A2 June 2024 Paper 1 Q10
10
(a) Write down the first three terms of the Maclaurin series for \(\ln(1 + x^3)\). [1]
(b) Use these three terms to show that \(\ln(1.125) \approx \dfrac{n}{1536}\), where \(n\) is an integer to be determined. [3]
(c) Charlie uses the same first three terms of the series to approximate \(\ln 9\) and gets an answer of 147, correct to 3 significant figures. However, \(\ln 9 = 2.20\) correct to 3 significant figures.
Explain Charlie’s error. [2]
Explain Charlie’s error. [2]
| Scheme | Marks | AO |
|---|---|---|
| \(x^3 - \frac{x^6}{2} + \frac{x^9}{3}\) | B1 | 1.1 |
| [1] |
Notes
B1: Must be simplified
| Scheme | Marks | AO |
|---|---|---|
| let \(x = \frac{1}{2}\) | M1 | 3.1a |
| \([\ln 1.125 \approx] \left(\frac{1}{2}\right)^3 - \frac{1}{2}\left(\frac{1}{2}\right)^6 + \frac{1}{3}\left(\frac{1}{2}\right)^9\) | A1 | 1.1 |
| \(= \dfrac{181}{1536} \quad\) [so \(n = 181\)] | A1 | 1.1 |
| [3] |
Notes
M1: Or \(x^3 = 0.125\). Must be seen, can be implied by explicit substitution.
A1: oe correct substitution. If \(x^3 = 0.125\) used then \(0.125 - \frac{0.125^2}{2} + \frac{0.125^3}{3}\). FT their three terms. Cannot be implied.
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| Charlie takes \(x = 2\) | M1 | 2.3 |
| The series only converges for \(-1 \lt x \leqslant 1\) | A1 | 2.4 |
| [2] |
Notes
M1: May be embedded
A1: See appendix. Do not award if spoilt by an incorrect statement.
Alternative method
| Scheme | Marks |
|---|---|
| Charlie takes \(x^3 = 8\) | M1 |
| The series only converges for \(-1 \lt x^3 \leqslant 1\) | A1 |
| [2] |
M1: May be embedded
A1: See appendix. Do not award if spoilt by an incorrect statement.
Appendix: exemplar responses for Q10c – A1 mark only; the M1 mark is not implied by any of the responses below and must be secured separately
| Response | Mark |
|---|---|
| [The series] converges/works only when \(-1 \lt x \leqslant 1\) | A1 |
| The series requires/works/converges when \(-1 \lt x \leqslant 1\) | A1 |
| The series converges only when \(-1 \lt x \lt 1\) | A0 |
| The series converges when \(-1 \leqslant x \leqslant 1\) | A0 |
| The series does not converge when \(x \gt 1\) | A1 |
| The series does not converge when \(x\) is greater than 1 | A1 |
| The series does not converge when \(x \geqslant 1\) | A0 |
| The series converges only when \(|x| \lt 1\) | A0 |
| The series does not converge when \(|x| \gt 1\) | A1 |
| The series does not converge when \(|x| \geqslant 1\) | A0 |
| The series does not converge when \(|x| \geqslant 1\) unless \(x = 1\) | A1 |
| \(x = 2\) is greater than 1 so outside the range for convergence | A1 |
| \(2 \geqslant 1\) | A0 |
| For series to converge \(x \lt 1\) | A0 |
| \(2 \gt 1\) so the series does not converge | A1 |
The following statements do not score by themselves but will not spoil an otherwise correct answer:
| The series converges when \(-1 \lt x \lt 1\) |
| The series converges when \(x\) is between \(-1\) and 1 |
| For the series to converge \(x\) must be between \(-1\) and 1 |
| The series converges when \(|x| \lt 1\) |
| \(2 \gt 1\) |
| The series does not converge when \(x = 2\) |