UPSC NDA II 2018 Mathematics Solved Paper

Concept:Finding the minimum of a composite function using derivatives and monotonicity.Explanation:Let $f(x) = \left[ x(x-1) + 1 \right]^{1/3}$ for $0 \le x \le 1$.Since the cube root function is increasing, $f(x)$ is minimum when the inner expression $g(x) = x(x-1) + 1 = x^2 - x + 1$ is minimum.Differentiate $g(x)$: $g'(x) = 2x - 1$.Set $g'(x) = 0$: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$.Find second derivative: $g''(x) = 2 > 0$, so $x = \frac{1}{2}$ gives a local minimum.Check endpoints: $g(0)=1$, $g(1)=1$, and $g(\frac{1}{2})=\frac{3}{4}$. Since $\frac{3}{4} < 1$, the minimum of $g(x)$ on $[0,1]$ is $\frac{3}{4}$.Thus the minimum of $f(x)$ is $\left( \frac{3}{4} \right)^{1/3}$.Answer:A. $\left( \frac{3}{4} \right)^{1/3}$