What is the number of solutions for $|\frac{4}{|x|}-2|=m$ for an $m\in \Bbb{R}$? For:


*

*$m=0$ we've got two solutions; these are the x-intercepts,

*$m>2$ we've got two solutions,

*$m<0$ we've got no solutions.


However, how many solutions are there for $m=2$?
If there were $4$ solutions for $m=2$, then $f(x)$ would intersect the horizontal asymptote.
 A: If $\left|\frac4{|x|}-2\right|=2$, then clearly $\frac4{|x|}$ must be $0$ or $4$. It can’t be $0$, so it must be $4$, and $|x|$ must be $1$. Thus, there are two solutions, $\pm 1$.
A: The equation is satisfied if
$$\frac{4}{|x|}-2=m\qquad \text{ or }\qquad 2-\frac{4}{|x|}=m$$
If $m=2$, these become
$$\frac{4}{|x|}=4\qquad \text{ or }\qquad -\frac{4}{|x|}=0$$
so $|x|=1$, i.e. $x=-1$ or $x=1$.
A: You say correctly that there is no solution for $m<0$.
Assume $m\ge0$ and set $t=4/|x|$, so the equation becomes
$$
|t-2|=m
$$
with the condition $t>0$. Since $m\ge0$, we can square and get an equivalent equation:
$$
t^2-4t+4-m^2=0
$$
Now use the “rule of signs”. If $4-m^2<0$, that is, $m>2$, the equation has one positive root and one negative root. The negative root is discarded, the positive one gives two solutions for $x$.
For $0<m<2$ the equation has two positive roots (the discriminant is positive), that produce four solutions in $x$.
It remains to consider the cases $m=0$ and $m=2$.
If $m=0$, we get $t=2$ and so $x=\pm2$.
If $m=2$, the equation is $t^2-4t=0$; since $t>0$, we obtain $t=4$ and so $x=\pm1$.
