| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for ln(trigonometric expressions) |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on Maclaurin series with clear scaffolding. Parts (a) and (b)(i) involve standard differentiation and series expansion techniques. The challenge lies in part (c) requiring insight to use logarithm properties and combine previous results, but the multi-part structure guides students through the solution. Harder than average A-level but routine for FP3 students. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{-\sin x + \cos x}{\cos x + \sin x}\) | B1 | Correct first derivative |
| \(\frac{d^2y}{dx^2} = \frac{(-\cos x - \sin x)(\cos x+\sin x)-(-\sin x+\cos x)^2}{(\cos x+\sin x)^2}\) | M1 | Quotient rule attempted |
| Numerator \(= -(\cos x+\sin x)^2 - (\cos x - \sin x)^2 = -2\) | M1 | Expanding numerator correctly |
| \(= \frac{-2}{(\cos x+\sin x)^2} = \frac{-2}{1+\sin 2x}\) | A1 | Using \((\cos x+\sin x)^2 = 1+\sin 2x\) to complete |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^3y}{dx^3} = \frac{4\cos 2x}{(1+\sin 2x)^2}\) | B1 | Correct third derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y(0)=0,\; y'(0)=1,\; y''(0)=-2,\; y'''(0)=4\) | B1 | Correct values at \(x=0\) |
| \(\ln(\cos x+\sin x) \approx x - x^2 + \frac{2}{3}x^3\) | M1A1 | Applying Maclaurin; correct terms shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln(\cos x - \sin x) \approx -x - x^2 - \frac{2}{3}x^3\) | B1 | Replace \(x\) with \(-x\) in (b)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln\!\left(\frac{\cos 2x}{e^{3x-1}}\right) = \ln(\cos 2x) - (3x-1)\) | M1 | Splitting the log correctly |
| \(\ln(\cos 2x) = \ln(\cos 2x + \sin 2x) + \ln(\cos 2x - \sin 2x)\), use (b)(i)+(b)(ii) with \(x\to 2x\) | M1 | Correct strategy using previous results |
| \(\ln(\cos 2x) \approx -2x^2 - \frac{8}{3}x^3 \cdots\) (leading terms) | A1 | Correct expansion |
| Final answer: \(1 - 3x - 2x^2 - \frac{8}{3}x^3\) (first three non-zero terms) | A1 | Correct final answer |
# Question 7:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-\sin x + \cos x}{\cos x + \sin x}$ | B1 | Correct first derivative |
| $\frac{d^2y}{dx^2} = \frac{(-\cos x - \sin x)(\cos x+\sin x)-(-\sin x+\cos x)^2}{(\cos x+\sin x)^2}$ | M1 | Quotient rule attempted |
| Numerator $= -(\cos x+\sin x)^2 - (\cos x - \sin x)^2 = -2$ | M1 | Expanding numerator correctly |
| $= \frac{-2}{(\cos x+\sin x)^2} = \frac{-2}{1+\sin 2x}$ | A1 | Using $(\cos x+\sin x)^2 = 1+\sin 2x$ to complete |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^3y}{dx^3} = \frac{4\cos 2x}{(1+\sin 2x)^2}$ | B1 | Correct third derivative |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y(0)=0,\; y'(0)=1,\; y''(0)=-2,\; y'''(0)=4$ | B1 | Correct values at $x=0$ |
| $\ln(\cos x+\sin x) \approx x - x^2 + \frac{2}{3}x^3$ | M1A1 | Applying Maclaurin; correct terms shown |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln(\cos x - \sin x) \approx -x - x^2 - \frac{2}{3}x^3$ | B1 | Replace $x$ with $-x$ in (b)(i) |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln\!\left(\frac{\cos 2x}{e^{3x-1}}\right) = \ln(\cos 2x) - (3x-1)$ | M1 | Splitting the log correctly |
| $\ln(\cos 2x) = \ln(\cos 2x + \sin 2x) + \ln(\cos 2x - \sin 2x)$, use (b)(i)+(b)(ii) with $x\to 2x$ | M1 | Correct strategy using previous results |
| $\ln(\cos 2x) \approx -2x^2 - \frac{8}{3}x^3 \cdots$ (leading terms) | A1 | Correct expansion |
| Final answer: $1 - 3x - 2x^2 - \frac{8}{3}x^3$ (first three non-zero terms) | A1 | Correct final answer |7
\begin{enumerate}[label=(\alph*)]
\item It is given that $y = \ln ( \cos x + \sin x )$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \frac { 2 } { 1 + \sin 2 x }$.
\item Find $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Hence use Maclaurin's theorem to show that the first three non-zero terms in the expansion, in ascending powers of $x$, of $\ln ( \cos x + \sin x )$ are $x - x ^ { 2 } + \frac { 2 } { 3 } x ^ { 3 }$.
\item Write down the first three non-zero terms in the expansion, in ascending powers of $x$, of $\ln ( \cos x - \sin x )$.
\end{enumerate}\item Hence find the first three non-zero terms in the expansion, in ascending powers of $x$, of $\ln \left( \frac { \cos 2 x } { \mathrm { e } ^ { 3 x - 1 } } \right)$.\\[0pt]
[4 marks]
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\hfill \mbox{\textit{AQA FP3 2014 Q7 [4]}}