JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the function \(f(x)=2 x^3-9 a x^2+12 a^2 x+1, a>0\) has a local maximum at \(\mathrm{x}=\alpha\) and a local minimum \(x=\alpha^2\), then \(\alpha\) and \(\alpha^2\) are the roots of the equation :
- A \(\mathrm{x}^2-6 \mathrm{x}+8=0\)
- B \(8 x^2+6 x-8=0\)
- C \(8 x^2-6 x+1=0\)
- D \(x^2+6 x+8=0\)
Answer & Solution
Correct Answer
(A) \(\mathrm{x}^2-6 \mathrm{x}+8=0\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}(x)=6 x^2-18 a x+12 a^2=0 \backslash_{\alpha^2}^\alpha\) \(\begin{aligned} & \alpha+\alpha^2=3 \mathrm{a} \& \alpha \times \alpha^2=2 \mathrm{a}^2 \\ & \quad \downarrow \\ & \left(\alpha+\alpha^2\right)^3=27 \mathrm{a}^3\end{aligned}\)…
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