| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Differential equation given |
| Difficulty | Challenging +1.2 This is a standard Further Maths Taylor series question requiring systematic differentiation of a differential equation to find higher derivatives, then constructing a series expansion about x=2. While it involves multiple steps and careful algebraic manipulation, the technique is routine for F2 students with no novel problem-solving required—just methodical application of the standard algorithm taught in this topic. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance Notes |
| \(2\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} - x\frac{dy}{dx} - y = 0\) | B1M1 | B1: \(2\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2}\) or equivalent correct terms; M1: attempt product rule on \(xy\), allow sign errors, need \(\pm x\frac{dy}{dx} \pm y\) |
| \(2\frac{d^4y}{dx^4} + \frac{d^3y}{dx^3} - x\frac{d^2y}{dx^2} - \frac{dy}{dx} - \frac{dy}{dx} = 0\) | M1 | Differentiates again to obtain expression containing fourth derivative including product rule on \(x\frac{dy}{dx}\) to give \(\pm x\frac{d^2y}{dx^2} \pm \frac{dy}{dx}\) |
| \(\frac{d^4y}{dx^4} = \frac{1}{2}\!\left(2\frac{dy}{dx} + x\frac{d^2y}{dx^2} - \frac{d^3y}{dx^3}\right)\) | A1 | If "1" is not dealt with correctly e.g. disappears at wrong time, withhold this mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance Notes |
| \(y''(2)=1,\ y'''(2)=1,\ y''''(2)=\frac{3}{2}\) | M1A1 | M1: attempt \(y''(2)\), \(y'''(2)\), \(y''''(2)\); A1: correct values |
| \(y = f(2)+(x-2)f'(2)+\frac{(x-2)^2}{2!}f''(2)+\frac{(x-2)^3}{3!}f'''(2)+\frac{(x-2)^4}{4!}f''''(2)\) | M1 | Attempt correct Taylor expansion with their values; allow terms to be "listed" |
| \(y = 1+(x-2)+\frac{(x-2)^2}{2}+\frac{(x-2)^3}{6}+\frac{(x-2)^4}{16}\) | A1 | Correct simplified expression |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance Notes |
| \(x=2.1 \Rightarrow y = 1+(0.1)+\frac{(0.1)^2}{2}+\frac{(0.1)^3}{6}+\frac{(0.1)^4}{16}\) | M1 | Substitutes \(x=2.1\) into expansion involving \((x-2)\) |
| \(y = 1.105\) only (not awrt) | A1 | cao; must follow correct expansion; incorrect answer with no working scores M0; correct answer following correct expansion scores M1A1 |
# Question 3(a):
| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| $2\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} - x\frac{dy}{dx} - y = 0$ | B1M1 | B1: $2\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2}$ or equivalent correct terms; M1: attempt product rule on $xy$, allow sign errors, need $\pm x\frac{dy}{dx} \pm y$ |
| $2\frac{d^4y}{dx^4} + \frac{d^3y}{dx^3} - x\frac{d^2y}{dx^2} - \frac{dy}{dx} - \frac{dy}{dx} = 0$ | M1 | Differentiates again to obtain expression containing fourth derivative including product rule on $x\frac{dy}{dx}$ to give $\pm x\frac{d^2y}{dx^2} \pm \frac{dy}{dx}$ |
| $\frac{d^4y}{dx^4} = \frac{1}{2}\!\left(2\frac{dy}{dx} + x\frac{d^2y}{dx^2} - \frac{d^3y}{dx^3}\right)$ | A1 | If "1" is not dealt with correctly e.g. disappears at wrong time, withhold this mark |
# Question 3(b):
| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| $y''(2)=1,\ y'''(2)=1,\ y''''(2)=\frac{3}{2}$ | M1A1 | M1: attempt $y''(2)$, $y'''(2)$, $y''''(2)$; A1: correct values |
| $y = f(2)+(x-2)f'(2)+\frac{(x-2)^2}{2!}f''(2)+\frac{(x-2)^3}{3!}f'''(2)+\frac{(x-2)^4}{4!}f''''(2)$ | M1 | Attempt correct Taylor expansion **with their values**; allow terms to be "listed" |
| $y = 1+(x-2)+\frac{(x-2)^2}{2}+\frac{(x-2)^3}{6}+\frac{(x-2)^4}{16}$ | A1 | Correct simplified expression |
# Question 3(c):
| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| $x=2.1 \Rightarrow y = 1+(0.1)+\frac{(0.1)^2}{2}+\frac{(0.1)^3}{6}+\frac{(0.1)^4}{16}$ | M1 | Substitutes $x=2.1$ into expansion involving $(x-2)$ |
| $y = 1.105$ **only** (not awrt) | A1 | cao; must follow correct expansion; incorrect answer with no working scores M0; correct answer following correct expansion scores M1A1 |
---3.
$$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} x } - x y = 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = \frac { 1 } { 2 } \left( a \frac { \mathrm {~d} y } { \mathrm {~d} x } + b x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + c \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } \right)$$
where $a , b$ and $c$ are constants to be found.
Given that $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ when $x = 2$
\item find a series solution for $y$ in ascending powers of ( $x - 2$ ), up to and including the term in $( x - 2 ) ^ { 4 }$. Write each term in its simplest form.
\item Use the solution to part (b) to find an approximate value for $y$ when $x = 2.1$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2018 Q3 [10]}}