If the function f(x)=2 x^3-9 \mathrm{ax}^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 :

Question

If the function f(x)=2 x^3-9 \mathrm{ax}^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 :

Accepted Answer

Answer: A. x^2-6 x+8=0 \begin{aligned} & f(x)=6 x^2-18 a x+12 a^2 \ & =6\left(x^2-3 a+2 a^2ight) \ & =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 & 	herefore & \alpha=2 \ & & \alpha^2=4 \end{array} 	herefore x^2-6 x+8=0 is the required quadratic equation.

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

Answer

Explanation

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