=0$.Answer:The minimum value occurs at $x = 0$, corresponding to option B.