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If the function $f(x)=2 x^3-9 x^2+12 \mathrm{a}^2 x+1, \mathrm{a}>0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$, then $\alpha$ and $\alpha^2$ are the roots of the equation :
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$x^2-6 x+8=0$
$\begin{aligned} & \alpha+\alpha^2=3 \mathrm{a} \& \alpha \times \alpha^2=2 \mathrm{a}^2 \\ & \quad \downarrow \\ & \left(\alpha+\alpha^2\right)^3=27 \mathrm{a}^3 \\ & \Rightarrow 2 \mathrm{a}^2+4 \mathrm{a}^4+3(3 \mathrm{a})\left(2 \mathrm{a}^2\right)=27 \mathrm{a}^3 \\ & \Rightarrow 2+4 \mathrm{a}^2+18 \mathrm{a}=27 \mathrm{a} \\ & \Rightarrow 4 \mathrm{a}^2-9 \mathrm{a}+2=0 \\ & \Rightarrow 4 \mathrm{a}^2-8 \mathrm{a}-\mathrm{a}+2=0 \\ & \Rightarrow(4 \mathrm{a}-1)(\mathrm{a}-2)=0 \Rightarrow \mathrm{a}=2 \\ & \text { so } 6 \mathrm{x}^2-36 \mathrm{x}+48=0\end{aligned}$
$\Rightarrow x^2-6 x+8=0$ ...(1)
If we take $\mathrm{a}=\frac{1}{4}$ then $\alpha=\frac{1}{2}$ which is not possible
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