Q242 / 270 · NDA 2015 (Phase 2) · Mathematics

Consider the function $\mathrm{f}(\mathrm{x})=\left(\frac{1}{\mathrm{x}}\right)^{2 \mathrm{x}^2}$, where $\mathrm{x} \gt 0$ The maximum value of the function is

  • A
    e
  • B
    $\mathrm{e}^{\frac{2}{\mathrm{e}}}$
  • Correct answer
    C
    $\mathrm{e}^{\frac{1}{\mathrm{e}}}$
  • D
    $\frac{1}{\mathrm{e}}$

Solution

$\begin{aligned} & f(x)=\left(\frac{1}{x}\right)^{2 x^2} \\ & f\left(e^{-\frac{1}{2}}\right)=\left(e^{\frac{1}{2}}\right)^{2 \times e^{-1}} \\ & =\left(e^{\frac{1}{2}}\right)^{\frac{2}{e}} \\ & =e^{\frac{1}{2} \times \frac{2}{e}} \\ & =e^{1 / e}\end{aligned}$

Hence, the correct answer is option: 'C'.

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