Pearson Edexcel International Advanced Level Mathematics: Proof, partial fractions and binomial series (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover Unit P4 sections 1.1 (proof by contradiction), 2.1 (partial fractions) and 4.1 (the binomial series for any rational n) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. P4 is listed as compulsory for IAL Mathematics and Pure Mathematics and is not part of IAS. Learn the methods first from the study guide and the revision notes. See also the course hub, the printable checklist and the free diagnostics.

Questions

1. Prove by contradiction that √7 is irrational. [5]

2. Prove by contradiction that there are no positive integers a and b such that a² − b² = 14. [5]

3. A student says that the proof of the infinity of primes shows that 2 × 3 × 5 × 7 × 11 × 13 × 17 + 1 is prime.

(a) Show that the student is wrong. [1] (b) Explain why this does not damage the proof. [3]

4. Express (13x² + 43x + 30)/((x − 2)(x + 4)(3x + 1)) in partial fractions. [4]

5. Express (9x² − 30x − 5)/((2x + 1)(x − 3)²) in the form A/(2x + 1) + B/(x − 3) + C/(x − 3)², finding A, B and C. [4]

6. Express (6x² − 26)/((x + 1)(2x − 3)) in the form A + B/(x + 1) + C/(2x − 3). [5]

7. Expand (1 + 4x)⁻³ in ascending powers of x up to and including x³, and state the values of x for which the expansion is valid. [4]

8. g(x) = (4 + 3x)^(−1/2).

(a) Find the first four terms of the binomial expansion of g(x) in ascending powers of x. [4] (b) State the range of x for which the expansion is valid. [1] (c) Use x = 0.04 to estimate 1/√4.12, giving your answer to 5 significant figures. [2]

9. f(x) = (5x + 4)/((1 − x)(1 + 2x)).

(a) Find constants A and B such that f(x) = A/(1 − x) + B/(1 + 2x). [3] (b) Hence expand f(x) in ascending powers of x up to and including x³. [3] (c) State the range of x for which your expansion is valid. [1]

10. f(x) = (6x² + x + 7)/((1 + x)(1 − 3x)).

(a) Find A, B and C such that f(x) = A + B/(1 + x) + C/(1 − 3x). [5] (b) Hence expand f(x) in ascending powers of x up to and including x³. [3] (c) State the range of x for which the expansion is valid. [1]

11. The expansion of (2 + kx)⁻³ in ascending powers of x begins 1/8 + px + qx² + …, where k and p are constants and |kx| < 2.

(a) Find p and q in terms of k. [3] (b) Given that p = −3/4, find the value of k. [2] (c) Find the coefficient of x² in the expansion of (1 + x)(2 + kx)⁻³. [2] (d) State the range of x for which this expansion is valid. [1]

Answers

1. Assume √7 is rational, so √7 = p/q with p and q positive integers with no common factor. [1] Then 7 = p²/q², so p² = 7q². [1] So 7 divides p². Since 7 is prime, 7 divides p. Write p = 7r. [1] Then 49r² = 7q², so q² = 7r², and in the same way 7 divides q. [1] Now 7 divides both p and q, which contradicts “no common factor”. So √7 is irrational. [1] [5] Examiner insight: Examiners look for the reason a prime divides p from p², not only the statement. Closing with the contradiction in words is part of a complete proof.

2. Assume positive integers a and b exist with a² − b² = 14. Then (a − b)(a + b) = 14. [1] (a + b) − (a − b) = 2b is even, so the two factors are both odd or both even. [1] If both are odd, their product is odd, but 14 is even. [1] If both are even, their product is a multiple of 4, but 14 is not. [1] Both cases are impossible, so no such integers exist. [1] [5] Examiner insight: A proof by cases must cover every case. If you say “the factors are 2 and 7” and stop, you have not dealt with the other factor pairs of 14.

3. (a) 2 × 3 × 5 × 7 × 11 × 13 × 17 + 1 = 510511 = 19 × 97 × 277, so it is not prime. [1] (b) The proof only claims that N has a prime factor. [1] Here the prime factors 19, 97 and 277 are not in the list 2, 3, 5, 7, 11, 13, 17, so that list was incomplete. [1] N does not have to be prime, so the argument still gives a prime outside any finite list. [1] [4] Examiner insight: A clear answer states what the proof really needs (a prime factor of N), not only that 510511 happens to be composite.

4. Write (13x² + 43x + 30)/((x − 2)(x + 4)(3x + 1)) = A/(x − 2) + B/(x + 4) + C/(3x + 1). [1] Then 13x² + 43x + 30 = A(x + 4)(3x + 1) + B(x − 2)(3x + 1) + C(x − 2)(x + 4). x = 2: 168 = 42A, so A = 4. [1] x = −4: 66 = 66B, so B = 1. [1] x = −1/3: 154/9 = −77C/9, so C = −2. [1] Result: 4/(x − 2) + 1/(x + 4) − 2/(3x + 1). [4] Examiner insight: Substituting the root of each factor is quick. A final check with an easy value such as x = 0 catches slips before you move on.

5. 9x² − 30x − 5 = A(x − 3)² + B(2x + 1)(x − 3) + C(2x + 1). [1] x = −1/2: 49/4 = 49A/4, so A = 1. [1] x = 3: −14 = 7C, so C = −2. [1] Coefficient of x²: 9 = A + 2B, so B = 4. [1] Result: 1/(2x + 1) + 4/(x − 3) − 2/(x − 3)². [4] Examiner insight: Show the substitution or coefficient comparison that gives B. A final answer with no supporting working leaves little to credit if one constant is wrong.

6. Divide: 2x² − x − 3 is the denominator, and 6x² − 26 = 3(2x² − x − 3) + (3x − 17). [1] So A = 3, and the remainder is (3x − 17)/((x + 1)(2x − 3)). [1] Write (3x − 17)/((x + 1)(2x − 3)) = B/(x + 1) + C/(2x − 3), so 3x − 17 = B(2x − 3) + C(x + 1). [1] x = −1: −20 = −5B, so B = 4. [1] x = 3/2: −25/2 = 5C/2, so C = −5. [1] Result: 3 + 4/(x + 1) − 5/(2x − 3). [5] Examiner insight: If you write the fraction as proper partial fractions only, with no constant, the working will not match the given form. Compare degrees before you start.

7. (1 + 4x)⁻³ = 1 + (−3)(4x) + (−3)(−4)/2² + (−3)(−4)(−5)/6³ [1] = 1 − 12x + 96x² [1] − 640x³ [1] Valid for |4x| < 1, so |x| < 1/4. [1] [4] Examiner insight: Put brackets round 4x before you square or cube it. Missing brackets are the usual reason the x² and x³ terms come out wrong.

8. (a) (4 + 3x)^(−1/2) = 4^(−1/2)(1 + 3x/4)^(−1/2) = (1/2)(1 + 3x/4)^(−1/2). [1] (1 + u)^(−1/2) = 1 − u/2 + 3u²/8 − 5u³/16 with u = 3x/4. [1] = 1 − 3x/8 + 27x²/128 − 135x³/1024. [1] Multiply by 1/2: 1/2 − 3x/16 + 27x²/256 − 135x³/2048. [1] (b) |3x/4| < 1, so |x| < 4/3. [1] (c) 4 + 3(0.04) = 4.12, so x = 0.04 is valid and g(0.04) = 1/√4.12. [1] Series: 0.5 − 0.0075 + 0.00016875 − 0.0000042 = 0.49266 to 5 significant figures. [1] [7] Examiner insight: Take out the constant 4^(−1/2) = 1/2 first. Multiplying it in at the end is easy to forget, so write it on the first line.

9. (a) 5x + 4 = A(1 + 2x) + B(1 − x). [1] x = 1: 9 = 3A, so A = 3. [1] x = −1/2: 3/2 = 3B/2, so B = 1. [1] (b) 3(1 − x)⁻¹ = 3(1 + x + x² + x³) and (1 + 2x)⁻¹ = 1 − 2x + 4x² − 8x³. [1] [1] Sum: 4 + x + 7x² − 5x³. [1] (c) (1 − x)⁻¹ needs |x| < 1 and (1 + 2x)⁻¹ needs |x| < 1/2, so the expansion is valid for |x| < 1/2. [1] [7] Examiner insight: The final range should be the narrower one. Quoting |x| < 1 would suggest that the second expansion has been ignored.

10. (a) The numerator and denominator both have degree 2, since (1 + x)(1 − 3x) = 1 − 2x − 3x². [1] 6x² + x + 7 = A(1 − 2x − 3x²) + B(1 − 3x) + C(1 + x). Coefficient of x²: 6 = −3A, so A = −2. [1] x = −1: 12 = 4B, so B = 3. [1] x = 1/3: 8 = 4C/3, so C = 6. [1] So f(x) = −2 + 3/(1 + x) + 6/(1 − 3x). [1] (b) 3(1 − x + x² − x³) + 6(1 + 3x + 9x² + 27x³). [1] [1] f(x) = −2 + 3 + 6 + … = 7 + 15x + 57x² + 159x³. [1] (c) The ranges are |x| < 1 and |3x| < 1. The expansion is valid for |x| < 1/3. [1] [9] Examiner insight: Do not forget that the constant −2 adds to the constant terms. The sum 7 comes from −2 + 3 + 6, not from 3 + 6.

11. (a) (2 + kx)⁻³ = 2⁻³(1 + kx/2)⁻³ = (1/8)(1 + kx/2)⁻³. [1] (1 + u)⁻³ = 1 − 3u + 6u² − … with u = kx/2, which gives 1 − 3kx/2 + 3k²x²/2. [1] So p = −3k/16 and q = 3k²/16. [1] (b) −3k/16 = −3/4, so k = 4. [1] [1] (c) With k = 4: q = 3(16)/16 = 3. The expansion is 1/8 − (3/4)x + 3x² + … [1] Coefficient of x² in (1 + x)(…): 3 + (−3/4) = 9/4. [1] (d) |kx/2| < 1 gives |2x| < 1, so |x| < 1/2. [1] [8] Examiner insight: Pick up x² terms from both products, here 1 × 3x² and x × (−3/4)x. Listing the contributions before adding them shows the method clearly.

Where marks are usually lost

• Starting a contradiction proof by assuming the statement you want to prove.
• Leaving out “no common factor” or “lowest terms” in the √7 type of proof.
• Using a proof by cases without listing all the cases.
• Writing only one fraction for a repeated linear factor.
• Skipping the division when the numerator degree equals the denominator degree.
• Forgetting to take out the constant before using the binomial series.
• Not using brackets when substituting u, so signs and powers go wrong.
• Quoting |x| < 1 instead of the range for the actual bracket.
• Taking the wider range of two expansions instead of the narrower one.

Next steps

• Review the revision notes for this unit.
• Re-read the study guide for any section you found hard.
• Return to the course hub and tick off the printable checklist.
• Take the free 10-minute diagnostics to see what to revise next.
• Book a free trial class if you would like help from a teacher.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited. Unit P4: Pure Mathematics 4, sections 1.1, 2.1 and 4.1.