The interval, in which the function f(x)=3/x+x/3 is strictly decreasi…

The correct answer is Option (3) → $(-3,0)∪(0,3)$

We are given

$f(x) = 3/x + x/3$

To find where the function is strictly decreasing, we check the derivative.

Step 1: Differentiate

$f'(x)$ = derivative of $3/x$ + derivative of $x/3$

$= -3/x^2 + 1/3$

Step 2: For decreasing, $f'(x) < 0$

So,

$1/3-3/x^2 < 0$

Multiply both sides by $3x^2$ (positive for $x ≠ 0$, so inequality sign does not change):

$x^2-9<0$

$x^2 <9$

$-3 < x < 3$

But $x ≠ 0$ because the function is not defined at $x = 0$.

Step 3: Split the interval

So the function is strictly decreasing on

$(-3, 0) ∪ (0,3)$