The maximum value of ≤ft(1/x)^x is

Correct Answer is : e1ee¹/e}ee1

Correct Answer is : e1ee¹/e}ee1

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KCET 2016 Math Solved Paper

Question : 33 of 60

Marks: +1, -0

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

Solution:

Let

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

Taking log on both the sides, we get

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

\log y = -x \log x

→ (1)

Now, differentiating Eq. (1) with respect to x, we get

\frac{1}{y} \frac{dy}{dx} = -\left( x \cdot \frac{1}{x} + \log x \right)

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

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

→ (2)

\Rightarrow \frac{dy}{dx} = -\left(\frac{1}{x}\right)^x (1+\log x)

Now,

\frac{dy}{dx}=0

-y(1+\log x)=0

\Rightarrow \log x = -1 \Rightarrow x = e^{-1}

Now, differentiating Eq. (2) with respect to x, we get

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

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

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

x = e^{-1}

, we get

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

\Rightarrow \left. \frac{d^2 y}{dx^2} \right|_{x=e^{-1}} < 0

Therefore, maximum value of y is,

y_{\max} = e^{\frac{1}{e}}

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