JAMIA 2021 — Mathematics PYQ

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Question

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

Maximum value of $(\frac{1}{x})^{x}$ is

Choose the correct answer:

Explanation

1. Let the function be:

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

2. Take natural log ( $\ln$ ) on both sides:

\ln y = \ln(x^{-x})

\ln y = -x \ln x

3. Differentiate with respect to $x$ :

\frac{1}{y} \frac{dy}{dx} = - \left[ x \cdot \frac{1}{x} + \ln x \cdot 1 \right]

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

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

4. Set $\frac{dy}{dx} = 0$ for maximum value:

-(1 + \ln x) = 0

\ln x = -1

x = e^{-1} = \frac{1}{e}

5. Substitute $x = \frac{1}{e}$ into the original function:

y_{max} = \left(\frac{1}{1/e}\right)^{1/e}

y_{max} = e^{1/e}

Final Answer:

e^{1/e}