| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Exponential or trigonometric base functions |
| Difficulty | Challenging +1.2 This is a structured Further Maths question requiring systematic differentiation to find a pattern, then using it to generate Maclaurin series terms. Part (a) guides students to discover the recurrence relation (k=8), and part (b) is computational but lengthy. While it requires careful algebra and organization across 6 derivatives, the approach is methodical rather than requiring novel insight—harder than average due to Further Maths content and computational demand, but not exceptionally challenging. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = e^x \sin x \Rightarrow \frac{dy}{dx} = e^x \sin x + e^x \cos x\) \(\Rightarrow \frac{d^2y}{dx^2} = e^x \sin x + e^x \cos x + e^x \cos x - e^x \sin x = 2e^x \cos x\) | M1 | Applies product rule on \(y\) and \(y'\) to reach at least the second derivative. May work in terms of \(x\) or \(y\) or combination. |
| \(\frac{d^3y}{dx^3} = 2e^x \cos x - 2e^x \sin x\) \(\frac{d^4y}{dx^4} = 2e^x \cos x - 2e^x \sin x - 2e^x \cos x - 2e^x \sin x\) | dM1 | Continues differentiation to obtain an expression for the 4th derivative. |
| \(\frac{d^4y}{dx^4} = -4e^x \sin x = -4y\) | A1 | Any correct fourth derivative expression. Must be correctly identified as the fourth derivative. |
| \(\frac{d^4y}{dx^4} = -4y \Rightarrow \frac{d^5y}{dx^5} = -4\frac{dy}{dx} \Rightarrow \frac{d^6y}{dx^6} = -4\frac{d^2y}{dx^2}\) | A1 | Completes the process to a correct expression, or for \(k = -4\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y)_0 = 0, \left(\frac{dy}{dx}\right)_0 = 1, \left(\frac{d^2y}{dx^2}\right)_0 = 2, \left(\frac{d^3y}{dx^3}\right)_0 = 2, \left(\frac{d^4y}{dx^4}\right)_0 = 0, \left(\frac{d^5y}{dx^5}\right)_0 = -4, \left(\frac{d^6y}{dx^6}\right)_0 = -8\) | M1 | Attempts to find values for all 6 derivatives at \(x=0\), condone one missing as a slip. |
| \(y = 0 + x + \frac{x^2}{2} \times 2 + \frac{x^3}{3!} \times 2 + 0 - \frac{x^5}{5!} \times 4 - \frac{x^6}{6!} \times 8 + \ldots\) | M1 | Applies Maclaurin's theorem correctly with their values, allowing one slip in the factorials, up to the term in \(x^6\). |
| \(y = x + x^2 + \frac{x^3}{3} - \frac{x^5}{30} - \frac{x^6}{90} + \ldots\) | A1 | Correct expansion, condone missing "\(y=\)". |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^x \sin x \Rightarrow \frac{dy}{dx} = e^x \sin x + e^x \cos x$ $\Rightarrow \frac{d^2y}{dx^2} = e^x \sin x + e^x \cos x + e^x \cos x - e^x \sin x = 2e^x \cos x$ | M1 | Applies product rule on $y$ and $y'$ to reach at least the second derivative. May work in terms of $x$ or $y$ or combination. |
| $\frac{d^3y}{dx^3} = 2e^x \cos x - 2e^x \sin x$ $\frac{d^4y}{dx^4} = 2e^x \cos x - 2e^x \sin x - 2e^x \cos x - 2e^x \sin x$ | dM1 | Continues differentiation to obtain an expression for the 4th derivative. |
| $\frac{d^4y}{dx^4} = -4e^x \sin x = -4y$ | A1 | Any correct fourth derivative expression. Must be correctly identified as the fourth derivative. |
| $\frac{d^4y}{dx^4} = -4y \Rightarrow \frac{d^5y}{dx^5} = -4\frac{dy}{dx} \Rightarrow \frac{d^6y}{dx^6} = -4\frac{d^2y}{dx^2}$ | A1 | Completes the process to a correct expression, or for $k = -4$. |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y)_0 = 0, \left(\frac{dy}{dx}\right)_0 = 1, \left(\frac{d^2y}{dx^2}\right)_0 = 2, \left(\frac{d^3y}{dx^3}\right)_0 = 2, \left(\frac{d^4y}{dx^4}\right)_0 = 0, \left(\frac{d^5y}{dx^5}\right)_0 = -4, \left(\frac{d^6y}{dx^6}\right)_0 = -8$ | M1 | Attempts to find values for all 6 derivatives at $x=0$, condone one missing as a slip. |
| $y = 0 + x + \frac{x^2}{2} \times 2 + \frac{x^3}{3!} \times 2 + 0 - \frac{x^5}{5!} \times 4 - \frac{x^6}{6!} \times 8 + \ldots$ | M1 | Applies Maclaurin's theorem correctly with their values, allowing one slip in the factorials, up to the term in $x^6$. |
| $y = x + x^2 + \frac{x^3}{3} - \frac{x^5}{30} - \frac{x^6}{90} + \ldots$ | A1 | Correct expansion, condone missing "$y=$". |
---\begin{enumerate}
\item Given that $y = \mathrm { e } ^ { x } \sin x$\\
(a) show that
\end{enumerate}
$$\frac { \mathrm { d } ^ { 6 } y } { \mathrm {~d} x ^ { 6 } } = k \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$
where $k$ is a constant to be determined.\\
(b) Hence determine the first 5 non-zero terms in the Maclaurin series expansion for $y$, giving each coefficient in simplest form.
\hfill \mbox{\textit{Edexcel F2 2024 Q7 [7]}}