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Basics & Logarithms: Number Negative Integer Solutions Equation

JEE Maths question with a full step-by-step solution.

Question
Number of negative integer solutions of the equation ∣x+1∣−∣x∣+3∣x−1∣−2∣x−2∣=x+2|x+1| - |x| + 3|x-1| - 2|x-2| = x+2 is
Solution
Answer: 1
Case 1: x≥2x \geq 2: (x+1)−x+3(x−1)−2(x−2)=x+2  ⟹  x+2=x+2(x+1) - x + 3(x-1) - 2(x-2) = x+2 \implies x+2 = x+2 True for all x≥2x \geq 2. No negative integers here. Case 2: 1≤x<21 \leq x < 2: (x+1)−x+3(x−1)−2(2−x)=x+2  ⟹  5x−6=x+2  ⟹  x=2(x+1) - x + 3(x-1) - 2(2-x) = x+2 \implies 5x-6 = x+2 \implies x = 2 Not in [1,2)[1,2). No solution. Case 3: 0≤x<10 \leq x < 1: (x+1)−x+3(1−x)−2(2−x)=x+2  ⟹  −x=x+2  ⟹  x=−1(x+1) - x + 3(1-x) - 2(2-x) = x+2 \implies -x = x+2 \implies x = -1 Not in [0,1)[0,1). No solution. Case 4: −1≤x<0-1 \leq x < 0: (x+1)+x+3(1−x)−2(2−x)=x+2  ⟹  x+0=x+2  ⟹  0=2(x+1) + x + 3(1-x) - 2(2-x) = x+2 \implies x+0 = x+2 \implies 0 = 2 Contradiction. No solution. Case 5: x<−1x < -1: −(x+1)+x+3(1−x)−2(2−x)=x+2  ⟹  −x−2=x+2  ⟹  x=−2-(x+1) + x + 3(1-x) - 2(2-x) = x+2 \implies -x-2 = x+2 \implies x = -2 x=−2<−1x = -2 < -1. Valid. Verify: ∣−1∣−∣−2∣+3∣−3∣−2∣−4∣=1−2+9−8=0=−2+2|-1|-|-2|+3|-3|-2|-4| = 1-2+9-8 = 0 = -2+2. Confirmed. Answer: 1
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