| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for ln(a+bx) |
| Difficulty | Standard +0.3 This is a structured, step-by-step Maclaurin series question with clear guidance through each part. Part (a) involves routine differentiation of ln(4+3x), part (b) applies the standard Maclaurin formula, part (c) is immediate by substitution, and part (d) uses logarithm laws and discards higher-order terms. While it's Further Maths content, the question requires only methodical application of taught techniques with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.07l 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 |
|---|---|---|
| \(\frac{dy}{dx} = \frac{1}{4+3x} \times 3\) | M1 | Chain rule |
| \(\frac{d^2y}{dx^2} = -3(4+3x)^{-2} \times 3 = -9(4+3x)^{-2}\) | M1A1 | M1 for quotient (PI) or chain rule used; 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| First three terms: \(\ln 4 + \frac{3}{4}x - \frac{9}{32}x^2\) | M1, A1F | Clear attempt to use Maclaurin's theorem with numerical values for \(y'(0)\) and \(y''(0)\) are \(\ne 0\). Accept 1.38(6..) for ln4; 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ln(4-3x) = \ln 4 - \frac{3}{4}x - \frac{9}{32}x^2\) | B1F | ft \(x \to -x\) in c's answer to (b); 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\approx \frac{3}{2}x\) | M1, A1 | CSO AG; 2 marks total |
**(a)**
$\frac{dy}{dx} = \frac{1}{4+3x} \times 3$ | M1 | Chain rule
$\frac{d^2y}{dx^2} = -3(4+3x)^{-2} \times 3 = -9(4+3x)^{-2}$ | M1A1 | M1 for quotient (PI) or chain rule used; 3 marks total
**(b)**
$\ln(4+3x) = \ln 4 + y'(0) + y''(0) \cdot \frac{1}{2}x^2 + \ldots$
First three terms: $\ln 4 + \frac{3}{4}x - \frac{9}{32}x^2$ | M1, A1F | Clear attempt to use Maclaurin's theorem with numerical values for $y'(0)$ and $y''(0)$ are $\ne 0$. Accept 1.38(6..) for ln4; 2 marks total
**(c)**
$\ln(4-3x) = \ln 4 - \frac{3}{4}x - \frac{9}{32}x^2$ | B1F | ft $x \to -x$ in c's answer to (b); 1 mark total
**(d)**
$\ln\left(\frac{4+3x}{4-3x}\right) = \ln(4+3x) - \ln(4-3x)$
$\approx \ln 4 + \frac{3}{4}x - \frac{9}{32}x^2 - \ln 4 + \frac{3}{4}x + \frac{9}{32}x^2$
$\approx \frac{3}{2}x$ | M1, A1 | CSO AG; 2 marks total
**Total: 8 marks**
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\begin{enumerate}[label=(\alph*)]
\item Given that $y = \ln ( 4 + 3 x )$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Hence, by using Maclaurin's theorem, find the first three terms in the expansion, in ascending powers of $x$, of $\ln ( 4 + 3 x )$.
\item Write down the first three terms in the expansion, in ascending powers of $x$, of $\ln ( 4 - 3 x )$.
\item Show that, for small values of $x$,
$$\ln \left( \frac { 4 + 3 x } { 4 - 3 x } \right) \approx \frac { 3 } { 2 } x$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2010 Q2 [8]}}