Let the function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{3}+\frac{3}{\mathrm{x}}+3,\mathrm{x}\neq 0$ be strictly increasing in $\left(-\infty ,{\alpha }_{1}\right)\cup \left({\alpha }_{2},\infty \right)$ and strictly decreasing in $\left({\alpha }_{3},{\alpha }_{4}\right)\cup \left({\alpha }_{4},{\alpha }_{5}\right)$. Then $\sum _{i=1}^{5}{\alpha }_{i}^{2}$ is equal to :- [JEE Main 2025, 8 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: d) $36$. $f(x)=\frac{x}{3}+\frac{3}{x}+3,\ x\neq 0$ $f'(x)=\frac{1}{3}-\frac{3}{x^2}=0\Rightarrow x=\pm 3$ $f'(x)=\frac{x^2-3}{3x^2}$ $f'(x)>0\ \forall(-\infty,-3)\cup(3,\infty)\rightarrow$ increasing $f'(x)<0\ \forall(-3,0)\cup(0,3)\rightarrow$ decreasing $\sum_{i=1}^{5}\alpha_i^2=(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2=36$

Let the function \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{3}+\frac{3}{\mathrm{x}}+3,\mathrm{x}\neq 0\) be strictly inc…

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