JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 3)
The local maximum value of the function $$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$, x > 0, is
$${\left( {2\sqrt e } \right)^{{1 \over e}}}$$
$${\left( {{4 \over {\sqrt e }}} \right)^{{e \over 4}}}$$
$${(e)^{{2 \over e}}}$$
1
Explanation
$$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$ ; x > 0
$$\ln f(x) = {x^2}(\ln 2 - \ln x)$$
$$f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\} $$
$$f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}$$
$$g(x) = 2{\ln ^2} - 2\ln x - 1$$
$$ = \ln {4 \over {{x^2}}} - 1 = 0 \Rightarrow x = {2 \over {\sqrt e }}$$
_26th_August_Evening_Shift_en_3_2.png)
$$LM = {2 \over {\sqrt e }}$$
Local maximum value = $${\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}$$
$$\ln f(x) = {x^2}(\ln 2 - \ln x)$$
$$f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\} $$
$$f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}$$
$$g(x) = 2{\ln ^2} - 2\ln x - 1$$
$$ = \ln {4 \over {{x^2}}} - 1 = 0 \Rightarrow x = {2 \over {\sqrt e }}$$
$$LM = {2 \over {\sqrt e }}$$
Local maximum value = $${\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}$$
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