The number of distinct real values of x, satisfying the equation max{x,2}−min{x,2}=|x+2|−|x−2|, is

Question

UPSC Quizroom · Accepted Answer

We need to solve:

max{x, 2} − min{x, 2} = |x + 2| − |x − 2|

Step 1: Simplify left-hand side

max{x, 2} − min{x, 2} = |x − 2|

Equation becomes:

|x − 2| = |x + 2| − |x − 2|

Add |x − 2| to both sides:

2|x − 2| = |x + 2| → |x − 2| = (1/2) |x + 2|

Step 2: Solve the absolute value equation

Case 1: x ≥ 2

|x − 2| = x − 2

|x + 2| = x + 2

Equation: x − 2 = (1/2)(x + 2) → x = 6 ✅ valid

Case 2: x < 2

Subcase 2a: x ≥ −2 → |x + 2| = x + 2, |x − 2| = 2 − x

Equation: 2 − x = (1/2)(x + 2) → x = 2/3 ✅ valid

Subcase 2b: x < −2 → |x + 2| = −(x + 2), |x − 2| = 2 − x

Equation: 2 − x = (1/2)(−x − 2) → x = 6 ❌ not valid

Step 3: Collect valid solutions

x = 6, x = 2/3

Number of distinct real values: 2

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