🧮 Algebra Absolute Value Piecewise (12 Focused Problems)
Recommended: 30–40 minutes. No calculator.
Problems
1.
Tags: Basic Absolute Value · Easy · source: Original (AMC-style)
What is $|5|$?
A) $-5$
B) $0$
C) $5$
D) $10$
E) $25$
Answer & Solution
Answer: C
$|5| = 5$ since 5 is positive.
2.
Tags: Basic Absolute Value · Easy · source: Original (AMC-style)
What is $|-7|$?
A) $-7$
B) $0$
C) $7$
D) $14$
E) $49$
Answer & Solution
Answer: C
$|-7| = 7$ since the absolute value of a negative number is its positive value.
3.
Tags: Absolute Value Expression · Easy · source: Original (AMC-style)
What is $|3 - 8|$?
A) $-5$
B) $0$
C) $5$
D) $11$
E) $24$
Answer & Solution
Answer: C
$|3 - 8| = |-5| = 5$.
4.
Tags: Absolute Value Expression · Easy · source: Original (AMC-style)
What is $|2x - 6|$ when $x = 4$?
A) $2$
B) $4$
C) $6$
D) $8$
E) $10$
Answer & Solution
Answer: A
When $x = 4$: $|2(4) - 6| = |8 - 6| = |2| = 2$.
5.
Tags: Absolute Value Equation · Medium · source: Original (AMC-style)
How many solutions does $|x - 3| = 5$ have?
A) $0$
B) $1$
C) $2$
D) $3$
E) Infinitely many
Answer & Solution
Answer: C
We have $x - 3 = 5$ or $x - 3 = -5$, giving $x = 8$ or $x = -2$. So there are 2 solutions.
6.
Tags: Absolute Value Equation · Medium · source: Original (AMC-style)
Solve $|2x + 1| = 7$.
A) $x = 3$ only
B) $x = -4$ only
C) $x = 3$ or $x = -4$
D) $x = 3$ or $x = 4$
E) No solution
Answer & Solution
Answer: C
We have $2x + 1 = 7$ or $2x + 1 = -7$. So $2x = 6$ or $2x = -8$, giving $x = 3$ or $x = -4$.
7.
Tags: Absolute Value Inequality · Medium · source: Original (AMC-style)
For what values of $x$ is $|x - 2| < 3$?
A) $-1 < x < 5$
B) $x < -1$ or $x > 5$
C) $-5 < x < 1$
D) $x < -5$ or $x > 1$
E) All real numbers
Answer & Solution
Answer: A
We have $-3 < x - 2 < 3$. Adding 2: $-1 < x < 5$.
8.
Tags: Piecewise Function · Medium · source: Original (AMC-style)
If $f(x) = \begin{cases} x + 1 & \text{if } x < 0 \ x - 1 & \text{if } x \geq 0 \end{cases}$, what is $f(-2) + f(3)$?
A) $1$
B) $2$
C) $3$
D) $4$
E) $5$
Answer & Solution
Answer: A
Since $-2 < 0$: $f(-2) = -2 + 1 = -1$. Since $3 \geq 0$: $f(3) = 3 - 1 = 2$. So $f(-2) + f(3) = -1 + 2 = 1$.
9.
Tags: Absolute Value with Parameters · Hard · source: Original (AMC-style)
How many integer solutions does $|x - 1| + |x - 3| = 2$ have?
A) $0$
B) $1$
C) $2$
D) $3$
E) Infinitely many
Answer & Solution
Answer: E
For $x \leq 1$: $|x-1| + |x-3| = (1-x) + (3-x) = 4-2x = 2$, so $x = 1$. For $1 < x < 3$: $|x-1| + |x-3| = (x-1) + (3-x) = 2$, which is always true. For $x \geq 3$: $|x-1| + |x-3| = (x-1) + (x-3) = 2x-4 = 2$, so $x = 3$. Therefore, all $x$ with $1 \leq x \leq 3$ are solutions, giving infinitely many integer solutions.
10.
Tags: Complex Absolute Value · Hard · source: Original (AMC-style)
If $|x - 2| = |x + 4|$, what is $x$?
A) $-1$
B) $0$
C) $1$
D) $2$
E) $3$
Answer & Solution
Answer: A
We have $x - 2 = x + 4$ or $x - 2 = -(x + 4)$. The first gives $0 = 6$, which is false. The second gives $x - 2 = -x - 4$, so $2x = -2$ and $x = -1$.
11.
Tags: Piecewise with Absolute Value · Hard · source: Original (AMC-style)
If $f(x) = \begin{cases} |x| & \text{if } x < 0 \ x^2 & \text{if } x \geq 0 \end{cases}$, what is $f(-3) + f(2)$?
A) $5$
B) $7$
C) $9$
D) $11$
E) $13$
Answer & Solution
Answer: D
Since $-3 < 0$: $f(-3) = |-3| = 3$. Since $2 \geq 0$: $f(2) = 2^2 = 4$. So $f(-3) + f(2) = 3 + 4 = 7$.
12.
Tags: Advanced Absolute Value · Hard · source: Original (AMC-style)
How many solutions does $|x - 1| + |x - 2| + |x - 3| = 2$ have?
A) $0$
B) $1$
C) $2$
D) $3$
E) Infinitely many
Answer & Solution
Answer: A
The minimum value of $|x-1| + |x-2| + |x-3|$ occurs when $x = 2$ (the median), giving $|2-1| + |2-2| + |2-3| = 1 + 0 + 1 = 2$. So there is exactly one solution: $x = 2$.
Answer Key
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ans | C | C | C | A | C | C | A | A | E | A | D | A |