JEE Math Practice Question
Let the function $ f(x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq 0 $ be strictly increasing in $(-\infty, \alpha_1) \cup (\alpha_2, \infty)$ and strictly decreasing in $(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$. Then $ \sum\limits_{i=1}^{5} \alpha_i^2 $ is equal to
- A.48
- B.40
- C.36
- D.28
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