UPSC NDA I 2016 Mathematics Solved Paper

Concept:Solve $f(x) = -1$ by substituting $t = x^{1/3}$ to transform the equation into a cubic, then analyze its roots using calculus.Explanation:Set $f(x) = -1$: $\frac{27(x^{2/3} - x)}{4} = -1$.Multiply by 4: $27(x^{2/3} - x) = -4$.Let $t = x^{1/3}$, so $x = t^3$ and $x^{2/3} = t^2$.Substitute: $27(t^2 - t^3) = -4$.Simplify: $t^2 - t^3 = -\frac{4}{27}$.Rearrange: $27t^2 - 27t^3 = -4$ → $27t^3 - 27t^2 - 4 = 0$.Define $g(t) = 27t^3 - 27t^2 - 4$. Differentiate: $g'(t) = 81t^2 - 54t = 27t(3t - 2)$.Set $g'(t) = 0$: $t = 0$ or $t = \frac{2}{3}$.Evaluate $g$ at these critical points: $g(0) = -4$, $g\left(\frac{2}{3}\right) = 27\cdot\frac{8}{27} - 27\cdot\frac{4}{9} - 4 = 8 - 12 - 4 = -8$. Both are negative.Since the leading coefficient (coefficient of $t^3$) is positive ($27 > 0$), the cubic $g(t)$ goes to $+\infty$ as $t \to +\infty$ and $-\infty$ as $t \to -\infty$. With both turning points below the x-axis, the curve crosses the x-axis exactly once. Hence $g(t) = 0$ has exactly one real root, and that root is positive (because $g(0) = -4 < 0$ and eventually positive for large $t$).Thus $t$ has one positive solution, meaning $x = t^3$ has one corresponding solution.Answer:One solution.