JEE Main 2025 Apr 8 Shift 2, Mathematics Q3: Monotonocity

JEE Main2025Apr 8, Shift 2Mathematics

Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3$ , $x \neq = 0$ be strictly increasing in $(- \infty, α_{1}) \cup (α_{2}, \infty)$ and strictly decreasing in $(α_{3}, α_{4}) \cup (α_{4}, α_{5})$ . Then $i = 1 \sum 5 α_{i}^{2}$ is equal to:

A
$48$
B
$28$
C
$40$
D
$36$

Solution

Differentiating gives $f^{'} (x) = \frac{1}{3} - \frac{3}{x ^{2}} = \frac{x ^{2} - 9}{3 x ^{2}}$ .

The sign of $f^{'} (x)$ is controlled by the numerator $x^{2} - 9$ : positive outside $[- 3, 3]$ and negative on $(- 3, 3)$ , excluding $x = 0$ where $f$ is undefined.

So $f$ is strictly increasing on $(- \infty, - 3) \cup (3, \infty)$ , giving $α_{1} = - 3$ and $α_{2} = 3$ .

$f$ is strictly decreasing on $(- 3, 0) \cup (0, 3)$ , so the three boundary values in that region are $α_{3} = - 3$ , $α_{4} = 0$ , $α_{5} = 3$ .

Summing the squares: $(- 3)^{2} + 3^{2} + (- 3)^{2} + 0^{2} + 3^{2} = 9 + 9 + 9 + 0 + 9 = 36$ , matching option (4).

Concept behind this question

MonotonocityNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 8 Shift 2, Mathematics Q3: Monotonocity

Concept behind this question

More previous year questions on Monotonocity

Practice more Mathematics