OxfordAQA International AS and A-level Mathematics: Binomial series for rational powers and series of rational functions (9660) – 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 specification section P2.2 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: the binomial series for any rational power and series of rational functions. The specification places this content in International A-level (A2) Unit P2 (Pure Maths); it is not examined at International AS. Read the study guide first, or use the revision notes to refresh the methods. The course hub is OxfordAQA A-level Mathematics, the printable checklist tracks each outcome, and the free 10-minute diagnostics show where to focus.

Questions

1. Expand (1 − x)^(−1/3) in ascending powers of x up to and including the term in x³. [3] State the values of x for which the expansion is valid. [1]

2. Expand (4 + x)^(−1/2) in ascending powers of x up to and including the term in x³, and state the range of x for which the expansion is valid. [5]

3. (a) Expand (27 + 2x)^(1/3) in ascending powers of x up to and including the term in x². [3] (b) State the range of x for which the expansion is valid. [1] (c) Use x = 0.2 in your expansion to estimate the cube root of 27.4, giving your answer to 6 significant figures. [2]

4. Find the coefficient of x³ in the expansion of (2 − x)(1 + 3x)⁻². [4]

5. In the expansion of (1 + ax)ⁿ, the first three terms are 1 − 6x + 54x² + … Find the values of n and a [4], and state the range of x for which the expansion is valid. [1]

6. A student writes (3 + 2x)⁻¹ = 1 − 2x + 4x² − … and says this is valid for |x| < 1/2. Explain the error, give the correct expansion up to the term in x², and state the correct range of validity. [4]

7. Let f(x) = (5 − 7x) / ((1 + x)(1 − 3x)). (a) Express f(x) in partial fractions. [3] (b) Hence expand f(x) in ascending powers of x up to and including the term in x³. [4] (c) State the range of x for which your expansion is valid. [1]

8. Let g(x) = (6 − 6x − 3x²) / ((1 − 2x)(1 + x)²). (a) Find constants A, B and C such that g(x) = A/(1 − 2x) + B/(1 + x) + C/(1 + x)². [4] (b) Hence find the expansion of g(x) up to and including the term in x³. [4] (c) State the range of x for which the expansion is valid. [1]

9. Show that, for small x, (1 + x)^(1/2) (1 − x)^(−1/2) = 1 + x + x²/2 + x³/2 + … [6]

10. Use the first three terms of the expansion of (1 − x)^(−1/2), with x = 0.04, to estimate 1/√0.96. [3] The value of 1/√0.96 is 1.0206207 to 7 decimal places. Find the absolute error of your estimate, to 2 significant figures. [1]

11. Let h(x) = (21 − 5x + 4x²) / ((1 + 2x)(3 − x)²). (a) Find constants A, B and C such that h(x) = A/(1 + 2x) + B/(3 − x) + C/(3 − x)². [4] (b) Hence expand h(x) in ascending powers of x up to and including the term in x². [5] (c) Explain why the expansion is valid for |x| < 1/2 and not for a larger range. [2]

12. Expand (1 − x)/(4 + 3x)² in ascending powers of x up to and including the term in x³, and state the range of x for which the expansion is valid. [6]

Answers

1. With u = −x and n = −1/3: (1 + u)^(−1/3) = 1 − u/3 + (−1/3)(−4/3)/2 u² + (−1/3)(−4/3)(−7/3)/6 u³ [1]. Substituting gives 1 + x/3 + 2x²/9 + 14x³/81 + … [1] with the x³ term correct [1]. The expansion needs |−x| < 1, so |x| < 1 [1]. [4] Examiner insight: Write the generic expansion with n and u substituted before simplifying; a sign slip in the x³ coefficient is easier to trace that way.

2. (4 + x)^(−1/2) = 4^(−1/2)(1 + x/4)^(−1/2) = (1/2)(1 + x/4)^(−1/2) [1]. (1 + u)^(−1/2) = 1 − u/2 + 3u²/8 − 5u³/16 [1], and with u = x/4 this is 1 − x/8 + 3x²/128 − 5x³/1024 [1]. Multiply by 1/2: 1/2 − x/16 + 3x²/256 − 5x³/2048 + … [1]. Validity: |x/4| < 1, so |x| < 4 [1]. [5] Examiner insight: Show the factor 4^(−1/2) = 1/2 as its own line; omitting it is an easy way to lose an otherwise correct answer.

3. (a) (27 + 2x)^(1/3) = 3(1 + 2x/27)^(1/3) [1]. (1 + u)^(1/3) = 1 + u/3 − u²/9 + …, and with u = 2x/27 this is 1 + 2x/81 − 4x²/6561 [1]. Multiply by 3: 3 + 2x/27 − 4x²/2187 + … [1]. [3] (b) |2x/27| < 1, so |x| < 27/2 [1]. [1] (c) x = 0.2 gives 27 + 0.4 = 27.4, which is within the range [1]. Then 3 + 0.0148148 − 0.0000732 = 3.0147416, so the estimate is 3.01474 [1]. [2] Examiner insight: In an estimate question, say which x you substitute and why it makes the expression the required number. Check that it lies inside the validity range.

4. (1 + 3x)⁻² = 1 − 2(3x) + 3(3x)² − 4(3x)³ [1] = 1 − 6x + 27x² − 108x³ [1]. Coefficient of x³ in the product: 2(−108) + (−1)(27) [1] = −243 [1]. [4] Examiner insight: List the pairs of terms that give x³ before multiplying out.

5. The x coefficient gives na = −6 [1]. The x² coefficient gives n(n − 1)a²/2 = 54 [1]. Substitute a = −6/n: 18(n − 1)/n = 54, so n − 1 = 3n and n = −1/2 [1]. Then a = 12 [1]. Validity: |12x| < 1, so |x| < 1/12 [1]. [5] Examiner insight: Eliminate one unknown first, rather than guessing values. The validity range depends on the value of a you found, so a wrong a also costs you the final answer.

6. The student has expanded (1 + 2x)⁻¹, not (3 + 2x)⁻¹: the 3 was never taken out [1]. (3 + 2x)⁻¹ = (1/3)(1 + 2x/3)⁻¹ = (1/3)(1 − 2x/3 + 4x²/9) [1], giving 1/3 − 2x/9 + 4x²/27 + … [1]. Validity: |2x/3| < 1, so |x| < 3/2 [1]. [4] Examiner insight: Name the specific slip, not just that the answer is wrong.

7. (a) 5 − 7x = A(1 − 3x) + B(1 + x) [1]. Put x = −1: 12 = 4A, so A = 3 [1]. Put x = 1/3: 8/3 = (4/3)B, so B = 2 [1]. f(x) = 3/(1 + x) + 2/(1 − 3x). [3] (b) 3/(1 + x) = 3 − 3x + 3x² − 3x³ [1]. 2/(1 − 3x) = 2 + 6x + 18x² + 54x³ [1]. Add the like terms [1]: 5 + 3x + 21x² + 51x³ + … [1]. [4] (c) The parts need |x| < 1 and |3x| < 1. Both hold when |x| < 1/3 [1]. [1] Examiner insight: Check the partial fractions by putting x = 0: 3 + 2 = 5, which matches f(0). This catches a wrong constant before you expand.

8. (a) 6 − 6x − 3x² = A(1 + x)² + B(1 − 2x)(1 + x) + C(1 − 2x) [1]. Put x = 1/2: 9/4 = A(9/4), so A = 1 [1]. Put x = −1: 9 = 3C, so C = 3 [1]. Compare x²: −3 = A − 2B, so B = 2 [1]. [4] (b) 1/(1 − 2x) = 1 + 2x + 4x² + 8x³ [1]. 2/(1 + x) = 2 − 2x + 2x² − 2x³ [1]. 3/(1 + x)² = 3 − 6x + 9x² − 12x³ [1]. Add: 6 − 6x + 15x² − 6x³ + … [1]. [4] (c) The parts need |2x| < 1 and |x| < 1. Both hold when |x| < 1/2 [1]. [1] Examiner insight: A squared factor needs both a single and a squared term in the partial-fraction form; writing only one cannot fit the numerator.

9. (1 + x)^(1/2) = 1 + x/2 + (1/2)(−1/2)/2 x² + (1/2)(−1/2)(−3/2)/6 x³ [1] = 1 + x/2 − x²/8 + x³/16 [1]. (1 − x)^(−1/2) = 1 + x/2 + 3x²/8 + 5x³/16 [1] with the signs correct [1]. Multiply the series and collect terms: x: 1/2 + 1/2 = 1 [1] x²: 3/8 + 1/4 − 1/8 = 1/2 x³: 5/16 + 3/16 − 1/16 + 1/16 = 1/2 The product is 1 + x + x²/2 + x³/2 + … as required [1]. [6] Examiner insight: On a “show that” question, give each coefficient calculation. A stated result with no products shown gives the reader nothing to credit.

10. (1 − x)^(−1/2) = 1 + x/2 + 3x²/8 + … [1]. With x = 0.04, 1/√0.96 = (1 − 0.04)^(−1/2) [1], so the estimate is 1 + 0.02 + 3(0.0016)/8 = 1.0206 [1]. Absolute error = 1.0206207 − 1.0206 = 0.0000207, which is 2.1 × 10⁻⁵ to 2 s.f. [1]. [4] Examiner insight: State that 1 − x = 0.96, so the choice of x is visible.

11. (a) 21 − 5x + 4x² = A(3 − x)² + B(1 + 2x)(3 − x) + C(1 + 2x) [1]. Put x = −1/2: 24.5 = A(3.5)², so A = 2 [1]. Put x = 3: 42 = 7C, so C = 6 [1]. Compare x²: 4 = A − 2B, so B = −1 [1]. [4] (b) 2/(1 + 2x) = 2 − 4x + 8x² [1]. −1/(3 − x) = −(1/3)(1 − x/3)⁻¹ = −1/3 − x/9 − x²/27 [1]. 6/(3 − x)² = (2/3)(1 − x/3)⁻² = 2/3 + 4x/9 + 2x²/9 [1], using 1 + 2u + 3u² with u = x/3 [1]. Add: constant 2 − 1/3 + 2/3 = 7/3; x: −4 − 1/9 + 4/9 = −11/3; x²: 8 − 1/27 + 6/27 = 221/27. h(x) = 7/3 − 11x/3 + 221x²/27 + … [1] [5] (c) The first part needs |2x| < 1, so |x| < 1/2 [1]. The others need |x| < 3. The series is valid only where all parts are valid, so the narrower condition |x| < 1/2 applies [1]. [2] Examiner insight: For a “why” question, state both conditions and say which is stricter.

12. (4 + 3x)⁻² = (1/16)(1 + 3x/4)⁻² [1]. With u = 3x/4: 1 − 2u + 3u² − 4u³ = 1 − 3x/2 + 27x²/16 − 27x³/16 [1] and [1] for the correct terms. Multiply by (1 − x)/16 and collect terms [1]: 1/16 − 5x/32 + 51x²/256 − 27x³/128 + … [1] Validity: |3x/4| < 1, so |x| < 4/3 [1]. [6] Examiner insight: The factor (1 − x) is a polynomial and gives no restriction, so the range comes only from the binomial part.

Where marks are usually lost

• Writing the series for (a + bx)ⁿ without taking a out first, or taking it out and forgetting aⁿ.
• Losing a negative sign when u is negative, especially in the u² and u³ terms.
• Stating |x| < 1 for every expansion instead of solving |u| < 1.
• Using one fraction for a repeated factor in partial fractions.
• Not checking the constants, for example by putting x = 0 into both sides.
• Estimating with an x outside the range of validity.
• Giving the range for one partial fraction instead of the narrowest range.
• Stopping at the wrong power, or missing cross-terms in a product.

Next steps

• Re-read the revision notes.
• Return to the study guide.
• Go to the course hub and the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section P2.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA.