| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Deduce related series from given series |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring standard Maclaurin series techniques. Part (i) involves routine differentiation and substitution (though ln(2+x) requires the chain rule). Part (ii) is pattern recognition (replacing x with -x) followed by simple logarithm laws and algebraic cancellation. While it's Further Maths content, the execution is mechanical with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| (i) EITHER: If \(f(x) = \ln(x+2)\), then \(f'(x) = \frac{1}{2+x}\) | M1 | For at least one differentiation attempt |
| and \(f''(x) = -\frac{1}{(2+x)^2}\) | A1 | For correct first and second derivatives |
| \(f(0) = \ln 2\), \(f'(0) = \frac{1}{2}\), \(f''(0) = -\frac{1}{4}\) | A1√ | For all three evaluations correct |
| Hence \(\ln(x+2) = \ln 2 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots\) | A1 | For three correct terms |
| OR: \(\ln(2+x) = \ln\left[2\left(1+\frac{x}{2}\right)\right]\) | M1 | For factoring in this way |
| \(= \ln 2 + \ln\left(1+\frac{1}{2}x\right)\) | A1 | For using relevant log law correctly |
| \(= \ln 2 + \frac{1}{2}x - \frac{(\frac{1}{2}x)^2}{2} + \ldots\) | M1 | For use of standard series expansion |
| \(= \ln 2 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots\) | A1 | For three correct terms |
| \(-\) | 4 | |
| (ii) \(\ln(2-x) = \ln 2 - \frac{1}{2}x - \frac{1}{8}x^2\) | B1√ | For replacing \(x\) by \(-x\) |
| \(\ln\left(\frac{2+x}{2-x}\right) = \left(\ln 2 + \frac{1}{2}x - \frac{1}{8}x^2\right) - \left(\ln 2 - \frac{1}{2}x - \frac{1}{8}x^2\right)\) | M1 | For subtracting the two series |
| \(= x\), as required | A1 | For showing given answer correctly |
| \(-\) | 3 |
(i) **EITHER:** If $f(x) = \ln(x+2)$, then $f'(x) = \frac{1}{2+x}$ | M1 | For at least one differentiation attempt |
and $f''(x) = -\frac{1}{(2+x)^2}$ | A1 | For correct first and second derivatives |
$f(0) = \ln 2$, $f'(0) = \frac{1}{2}$, $f''(0) = -\frac{1}{4}$ | A1√ | For all three evaluations correct |
Hence $\ln(x+2) = \ln 2 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots$ | A1 | For three correct terms |
**OR:** $\ln(2+x) = \ln\left[2\left(1+\frac{x}{2}\right)\right]$ | M1 | For factoring in this way |
$= \ln 2 + \ln\left(1+\frac{1}{2}x\right)$ | A1 | For using relevant log law correctly |
$= \ln 2 + \frac{1}{2}x - \frac{(\frac{1}{2}x)^2}{2} + \ldots$ | M1 | For use of standard series expansion |
$= \ln 2 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots$ | A1 | For three correct terms |
$-$ | **4** | |
(ii) $\ln(2-x) = \ln 2 - \frac{1}{2}x - \frac{1}{8}x^2$ | B1√ | For replacing $x$ by $-x$ |
$\ln\left(\frac{2+x}{2-x}\right) = \left(\ln 2 + \frac{1}{2}x - \frac{1}{8}x^2\right) - \left(\ln 2 - \frac{1}{2}x - \frac{1}{8}x^2\right)$ | M1 | For subtracting the two series |
$= x$, as required | A1 | For showing given answer correctly |
$-$ | **3** | |
---3 (i) Find the first three terms of the Maclaurin series for $\ln ( 2 + x )$.\\
(ii) Write down the first three terms of the series for $\ln ( 2 - x )$, and hence show that, if $x$ is small, then
$$\ln \left( \frac { 2 + x } { 2 - x } \right) \approx x$$
\hfill \mbox{\textit{OCR FP2 Q3 [7]}}