The maximum value of the function f(x)=≤ft(1/x)^x is

Correct Answer is : e1ee¹/e}ee1

Correct Answer is : e1ee¹/e}ee1

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CGPET 2022 Solved Paper

Section: Mathematics

Question : 122 of 150

Marks: +1, -0

f(x)=\left(\frac{1}{x}\right)^x

Solution:

y=\left(\frac{1}{x}\right)^x=x^{-x}

\Rightarrow \log y = -x \log x

\Rightarrow \frac{1}{y} \frac{dy}{dx} = -(1+\log x)

\Rightarrow \frac{dy}{dx} = -y(1+\log x)

\Rightarrow \frac{d^2 y}{dx^2} = \frac{-dy}{dx}(1+\log x) - \frac{y}{x} = y(1+\log x)^2 - \frac{y}{x}

= x^{-x}(1+\log x)^2 - \frac{x^{-x}}{x}

= x^{-x}(1+\log x)^2 - x^{-x-1}

At points of local maximum and local minimum, we must have

\frac{dy}{dx} = 0

-y(1+\log x) = 0

\Rightarrow 1+\log x = 0

\Rightarrow \log x = -1

\Rightarrow x = e^{-1}

\Rightarrow \left(\frac{d^2 y}{dx^2}\right)_{x=e^{-1}} = \left(\frac{1}{e}\right)^{-\frac{1}{e}} \left(1+\log \frac{1}{e}\right)^2 - \left(\frac{1}{e}\right)^{-\frac{1}{e}-1}

= (e^{-1})^{-\frac{1}{e}} (1-\log e)^2 - (e^{-1})^{-\frac{1}{e}-1}

= -e^{\frac{1}{e}+1} < 0

So,

x=\frac{1}{e}

is a point of local maximum. The local maximum value of

y

is obtained by putting

x=\frac{1}{e}

y

and is equal to

e^{\frac{1}{e}}

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