Let the function f(x)=x/3+3/x+3, x neq 0 be strictly increasing in (-∞, α 1 ) cup (α 2 , ∞ ) and strictly decreasing in (α 3 , α 4 ) cup (α 4 , α 5 ) . Then sum i=1 5 α i 2 is equal to

Mathematics · Application of Derivatives

JEE Main 2025 — 8 April, Evening Shift — Question 32

Let the function $f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0$ 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

Option A:
48
Option B:
40
Option C:
36
Correct
Option D:
28

Answer: C

Step-by-step solution

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

$f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^{2}}=\left(\frac{x^{2}-9}{3 x^{2}}\right)=\frac{(x-3)(x+3)}{3 x^{2}}$

$\Rightarrow f^{\prime}(x)>0 \forall x \in(-\infty,-3) \cup(3, \infty)$

$f^{\prime}(x)<0 \forall x \in(-3,0) \cup(0,3)$

$\Rightarrow \alpha_{1}=-3, \alpha_{2}=3, \alpha_{3}=-3, \alpha_{4}=0, \alpha_{5}=3$

$\Rightarrow \sum_{i=1}^{5} \alpha_{i}^{2}=(-3)^{2}+\left(3^{2}\right)+0^{2}+(-3)^{2}+(3)^{2}$

=4(9)=36

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives
Topic: Monotonicity