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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
MCQ
The maximum value of $\text{x}^\frac{1}{\text{x}}, \text{x}>0$ is.
  • ✓
    $\text{e}^\frac{1}{\text{e}}$
  • B
    $(\frac{1}{\text{e}})^\text{e}$
  • C
    $1$
  • D
    none of these.
✓

Answer

Correct option: A.
$\text{e}^\frac{1}{\text{e}}$
We have $\text{f}(\text{x})=\text{x}^\frac{1}{\text{x}}$
Taking log on both side, we get
$\log\text{f}(\text{x})=\frac{1}{\text{x}}\ \log\text{x}$
Differentitating $\text{w.r.t. x},$ we get
$\frac{1}{\text{f}(\text{x})}\text{f}\ '(\text{x})=\frac{-1}{\text{x}^{2}}\ \log\text{x}+\frac{1}{\text{x}^{2}}$
$\Rightarrow\text{f}\ '(\text{x})=\text{f}'(\text{x})\frac{1}{\text{x}^{2}}\ (1-\log\text{x})$
$\Rightarrow\text{f}\ '(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big) ...(\text{i})$
$\Rightarrow\text{f}\ '(\text{x})=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x})$
For a local maximum or a local minima, we must have $f\ '(x) = 0$
$=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x}) =0$
$\Rightarrow \log\text{x}=1$
$\therefore \text{x}=\text{e}$
Now, $\text{f}\ ''(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$
$=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$
At $x = e$
$\text{f}\ ''(\text{e})=\text{e}^\frac{1}{\text{e}}\Big(\frac{1}{\text{e}^{2}}-\frac{1}{\text{e}^{2}}\log\text{e}\Big)^{2}+\text{e}^\frac{1}{\text{e}}\Big(\frac{-3}{\text{e}^{3}}+\frac{2}{\text{e}^{3}}\log\text{e}\Big)$
$=-\text{e}^\frac{1}{\text{e}}(\frac{1}{\text{e}^{3}})<0$
So$, x = e$ is a point of local maximum.
Thus, the maximum value is given by
$\text{f}(\text{e})=\text{e}^\frac{1}{\text{e}}$

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