Let the function be strictly increasing in and strictly decreasing in . Then is equal to [2025]

Question

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 $\sum_{i = 1}^{5} α_{i}^{2}$ is equal to [2025]

Smart Achievers · Accepted Answer

(3)

We have,

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

$\Rightarrow f' (x) = \frac{1}{3} - \frac{3}{x^{2}}$

For critical points,

$f' (x) = 0 \Rightarrow \frac{1}{3} - \frac{3}{x^{2}} = 0$

$\Rightarrow x^{2} = 9 \Rightarrow x = \pm 3$

Now, $f' (x) > 0 \forall x \in (- \infty, - 3) \cup (3, \infty)$

and $f' (x) < 0 \forall x \in (- 3, 0) \cup (0, 3)$

$∴$ f(x) is increasing in $(- \infty, - 3) \cup (3, \infty)$ and decreasing in $(- 3, 0) \cup (0, 3)$

$∴ α_{1} = - 3, α_{2} = 3, α_{3} = - 3, α_{4} = 0, α_{5} = 3$

$∴ \sum_{i = 1}^{5} α_{i}^{2} = 9 + 9 + 9 + 0 + 9 = 36$ .

Solution

Q.
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 $\sum_{i = 1}^{5} α_{i}^{2}$ is equal to [2025]

1 40

2 28

3 36

4 48

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