JEE Advanced Maths: Inequalities questions with solutions

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

1. x in (-inf, -2) U (-2, -1) U (1, inf)
2. x in (-1, 1)
3. x in (-2, 1)
4. x in (-inf, -1) U (1, inf), x != -2

Answer: x in (-inf, -2) U (-2, -1) U (1, inf)

The factor (x+2)² is non-negative and equals zero only at x = -2 (excluded since we need strict <0, but at x=-2 the expression is 0, not <0, so x=-2 is excluded). Rewriting: ((x-1)(x+2)²)/(-(x+1)) < 0 is equivalent to ((x-1)(x+2)²)/(x+1) > 0. Sign analysis of (x-1)/(x+1) (with (x+2)² > 0 for x != -2) gives the solution x < -1 or x > 1, excluding x = -2 where the expression is zero. So x in (-inf,-2) U (-2,-1) U (1, inf).

Q2. Solve the inequality over the real numbers: ((x - 1)² * (x + 1)³) / (x⁴ * (x - 2)) <= 0.

1. x in [-1, 0) U (0, 1] U (1, 2), i.e. [-1, 2) excluding 0 (with x = 1 allowed); equivalently -1 <= x < 2, x != 0
2. x in (-infinity, -1] U (2, infinity)
3. x in (-1, 2)
4. x in [-1, 2]

Answer: x in [-1, 0) U (0, 1] U (1, 2), i.e. [-1, 2) excluding 0 (with x = 1 allowed); equivalently -1 <= x < 2, x != 0

Domain excludes x = 0 and x = 2 (denominator zero). The factor (x - 1)² >= 0 and x⁴ > 0 (for x != 0), so the sign of the whole expression matches the sign of (x + 1)³/(x - 2), i.e. of (x + 1)/(x - 2). We need (x+1)/(x-2) <= 0, which holds for -1 <= x < 2. Also the expression equals 0 at x = 1 (from (x-1)²) and at x = -1, both allowed. Removing x = 0 (not in domain), the solution is -1 <= x < 2 with x != 0.

Q3. Solve the inequality over the reals: ((2 - x²)*(x - 3)³) / ((x + 1)*(x² - 3x - 4)) >= 0.

1. [-sqrt(2), -1) U (-1, sqrt(2)] U [3, 4)
2. [-sqrt(2), sqrt(2)] U [3, 4]
3. (-1, sqrt(2)] U [3, 4)
4. [-sqrt(2), -1) U (3, 4)

Answer: [-sqrt(2), -1) U (-1, sqrt(2)] U [3, 4)

Denominator = (x+1)(x-4)(x+1) = (x+1)²*(x-4). Since (x+1)² > 0 for x != -1, the inequality reduces to (2 - x²)*(x - 3)³/(x - 4) >= 0 with x != -1, 4. Critical points: x = -sqrt(2), sqrt(2), 3, 4. Sign analysis shows the expression is >= 0 on [-sqrt(2), sqrt(2)] and [3, 4), with x = -1 excluded (it lies in (-sqrt(2), sqrt(2))) and x = 4 excluded. Zeros at +-sqrt(2) and 3 are included.

Q4. Solve the inequality over the real numbers: x/(x + 1) > 2. Which option is the solution set?

1. (-2, -1)
2. (-1, infinity)
3. (-infinity, -2)
4. (-2, -1]

Answer: (-2, -1)

Subtract 2 and form one fraction, then do a sign analysis, excluding x = -1 where the expression is undefined.

Q5. For f(x) = (x³ + 4x² - 11x - 30)/(x² - 6x - 7), determine the set of x for which f(x) > 0.

1. x in (-5, -2) U (-1, 3) U (7, inf)
2. x in (-inf, -5) U (-2, -1) U (3, 7)
3. x in (-5, -1) U (3, 7)
4. x in (-inf, -5) U (-1, 7)

Answer: x in (-5, -2) U (-1, 3) U (7, inf)

Numerator x³ + 4x² - 11x - 30 factors as (x+5)(x+2)(x-3) (roots -5, -2, 3). Denominator x² - 6x - 7 = (x-7)(x+1) (roots 7, -1). Critical points in order: -5, -2, -1, 3, 7. A sign chart of (x+5)(x+2)(x-3)/((x-7)(x+1)) gives f(x) > 0 on (-5, -2) U (-1, 3) U (7, inf).

Q6. Solve for real x: 1/(x + 1) + 2/(x + 2) > 3/(x + 3).

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

Answer: x in (-infinity, -3) U (-2, -3/2) U (-1, infinity)

Move the right side over, combine to a single fraction, find its numerator roots and the excluded points, then use a sign chart across the critical values.

Q7. Solve over the real numbers: (x - 1)² (x + 1)³ (x - 4) < 0. Give the solution set.

1. (-1, 1) U (1, 4)
2. (-infinity, -1) U (4, infinity)
3. (-1, 4)
4. (1, 4) only

Answer: (-1, 1) U (1, 4)

Zeros: x = -1 (odd mult 3), x = 1 (even mult 2), x = 4 (odd mult 1). (x-1)² >= 0, zero only at x = 1, so x = 1 is excluded. Sign of (x+1)³ (x-4) is negative for -1 < x < 4. Multiplied by the positive even factor, the product is < 0 on (-1, 4) except at x = 1 where it is 0. Endpoints -1, 4 give 0 (excluded for strict <). Solution = (-1, 1) U (1, 4).