MCQ
Maximum value of ${\left( {{1 \over x}} \right)^x}$ is
- A${(e)^e}$
- ✓${(e)^{1/e}}$
- C${(e)^{ - e}}$
- D${\left( {{1 \over e}} \right)^e}$
==> $f'(x) = {\left( {\frac{1}{x}} \right)^x}\left( {\log \frac{1}{x} - 1} \right)$
$f'(x) = 0 \Rightarrow \log \frac{1}{x} = 1 = \log e$
$\Rightarrow \frac{1}{x} = e \Rightarrow x = \frac{1}{e}$
Therefore maximum value of function is ${e^{1/e}}$.
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$f\left( x \right) = \int_1^x {\left\{ {2\left( {t - 1} \right){{\left( {t - 2} \right)}^3} + 3{{\left( {t - 1} \right)}^2}{{\left( {t - 2} \right)}^2}} \right\}} dt$ is maximum when $x$ is equal to