The function f(x) = 9x^2 - 2log x , x gt 0 , is strictly decreasing i…

The correct answer is Option (3) → \(\left(0, \displaystyle\frac{1}{3}\right)\)

Differentiating:

    \( \displaystyle f'(x) = 18x - \frac{2}{x} \)

    \( \displaystyle = \frac{2(9x^{2} - 1)}{x} = \frac{2(3x - 1)(3x + 1)}{x} \)

For x > 0, the factors x and 3x + 1 are positive, so the sign of f′(x) is the sign of 3x − 1:

    \( f'(x) \lt 0 \) for \( \displaystyle 0 \lt x \lt \frac{1}{3} \)
    \( f'(x) \gt 0 \) for \( \displaystyle x \gt \frac{1}{3} \)

So f is strictly decreasing in \(\left(0, \frac{1}{3}\right)\) and strictly increasing in \(\left(\frac{1}{3}, \infty\right)\).