Solve \(|x| + |x - 2| = 2\)

Step 1 (Geometric Insight): By the triangle inequality, |x| + |x - 2| ≥ |(x) - (x-2)| = |2| = 2. Equality holds when x and (x-2) have the same sign relative to 0, meaning x lies between 0 and 2. Step 2 (Verification by Cases): • If x 2 ✗ • If 0 ≤ x ≤ 2: |x| + |x - 2| = x + (2-x) = 2 ✓ • If x > 2: |x| + |x - 2| = x + (x-2) = 2x-2 > 2 ✗ Step 3: The solution set is the entire interval [0, 2]. ∴ Answer: A (x ∈ [0, 2])

Basic Mathematics & Logarithm

Modulus Equations

Grade 11

Question:

Solve \(|x| + |x - 2| = 2\)

\(0 \leq x \leq 2\)

\(x = 0\) or \(x = 2\)

\(x \in (-\infty, 0] \cup [2, \infty)\)

\(x \in (0, 2)\)

Step-by-Step Solution

Key Concept: The equation |x| + |x - 2| = 2 represents the sum of distances from x to points 0 and 2 on a number line. Since these points are distance 2 apart, the sum equals 2 only when x lies on the line segment connecting them.

Step 1 (Geometric Insight): By the triangle inequality, |x| + |x - 2| ≥ |(x) - (x-2)| = |2| = 2. Equality holds when x and (x-2) have the same sign relative to 0, meaning x lies between 0 and 2.Step 2 (Verification by Cases):• If x < 0: |x| + |x - 2| = -x + (2-x) = 2-2x > 2 ✗• If 0 ≤ x ≤ 2: |x| + |x - 2| = x + (2-x) = 2 ✓• If x > 2: |x| + |x - 2| = x + (x-2) = 2x-2 > 2 ✗Step 3: The solution set is the entire interval [0, 2].∴ Answer: A (x ∈ [0, 2])

Correct Answer: A

Question:

Step-by-Step Solution

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