Simple Absolute Value Inequalities
\( 1. \quad |\; f(x) | \; \leq \; A \) For a function \(f(x)\), where \( A \geq 0\):
$$ -A \;\leq \;f(x) \; \leq \; A $$
\( 2. \quad |\; f(x) | \; \geq \; A \) For a function \(f(x)\): \(f(x) \geq A \) or \( f(x) \leq -A \).
Examples:
\( \bullet \quad \displaystyle {\left |\frac{2x \ – \ 1}{3} \right|} \leq 1 \)
\[\Rightarrow -1 \leq \frac{2x \ – \ 1}{3} \leq 1 \]
\[ \Rightarrow -3 \leq 2x \ – \ 1 \leq 3 \]
\[ \Rightarrow -3+1 \leq 2x \ – \ 1 \leq 3 +1 \]
\[ \Rightarrow -2 \leq 2x \leq 4 \]
\[ \Rightarrow -1 \leq x \leq 2 \]
\( \bullet \quad |3x\ – \ 1 | > 4 \)
\[\Rightarrow 3x \ – \ 1 > 4 \quad \text{or} \quad 3x \ – \ 1 < -4 \]
\[\Rightarrow 3x > 4+1 \quad \text{or} \quad 3x < -4+1 \]
\[\Rightarrow x > \frac{5}{3} \quad \text{or} \quad x < -1 \quad\quad\quad\quad \]
Question 1
How many integer values of x satisfy the inequality ? \[ \frac{|2x \ – \ 2 |\;+ \; | 1 \ – \ x| }{1 \;+\; |x \ – \ 1|} <2 \]
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
Since \( |1 \ – \ x | = |x \ – \ 1| \), we have:
\[ \frac{|2x \ – \ 2| + | 1 \ – \ x |}{1 \;+\; |x \ – \ 1|} < 2 \Rightarrow \frac{|2 (x \ – \ 1)| + | x \ – \ 1|}{1+ |x \ – \ 1|} < 2 \]
Since \( 1 + |x \ – \ 1 | > 0 \):
\[ \Rightarrow \frac{ 3 | x \ – \ 1 |}{ 1 + |x \ – \ 1 |} < 2 \]
\[ \Rightarrow 3\; | x \ – \ 1 | < 2 \;(1+ | x \ – \ 1 | ) \]
\[ \Rightarrow |x \ – \ 1 | < 2 \]
\[ \Rightarrow -2 < x \ – \ 1 < 2 \]
\[ \Rightarrow -1 < x < 3 \]
Thus, the possible integer values for x are \( 0, 1, 2 \), which gives a total of 3 values.
\(\textbf{Answer: C} \)
Question 2
How many integer values of x satisfy the inequality ? \[ | \; |x| – \; x \; | + x \; < \; 3 \]
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
For \( x > 0 \), since \( |x | = x \):
\[ | \; |x| – \; x \; | + x \; < \; 3 \Rightarrow | x – x | + x < 3\]
\[ \Rightarrow x < 3 \]
Since \( x > 0 \), this yields the interval \( 0 \leq x < 3 \).
For \( x < 0 \), since \( |x | = -x \):
\[ | \; |x| \ – \ \; x \; | + x \; < \; 3 \Rightarrow | -x \ – \ x | + x < 3\]
\[ \Rightarrow |-2x| + x < 3 \]
Since \( x < 0 \), \( -2x > 0 \), so \( |-2x| = -2x \):
\[ \Rightarrow -2x + x < 3 \]
\[ \Rightarrow -x < 3 \]
\[ \Rightarrow x > -3 \]
Since \( x < 0 \), this yields the interval \( -3 < x < 0 \).
Combining both cases, we get \( -3 < x < 3 \). Therefore, the integer values x can take are \( -2, -1, 0, 1, 2 \), which gives a total of 5 values.
\(\textbf{Answer: E} \)
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