Free Algebra questions for A Level Mathematics (9709), part of Pure Mathematics 3. Each question has the answer and the marking points an examiner would look for.
Generate a full Algebra worksheet →All Pure Mathematics 3 questionsSolve the inequality |2x − 4| > |x − 4|.
| Part | Answer and marking points | Marks |
|---|---|---|
| x < 0 or x > 83 square both sides or consider non-modular equations M1 critical values 0 and 83 A1 correct method for the inequality (outer region) M1 x < 0, x > 83 A1 | 4 |
In this question m is a real constant.
Solve the inequality |2x − 3| < 1 − ½x.
Find the set of values of m for which the equation |2x − 3| = m|x| + 1 has exactly two distinct real roots.
Find the value of m for which the equation has two roots whose product is −1, and find these roots exactly.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 43 < x < 85 solves 3 − 2x = 1 − ½x or 2x − 3 = 1 − ½x M1 critical values 43 and 85 A1 43 < x < 85 A1 | 3 |
| (b) | −23 < m < 2 or m > 2 considers the three intervals x < 0, 0 ≤ x < 32, x ≥ 32 (or graphs of y = |2x − 3| and y = m|x| + 1) M1 x = 22 − m, valid (negative) only when m > 2 A1 x = 2m + 2, valid when m > −23 A1 x = 42 − m, valid when −23 ≤ m < 2 A1 m = 2 gives only x = ½; m = −23 gives only x = 32 M1 −23 < m < 2 or m > 2 A1 | 6 |
| (c) | m = 2√2; x = √2 − 1 and −(√2 + 1) product negative so m > 2: 2m + 2 × 22 − m = −1 M1 m2 = 8, m = 2√2 A1 x = √2 − 1 and −1 − √2 A1 | 3 |
Express −3x + 1(x + 1)(x2 + 1) in partial fractions.
| Part | Answer and marking points | Marks |
|---|---|---|
| 2x + 1 + −2x − 1x2 + 1 correct form A/(x + 1) + (Bx + C)/(x2 + 1) B1 equate numerators/substitute x = −1 M1 A = 2 A1 B = −2 A1 C = −1 A1 | 5 |
Let f(x) = −15x − 5(1 − 3x)(3 + x).
Express f(x) in partial fractions.
Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x2.
State the set of values of x for which the expansion is valid.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | −31 − 3x + 43 + x correct form B1 method to find constants M1 A = −3, B = 4 A1 | 3 |
| (b) | −53 − 859x − 72527x2 (1 − px)−1 expanded M1 (q + x)−1 written as 1q(1 + xq)−1 and expanded M1 one expansion correct A1 second expansion correct A1 −53 − 859x − 72527x2 A1 | 5 |
| (c) | |x| < 13 |x| < 13 B1 | 1 |
The polynomial p(x) is defined by p(x) = 2x3 + ax2 + bx − 2, where a and b are constants. It is given that (x − 2) is a factor of p(x) and that the remainder is 0 when p(x) is divided by (x + 1).
Find the values of a and b.
Hence factorise p(x) completely.
Hence solve p(3y) = 0, giving your answers correct to 3 significant figures where appropriate.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | a = −1, b = −5 p(2) = 0 substituted M1 p(−1) = 0 substituted M1 two correct linear equations A1 solve simultaneously M1 a = −1, b = −5 A1 | 5 |
| (b) | (x − 2)(2x + 1)(x + 1) divide by (x − 2) to obtain the quadratic factor M1 complete factorisation A1 | 2 |
| (c) | y = 0.631 (3y > 0 so negative root rejected) 3y = positive root(s) M1 logarithms used correctly M1 0.631, and negative root rejected A1 | 3 |
Solve the inequality |3x − 4| > |x − 4|.
| Part | Answer and marking points | Marks |
|---|---|---|
| x < 0 or x > 2 square both sides or consider non-modular equations M1 critical values 0 and 2 A1 correct method for the inequality (outer region) M1 x < 0, x > 2 A1 | 4 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in