Q241 / 270 · NDA 2015 (Phase 2) · Mathematics

Consider the function $\mathrm{f}(\mathrm{x})=\left(\frac{1}{\mathrm{x}}\right)^{2 \mathrm{x}^2}$, where $\mathrm{x} \gt 0$ At what value of $x$ does the function attain maximum value?

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

Solution

$\begin{aligned} & f(x)=\left(\frac{1}{x}\right)^{2 x^2}=y \text { (say) } \\ & \log y=2 x^2 \log \left(\frac{1}{x}\right) \\ & \frac{1}{y} \frac{d y}{d x}=2\left[2 x \cdot \log \left(\frac{1}{x}\right)+x^2 \cdot \frac{1}{1 / x} \cdot\left(\frac{-1}{x^2}\right)\right] \\ & \frac{d y}{d x}=2 y\left[2 x \log \left(\frac{1}{x}\right)-x\right] \end{aligned}$ For $\max$. or $\min$. value $\frac{\mathrm{dy}}{\mathrm{dx}}=0$ $\begin{aligned} & 2 x \log \left(\frac{1}{x}\right)-x=0 \\ & x\left[2 \log \left(\frac{1}{x}\right)-1\right]=0 \\ & \because x \neq 0 \\ & \Rightarrow \quad 2 \log \left(\frac{1}{x}\right)=1 \\ & \Rightarrow \quad \frac{1}{x}=e^{\frac{1}{2}} \end{aligned}$ $\begin{aligned} & \Rightarrow \\ & \Rightarrow=e^{-\frac{1}{2}} \\ & \text {Again } \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left\{2 \mathrm{xy}\left[\log \left(\frac{1}{\mathrm{x}}\right)-1\right]\right\} \\ & =(2 \mathrm{xy})^{\prime}\left(\log \frac{1}{\mathrm{x}}-1\right)+2 \mathrm{xy}\left[\mathrm{x} \cdot\left(-\frac{1}{\mathrm{x}^2}\right)\right]\end{aligned}$ $\begin{aligned} & =(2 x y)^{\prime}\left(\log \frac{1}{x}-1\right)-2 y \\ & =2\left[\left(y+x y^1\right)\left(\log \frac{1}{x}-1\right)-y\right] \lt 0 \end{aligned}$ So at $\mathrm{x}=\mathrm{e}^{-1 / 2}$ function is maximum.

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

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