JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 17)
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 :
$$x^2-6 x+8=0$$
$$8 x^2-6 x+1=0$$
$$8 x^2+6 x-1=0$$
$$x^2+6 x+8=0$$
Explanation
$$\begin{aligned} & f(x)=6 x^2-18 a x+12 a^2 \\ & =6\left(x^2-3 a+2 a^2\right) \\ & =6(x-a)(x-2 a)=0 \\ & x=a, 2 a \end{aligned}$$
$$a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2$$
$$\begin{array}{lll} a>0 & \therefore & \alpha=2 \\ & & \alpha^2=4 \end{array}$$
$$\therefore x^2-6 x+8=0$$ is the required quadratic equation.
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