The interval, in which the function \(f(x)=\displaystyle\frac{3}{x}+\frac{x}{3}\) is strictly decreasing, is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-3,0)\cup(0,3)\)
The function is defined for all x ≠ 0. Differentiating:
\( \displaystyle f'(x) = -\frac{3}{x^2} + \frac{1}{3} \)
\( \displaystyle = \frac{x^2 - 9}{3x^2} \)
Since \(3x^2 \gt 0\) for x ≠ 0, f is strictly decreasing where f′(x) < 0, that is, where
\(x^2 - 9 \lt 0\)
\(-3 \lt x \lt 3,\quad x \ne 0\)
So f is strictly decreasing on each of the intervals (−3, 0) and (0, 3), which is how the union in Option (3) is to be read. (Taken as one set, the union is not an interval of decrease: f(−1) = −10/3 is less than f(1) = 10/3, because f is undefined at 0 and jumps there.)
Why the other options are incorrect
Option (1) \((-\infty,-3)\cup(3,\infty)\) — Incorrect. Here \(x^2 \gt 9\), so f′(x) > 0: f is strictly increasing on these intervals.
Option (2) \((-3,3)\) — Incorrect. This interval contains x = 0, where f is not defined.
Option (4) \(\mathrm{R}-\{0\}\) — Incorrect. For |x| > 3, f′(x) > 0, so f is not decreasing on the whole of R − {0}.