UPSC NDA I 2024 Mathematics Paper

Concept:Given $f(x) = \frac{x}{\ln x}$ for $x>1$, evaluate the three statements using derivatives and critical point analysis.Explanation:First, compute the first derivative using the quotient rule:$f'(x) = \frac{(\ln x)(1) - x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2}$.Set $f'(x) = 0$ to find critical points: $\ln x - 1 = 0 \Rightarrow \ln x = 1 \Rightarrow x = e$.Now compute the second derivative:$f''(x) = \frac{d}{dx}\left( \frac{\ln x - 1}{(\ln x)^2} \right) = \frac{1}{x(\ln x)^2} - \frac{2(\ln x - 1)}{x(\ln x)^3}$.Simplify: $f''(x) = \frac{\ln x - 2(\ln x - 1)}{x(\ln x)^3} = \frac{2 - \ln x}{x(\ln x)^3}$? Wait, let's re-derive accurately.Actually, using the product rule on $f'(x) = (\ln x - 1)(\ln x)^{-2}$:$f''(x) = \frac{1}{x}(\ln x)^{-2} + (\ln x - 1)(-2)(\ln x)^{-3}\frac{1}{x} = \frac{1}{x(\ln x)^2} - \frac{2(\ln x - 1)}{x(\ln x)^3}$.Combine over common denominator $x(\ln x)^3$:$f''(x) = \frac{\ln x - 2(\ln x - 1)}{x(\ln x)^3} = \frac{2 - \ln x}{x(\ln x)^3}$.Evaluate $f''(e)$: $\ln e = 1$, so $f''(e) = \frac{2 - 1}{e \cdot 1^3} = \frac{1}{e}$. Statement 1 is correct.Since $f''(e) > 0$, $x=e$ gives a local minimum. Statement 2 is correct.Compute $f(e)$: $f(e) = \frac{e}{\ln e} = \frac{e}{1} = e$. The local minimum value is $e$. Statement 3 is correct.All three statements are correct.Answer:Statements 1, 2, and 3 are correct. The correct option is D.