JEE Advanced Maths: Modulus Equations questions with solutions

Q1. Solve the equation |3x + 5| + |4x + 7| = 12 for real x.

1. x = 0 and x = -24/7
2. x = 0 only
3. x = -24/7 only
4. no real solution

Answer: x = 0 and x = -24/7

Critical points: x = -5/3 (≈ -1.667) and x = -7/4 (= -1.75). Region x >= -5/3: (3x+5)+(4x+7) = 7x + 12 = 12 => x = 0 (valid). Region -7/4 <= x < -5/3: -(3x+5)+(4x+7) = x + 2 = 12 => x = 10 (reject). Region x < -7/4: -(3x+5)-(4x+7) = -7x - 12 = 12 => x = -24/7 ≈ -3.43 (valid). Solutions: x = 0 and x = -24/7.