Let the function $$f(x) = \frac{x}{3} + \frac{3}{x} + 3$$, $$x \ne 0$$ be strictly increasing in $$(-\infty, \alpha_1) \cup (\alpha_2, \infty)$$ and strictly decreasing in $$(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$$. Then $$\sum_{i=1}^{5} \alpha_i^2$$ is equal to :

The given function is $$f(x)=\frac{x}{3}+\frac{3}{x}+3,\;x\neq 0$$.

Step 1: Find the derivative and its critical points
Using the rule $$\frac{d}{dx}\left(\frac{k}{x}\right)=-\frac{k}{x^{2}}$$, we get
$$f'(x)=\frac{1}{3}-\frac{3}{x^{2}}.$$ Set $$f'(x)=0$$ to locate stationary points:
$$\frac{1}{3}-\frac{3}{x^{2}}=0 \;\Longrightarrow\; \frac{1}{3}=\frac{3}{x^{2}} \;\Longrightarrow\; x^{2}=9 \;\Longrightarrow\; x=\pm 3.$$ Thus the only stationary points are $$x=-3$$ and $$x=3$$. In addition, $$f(x)$$ is not defined at $$x=0$$, so $$x=0$$ also separates intervals.

Step 2: Sign of $$f'(x)$$ in each interval
We test one value from each of the four regions determined by $$x=-3,0,3$$.

For $$x\lt -3$$, take $$x=-4$$:
$$f'(-4)=\frac{1}{3}-\frac{3}{16}\approx 0.145\gt 0,$$ so $$f'(x)\gt 0$$ in $$(-\infty,-3)$$ ⇒ increasing.

For $$-3\lt x\lt 0$$, take $$x=-1$$:
$$f'(-1)=\frac{1}{3}-3=-\frac{8}{3}\lt 0,$$ so $$f'(x)\lt 0$$ in $$(-3,0)$$ ⇒ decreasing.

For $$0\lt x\lt 3$$, take $$x=1$$:
$$f'(1)=\frac{1}{3}-3=-\frac{8}{3}\lt 0,$$ so $$f'(x)\lt 0$$ in $$(0,3)$$ ⇒ decreasing.

For $$x\gt 3$$, take $$x=4$$:
$$f'(4)=\frac{1}{3}-\frac{3}{16}\approx 0.146\gt 0,$$ so $$f'(x)\gt 0$$ in $$(3,\infty)$$ ⇒ increasing.

Step 3: Match the intervals with the symbols $$\alpha_i$$
From the sign chart, the intervals of monotonicity are
• Increasing: $$(-\infty,-3)$$ and $$(3,\infty)$$.
• Decreasing: $$(-3,0)$$ and $$(0,3)$$.

The problem states that
  • $$f(x)$$ is strictly increasing in $$(-\infty,\alpha_1)\cup(\alpha_2,\infty)$$,
  • $$f(x)$$ is strictly decreasing in $$(\alpha_3,\alpha_4)\cup(\alpha_4,\alpha_5)$$.

Comparing, we can assign
$$\alpha_1=-3,\qquad \alpha_2=3,$$
$$\alpha_3=-3,\qquad \alpha_4=0,\qquad \alpha_5=3.$$ (The same numerical value may occur for different indices; only the positions in the description matter.)

Step 4: Compute the required sum
$$\sum_{i=1}^{5}\alpha_i^{2} =\alpha_1^{2}+\alpha_2^{2}+\alpha_3^{2}+\alpha_4^{2}+\alpha_5^{2}$$ $$=(-3)^{2}+3^{2}+(-3)^{2}+0^{2}+3^{2}$$ $$=9+9+9+0+9$$ $$=36.$$

Answer: $$\sum_{i=1}^{5}\alpha_i^{2}=36$$, which corresponds to Option D.