Let the function f(x)=(x)/(3)+(3)/(x)+3, x ≠ 0 be strictly increasing in (-∞, α_1) U(α_2, ∞) and strictly decreasing in (α_3, α_4) U(α_4, α_5) . Then ∑_ i=1 ^5 α_ i ^2 is equal to :-

Correct answer: D. 36 aligned & f(x)=(x)/(3)+(3)/(x)+3, x ≠ 0 & f^ (x)=(1)/(3)-(3)/(x^2)=0 x= ± 3 & f^ (x)=(x^2-3)/(3 x^2) aligned aligned & f^ (x) > 0 (-∞,-3) (3, ∞) arrow increasing & f^ (x) < 0 (-3,0) (0,3) arrow decreasing aligned ∑_ i=1 ^5 α_i^2=(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2 =36

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Application of Derivatives

Let the function f(x)=(x)/(3)+(3)/(x)+3, x ≠ 0 be strictly...

Let the function

f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0

be strictly increasing in \left(-\infty, \alpha_1\right) \mathrm{U}\left(\alpha_2, \infty\right) and strictly decreasing in \left(\alpha_3, \alpha_4\right) \mathrm{U}\left(\alpha_4, \alpha_5\right)

. Then

\sum_{\mathrm{i}=1}^5 \alpha_{\mathrm{i}}^2 is equal to :-

Options

Correct answer

D. 36

Step-by-step solution

\begin{aligned} & f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0 \\ & f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^2}=0 \quad \Rightarrow x= \pm 3 \\ & f^{\prime}(x)=\frac{x^2-3}{3 x^2}\end{aligned}

\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x}) > 0 \forall(-\infty,-3) \cup(3, \infty) \rightarrow \text{ increasing } \\ & \mathrm{f}^{\prime}(\mathrm{x}) < 0 \forall(-3,0) \cup(0,3) \rightarrow \text{ decreasing }\end{aligned}

\sum_{i=1}^5 \alpha_i^2=(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2 =36$

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