MCQ
વિધેય $f(x)=\left(\frac{2}{x}\right)^{x^{2}}, x>0$ ની સ્થાનીય મહતમ કિમંત મેળવો.
- A$(2 \sqrt{\mathrm{e}})^{\frac{1}{\mathrm{e}}}$
- B$\left(\frac{4}{\sqrt{\mathrm{e}}}\right)^{\frac{\mathrm{e}}{4}}$
- C$(\mathrm{e})^{\frac{2}{\mathrm{e}}}$
- D$1$
$\ell n f(x)=x^{2}(\ell \ln 2-\ell n x)$
$f^{\prime}(x)=f(x)\{-x+(\ell n 2-\ell n x) 2 x\}$
$f^{\prime}(x)=\underbrace{f(x)}_{+} \cdot \underbrace{x}_{+} \underbrace{(2 \ell n 2-2 \ell n x-1)}_{g(x)}$
$g(x)=2 \ell n^{2}-2 \ell n x-1$
$=\ell n \frac{4}{x^{2}}-1=0 \Rightarrow x=\frac{2}{\sqrt{e}}$
$\mathrm{LM}=\frac{2}{\sqrt{\mathrm{e}}}$
Local maximum value $=\left(\frac{2}{2 / \sqrt{\mathrm{e}}}\right)^{\frac{4}{\mathrm{e}}} \Rightarrow \mathrm{e}^{\frac{2}{\mathrm{e}}}$
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