Question

Let $f\left(x\right)=x^3-3x^2+2x$. If the equation $f\left(x\right)=k$ has exactly one positive and one negative solution, then the value of $k$ equals to

• A. $-\frac{2\sqrt{3}}{9}$
• B. $-\frac{2}{9}$
• C. $\frac{2}{3\sqrt{3}}$
• D. $\frac{1}{3\sqrt{3}}$

Answer 1

The correct option is A $-\frac{2\sqrt{3}}{9}$

$f\left(x\right)=x^3-3x^2+2x$
$f^{\prime }\left(x\right)=3x^3-6x+2=0$ for min. or max.
$\Rightarrow x=\frac{6\pm \sqrt{36-24}}{6}=\frac{3\pm \sqrt{3}}{3}$
$\Rightarrow x_1=\frac{3-\sqrt{3}}{3}=1-\frac{1}{\sqrt{3}},x_2=\frac{3+\sqrt{3}}{3}=1+\frac{1}{\sqrt{3}}$
From graph of $f\left(x\right)$ it is clear that for equation $f\left(x\right)=k$ to have one negative
and one positive root $k=f\left(x_2\right)=x_2(x_2-1)(x_2-2)=-\left(\frac{1}{\sqrt{3}}\right).(1-\frac{1}{3})$
$\therefore k=-\frac{2\sqrt{3}}{9}$