UPSC NDA 2 2016 Mathematics Solved Paper

Concept:Use the first derivative test to determine local extrema when the second derivative test is inconclusive.Explanation:Given $f(x) = x^{3} \sin x$.Compute the first derivative: $f'(x) = 3x^{2} \sin x + x^{3} \cos x$.Set $f'(x)=0$ to find critical points. At $x=0$, $f'(0)=0$, so $x=0$ is a critical point.Compute the second derivative: $f''(x) = 6x \sin x + 6x^{2} \cos x - x^{3} \sin x$.At $x=0$, $f''(0)=0$, so the second derivative test fails.Apply the first derivative test. Examine the sign of $f'(x)$ for values near zero.For small $x$, use the approximations $\sin x \approx x$ and $\cos x \approx 1$.Then $f'(x) \approx 3x^{2} \cdot x + x^{3} \cdot 1 = 4x^{3}$.For $x>0$ (small positive), $4x^{3}>0$; for $x<0$ (small negative), $4x^{3}<0$.Thus $f'(x)$ changes from negative to positive at $x=0$, indicating a local minimum.Also check the function value: $f(0)=0$, and for any $x \neq 0$ near zero, $f(x) = x^{3}\sin x >0$ (since both factors have the same sign), so $f(x) \ge f(0)$ locally.Answer:The function $f(x)=x^{3}\sin x$ has a local minimum at $x=0$. Hence option B is correct.