In which of the following interval the function f(x) = x^x, x gt 0 is…

The correct answer is Option (2) → \(\left(\frac{1}{e}, \infty\right)\)

Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Option (1), Option (2) and Option (3) are all correct, as explained below.

By logarithmic differentiation, with \(y = x^x\):

    \(\log y = x \log x\)

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

    \(f'(x) = x^x(1 + \log x)\)

Since \(x^x \gt 0\) for \(x \gt 0\), \(f'(x) \gt 0\) exactly when

    \(\log x \gt -1\)
\( x \gt \frac{1}{e}\)

So f is strictly increasing on \(\left(\frac{1}{e}, \infty\right)\) and strictly decreasing on \(\left(0, \frac{1}{e}\right)\).

Option (1) \([e, \infty)\) — Should also be correct, though not the answer given by NTA. f is strictly increasing for every x bigger than \(\frac{1}{e} \approx 0.37\). Every number in \([e, \infty)\) is at least \(e \approx 2.72\), so f is strictly increasing there too.

Option (2) \(\left(\frac{1}{e}, \infty\right)\) — Correct. \(f'(x) \gt 0\) exactly for \(x \gt \frac{1}{e}\); this is the largest such interval.

Option (3) \([1, \infty)\) — Should also be correct, though not the answer given by NTA. Every number in \([1, \infty)\) is bigger than \(\frac{1}{e} \approx 0.37\), so f is strictly increasing there too.

Option (4) \([0, \infty)\) — Incorrect. f is not defined at 0, and it decreases on \(\left(0, \frac{1}{e}\right)\).

Advice for the exam: When more than one option is correct, mark the one that answers the question most completely, here the largest interval on which f is strictly increasing. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.