If the function f(x)=2 x^3-9 x^2+12 a^2 x+1, a>0 has a local maximum at x=α and a local minimum at x=α^2, then α and α^2 are the roots of the…

$\begin{aligned} & \alpha+\alpha^2=3 \mathrm{a} \& \alpha \times \alpha^2=2 \mathrm{a}^2 \\ & \quad \downarrow \\ & \left(\alpha+\alpha^2\right)^3=27 \mathrm{a}^3 \\ & \Rightarrow 2 \mathrm{a}^2+4 \mathrm{a}^4+3(3 \mathrm{a})\left(2 \mathrm{a}^2\right)=27 \mathrm{a}^3 \\ & \Rightarrow 2+4 \mathrm{a}^2+18 \mathrm{a}=27 \mathrm{a} \\ & \Rightarrow 4 \mathrm{a}^2-9 \mathrm{a}+2=0 \\ & \Rightarrow 4 \mathrm{a}^2-8 \mathrm{a}-\mathrm{a}+2=0 \\ & \Rightarrow(4 \mathrm{a}-1)(\mathrm{a}-2)=0 \Rightarrow \mathrm{a}=2 \\ & \text { so } 6 \mathrm{x}^2-36 \mathrm{x}+48=0\end{aligned}$​α+α2=3a&α×α2=2a2↓(α+α2)3=27a3⇒2a2+4a4+3(3a)(2a2)=27a3⇒2+4a2+18a=27a⇒4a2−9a+2=0⇒4a2−8a−a+2=0⇒(4a−1)(a−2)=0⇒a=2 so 6x2−36x+48=0​ $\Rightarrow x^2-6 x+8=0$⇒x2−6x+8=0 ...(1) If we take $\mathrm{a}=\frac{1}{4}$a=41​ then $\alpha=\frac{1}{2}$α=21​ which is not possible