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 $k$ is equal 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 $-\frac{2\sqrt{3}}{9}$

$f\left(x\right)=x(x^2-3x+2)$
$\Rightarrow f\left(x\right)=x(x-2)(x-1)$
Graph of $y=f\left(x\right):$


We have $f^{\prime }\left(x\right)=3x^2-6x+2=0$
$\Rightarrow x=\frac{6\pm \sqrt{36-24}}{6}$
$\Rightarrow x=\frac{3\pm \sqrt{3}}{3}$
$\Rightarrow x=1-\frac{1}{\sqrt{3}},1+\frac{1}{\sqrt{3}}$

If we consider $x=1-\frac{1}{\sqrt{3}},$ then there are two positive roots.
$\therefore k=f(1+\frac{1}{\sqrt{3}})$
$\Rightarrow k=(1+\frac{1}{\sqrt{3}})(1+\frac{1}{\sqrt{3}}-2)(1+\frac{1}{\sqrt{3}}-1)$
$\Rightarrow k=(1+\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}-1)\left(\frac{1}{\sqrt{3}}\right)$
$\Rightarrow k=(\frac{1}{3}-1)\left(\frac{1}{\sqrt{3}}\right)$
$\therefore k=-\frac{2\sqrt{3}}{9}$