Let the function f(x)= x 3 +3 x +3, x 0 be strictly increasing in… - Vidyadip

MCQ

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

A
48
B
28
C
40
✓
36

✓

Answer

Correct option: D.

36

(D) 36
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{3}+\frac{3}{\mathrm{x}}+3, \mathrm{x} \neq 0$
$f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^{2}}=0 \quad \Rightarrow x= \pm 3$
$f^{\prime}(x)=\frac{x^{2}-3}{3 x^{2}}$
$\mathrm{f}^{\prime}(\mathrm{x})>0 \forall(-\infty,-3) \cup(3, \infty) \rightarrow$ increasing
$\mathrm{f}^{\prime}(\mathrm{x})<0 \forall(-3,0) \cup(0,3) \rightarrow$ decreasing
$\sum_{i=1}^{5} \alpha_{i}^{2}=(-3)^{2}+(3)^{2}+(-3)^{2}+(0)^{2}+(3)^{2}$ $=36$

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