Match List – I with List – II.
| List – I | List – II | ||
| (A) | \(x^x\) has a stationary point at x equal to | (I) | \(e\) |
| (B) | For x > 0, minimum value of \(x^x\) | (II) | \(\displaystyle\frac{1}{e}\) |
| (C) | The greatest value of \(\left(\displaystyle\frac{1}{x}\right)^x\) | (III) | \(e^{\frac{1}{e}}\) |
| (D) | The stationary point of \(\displaystyle\frac{\log x}{x}\) for x, where x > 0, is | (IV) | \(e^{\frac{-1}{e}}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
A stationary point is where the derivative is zero. For \(y = x^x\), use logarithmic differentiation:
\(\log y = x\log x\)
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \log x + 1, \quad \frac{dy}{dx} = x^x(1 + \log x) \)
(A) \(x^x\) has a stationary point at x equal to
This matches (II) \(\frac{1}{e}\):
\(1 + \log x = 0\), so \(\log x = -1\), \(x = \frac{1}{e}\)
(B) For x > 0, minimum value of \(x^x\)
This matches (IV) \(e^{-1/e}\). Since \(x^x \gt 0\), dy/dx has the sign of 1 + log x: negative for \(x \lt \frac{1}{e}\), positive for \(x \gt \frac{1}{e}\). So \(x = \frac{1}{e}\) gives the minimum:
\( \displaystyle \left(\frac{1}{e}\right)^{1/e} = e^{-1/e} \)
(C) The greatest value of \(\left(\frac{1}{x}\right)^x\)
This matches (III) \(e^{1/e}\). Here \(\left(\frac{1}{x}\right)^x = \frac{1}{x^x}\), which is greatest where \(x^x\) is least, that is at \(x = \frac{1}{e}\):
\( \displaystyle \frac{1}{e^{-1/e}} = e^{1/e} \)
(D) The stationary point of \(\frac{\log x}{x}\) for x, where x > 0, is
This matches (I) e. By the quotient rule:
\( \displaystyle \frac{d}{dx}\left(\frac{\log x}{x}\right) = \frac{x\cdot\frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2} \)
\(1 - \log x = 0\), so \(x = e\)