Question

∣x−2∣−1 ∣x+2∣−2 ​ ≤0

Answer 1

(-4, 0) ∪ [1, 3]

Answer 2

To solve the inequality $\frac{|x-2|-1}{|x+2|-2} \leq 0$, we need to find the values of $x$ for which the numerator and denominator have opposite signs, or the numerator is zero (provided the denominator is not zero).

Step 1: Find the critical points where the numerator or denominator becomes zero.

• Numerator: $|x-2|-1 = 0$

  $|x-2| = 1$

  This implies $x-2 = 1$ or $x-2 = -1$.

  So, $x = 3$ or $x = 1$.

  These points make the numerator zero, so they are potential solutions if the denominator is non-zero.

• Denominator: $|x+2|-2 = 0$

  $|x+2| = 2$

  This implies $x+2 = 2$ or $x+2 = -2$.

  So, $x = 0$ or $x = -4$.

  These points make the denominator zero, so they must be excluded from the solution set.

Step 2: Plot the critical points on a number line.

The critical points, in increasing order, are $-4, 0, 1, 3$. These points divide the number line into several intervals: $(-\infty, -4)$, $(-4, 0)$, $(0, 1)$, $(1, 3)$, $(3, \infty)$.

Step 3: Test a value in each interval to determine the sign of the expression.

• Interval 1: $x < -4$ (e.g., $x=-5$)

  Numerator: $|-5-2|-1 = |-7|-1 = 7-1 = 6$ (Positive)

  Denominator: $|-5+2|-2 = |-3|-2 = 3-2 = 1$ (Positive)

  Expression: $\frac{+}{+} = +$. Not $\leq 0$.

• Interval 2: $-4 < x < 0$ (e.g., $x=-2$)

  Numerator: $|-2-2|-1 = |-4|-1 = 4-1 = 3$ (Positive)

  Denominator: $|-2+2|-2 = |0|-2 = 0-2 = -2$ (Negative)

  Expression: $\frac{+}{-} = -$. Is $\leq 0$. So, $(-4, 0)$ is part of the solution. (Note: $x=-4$ and $x=0$ are excluded because the denominator is zero).

• Interval 3: $0 < x < 1$ (e.g., $x=0.5$)

  Numerator: $|0.5-2|-1 = |-1.5|-1 = 1.5-1 = 0.5$ (Positive)

  Denominator: $|0.5+2|-2 = |2.5|-2 = 2.5-2 = 0.5$ (Positive)

  Expression: $\frac{+}{+} = +$. Not $\leq 0$.

• Interval 4: $1 < x < 3$ (e.g., $x=2$)

  Numerator: $|2-2|-1 = |0|-1 = 0-1 = -1$ (Negative)

  Denominator: $|2+2|-2 = |4|-2 = 4-2 = 2$ (Positive)

  Expression: $\frac{-}{+} = -$. Is $\leq 0$. So, $(1, 3)$ is part of the solution.

• Interval 5: $x > 3$ (e.g., $x=4$)

  Numerator: $|4-2|-1 = |2|-1 = 2-1 = 1$ (Positive)

  Denominator: $|4+2|-2 = |6|-2 = 6-2 = 4$ (Positive)

  Expression: $\frac{+}{+} = +$. Not $\leq 0$.

Step 4: Check the critical points themselves.

• At $x = -4$: Denominator is $0$. Expression is undefined. Exclude $x=-4$.
• At $x = 0$: Denominator is $0$. Expression is undefined. Exclude $x=0$.
• At $x = 1$: Numerator is $0$, denominator is $|1+2|-2 = 3-2 = 1 \neq 0$. Expression is $\frac{0}{1} = 0$. Since $0 \leq 0$, $x=1$ is included.
• At $x = 3$: Numerator is $0$, denominator is $|3+2|-2 = 5-2 = 3 \neq 0$. Expression is $\frac{0}{3} = 0$. Since $0 \leq 0$, $x=3$ is included.

Step 5: Combine the results.

The intervals where the inequality holds are $(-4, 0)$ and $(1, 3)$.

The points where the inequality holds are $x=1$ and $x=3$.

Combining these, the solution set is $(-4, 0) \cup [1, 3]$.