If the function f(x)=2x^3-9ax^2+12a^2x+1,a0 has a local maximum at x= and a local minimum…

Question

If the function $f(x)=2{x}^{3}-9{ax}^{2}+12{a}^{2}x+1,a>0$ 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 :

Accepted Answer

$$\begin{aligned}
& f(x)=6 x^2-18 a x+12 a^2 \
& =6\left(x^2-3 a+2 a^2\right) \
& =6(x-a)(x-2 a)=0 \
& x=a, 2 a
\end{aligned}$$

$$a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2$$

$$\[\begin{array}{lll} a>0 & \therefore & \alpha=2 \ & \alpha^2=4 \end{array}$$\]

$$\therefore x^2-6 x+8=0$$ is the required quadratic equation.

Solutions · 1