UPSC NDA II 2015 Mathematics Paper

Concept:A function $f(x)$ is monotonically increasing if its derivative $f'(x) > 0$.We use the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.Explanation:Given $f(x) = \frac{x^2}{e^x}$.Compute $f'(x)$ using the quotient rule:$f'(x) = \frac{(2x)(e^x) - (x^2)(e^x)}{(e^x)^2}$.Simplify: $f'(x) = \frac{2x e^x - x^2 e^x}{e^{2x}} = \frac{2x - x^2}{e^x}$.For monotonic increase, $f'(x) > 0$:$\frac{2x - x^2}{e^x} > 0$.Since $e^x > 0$ for all real $x$, we only need $2x - x^2 > 0$.Factor: $x(2 - x) > 0$.Solve the inequality: The product is positive when both factors have the same sign.Case 1: $x > 0$ and $2 - x > 0$ → $x > 0$ and $x < 2$ → $0 < x < 2$.Case 2: $x < 0$ and $2 - x < 0$ → $x < 0$ and $x > 2$ — impossible.Thus, only $0 < x < 2$ satisfies the condition.Answer:Option C: $0 < x < 2$.