Functions

Take logs: $y = x^{1/ x} ⟹ ln y = \frac{l n x}{x}$ .
Differentiate: $\frac{1}{y} y^{'} = \frac{1 \cdot x - l n x \cdot 1}{x ^{2}} = \frac{1 - l n x}{x ^{2}}$ .
Find Critical Point: $y^{'} = 0 ⟹ 1 - ln x = 0 ⟹ ln x = 1 ⟹ x = e$ .
Check sign change: For $x < e, 1 - ln x > 0$ (Increasing). For $x > e, 1 - ln x < 0$ (Decreasing). So Max occurs at $x = e$ .
Max Value: $f (e) = e^{1/ e}$ .

This medium difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of functions. It appeared in the 2025 exam.

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