| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question on hyperbolic functions requiring proof from definitions, equation solving, and integration. While it involves several techniques (double angle formulas, substitution, integration), each part follows standard FP2 methods without requiring novel insight. The proof and equation solving are routine for Further Maths students, and the final integral uses a standard hyperbolic substitution. It's harder than typical A-level Core questions due to the Further Maths content, but represents standard textbook exercises for this module. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sinh x = \frac{e^x-e^{-x}}{2}\), \(\cosh x = \frac{e^x+e^{-x}}{2}\) stated | B1 | |
| \(1+2\sinh^2 x = 1+2\cdot\frac{(e^x-e^{-x})^2}{4}\) | M1 | |
| \(= 1+\frac{e^{2x}-2+e^{-2x}}{2} = \frac{e^{2x}+e^{-2x}}{2} = \cosh 2x\) | A1 | Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\cosh 2x + \sinh x = 5 \Rightarrow 2(1+2\sinh^2 x)+\sinh x = 5\) | M1 | Using result from (i) |
| \(4\sinh^2 x + \sinh x - 3 = 0\) | A1 | |
| \((4\sinh x - 3)(\sinh x + 1) = 0\) | M1 | |
| \(\sinh x = \frac{3}{4}\) or \(\sinh x = -1\) | A1 | |
| \(x = \ln(\frac{3}{4}+\sqrt{\frac{9}{16}+1}) = \ln(\frac{3+5}{4}) = \ln 2\) | M1 | Using \(\sinh^{-1}\) formula |
| \(x = \ln(-1+\sqrt{2}) = -\ln(1+\sqrt{2})\) | A1 | Both answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\ln 3}\sinh^2 x\,dx = \int_0^{\ln 3}\frac{\cosh 2x - 1}{2}\,dx\) | M1 | |
| \(= \left[\frac{\sinh 2x}{4} - \frac{x}{2}\right]_0^{\ln 3}\) | A1 | |
| \(\sinh(2\ln 3) = \frac{e^{2\ln3}-e^{-2\ln3}}{2} = \frac{9-\frac{1}{9}}{2} = \frac{40}{9}\) | M1 A1 | |
| \(= \frac{40}{36} - \frac{\ln 3}{2} = \frac{10}{9} - \frac{1}{2}\ln 3\) | A1 | Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(x = 3\cosh u\), \(dx = 3\sinh u\, du\) | M1 | Substitution |
| \(\int 3\sinh u \cdot 3\sinh u\, du = 9\int\sinh^2 u\, du\) | A1 | |
| \(= \frac{9}{2}\left[\sinh u \cosh u - u\right]\) with correct limits | M1 | |
| Exact answer: \(\frac{9}{2}\left(\frac{5\cdot4}{9} - \frac{8}{9} - \cosh^{-1}\frac{5}{3}+\cosh^{-1}1\right)\)... | A1 | |
| \(= 8 - \frac{9}{2}\ln 3\) (or equivalent exact form) |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sinh x = \frac{e^x-e^{-x}}{2}$, $\cosh x = \frac{e^x+e^{-x}}{2}$ stated | B1 | |
| $1+2\sinh^2 x = 1+2\cdot\frac{(e^x-e^{-x})^2}{4}$ | M1 | |
| $= 1+\frac{e^{2x}-2+e^{-2x}}{2} = \frac{e^{2x}+e^{-2x}}{2} = \cosh 2x$ | A1 | Completion |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\cosh 2x + \sinh x = 5 \Rightarrow 2(1+2\sinh^2 x)+\sinh x = 5$ | M1 | Using result from (i) |
| $4\sinh^2 x + \sinh x - 3 = 0$ | A1 | |
| $(4\sinh x - 3)(\sinh x + 1) = 0$ | M1 | |
| $\sinh x = \frac{3}{4}$ or $\sinh x = -1$ | A1 | |
| $x = \ln(\frac{3}{4}+\sqrt{\frac{9}{16}+1}) = \ln(\frac{3+5}{4}) = \ln 2$ | M1 | Using $\sinh^{-1}$ formula |
| $x = \ln(-1+\sqrt{2}) = -\ln(1+\sqrt{2})$ | A1 | Both answers |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\ln 3}\sinh^2 x\,dx = \int_0^{\ln 3}\frac{\cosh 2x - 1}{2}\,dx$ | M1 | |
| $= \left[\frac{\sinh 2x}{4} - \frac{x}{2}\right]_0^{\ln 3}$ | A1 | |
| $\sinh(2\ln 3) = \frac{e^{2\ln3}-e^{-2\ln3}}{2} = \frac{9-\frac{1}{9}}{2} = \frac{40}{9}$ | M1 A1 | |
| $= \frac{40}{36} - \frac{\ln 3}{2} = \frac{10}{9} - \frac{1}{2}\ln 3$ | A1 | Completion |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $x = 3\cosh u$, $dx = 3\sinh u\, du$ | M1 | Substitution |
| $\int 3\sinh u \cdot 3\sinh u\, du = 9\int\sinh^2 u\, du$ | A1 | |
| $= \frac{9}{2}\left[\sinh u \cosh u - u\right]$ with correct limits | M1 | |
| Exact answer: $\frac{9}{2}\left(\frac{5\cdot4}{9} - \frac{8}{9} - \cosh^{-1}\frac{5}{3}+\cosh^{-1}1\right)$... | A1 | |
| $= 8 - \frac{9}{2}\ln 3$ (or equivalent exact form) | | |
---4 (i) Starting from the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials, prove that
$$1 + 2 \sinh ^ { 2 } x = \cosh 2 x$$
(ii) Solve the equation
$$2 \cosh 2 x + \sinh x = 5 ,$$
giving the answers in an exact logarithmic form.\\
(iii) Show that $\int _ { 0 } ^ { \ln 3 } \sinh ^ { 2 } x \mathrm {~d} x = \frac { 10 } { 9 } - \frac { 1 } { 2 } \ln 3$.\\
(iv) Find the exact value of $\int _ { 3 } ^ { 5 } \sqrt { x ^ { 2 } - 9 } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR MEI FP2 2006 Q4 [18]}}