| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Hyperbolic power functions |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on Maclaurin series with hyperbolic functions. Part (a) requires systematic differentiation using product rule (4 marks of calculation), part (b) applies the differential equation to find series terms (standard technique once the pattern is established), and part (c) requires recognizing the pattern in coefficients. While it involves multiple steps and Further Maths content, the question is highly scaffolded with clear signposting, making it moderately above average difficulty but not requiring significant novel insight. |
| Spec | 4.07d Differentiate/integrate: hyperbolic functions4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \sin x\cosh x + \cos x\sinh x\) | M1 | Uses product rule, attempts first derivative |
| \(\frac{d^2y}{dx^2} = \cos x\cosh x + \sin x\sinh x + \cos x\cosh x - \sin x\sinh x = 2\cos x\cosh x\) | M1 | Second application of product rule |
| \(\frac{d^3y}{dx^3} = 2\cos x\sinh x - 2\sin x\cosh x\) | M1 | Correct method for third derivative |
| \(\frac{d^4y}{dx^4} = -4\sinh x\sin x = -4y\) | A1* | Obtains correct 4th derivative and links back to \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\left(\frac{d^2y}{dx^2}\right)_0=2\), \(\left(\frac{d^6y}{dx^6}\right)_0=-8\), \(\left(\frac{d^{10}y}{dx^{10}}\right)_0=32\) | B1 | Makes connection with part (a), finds correct non-zero values |
| Uses \(y = y_0 + xy_0' + \frac{x^2}{2!}y_0'' + \frac{x^3}{3!}y_0''' + \ldots\) with values | M1 | Correct attempt at Maclaurin series with values |
| \(= \frac{x^2}{2!}(2)+\frac{x^6}{6!}(-8)+\frac{x^{10}}{10!}(32)\) | A1 | Correct expression unsimplified |
| \(= x^2 - \frac{x^6}{90}+\frac{x^{10}}{113400}\) | A1 | Correct expression simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2(-4)^{n-1}\frac{x^{4n-2}}{(4n-2)!}\) | M1 A1 | Generalising, dealing with signs, powers and factorials; correct expression |
## Question 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \sin x\cosh x + \cos x\sinh x$ | M1 | Uses product rule, attempts first derivative |
| $\frac{d^2y}{dx^2} = \cos x\cosh x + \sin x\sinh x + \cos x\cosh x - \sin x\sinh x = 2\cos x\cosh x$ | M1 | Second application of product rule |
| $\frac{d^3y}{dx^3} = 2\cos x\sinh x - 2\sin x\cosh x$ | M1 | Correct method for third derivative |
| $\frac{d^4y}{dx^4} = -4\sinh x\sin x = -4y$ | A1* | Obtains correct 4th derivative and links back to $y$ |
## Question 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left(\frac{d^2y}{dx^2}\right)_0=2$, $\left(\frac{d^6y}{dx^6}\right)_0=-8$, $\left(\frac{d^{10}y}{dx^{10}}\right)_0=32$ | B1 | Makes connection with part (a), finds correct non-zero values |
| Uses $y = y_0 + xy_0' + \frac{x^2}{2!}y_0'' + \frac{x^3}{3!}y_0''' + \ldots$ with values | M1 | Correct attempt at Maclaurin series with values |
| $= \frac{x^2}{2!}(2)+\frac{x^6}{6!}(-8)+\frac{x^{10}}{10!}(32)$ | A1 | Correct expression unsimplified |
| $= x^2 - \frac{x^6}{90}+\frac{x^{10}}{113400}$ | A1 | Correct expression simplified |
## Question 5(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2(-4)^{n-1}\frac{x^{4n-2}}{(4n-2)!}$ | M1 A1 | Generalising, dealing with signs, powers and factorials; correct expression |5.
$$y = \sin x \sinh x$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = - 4 y$
\item Hence find the first three non-zero terms of the Maclaurin series for $y$, giving each coefficient in its simplest form.
\item Find an expression for the $n$th non-zero term of the Maclaurin series for $y$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP2 Q5 [10]}}