If the coordinates of one of the two points on the curve  $y=x^4-2x^2-x$ that have a common tangent line, can be $(\alpha ,\beta )$ , then the value of  $|\left|\alpha \right|-\left|\beta \right||=$

$$\begin{gathered} y-(a^4-2a^2-a)=(4a^3-4a-1)(x-a) \\ \Rightarrow y=(4a^3-4a-1)x+(a^4-2a^2-a)-a(4a^3-4a-1) \\ \therefore y=(4a^3-4a-1)x-3a^4+2a^2\mathrm{....}\left(1\right) \end{gathered}$$

!!ly     tangent at Q is

$y=(4b^3-4b-1)x-3b^4+2b^2\mathrm{....}\left(2\right)$

Given (1) and (2) is same line and hence
$4a^3-4a-1=4b^3-4b-1$   (equating slope)
i.e.  $a^3-b^3=a-b\mathrm{...}\left(3\right)$ and   $-3a^4+2a^2=-3b^4+2b^2$ (equating constant term)
i.e  $2(a^2-b^2)=3(a^4-b^4)\mathrm{...}\left(4\right)$
$$\begin{gathered} \left(3\right) \Rightarrow (a-b)(a^2+ab+b^2)=(a-b)(a\ne b) \\ \therefore a^2+ab+b^2=\mathrm{1...}\left(5\right) \\ \left(4\right) \Rightarrow 2(a^2-b^2)=3(a^2-b^2)(a^2+b^2) \\ \therefore a^2=b^2 or a^2+b^2=\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \therefore a^2+\frac{1}{9a^2}=\frac{2}{3}\Rightarrow \frac{9a^4+1}{9a^2}=\frac{2}{3} \Rightarrow 9a^4-6a^2+1=0\Rightarrow {(3a^2-1)}^2=0 \\ \therefore a^2=\frac{1}{3} \Rightarrow b^2=\frac{1}{3} \end{gathered}$$    

$\therefore a^2=b^2$  which is not possible
$\Rightarrow$ this case is not possible