| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Express in form R cosh(x±α) or R sinh(x±α) |
| Difficulty | Standard +0.3 This is a structured multi-part question on hyperbolic functions that guides students through standard results. Parts (i) and (ii) are routine proofs using definitions, part (iii) is a standard R cosh(x±α) form problem solvable by comparing coefficients, and part (iv) follows directly. While it's Further Maths content (inherently harder), the question is methodical with clear scaffolding and uses well-established techniques, making it slightly easier than average overall. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.07n Stationary points: find maxima, minima using derivatives4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Get \(\frac{1}{2}(e^{\ln a} + e^{-\ln a})\) | M1 | |
| Use \(e^{\ln a} = a\) and \(e^{-\ln a} = \frac{1}{a}\) | M1 | |
| Clearly derive AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reasonable attempt to multiply out their attempts at exponential definitions of \(\cosh\) and \(\sinh\) | M1 | 4 terms in each |
| Correct expansion seen as \(e^{(x+y)}\) etc. | A1 | |
| Clearly tidy to AG | A1 | With \(e^{-(x-y)}\) seen or implied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(x = y\) and \(\cosh 0 = 1\) to get AG | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to expand and equate coefficients | M1 | \(13 = R\cosh\ln a = \frac{R(a^2+1)}{2a}\); \(5 = R\sinh\ln a = \frac{R(a^2-1)}{2a}\) |
| Attempt to eliminate \(R\) (or \(a\)) to set up solvable equation in \(a\) (or \(R\)) | M1 | SC: If exponential definitions used, \(8e^x + 18e^{-x} = Re^x/a + Rae^{-x}\) and same scheme follows |
| Get \(a = \frac{3}{2}\) (or \(R = 12\)) | A1 | |
| Replace for \(a\) (or \(R\)) in relevant equation to set up solvable equation in \(R\) (or \(a\)) | M1 | |
| Get \(R = 12\) (or \(a = \frac{3}{2}\)) | A1 | Ignore if \(a = \frac{2}{3}\) also given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Quote/derive \((\ln\frac{3}{2},\ 12)\) | B1\(\sqrt{}\) | On their \(R\) and \(a\) |
| B1\(\sqrt{}\) |
## Question 8(i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Get $\frac{1}{2}(e^{\ln a} + e^{-\ln a})$ | M1 | |
| Use $e^{\ln a} = a$ and $e^{-\ln a} = \frac{1}{a}$ | M1 | |
| Clearly derive AG | A1 | |
---
## Question 8(i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reasonable attempt to multiply out their attempts at exponential definitions of $\cosh$ and $\sinh$ | M1 | 4 terms in each |
| Correct expansion seen as $e^{(x+y)}$ etc. | A1 | |
| Clearly tidy to AG | A1 | With $e^{-(x-y)}$ seen or implied |
---
## Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $x = y$ and $\cosh 0 = 1$ to get AG | B1 | |
---
## Question 8(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to expand and equate coefficients | M1 | $13 = R\cosh\ln a = \frac{R(a^2+1)}{2a}$; $5 = R\sinh\ln a = \frac{R(a^2-1)}{2a}$ |
| Attempt to eliminate $R$ (or $a$) to set up solvable equation in $a$ (or $R$) | M1 | **SC:** If exponential definitions used, $8e^x + 18e^{-x} = Re^x/a + Rae^{-x}$ and same scheme follows |
| Get $a = \frac{3}{2}$ (or $R = 12$) | A1 | |
| Replace for $a$ (or $R$) in relevant equation to set up solvable equation in $R$ (or $a$) | M1 | |
| Get $R = 12$ (or $a = \frac{3}{2}$) | A1 | Ignore if $a = \frac{2}{3}$ also given |
---
## Question 8(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Quote/derive $(\ln\frac{3}{2},\ 12)$ | B1$\sqrt{}$ | On their $R$ and $a$ |
| | B1$\sqrt{}$ | |
---8 (i) Using the definitions of $\sinh x$ and $\cosh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that
\begin{enumerate}[label=(\alph*)]
\item $\cosh ( \ln a ) \equiv \frac { a ^ { 2 } + 1 } { 2 a }$, where $a > 0$,
\item $\cosh x \cosh y - \sinh x \sinh y \equiv \cosh ( x - y )$.\\
(ii) Use part (i)(b) to show that $\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1$.\\
(iii) Given that $R > 0$ and $a > 1$, find $R$ and $a$ such that
$$13 \cosh x - 5 \sinh x \equiv R \cosh ( x - \ln a )$$
(iv) Hence write down the coordinates of the minimum point on the curve with equation $y = 13 \cosh x - 5 \sinh x$.
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2009 Q8 [14]}}