A2 June 2025 Paper 1 Q8
8 The function \(\mathrm{f}(x)\) is defined as \(\mathrm{f}(x) = \ln(1 + x)\), for \(x \gt -1\).
[You are not required to show this series for \(\ln(1 + x)\) converges for \(-1 \lt x \leqslant 1\).] [3]
| Scheme | Marks | AO |
|---|---|---|
| Check result for \(n = 1\): \(\mathrm{f}^{\prime}(x) = \frac{1}{1 + x} = \frac{(-1)^{1+1}(1 - 1)!}{(1 + x)^1}\) [so true for \(n = 1\)] | B1 | 2.1 |
| Assume true for \(n = k\): \(\mathrm{f}^{(k)}(x) = \dfrac{(-1)^{k+1}(k - 1)!}{(1 + x)^k}\) \(\mathrm{f}^{(k+1)}(x) = \dfrac{(-k)(-1)^{k+1}(k - 1)!}{(1 + x)^{k+1}}\) | M1 | 2.1 |
| \(= \dfrac{(-1)^{(k+1)+1}((k + 1) - 1)!}{(1 + x)^{k+1}}\) [which is the result for \(n = k + 1\)] | A1 | 2.2a |
| As true for \(n = 1\), and if true for \(n = k\) then true for \(n = k + 1\), true for all \(n\) | A1 | 2.4 |
| [4] |
Notes
B1: no more simplified in initial step than \(\frac{(-1)^2(0!)}{1 + x}\); condone missing brackets
M1: differentiating expression for \(\mathrm{f}^{(k)}(x)\); condone bracketing errors or a sign slip only
A1: condone \(\frac{(-1)^{(k+2)}k!}{(1 + x)^{k+1}}\). Do not condone bracketing errors. M1 step must be seen.
A1: www. \(\frac{(-1)^{(k+1)+1}((k + 1) - 1)!}{(1 + x)^{k+1}}\) must have been seen to score this mark. \(n = 1\) must have been considered. Cannot score A0A1.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}^{(1)}(0) = \frac{(-1)^2(0!)}{1},\; \mathrm{f}^{(2)}(0) = \frac{(-1)^3(1!)}{1^2},\) \(\mathrm{f}^{(3)}(0) = \frac{(-1)^4(2!)}{1^3},\; \mathrm{f}^{(n)}(0) = \frac{(-1)^{n+1}(n - 1)!}{1^n}\) | M1 | 3.1a |
| \(n^{\text{th}}\) term coefficient \(= \frac{(-1)^{n+1}(n - 1)!}{n!} = \frac{(-1)^{n+1}}{n}\) | M1 | 3.1 |
| \(\mathrm{f}(0) = \ln 1 = 0\) \(\mathrm{f}(x) = [0] + x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \ldots + \dfrac{(-1)^{n+1}x^n}{n} + \ldots\) | A1 | 2.2a |
| [3] |
Notes
M1: using result from (a) to find at least two derivatives (can include general term). May be embedded in coefficients. Must include powers of \(-1\) and factorials in each term; condone missing division by 1. Allow one slip.
M1: considering coefficient of general term and simplifying using \(\frac{(n - 1)!}{n!}\)
A1: AG complete argument, www. Must clearly be a sum of infinitely many terms. Must have considered \(\mathrm{f}(0) = 0\).
Alternative method
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \mathrm{f}(0) + \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n - 1)!\,x^n}{n!}\) | M1 |
| \(= \ln(1) + \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}\) | A1 |
| \(= [0] + x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \ldots + \dfrac{(-1)^{n+1}x^n}{n} + \ldots\) | A1 |
M1: use of Maclaurin series with result from (a) used for all terms other than \(\mathrm{f}(0)\)
A1: \(\frac{(n - 1)!}{n!} = \frac{1}{n}\) seen or used
A1: AG