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4PM1/01 Jun 2019, Question 10: Definite Integrals and Areas

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 14 marks

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The question, in short

Expand a binomial with a fractional negative index, find its validity range, build a series for a related rational function, determine k from a zero coefficient, then integrate to estimate a definite integral.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 3 marks

Write the binomial expansion of (1 + 2x²)^(−1/3) in ascending powers of x through the x⁶ term, giving each coefficient as an exact fraction.

  1. 1(1 + 2x²)^(−1/3) = 1 + (−1/3)(2x²) + ((−1/3)(−4/3)/2!)(2x²)² + ((−1/3)(−4/3)(−7/3)/3!)(2x²)³ + …
  2. 2= 1 − (2/3)x² + (2/9)(4x⁴) − (14/81)(8x⁶) + …
  3. 3= 1 − (2/3)x² + (8/9)x⁴ − (112/81)x⁶

Answer1 − (2/3)x² + (8/9)x⁴ − (112/81)x⁶

Part (b) · 1 mark

Give the set of x values for which the expansion in part (a) converges.

  1. 1Valid for |2x²| < 1
  2. 2|x| < 1/√2

Answer−1/√2 < x < 1/√2 (|x| < 1/√2)

Part (c) · 3 marks

Expand f(x) = (2 + kx²)/(1 + 2x²)^(1/3) in ascending powers of x through x⁶, with coefficients in terms of k.

  1. 1f(x) = (2 + kx²)(1 − (2/3)x² + (8/9)x⁴ − (112/81)x⁶ + …)
  2. 2= 2 − (4/3)x² + (16/9)x⁴ − (224/81)x⁶ + kx² − (2k/3)x⁴ + (8k/9)x⁶ + …
  3. 3= 2 + (k − 4/3)x² + (16/9 − 2k/3)x⁴ + (8k/9 − 224/81)x⁶

Answer2 + (k − 4/3)x² + (16/9 − 2k/3)x⁴ + (8k/9 − 224/81)x⁶

Part (d) · 2 marks

Using the fact that the x⁴ coefficient of f(x) equals zero, determine k.

  1. 116/9 − 2k/3 = 0
  2. 2k = 8/3

Answerk = 8/3

Part (e) · 5 marks

Using your series, integrate algebraically from 0 to 0.5 to estimate the integral of f(x), correct to 4 decimal places.

  1. 1With k = 8/3: f(x) ≈ 2 + (4/3)x² − (32/81)x⁶
  2. 2∫ = [2x + 4x³/9 − 32x⁷/567] from 0 to 0.5
  3. 3= 1 + 4(0.5)³/9 − 32(0.5)⁷/567
  4. 4= 1.0551 (4 d.p.)

Answer1.0551

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