| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Implicit differential equation series solution |
| Difficulty | Challenging +1.8 This question requires systematic differentiation of an implicit differential equation, careful substitution of given conditions, and construction of a Taylor series. While methodical, it demands multiple derivatives, algebraic manipulation with trigonometric values, and careful bookkeeping across three connected parts—significantly above average difficulty but follows a clear algorithmic path once the approach is understood. |
| Spec | 4.08a Maclaurin series: find series for function4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\!\left(\frac{\pi}{4}\right)\frac{d^2y}{dx^2} + (1)^2(0) + \sin\!\left(\frac{\pi}{4}\right) = 0 \Rightarrow \frac{d^2y}{dx^2} = \ldots\) | M1 | Substitutes \(x=\frac{\pi}{4},\ y=1,\ \frac{dy}{dx}=0\) into DE and rearranges |
| \(\dfrac{d^2y}{dx^2} = -1 < 0\), therefore a (local) maximum | A1 | Must state \(< 0\) and conclusion; no contradictory statements |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiates to form: \(\pm\sin x\dfrac{d^2y}{dx^2} + \cos x\dfrac{d^3y}{dx^3} + \alpha y\!\left(\dfrac{dy}{dx}\right)^{1\text{ or }2} + y^2\dfrac{d^2y}{dx^2} \pm \cos x = 0\) | M1 | Correct differentiation form achieved |
| \(-\sin x\dfrac{d^2y}{dx^2} + \cos x\dfrac{d^3y}{dx^3} + 2\!\left(\dfrac{dy}{dx}\right)^2 + y^2\dfrac{d^2y}{dx^2} + \cos x = 0\) | A1 | Correct full differentiation |
| Substitutes \(x=\frac{\pi}{4},\ y=1,\ \frac{dy}{dx}=0,\ \frac{d^2y}{dx^2}=-1\) and rearranges for \(\dfrac{d^3y}{dx^3}\) | M1 | Correct substitution of known values |
| \(\dfrac{d^3y}{dx^3} = \sqrt{2}-2\) | A1* | From correct working with at least one unsimplified intermediate line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y=)1 + \left(x-\dfrac{\pi}{4}\right)(0) + \dfrac{\left(x-\frac{\pi}{4}\right)^2}{2!}(-1) + \dfrac{\left(x-\frac{\pi}{4}\right)^3}{3!}(\sqrt{2}-2)+\ldots\) | M1 | Taylor series with correct values for \(y_{\pi/4}=1,\ \frac{dy}{dx}=0,\ \frac{d^3y}{dx^3}=\sqrt{2}-2\) and their \(\frac{d^2y}{dx^2}\) |
| \((y=)1 - \dfrac{\left(x-\frac{\pi}{4}\right)^2}{2} + \dfrac{\left(x-\frac{\pi}{4}\right)^3(\sqrt{2}-2)}{6}+\ldots\) | A1 | Correct simplified expansion |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\!\left(\frac{\pi}{4}\right)\frac{d^2y}{dx^2} + (1)^2(0) + \sin\!\left(\frac{\pi}{4}\right) = 0 \Rightarrow \frac{d^2y}{dx^2} = \ldots$ | M1 | Substitutes $x=\frac{\pi}{4},\ y=1,\ \frac{dy}{dx}=0$ into DE and rearranges |
| $\dfrac{d^2y}{dx^2} = -1 < 0$, therefore a (local) maximum | A1 | Must state $< 0$ and conclusion; no contradictory statements |
---
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates to form: $\pm\sin x\dfrac{d^2y}{dx^2} + \cos x\dfrac{d^3y}{dx^3} + \alpha y\!\left(\dfrac{dy}{dx}\right)^{1\text{ or }2} + y^2\dfrac{d^2y}{dx^2} \pm \cos x = 0$ | M1 | Correct differentiation form achieved |
| $-\sin x\dfrac{d^2y}{dx^2} + \cos x\dfrac{d^3y}{dx^3} + 2\!\left(\dfrac{dy}{dx}\right)^2 + y^2\dfrac{d^2y}{dx^2} + \cos x = 0$ | A1 | Correct full differentiation |
| Substitutes $x=\frac{\pi}{4},\ y=1,\ \frac{dy}{dx}=0,\ \frac{d^2y}{dx^2}=-1$ and rearranges for $\dfrac{d^3y}{dx^3}$ | M1 | Correct substitution of known values |
| $\dfrac{d^3y}{dx^3} = \sqrt{2}-2$ | A1* | From correct working with at least one unsimplified intermediate line |
---
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y=)1 + \left(x-\dfrac{\pi}{4}\right)(0) + \dfrac{\left(x-\frac{\pi}{4}\right)^2}{2!}(-1) + \dfrac{\left(x-\frac{\pi}{4}\right)^3}{3!}(\sqrt{2}-2)+\ldots$ | M1 | Taylor series with correct values for $y_{\pi/4}=1,\ \frac{dy}{dx}=0,\ \frac{d^3y}{dx^3}=\sqrt{2}-2$ and their $\frac{d^2y}{dx^2}$ |
| $(y=)1 - \dfrac{\left(x-\frac{\pi}{4}\right)^2}{2} + \dfrac{\left(x-\frac{\pi}{4}\right)^3(\sqrt{2}-2)}{6}+\ldots$ | A1 | Correct simplified expansion |
---4.
$$\left[ \begin{array} { l }
\text { The Taylor series expansion of } \mathrm { f } ( x ) \text { about } x = a \text { is given by } \\
\mathrm { f } ( x ) = \mathrm { f } ( a ) + ( x - a ) \mathrm { f } ^ { \prime } ( a ) + \frac { ( x - a ) ^ { 2 } } { 2 ! } \mathrm { f } ^ { \prime \prime } ( a ) + \ldots + \frac { ( x - a ) ^ { r } } { r ! } \mathrm { f } ^ { ( r ) } ( a ) + \ldots
\end{array} \right]$$
The curve with equation $y = \mathrm { f } ( x )$ satisfies the differential equation
$$\cos x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \sin x = 0$$
Given that $\left( \frac { \pi } { 4 } , 1 \right)$ is a stationary point of the curve,
\begin{enumerate}[label=(\alph*)]
\item determine the nature of this stationary point, giving a reason for your answer.
\item Show that $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \sqrt { 2 } - 2$ at this stationary point.
\item Hence determine a series solution for $y$, in ascending powers of $\left( x - \frac { \pi } { 4 } \right)$ up to and including the term in $\left( x - \frac { \pi } { 4 } \right) ^ { 3 }$, giving each coefficient in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2024 Q4 [8]}}