The maximum value of \(\left(\displaystyle\frac{1}{x}\right)^x\) for \(x \gt 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(e^{1/e}\)
Let
\( \displaystyle y = \left(\frac{1}{x}\right)^x = x^{-x},\quad x \gt 0 \)
The variable is in both the base and the power, so logarithms are taken first (logarithmic differentiation):
\(\log y = -x\log x\)
\( \displaystyle \frac{1}{y}\,\frac{dy}{dx} = -(\log x + 1) \)
\( \displaystyle \frac{dy}{dx} = -y(\log x + 1) \)
Since \(y \gt 0\), \(\frac{dy}{dx} = 0\) when
\(\log x = -1\)
\( \displaystyle x = \frac{1}{e} \)
By the first derivative test: for \(x \lt \frac{1}{e}\), \(\log x + 1 \lt 0\), so \(\frac{dy}{dx} \gt 0\); for \(x \gt \frac{1}{e}\), \(\frac{dy}{dx} \lt 0\). So y rises and then falls, and \(x = \frac{1}{e}\) gives the maximum:
\( \displaystyle y_{\max} = \left(\frac{1}{1/e}\right)^{1/e} = e^{1/e} \)
NCERT Exemplar:
Application of Derivatives
