Question

∣x−2∣−2 ∣x-2∣−1 ​ ≤0

Answer 1

[0, 1) ∪ (3, 4]

Answer 2

To solve the inequality $\frac{|x-2|-2}{|x-2|-1} \leq 0$, we first simplify it by substituting $y = |x-2|$.
The inequality becomes: $\frac{y-2}{y-1} \leq 0$ For a rational expression $\frac{A}{B} \leq 0$, we need $A$ and $B$ to have opposite signs, or $A=0$ (provided $B \neq 0$).

The critical points are the values of $y$ where the numerator or denominator is zero:

1. Numerator $y-2 = 0 \implies y=2$.
2. Denominator $y-1 = 0 \implies y=1$.

We can use a sign analysis (or number line method) for the expression $\frac{y-2}{y-1}$:

• Case 1: $y < 1$
  Let $y=0$. $\frac{0-2}{0-1} = \frac{-2}{-1} = 2$. Since $2 \not\leq 0$, this interval is not part of the solution.

• Case 2: $1 < y < 2$
  Let $y=1.5$. $\frac{1.5-2}{1.5-1} = \frac{-0.5}{0.5} = -1$. Since $-1 \leq 0$, this interval is part of the solution.

• Case 3: $y > 2$
  Let $y=3$. $\frac{3-2}{3-1} = \frac{1}{2}$. Since $\frac{1}{2} \not\leq 0$, this interval is not part of the solution.

Also, we must consider the endpoints:

• If $y=1$, the denominator becomes zero, so the expression is undefined. Thus $y \neq 1$.
• If $y=2$, the numerator becomes zero, so $\frac{2-2}{2-1} = \frac{0}{1} = 0$. Since $0 \leq 0$, $y=2$ is included in the solution.

Combining these, the solution for $y$ is $1 < y \leq 2$.

Now, substitute back $y = |x-2|$: $1 < |x-2| \leq 2$ This compound inequality can be broken down into two separate inequalities:

1. $|x-2| > 1$
2. $|x-2| \leq 2$

Let's solve each part:

Part 1: $|x-2| > 1$
This inequality holds if $x-2 > 1$ OR $x-2 < -1$.

• $x-2 > 1 \implies x > 3$
• $x-2 < -1 \implies x < 1$

So, the solution for Part 1 is $x \in (-\infty, 1) \cup (3, \infty)$.

Part 2: $|x-2| \leq 2$
This inequality holds if $-2 \leq x-2 \leq 2$.
Add 2 to all parts of the inequality:

• $-2+2 \leq x-2+2 \leq 2+2$
• $0 \leq x \leq 4$

So, the solution for Part 2 is $x \in [0, 4]$.

Finally, we need to find the intersection of the solutions from Part 1 and Part 2, because both conditions must be satisfied simultaneously: $x \in ((-\infty, 1) \cup (3, \infty)) \cap [0, 4]$ Let's find the intersection for each part of the union:

• Intersection of $(-\infty, 1)$ and $[0, 4]$ is $[0, 1)$.
• Intersection of $(3, \infty)$ and $[0, 4]$ is $(3, 4]$.

Combining these two intervals, the final solution set for $x$ is: $x \in [0, 1) \cup (3, 4]$