JEE Main Maths: Theory of Equations / Inequalities questions with solutions

Q1. Solve the inequality over the real numbers: ((2x - 1)(x - 1)² (x - 2)³) / (x - 4)⁴ > 0.

1. x in (-infinity, 1/2) U (2, infinity), excluding x = 1 and x = 4
2. x in (1/2, 2)
3. x in (1/2, 4) only
4. x in (-infinity, 1/2) U (1, 2)

Answer: x in (-infinity, 1/2) U (2, infinity), excluding x = 1 and x = 4

Factors raised to even powers never flip the sign, but create excluded points (zeros or undefined). Only the odd-power factors determine where the whole expression is positive.

Q2. Solve for real x: |x| - 2|x + 1| + 3|x + 2| = 0.

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

Answer: x = -2 only

An absolute-value equation is solved piecewise across the points where each modulus changes sign; a candidate root counts only if it falls within the interval used to derive it.

Q3. Solve the inequality over the real numbers: [ (x - 2)² * (1 - x) * (x - 3)³ * (x - 4)² ] / (x + 1) <= 0.

1. x in (-1, 1] union {2} union [3, infinity)
2. x in (-infinity, -1) union [1, 3]
3. x in (-1, 1] union [3, 4]
4. x in [1, 3] union {4}

Answer: x in (-1, 1] union {2} union [3, infinity)

Even-power factors only create zeros (which satisfy <= 0) and never change sign. The odd factors (x-3)³ and (1-x) plus the linear denominator (x+1) determine the sign intervals. x = -1 is excluded (denominator zero). x = 2 is an isolated solution (a squared factor vanishes there).