Solutions of inequalities (graphical/analytical) practice questions.

Determine the solution set of x² − 5x + 6 < 0.

• x < 2 or x > 3
• 2 < x < 3
• 2 ≤ x ≤ 3
• x < −3 or x > −2

Answer: 2 < x < 3

Factorise: (x − 2)(x − 3) < 0. The upward parabola is negative strictly between its roots, so 2 < x < 3; the endpoints are excluded because the inequality is strict.

Determine the solution set of (x + 3)/(x − 1) < 0.

• −3 < x < 1
• x < −3 or x > 1
• −3 ≤ x ≤ 1
• x < 1

Answer: −3 < x < 1

A quotient is negative exactly when numerator and denominator have opposite signs, which happens between the critical values −3 and 1. Both endpoints are excluded: −3 gives 0 (not < 0) and 1 makes the expression undefined.

Solve the inequality x/(x − 2) ≤ 3.

• 2 < x ≤ 3
• x ≤ 3, x ≠ 2
• x < 2 or x ≥ 3
• x ≤ 2 or x ≥ 3

Answer: x < 2 or x ≥ 3

Bring 3 to the left: (6 − 2x)/(x − 2) ≤ 0, with critical values x = 3 (zero) and x = 2 (undefined). A sign table gives x < 2 or x ≥ 3; x = 2 must be excluded, x = 3 included.

Determine how many integers satisfy x² ≤ 4x.

• 3
• 5
• 4
• 6

Answer: 5

Rearrange to x(x − 4) ≤ 0, giving 0 ≤ x ≤ 4. The integers 0, 1, 2, 3, 4 satisfy it — five in total; dividing by x loses the solution x = 0 and is invalid for x < 0.

Determine all values of k for which x² + kx + 9 > 0 for every real x.

• k < −6 or k > 6
• −6 ≤ k ≤ 6
• k < 6
• −6 < k < 6

Answer: −6 < k < 6

An upward parabola is positive for all x exactly when it has no real roots, so the discriminant must satisfy k² − 36 < 0, giving −6 < k < 6. At k = ±6 the parabola touches the x-axis, where the value 0 is not > 0.

Solve 2/(x + 1) > 1, where x ≠ −1.

• x < 1
• −1 < x < 1
• x > 1
• x < −1 or x > 1

Answer: −1 < x < 1

Rearranging gives (1 − x)/(x + 1) > 0, which requires numerator and denominator to have the same sign, so −1 < x < 1. Multiplying directly by (x + 1) is invalid because its sign is unknown.