| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| 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 requiring chain rule differentiation of hyperbolic functions, repeated differentiation following a pattern, and Maclaurin series construction. While it involves multiple steps and Further Maths content (making it inherently harder than standard A-level), the question is highly scaffolded with part (a)(i) shown and a clear algorithmic path. The pattern recognition for the fourth derivative and evaluation at x=0 are routine for FM students who have practiced Taylor series. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \ldots\cosh^{n-1}x\sinh x\) | M1 | 1.1b – Correct form of first derivative |
| \(\frac{dy}{dx} = n\cosh^{n-1}x\sinh x\); \(\frac{d^2y}{dx^2} = n(n-1)\cosh^{n-2}x\sinh^2 x + n\cosh^n x\) | A1 | 2.1 – Correct derivatives |
| \(\frac{d^2y}{dx^2} = n(n-1)\cosh^{n-2}x(\cosh^2 x - 1) + n\cosh^n x\) | M1 | 2.1 – Uses \(\sinh^2 x = \cosh^2 x - 1\) |
| \(\frac{d^2y}{dx^2} = n^2\cosh^n x - n(n-1)\cosh^{n-2}x\) | A1* | 1.1b – Correct result (cso) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^3y}{dx^3} = \ldots\cosh^{n-1}x\sinh x - \ldots\cosh^{n-3}x\sinh x\); \(\frac{d^4y}{dx^4} = \ldots\cosh^{n-2}x\sinh^2 x + \ldots\cosh^n x - \ldots\cosh^{n-4}x\sinh^2 x - \ldots\) | M1 | 1.1b |
| \(\frac{d^3y}{dx^3} = n^3\cosh^{n-1}x\sinh x - n(n-1)(n-2)\cosh^{n-3}x\sinh x\); \(\frac{d^4y}{dx^4} = n^3(n-1)\cosh^{n-2}x\sinh^2 x + n^3\cosh^n x - n(n-1)(n-2)(n-3)\cosh^{n-4}x\sinh^2 x - n(n-1)(n-2)\cosh^{n-2}x\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(\frac{d^2y}{dx^2} = n^2y - n(n-1)\cosh^{n-2}x\) leading to \(\frac{d^3y}{dx^3} = n^2\frac{dy}{dx} - \ldots\cosh^{n-3}x\sinh x\); \(\frac{d^4y}{dx^4} = n^2\frac{d^2y}{dx^2} - \ldots\cosh^{n-4}x\sinh^2 x - \ldots\cosh^{n-2}x\) | M1 | 1.1b |
| \(\frac{d^3y}{dx^3} = n^2\frac{dy}{dx} - n(n-1)(n-2)\cosh^{n-3}x\sinh x\); \(\frac{d^4y}{dx^4} = n^2\frac{d^2y}{dx^2} - n(n-1)(n-2)(n-3)\cosh^{n-4}x\sinh^2 x - n(n-1)(n-2)\cosh^{n-2}x\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^4y}{dx^4} = n^3(n-1)\left(\frac{e^x+e^{-x}}{2}\right)^{n-2}\left(\frac{e^x-e^{-x}}{2}\right)^2 + n^3\left(\frac{e^x+e^{-x}}{2}\right)^{n-2} - n(n-1)(n-2)(n-3)\left(\frac{e^x+e^{-x}}{2}\right)^{n-4}\left(\frac{e^x-e^{-x}}{2}\right)^2 - n(n-1)(n-2)\left(\frac{e^x+e^{-x}}{2}\right)^{n-2}\) | M1 | Uses chain rule and product rule to find third and fourth derivatives of required form, condone sign slips |
| Correct fourth derivative (does not need to be simplified) | A1 | ISW |
| Answer | Marks | Guidance |
|---|---|---|
| \(y=1,\ y'=0,\ y''=n^2-n(n-1),\ y^{(3)}=0,\ y^{(4)}=n^3-n(n-1)(n-2)\) | M1 | Attempts evaluation of all four derivatives at \(x=0\) and applies Maclaurin formula. Note \(y^{(1)}(0)=0\) and \(y^{(3)}(0)=0\) may be implied. If \(y^{(3)}(0)\neq 0\), allow this mark for first 3 non-zero terms |
| Uses values in expansion \(y = y(0) + xy'(0) + \frac{x^2}{2!}y''(0) + \frac{x^3}{3!}y^{(3)}(0) + \frac{x^4}{4!}y^{(4)}(0)+\ldots\) | ||
| \(y = 1 + \frac{nx^2}{2} + \frac{(3n^2-2n)x^4}{24} + \ldots\) cso | A1 2.5 | Correct simplified expansion from correct derivatives |
## Question 9:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \ldots\cosh^{n-1}x\sinh x$ | M1 | 1.1b – Correct form of first derivative |
| $\frac{dy}{dx} = n\cosh^{n-1}x\sinh x$; $\frac{d^2y}{dx^2} = n(n-1)\cosh^{n-2}x\sinh^2 x + n\cosh^n x$ | A1 | 2.1 – Correct derivatives |
| $\frac{d^2y}{dx^2} = n(n-1)\cosh^{n-2}x(\cosh^2 x - 1) + n\cosh^n x$ | M1 | 2.1 – Uses $\sinh^2 x = \cosh^2 x - 1$ |
| $\frac{d^2y}{dx^2} = n^2\cosh^n x - n(n-1)\cosh^{n-2}x$ | A1* | 1.1b – Correct result (cso) |
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### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^3y}{dx^3} = \ldots\cosh^{n-1}x\sinh x - \ldots\cosh^{n-3}x\sinh x$; $\frac{d^4y}{dx^4} = \ldots\cosh^{n-2}x\sinh^2 x + \ldots\cosh^n x - \ldots\cosh^{n-4}x\sinh^2 x - \ldots$ | M1 | 1.1b |
| $\frac{d^3y}{dx^3} = n^3\cosh^{n-1}x\sinh x - n(n-1)(n-2)\cosh^{n-3}x\sinh x$; $\frac{d^4y}{dx^4} = n^3(n-1)\cosh^{n-2}x\sinh^2 x + n^3\cosh^n x - n(n-1)(n-2)(n-3)\cosh^{n-4}x\sinh^2 x - n(n-1)(n-2)\cosh^{n-2}x$ | A1 | 1.1b |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $\frac{d^2y}{dx^2} = n^2y - n(n-1)\cosh^{n-2}x$ leading to $\frac{d^3y}{dx^3} = n^2\frac{dy}{dx} - \ldots\cosh^{n-3}x\sinh x$; $\frac{d^4y}{dx^4} = n^2\frac{d^2y}{dx^2} - \ldots\cosh^{n-4}x\sinh^2 x - \ldots\cosh^{n-2}x$ | M1 | 1.1b |
| $\frac{d^3y}{dx^3} = n^2\frac{dy}{dx} - n(n-1)(n-2)\cosh^{n-3}x\sinh x$; $\frac{d^4y}{dx^4} = n^2\frac{d^2y}{dx^2} - n(n-1)(n-2)(n-3)\cosh^{n-4}x\sinh^2 x - n(n-1)(n-2)\cosh^{n-2}x$ | A1 | 1.1b |
## Question (a)(ii):
**Fourth derivative:**
$\frac{d^4y}{dx^4} = n^3(n-1)\left(\frac{e^x+e^{-x}}{2}\right)^{n-2}\left(\frac{e^x-e^{-x}}{2}\right)^2 + n^3\left(\frac{e^x+e^{-x}}{2}\right)^{n-2} - n(n-1)(n-2)(n-3)\left(\frac{e^x+e^{-x}}{2}\right)^{n-4}\left(\frac{e^x-e^{-x}}{2}\right)^2 - n(n-1)(n-2)\left(\frac{e^x+e^{-x}}{2}\right)^{n-2}$ | M1 | Uses chain rule and product rule to find third and fourth derivatives of required form, condone sign slips |
Correct fourth derivative (does not need to be simplified) | A1 | ISW |
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## Question (b):
**Values at $x = 0$:**
$y=1,\ y'=0,\ y''=n^2-n(n-1),\ y^{(3)}=0,\ y^{(4)}=n^3-n(n-1)(n-2)$ | M1 | Attempts evaluation of all four derivatives at $x=0$ and applies Maclaurin formula. Note $y^{(1)}(0)=0$ and $y^{(3)}(0)=0$ may be implied. If $y^{(3)}(0)\neq 0$, allow this mark for first 3 non-zero terms |
Uses values in expansion $y = y(0) + xy'(0) + \frac{x^2}{2!}y''(0) + \frac{x^3}{3!}y^{(3)}(0) + \frac{x^4}{4!}y^{(4)}(0)+\ldots$ | | |
$y = 1 + \frac{nx^2}{2} + \frac{(3n^2-2n)x^4}{24} + \ldots$ cso | A1 2.5 | Correct simplified expansion from correct derivatives |
**(2 marks)**9.
$$y = \cosh ^ { n } x \quad n \geqslant 5$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = n ^ { 2 } \cosh ^ { n } x - n ( n - 1 ) \cosh ^ { n - 2 } x$$
\item Determine an expression for $\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }$
\end{enumerate}\item Hence determine the first three non-zero terms of the Maclaurin series for $y$, giving each coefficient in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP2 2022 Q9 [8]}}