JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let the function \(f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0\) be strictly increasing in \(\left(-\infty, \alpha_1\right) \mathrm{U}\left(\alpha_2, \infty\right)\) and strictly decreasing in \(\left(\alpha_3, \alpha_4\right) \mathrm{U}\left(\alpha_4, \alpha_5\right)\). Then \(\sum_{\mathrm{i}=1}^5 \alpha_{\mathrm{i}}^2\) is equal to :-
- A 48
- B 28
- C 40
- D 36
Answer & Solution
Correct Answer
(D) 36
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0 \\ & f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^2}=0 \quad \Rightarrow x= \pm 3 \\ & f^{\prime}(x)=\frac{x^2-3}{3 x^2}\end{aligned}\)…
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