| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for ln(a+bx) |
| Difficulty | Standard +0.3 This is a straightforward application of the Maclaurin series formula requiring routine differentiation of ln(3+x) and substitution of x=0. The question guides students through each step explicitly, making it slightly easier than average for Further Maths, though the topic itself (Taylor series) is more advanced than standard A-level content. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(0) = \ln 3\) | B1 | |
| \(f'(0) = \frac{1}{3}\) | B1 | |
| \(f''(0) = -\frac{1}{9}\) | B1 | Clearly derived |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reasonable attempt at Maclaurin series | M1 | Form \(\ln 3 + ax + bx^2\), with \(a,b\) related to \(f'\), \(f''\) |
| \(f(x) = \ln 3 + \frac{1}{3}x - \frac{1}{18}x^2\) | A1 | On their values off \(f'\) and \(f''\) |
| SR | Use \(\ln(3+x) = \ln 3 + \ln(1 + \frac{1}{3}x)\); M1 use formulae book to get \(\ln 3 + \frac{1}{3}x - \frac{1}{18}x^2\) | |
| A1 |
## Question 1:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0) = \ln 3$ | B1 | |
| $f'(0) = \frac{1}{3}$ | B1 | |
| $f''(0) = -\frac{1}{9}$ | B1 | Clearly derived |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reasonable attempt at Maclaurin series | M1 | Form $\ln 3 + ax + bx^2$, with $a,b$ related to $f'$, $f''$ |
| $f(x) = \ln 3 + \frac{1}{3}x - \frac{1}{18}x^2$ | A1 | On their values off $f'$ and $f''$ |
| | SR | Use $\ln(3+x) = \ln 3 + \ln(1 + \frac{1}{3}x)$; M1 use formulae book to get $\ln 3 + \frac{1}{3}x - \frac{1}{18}x^2$ |
| | A1 | |
---1 It is given that $\mathrm { f } ( x ) = \ln ( 3 + x )$.\\
(i) Find the exact values of $f ( 0 )$ and $f ^ { \prime } ( 0 )$, and show that $f ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 9 }$.\\
(ii) Hence write down the first three terms of the Maclaurin series for $\mathrm { f } ( x )$, given that $- 3 < x \leqslant 3$.
\hfill \mbox{\textit{OCR FP2 2007 Q1 [5]}}