**Question:**

Solve each of the following in equations and represent the solution set on the number line.

$\frac{|x|-1}{|x|-2} \geq 0, x \in$ R. $-\{-2,2\}$

**Solution:**

Given:

$\frac{|x|-1}{|x|-2} \geq 0, x \in$ R. $-\{-2,2\}$

Intervals of $|x|$ :

$x \geq 0,|x|=x$ and $x<0,|x|=-x$

Domain of $\frac{|x|-1}{|x|-2} \geq 0$

$\frac{|x|-1}{|x|-2}$ is not defined for $x=-2$ and $x=2$

Therefore, Domain: x < -2 or -2 < x < 2 or x > 2

Combining intervals with domain:

$x<-2,-2

For $x<-2$ :

$\frac{|x|-1}{|x|-2}=\frac{-x-1}{-x-2}$

$\frac{-x-1}{-x-2} \geq 0$

Signs of $-x-1$ :

$-x-1=0 \rightarrow x=-1$

(Adding 1 to both the sides and then dividing by -1 on both the sides)

$-x-1>0 \rightarrow x<-1$

(Adding 1 to both the sides and then multiplying by -1 on both the sides)

$-x-1<0 \rightarrow x>-1$

(Adding 1 to both the sides and then multiplying by -1 on both the sides)

Signs of $-x-2$ :

$-x-2=0 \rightarrow x=-2$

(Adding 2 to both the sides and then dividing by -1 on both the sides)

$-x-2>0 \rightarrow x<-2$

(Adding 2 to both the sides and then multiplying by -1 on both the sides)

$-x-2<0 \rightarrow x>-2$

(Adding 2 to both the sides and then multiplying by -1 on both the sides)

Intervals satisfying the required condition: ≥ 0

$x<-2$ or $x=-1$ or $x>-1$

Merging overlapping intervals:

$x<-2$ or $x \geq-1$

Combining the intervals:

$x<-2$ or $x \geq-1$ and $x<-2$

Merging overlapping intervals:

$x<-2$

Merging overlapping intervals

$x<-2$

Similarly, for $-2

$\frac{|x|-1}{|x|-2}=\frac{-x-1}{-x-2}$

$\frac{-x-1}{-x-2} \geq 0$

Therefore,

Intervals satisfying the required condition: ≥ 0

$x<-2$ or $x=-1$ or $x>-1$

Merging overlapping intervals:

$x<-2$ or $x \geq-1$

Combining the intervals:

$x<-2$ or $x \geq-1$ and $-2

Merging overlapping intervals:

$-1 \leq x<0$

For 0 ≤ x < 2,

$\frac{|x|-1}{|x|-2}=\frac{x-1}{x-2}$

$\frac{x-1}{x-2} \geq 0$

Signs of x – 1:

$x-1=0 \rightarrow x=1$ (Adding 1 to both the sides)

$x-1>0 \rightarrow x>1$ (Adding 1 to both the sides)

$x-1<0 \rightarrow x<1$ (Adding 1 to both the sides)

Signs of x – 2:

$x-2=0 \rightarrow x=2$ (Adding 2 to both the sides)

$x-2<0 \rightarrow x<2$ (Adding 2 to both the sides)

$x-2>0 \rightarrow x>2$ (Adding 2 to both the sides)

At $x=2, \frac{x-1}{x-2}$ is not defined

Intervals satisfying the required condition: ≥ 0

$x<1$ or $x=1$ or $x>2$

Merging overlapping intervals:

x ≤ 1 or x > 2

Combining the intervals:

$x \leq 1$ or $x>2$ and $0 \leq x<2$

Merging overlapping intervals:

0 ≤ x ≤ 1

Similarly, for x > 2:

$\frac{|x|-1}{|x|-2}=\frac{x-1}{x-2}$

$\frac{x-1}{x-2} \geq 0$

Therefore,

Intervals satisfying the required condition: ≥ 0

$x<1$ or $x=1$ or $x>2$

Merging overlapping intervals:

$x \leq 1$ or $x>2$

Combining the intervals:

x ≤ 1 or x > 2 and x > 2

Merging overlapping intervals:

x > 2

Combining all the intervals:

$x<-2$ or $-1 \leq x<0$ or $0 \leq x \leq 1$ or $x>2$

Merging the overlapping intervals:

$x<-2$ or $-1 \leq x \leq 1$ or $x>2$

Therefore,

$x \in(-\infty,-2) \cup[-1,1] \cup(2, \infty)$