The function \(f(x) = x^2\log x\), \(x \gt 0\), is strictly decreasing in the interval
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left(0, \displaystyle\frac{1}{\sqrt{e}}\right)\)
By the product rule:
\( \displaystyle f'(x) = 2x\log x + x^2 \cdot \frac{1}{x} \)
\( = x(2\log x + 1) \)
Since x > 0, the sign of f’(x) is the sign of \(2\log x + 1\).
\( \displaystyle 2\log x + 1 \lt 0 \iff \log x \lt -\frac{1}{2} \iff x \lt \frac{1}{\sqrt{e}} \)
So f is strictly decreasing in \(\left(0, \frac{1}{\sqrt{e}}\right)\) and strictly increasing in \(\left(\frac{1}{\sqrt{e}}, \infty\right)\).
Why the other options are incorrect
Option (2) \(\left(\displaystyle\frac{1}{\sqrt{e}}, \infty\right)\) — Incorrect. Here \(2\log x + 1 \gt 0\), so f’(x) > 0 and f is increasing.
Option (3) \((0, 1)\) — Incorrect. This is where \(\log x \lt 0\); but f’(x) > 0 between \(\frac{1}{\sqrt{e}} \approx 0.61\) and 1.
Option (4) \(\left(\displaystyle\frac{1}{e}, 1\right)\) — Incorrect. This contains \(\frac{1}{\sqrt{e}} \approx 0.61\), and f increases after that point.