UPSC NDA 2 2017 Mathematics Solved Paper

Concept:The cubic function $f(x) = x(x-1)(x+1)$ has both a local maximum and a local minimum.Explanation:Simplify the function: $f(x) = x(x^2-1) = x^3 - x$.Find the first derivative: $f'(x) = 3x^2 - 1$.Set $f'(x) = 0$ to locate stationary points: $3x^2 - 1 = 0$ gives $x = \pm \frac{1}{\sqrt{3}}$.Compute the second derivative: $f''(x) = 6x$.For $x = \frac{1}{\sqrt{3}}$: $f''\left(\frac{1}{\sqrt{3}}\right) = \frac{6}{\sqrt{3}} > 0$, so this is a local minimum.Minimum value: $f\left(\frac{1}{\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}\right)^3 - \frac{1}{\sqrt{3}} = -\frac{2}{3\sqrt{3}}$.For $x = -\frac{1}{\sqrt{3}}$: $f''\left(-\frac{1}{\sqrt{3}}\right) = -\frac{6}{\sqrt{3}} < 0$, so this is a local maximum.Maximum value: $f\left(-\frac{1}{\sqrt{3}}\right) = \left(-\frac{1}{\sqrt{3}}\right)^3 + \frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}}$.Since $\frac{2}{3\sqrt{3}} > -\frac{2}{3\sqrt{3}}$, the local maximum value is larger than the local minimum value.Answer:Option A: The local maximum value is larger than local minimum value.