Question

If equation $x^3-x+a^2-a+\frac{2}{3\sqrt{3}}<0$ is true for atleast one positive $x$, then a belongs to

• A. $(0,1)$
• B. $(1,\infty )$
• C. $(2,3\sqrt{3})$
• D. none of these

Answer 1

The correct option is A $(0,1)$

$x^3-x<-a^2+a-\frac{2}{3\sqrt{3}}$
Let $f\left(x\right)=x^3-x$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-1=0$ for estrema.
$\Rightarrow x=\pm \frac{1}{\sqrt{3}}$
In positive region minimum value of $f\left(x\right)=\frac{1}{3\sqrt{3}}-\frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$
So for having at least one positive $x$ root of given equation,
$-a^2+a-\frac{2}{3\sqrt{3}}>-\frac{2}{3\sqrt{3}}$
$\Rightarrow a^2-a<0$

$\Rightarrow a\in (0,1)$