The problem is as follows:
Find the set:
$$K=\left\{x\in\mathbb{R}/\frac{2-|x||x-1|}{2-|x|}\leq 0\right\}$$
The choices given by my textbook are as follows:
$\begin{array}{ll} 1.&\langle -2,0\rangle\\ 2.&\langle -1,2\rangle\\ 3.&\langle -2,1]\\ 4.&\langle -2,2\rangle\\ \end{array}$
What I'm about to post below is the official solution according to my workbook, and it does not give any details or explanation of why each step was chosen this way.
Interval $1$: $x<0$, in $K$:
Hence:
$\frac{2-(-x)(-(x-1))}{2-(-x)}\leq 0$
$\frac{-x^2+x+2}{x+2}\leq 0$
$\frac{x^2-x-2}{x+2}\geq 0$
$\frac{(x-2)(x+1)}{(x+2)}\geq 0$
Therefore:
$x\in\langle -2,-1]\cup[2,+\infty\rangle \,\land\, x<0$
Therefore:
$x\in \langle -2,-1]$
Interval $2$: $0\leq x < 1$ in $K$
$\frac{2-x(1-x)}{2-x}\leq 0$
$\frac{x^2-x+2}{2-x}\leq 0$, you may cancel the positive factor $x^2-x+2$, $\forall x\in \mathbb{R}$
$\frac{1}{2-x}\leq 0$
$\frac{1}{x-2}\geq 0$
Therefore:
$x-2>0$
$x>2$
Notice: $x>2 \,\land\,0\leq x<1$
$x\in \varnothing$
Interval $3$: $x\geq 1$, in $K$:
$\frac{2-x(x-1)}{2-x}\leq 0$
$\frac{-x^2+x+2}{2-x}\leq 0$
$\frac{x^2-x-2}{x-2}\leq 0$
$\frac{(x-2)(x+1)}{(x-2)}\leq 0$
$x+1\leq 0$ ; $x\neq 2$
Therefore
$x\leq -1 \, \land \, x\geq 1$
$x\in \varnothing$
From joining the three intervals: We conclude
$(1)\cup (2) \cup (3)$
$K=\langle -2,-1]$
Thus the answer is choice $3$
This is where it ends the explanation given by the author. But I'm confused on exactly why did he used all these steps can someone explain me why?.
The thing is what I'm confused is why should I split the interval in three?. Is this the only method of solution for this problem?.
Since what I am struggling the most is the understanding of this method of solution it would help me if someone could explain this to me line by line and offer some alternate method of solution or something easier. As I feel that in this problem the author did made a mistake here:
$\frac{2-x(1-x)}{2-x}\leq 0$
and from then I believe that for such given interval ($0\leq x < 1$) the rest is void. It is also worthy to note that the author at times cancelled positive factors and others cancelled common terms. This is somewhat confusing.
Thus can someone help me please I'm really going in circles with this problem.
Remark that for a fraction we have the following characterization:
$$\frac ab\le 0\iff \begin{cases}b>0 &\text{and}& a\le 0 \\b<0 &\text{and}& a\ge 0\end{cases}\iff \begin{cases} |x|<2 &\text{and}& |x||x-1|\ge 2\\|x|>2 &\text{and}& |x||x-1|\le 2\end{cases}$$
We now have to solve the quadratics $f(x)=x(x-1)=\pm 2$
Notice that $f(x)$ is an upward parabola with roots $0$ and $1$ so the minimum is in the middle $f(\frac 12)=-\frac 14$ so $f(x)=-2$ has no solution.
We just need to solve $f(x)=2$ which has roots $-1$ and $2$. Thus $|f(x)|\le 2$ between the roots, and greater outside the roots.
Our problem is reduced to
$$\begin{cases} -2<x<2 &\text{and}& (x\le-1\text{ or }x\ge 2)\\(x<-2\text{ or }x>2) &\text{and}& -1\le x\le 2\end{cases}$$
The last step consist in aggregating these results (i.e. find the intersection of all these intervals), I let you check that the only subsisting interval is
$$-2<x\le -1$$
I hope this presentation is clearer than the one presented in your book. :-)