\[\begin{array}{1 1}(1)\bigg(-\frac{1}{3},-3\bigg)&(2)\bigg(\frac{1}{3},3\bigg)\\(3)\bigg(-\frac{1}{3},3\bigg)&(4)\bigg(\frac{1}{3},1\bigg)\end{array}\]

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Let $y=x^3-2x^2+3x+8$

Given $\large\frac{dy}{dt} $$=2 \large\frac{dx}{dt}$

$y= x^3-2x^2+3x+8$

$\large\frac{dy}{dx} $$=(3x^2-4x+3) $$\large\frac{dx}{dt}$

$2 \large\frac{dx}{dt} $$=(3x^2-4x+3) $$\large\frac{dx}{dt}$

$2= 3x^2-4x+3$

$3x^2-4x+1=0$

$3x^2-3x-x+1=0$

$3x(x-1) -1(x-1) =0$

$(3x-1) (x-1)=0$

$x=1$ or $x=\large\frac{1}{3}$

Hence 4 is the correct answer

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