| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Direct substitution into standard series |
| Difficulty | Moderate -0.3 This is a straightforward application of a standard Maclaurin series with direct substitution of x² for x, followed by recognizing the series at x=1/2. Part (i) requires only recall and substitution (no derivation), and part (ii) is a routine 'hence' question where the connection is clearly signposted. Slightly easier than average due to minimal problem-solving required. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Validity: \(-1 \leq x \leq 1\) or \( | x | \leq 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 4\ln\left(\frac{5}{4}\right)\) isw | M1, A1 | Sub \(x=\frac{1}{2}\) into their ans to (i); Single ln expression |
**(i)** $\ln 1+x = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$
$\Rightarrow \ln 1+x^2 = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} \ldots$
Validity: $-1 \leq x \leq 1$ or $|x| \leq 1$ | B1, B1, B1 | 2 or 3 terms correct unsimplified; All terms correct
**(ii)** Substitute $x = \frac{1}{2}$
$\Rightarrow \ln\left(\frac{5}{4}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2}\left(\frac{1}{2}\right)^4 + \frac{1}{3}\left(\frac{1}{2}\right)^6 - \frac{1}{4}\left(\frac{1}{2}\right)^8 + \ldots$
$= \frac{1}{4}\left(1 - \frac{1}{2}\left(\frac{1}{2}\right)^2 + \frac{1}{3}\left(\frac{1}{2}\right)^4 - \frac{1}{4}\left(\frac{1}{2}\right)^6 + \ldots\right)$
$\Rightarrow 1 - \frac{1}{2}\left(\frac{1}{2}\right)^2 + \frac{1}{3}\left(\frac{1}{2}\right)^4 - \frac{1}{4}\left(\frac{1}{2}\right)^6 + \ldots$
$= 4\ln\left(\frac{5}{4}\right)$ isw | M1, A1 | Sub $x=\frac{1}{2}$ into their ans to (i); Single ln expression
---2 It is given that $\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)$.\\
(i) Using the standard Maclaurin expansion for $\ln ( 1 + x )$, write down the first four terms in the expansion of $\mathrm { f } ( x )$, stating the set of values of $x$ for which the expansion is valid.\\
(ii) Hence find the exact value of
$$1 - \frac { 1 } { 2 } \left( \frac { 1 } { 2 } \right) ^ { 2 } + \frac { 1 } { 3 } \left( \frac { 1 } { 2 } \right) ^ { 4 } - \frac { 1 } { 4 } \left( \frac { 1 } { 2 } \right) ^ { 6 } + \ldots .$$
\hfill \mbox{\textit{OCR FP2 2014 Q2 [5]}}