Inequalities with absolute values
Task number: 2761
Solve the following inequalities over the domain of real numbers:
Variant 1
\(|x-1|< 3\)
Hint
Analyze the cases where the expression in the absolute value is positive and where it is negative.
Resolution
\(x-1=0 \longrightarrow x=1\)
- \(x\le 1: |x-1|=1-x,\ 1-x< 3 \longrightarrow x> -2 \longrightarrow x\in(-2{,}1\rangle\)
- \(x\ge 1: |x-1|=x-1,\ x-1< 3 \longrightarrow x< 4 \longrightarrow x\in\langle 1{,}4)\)
Result
The solution is \(x\in (-2{,}4)\).
Variant 2
\(|x+2|\ge 4\)
Resolution
- \(x\ge -2: x+2\ge 4 \longrightarrow x\ge 2 \longrightarrow x\in \langle 2,\infty)\)
- \(x\le -2: -x-2\ge 4 \longrightarrow x\le -6 \longrightarrow x\in (-\infty,-6\rangle \)
Result
The solution is \(x\in (-\infty,-6\rangle \cup \langle 2,\infty)\).
Variant 3
\(|5x-2|<x\)
Resolution
- \(x\le \frac25 \longrightarrow 2-5x < x \longrightarrow 6x>2 \longrightarrow x>\frac13 \longrightarrow x\in\left(\frac13,\frac25 \right\rangle \)
- \(x\ge \frac25 \longrightarrow 4x< 2 \longrightarrow x<\frac12 \longrightarrow x\in \left\langle \frac25,\frac12\right)\)
Result
The solution is \(x\in \left(\frac13,\frac12\right)\)
Variant 4
\(|x-1|<|x+5|\)
Resolution
- \(x\le -5: 1-x < -x-5 \longrightarrow 0< -4\), no solution.
- \(-5 \le x \le 1: 1-x < x+5 \longrightarrow 2x+4 > 0 \longrightarrow x> 2 \longrightarrow x\in (-2{,}1 \rangle\)
- \(x \ge 1: x-1< x+5 \longrightarrow -1<5 \longrightarrow x\in \langle 1, \infty )\)
Result
The solution is \(x\in (-2,\infty)\).
Variant 5
\(\left| \frac{x+1}{x-1}\right| \le 1\)
Hint
Do not forget about the case where the value of the expression is undefined.
Resolution
- \(x\le -1: \frac{-x-1}{-x+1}\le 1 \longrightarrow x+1 \ge x-1 \longrightarrow x\in (-\infty,-1\rangle\)
- \(-1\le x < 1: \frac{x+1}{-x+1}\le 1 \longrightarrow x+1 \le 1-x \longrightarrow x\le 0 \longrightarrow x\in \langle -1, 0\rangle\)
- \(x> 1: \frac{x+1}{x-1} \le 1 \longrightarrow x+1\le x-1 \longrightarrow 1\le -1\) which has no solution.
Result
The solution is \(x\in (-\infty, 0\rangle \).
Variant 6
\(|x^2+2x-3|\ge |x^2+3x-4|\)
Resolution
We first determine the values of \(x\) at which the expressions inside the absolute values change sign.
\(x^2+2x-3=0 \longrightarrow x_{1{,}2}=\frac{-2 \pm \sqrt{4+12}}{2}=-1\pm 2 \longrightarrow x_1=-3,\ x_2=1\)
\(x^2+3x-4=0 \longrightarrow x_{1{,}2}=\frac{-3 \pm \sqrt{9+16}}{2} \longrightarrow x_1=-4,\ x_2=1\)
- \(x \le -4: x^2+2x-3 \ge x^2 +3x -4 \longrightarrow x \le 1 \longrightarrow x\in (-\infty,-4\rangle \)
- \(-4\le x\le -3: x^2+2x-3 \ge -x^2 -3x +4 \longrightarrow 2x^2+5x-7\ge 0 \longrightarrow \\ (x-1)(2x+7)\ge 0 \longrightarrow 2x+7 \le 0 \longrightarrow x\in \left\langle -4,-\frac72 \right\rangle\)
- \(-3\le x \le 1: -x^2-2x+3 \ge -x^2 -3x +4 \longrightarrow x\ge 1 \longrightarrow x=1\)
- \(x \ge 1: x^2+2x-3 \ge x^2 +3x -4 \longrightarrow x\le 1 \longrightarrow x=1\)
Result
The solution is \(x\in \left(-\infty,-\frac72 \right\rangle \cup \{1\}\)
Variant 7
\(||x-2|+1|\le 5\)
Resolution
- \(x\le 2 \longrightarrow |x-2+1|=|3-x|> 0: 3-x\le 5 \longrightarrow x\ge -2\longrightarrow x\in \langle -2{,}2\rangle\)
- \(x\ge 2 \longrightarrow |x-2+1|=x-1>0: x-1\le 5 \longrightarrow x\le 6 \longrightarrow x\in\langle 2{,}6\rangle \)
Result
The solution is \(x\in \langle -2{,}6\rangle\).
Variant 8
\(|x^2-4x+3|\le |x^2-4|\)
Resolution
\(x^2-4x+3=0 \longrightarrow x_{1{,}2}=\frac{4\pm\sqrt{16-12}}{2}=2\pm 1\)
\(x^2-4=0 \longrightarrow x_{1{,}2}=\pm 2\)
- \(x\le -2: x^2-4x+3\le x^2-4 \longrightarrow -4x\le -7 \longrightarrow x\ge \frac74\) which has no solution
- \(-2\le x \le 1: x^2-4x+3\le -x^2+4 \longrightarrow 2x^2-4x-1\le 0\\ x_{1{,}2}=\frac{4\pm \sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}2 \longrightarrow x\in \left\langle 1-\frac{\sqrt{6}}2{,}1\right\rangle\)
- \(1\le x\le 2: -x^2+4x-3\le -x^2+4 \longrightarrow 4x\le 7 \longrightarrow x\le \frac74 \longrightarrow x\in \left\langle 1,\frac74\right\rangle\)
- \(2\le x\le 3: -x^2+4x-3\le x^2-4 \longrightarrow 2x^2-4x-1\ge 0 \longrightarrow x\in \left\langle 1+\frac{\sqrt{6}}2{,}3\right\rangle\)
- \(x\ge 3: x^2-4x+3\le x^2-4 \longrightarrow -4x\le -7 \longrightarrow x\ge \frac74 \longrightarrow x\in \langle 3,\infty)\)
Result
The solution is \(x\in \left\langle 1-\frac{\sqrt{6}}2,\frac74\right\rangle \cup \left\langle 1+\frac{\sqrt{6}}2,\infty\right)\).
