UPSC NDA I 2015 Mathematics Paper

Concept:The function $f(x) = \frac{x^{2} - 1}{x^{2} + 1}$ is a rational function where the numerator and denominator are both quadratic in $x$.Since $x^{2} \ge 0$ for all real $x$, the minimum of $f(x)$ occurs when $x^{2}$ is smallest, i.e., $x = 0$.Explanation:Rewrite $f(x)$ as: $f(x) = 1 - \frac{2}{x^{2} + 1}$.Here, $\frac{2}{x^{2} + 1}$ is always positive and becomes maximum when $x^{2} + 1$ is minimum, i.e., at $x = 0$.At $x = 0$, $\frac{2}{x^{2} + 1} = \frac{2}{0 + 1} = 2$.Thus $f(0) = 1 - 2 = -1$.For any other $x$, $x^{2} + 1 > 1$ so $\frac{2}{x^{2} + 1} < 2$, making $f(x) > -1$.Hence the minimum value of $f(x)$ is $-1$.Answer:The minimum value of $f(x)$ is $-1$, which corresponds to option C.