| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Explicit differential equation series solution |
| Difficulty | Challenging +1.8 This Further Maths question requires systematic differentiation using the chain rule multiple times to find higher derivatives, then applying Taylor series expansion around a non-zero point. While the techniques are standard for FM students, the multi-step nature (finding 4 derivatives, identifying coefficients A and B, then computing numerical values at x=-1, and constructing the series) combined with the algebraic manipulation required makes this significantly harder than average A-level questions but not exceptionally difficult for Further Maths Pure 2. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = y^2 - x \Rightarrow \frac{d^2y}{dx^2} = 2y\frac{dy}{dx} - 1\) | B1 | Correct expression for \(\frac{d^2y}{dx^2}\) |
| \(\frac{d^3y}{dx^3} = 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2\) | M1 A1 | M1: Applies product rule to obtain \(\frac{d^3y}{dx^3} = Ay\frac{d^2y}{dx^2} + \ldots\) or \(\frac{d^3y}{dx^3} = \ldots + B\left(\frac{dy}{dx}\right)^2\) where \(\ldots\) is non-zero. A1: Correct expression. |
| \(\frac{d^4y}{dx^4} = 2y\frac{d^3y}{dx^3} + 6\frac{dy}{dx}\frac{d^2y}{dx^2}\) | A1 | Correct expression for \(\frac{d^4y}{dx^4}\) or correct values for \(A\) and \(B\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((y)_{-1}=1,\ (y')_{-1}=2,\ (y'')_{-1}=3,\ (y''')_{-1}=14,\ (y'''')_{-1}=64\) | M1 | Attempts values up to at least the 3rd derivative using \(x=-1\) and \(y=1\). Condone slips provided intention is clear. |
| \((y=)1+2(x+1)+\frac{3(x+1)^2}{2}+\frac{14(x+1)^3}{3!}+\frac{64(x+1)^4}{4!}+\ldots\) | M1 | Correct application of Taylor series in powers of \((x+1)\). If expansion just written down with no formula quoted then it must be correct for their values. |
| \((y=)1+2(x+1)+\frac{3(x+1)^2}{2}+\frac{7(x+1)^3}{3}+\frac{8(x+1)^4}{3}+\ldots\) | A1 | Correct simplified expansion. The "\(y=\)" is not required. |
| Total part (b): 3 marks | Total Q4: 7 marks |
## Question 4:
### Part (a):
| $\frac{dy}{dx} = y^2 - x \Rightarrow \frac{d^2y}{dx^2} = 2y\frac{dy}{dx} - 1$ | B1 | Correct expression for $\frac{d^2y}{dx^2}$ |
| $\frac{d^3y}{dx^3} = 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2$ | M1 A1 | M1: Applies product rule to obtain $\frac{d^3y}{dx^3} = Ay\frac{d^2y}{dx^2} + \ldots$ or $\frac{d^3y}{dx^3} = \ldots + B\left(\frac{dy}{dx}\right)^2$ where $\ldots$ is non-zero. A1: Correct expression. |
| $\frac{d^4y}{dx^4} = 2y\frac{d^3y}{dx^3} + 6\frac{dy}{dx}\frac{d^2y}{dx^2}$ | A1 | Correct expression for $\frac{d^4y}{dx^4}$ or correct values for $A$ and $B$ |
**Note:** If $\frac{d^2y}{dx^2} = 2y\frac{dy}{dx}$ is obtained, allow recovery in (a) so B0M1A1A1 is possible.
**Total part (a): 4 marks**
### Part (b):
| $(y)_{-1}=1,\ (y')_{-1}=2,\ (y'')_{-1}=3,\ (y''')_{-1}=14,\ (y'''')_{-1}=64$ | M1 | Attempts values up to at least the 3rd derivative using $x=-1$ and $y=1$. Condone slips provided intention is clear. |
| $(y=)1+2(x+1)+\frac{3(x+1)^2}{2}+\frac{14(x+1)^3}{3!}+\frac{64(x+1)^4}{4!}+\ldots$ | M1 | Correct application of Taylor series in powers of $(x+1)$. If expansion just written down with no formula quoted then it must be correct for their values. |
| $(y=)1+2(x+1)+\frac{3(x+1)^2}{2}+\frac{7(x+1)^3}{3}+\frac{8(x+1)^4}{3}+\ldots$ | A1 | Correct simplified expansion. The "$y=$" is not required. |
**Total part (b): 3 marks | Total Q4: 7 marks**
---4.
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } - x$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = A y \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } + B \frac { \mathrm {~d} y } { \mathrm {~d} x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$
where $A$ and $B$ are integers to be determined.
Given that $y = 1$ at $x = - 1$
\item determine the Taylor series solution for $y$, in ascending powers of $( x + 1 )$ up to and including the term in $( x + 1 ) ^ { 4 }$, giving each coefficient in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2023 Q4 [7]}}