JEE Main Maths: Algebra (Inequalities) questions with solutions

Q1. Solve over the real numbers: (x² + 4x + 4)/(2x² - x - 1) < 0.

1. (-1/2, 1), x != -2
2. (-infinity, -2) U (-2, -1/2)
3. (-1/2, 1)
4. (-2, 1)

Answer: (-1/2, 1), x != -2

Numerator = (x+2)² >= 0, zero only at x = -2. Denominator = 2x² - x - 1 = (2x+1)(x-1), roots at x = -1/2 and x = 1. The fraction is < 0 only when numerator > 0 and denominator < 0. Numerator > 0 for x != -2. Denominator < 0 between its roots: -1/2 < x < 1. Exclude x = -2 (it lies outside this interval anyway, but in general the numerator zero must be excluded). So solution = (-1/2, 1) with x != -2; since -2 is not in (-1/2,1) the set is effectively (-1/2, 1). The form listed as the answer keeps the x != -2 condition explicit.

Q2. Solve the inequality x⁴ - 5x² + 4 < 0 over the real numbers.

1. (-2, -1) union (1, 2)
2. (-1, 1)
3. (-2, 2)
4. (-inf, -2) union (2, inf)

Answer: (-2, -1) union (1, 2)

Factor as (x² - 1)(x² - 4) = (x-1)(x+1)(x-2)(x+2). The product is negative between consecutive roots. With roots at -2, -1, 1, 2, the expression is negative on (-2, -1) and (1, 2).

Q3. Find all real x satisfying the double inequality 7 - 3x < x - 1 <= 2x + 9.

1. x > 2
2. -10 <= x < 2
3. x in (2, infinity) and x >= -10, i.e. x > 2
4. x <= -10

Answer: x > 2

Solve each part separately. The left inequality yields x > 2; the right yields x >= -10. The intersection of x > 2 and x >= -10 is x > 2.

Q4. Solve over the real numbers: (3x - 1)/(4x + 1) <= 0.

1. x in (-1/4, 1/3]
2. x in [-1/4, 1/3]
3. x in (-1/3, 1/4]
4. x in (-infinity, -1/4) U [1/3, infinity)

Answer: x in (-1/4, 1/3]

The expression is non-positive between the root of the numerator and the value making the denominator zero. Numerator zero at x = 1/3 is allowed (gives 0). Denominator zero at x = -1/4 is excluded. Sign analysis shows the fraction is negative on (-1/4, 1/3), giving the solution (-1/4, 1/3].