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) $36$. Sol: $\begin{matrix}f'(x)=\frac{1}{3}−\frac{3}{{x}^{2}} \ =\frac{{x}^{2}−9}{3{x}^{2}}=\frac{(x+3)(x−3)}{3{x}^{2}}\end{matrix}$ For strictly increasing ${\mathrm{f}}^{'}(\mathrm{x})>0\Rightarrow \mathrm{x}\in (-\infty ,-3)\cup (3,\infty )$ For strictly decreasing ${\mathrm{f}}^{'}(\mathrm{x}) \(∴{\alpha }_{1}=−3,{\alpha }_{2}=3\ {\alpha }_{3}=−3,{\alpha }_{4}=0\ {\alpha }_{5}=3$ $\sum _{i=1}^{5}{\alpha }_{i}^{2}=36$

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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