UPSC NDA I 2016 Mathematics Solved Paper

Concept:Solve an equation by substituting a variable to reduce it to a cubic, then factor to find real roots.Explanation:Set $f(x) = 1$: $\frac{27(x^{2/3} - x)}{4} = 1$.Multiply by $4$: $27(x^{2/3} - x) = 4$.Divide by $27$: $x^{2/3} - x = \frac{4}{27}$.Let $t = x^{1/3}$. Then $x = t^3$ and $x^{2/3} = t^2$.The equation becomes $t^2 - t^3 = \frac{4}{27}$.Multiply by $27$: $27t^2 - 27t^3 = 4$.Rearrange: $27t^3 - 27t^2 + 4 = 0$.Factor the cubic: $(3t - 2)^2 (3t + 1) = 0$.Solve for $t$: $t = \frac{2}{3}$ (double root) or $t = -\frac{1}{3}$.Back-substitute: $x = t^3$.For $t = \frac{2}{3}$: $x = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$.For $t = -\frac{1}{3}$: $x = \left(-\frac{1}{3}\right)^3 = -\frac{1}{27}$.Both $x$ values are real and satisfy the original equation, giving two distinct solutions.Answer:Two