JEE Advanced Maths: Modulus / Absolute Value Equations questions with solutions

Q1. Solve for x: |2x - 3| + |2x + 1| + |2x + 5| = 12.

1. x = -5/2 or x = 3/2
2. x = -3/2 or x = 3/2
3. x = -1/2 or x = 5/2
4. x = 0 or x = 1

Answer: x = -5/2 or x = 3/2

Critical points: 2x-3=0 => x=3/2; 2x+1=0 => x=-1/2; 2x+5=0 => x=-5/2. The minimum value of the sum (at the median point x=-1/2) is |−4|+0+|4| = 8 < 12, so two solutions lie in the outer regions. For x >= 3/2: (2x-3)+(2x+1)+(2x+5) = 6x+3 = 12 => x = 3/2. For x <= -5/2: -(2x-3)-(2x+1)-(2x+5) = -6x-3 = 12 => x = -5/2.

Q2. Solve for x: |x² - 4x + 3| = |x² - 5x + 4|.

1. x = 1 and x = 7/2
2. x = 1 only
3. x = 3 and x = 4
4. x = 7/2 only

Answer: x = 1 and x = 7/2

Factor: (x-1)(x-3) and (x-1)(x-4). |A|=|B| gives A=B or A=-B. Case A=B: (x-1)(x-3) = (x-1)(x-4) => (x-1)[(x-3)-(x-4)] = (x-1)(1) = 0 => x = 1. Case A=-B: (x-1)(x-3) = -(x-1)(x-4) => (x-1)[(x-3)+(x-4)] = (x-1)(2x-7) = 0 => x = 1 or x = 7/2. Combined solutions: x = 1 and x = 7/2.