| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring fluency with hyperbolic functions including inverse forms, integration via substitution, solving equations with double angle formulas, and differentiation. While the techniques are standard for FP2, the combination of four parts testing different aspects and requiring exact logarithmic answers elevates it above average difficulty. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}(e^x+e^{-x})=k\), \(e^{2x}-2ke^x+1=0\) | M1, M1 | or \(\cosh x+\sinh x=e^x\); or \(k \pm \sqrt{k^2-1}=e^x\) |
| \(e^x=\frac{2k\pm\sqrt{4k^2-4}}{2}=k\pm\sqrt{k^2-1}\) | A1 | One value sufficient |
| \(x=\ln(k+\sqrt{k^2-1})\) or \(\ln(k-\sqrt{k^2-1})\) | A1 | |
| \((k+\sqrt{k^2-1})(k-\sqrt{k^2-1})=k^2-(k^2-1)=1\) | M1 | or \(\cosh x\) is an even function (or equivalent) |
| \(\ln(k-\sqrt{k^2-1})=\ln\left(\frac{1}{k+\sqrt{k^2-1}}\right)=-\ln(k+\sqrt{k^2-1})\) | A1 ag | |
| \(x=\pm\ln(k+\sqrt{k^2-1})\) | ||
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_1^2 \frac{1}{\sqrt{4x^2-1}}dx = \left[\frac{1}{2}\text{arcosh}2x\right]_1^2\) | M1, A1 | For arcosh or \(\ln(2x+\sqrt{2x^2-\ldots})\) or any cosh substitution; For arcosh \(2x\) or \(2x=\cosh u\) or \(\ln(2x+\sqrt{4x^2-1})\) or \(\ln(x+\sqrt{x^2-\frac{1}{4}})\); For \(\frac{1}{2}\) or \(\frac{1}{4}du\) |
| \(= \frac{1}{2}(\text{arcosh}4-\text{arcosh}2) = \frac{1}{2}\left(\ln(4+\sqrt{15})-\ln(2+\sqrt{3})\right)\) | M1, A1 | Exact numerical logarithmic form |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(6\sinh x-2\sinh x\cosh x=0\), \(\cosh x=3\) (or \(\sinh x=0\)) | M1, M1 | Obtaining a value for \(\cosh x\) |
| \(x=0\) | B1 | |
| \(x=\pm\ln(3+\sqrt{8})\) | A1 | or \(x=\ln(3\pm\sqrt{8})\) |
| 4 | ||
| OR \(e^{4x}-6e^{3x}+6e^x-1=0\), \((e^{2x}-1)(e^{2x}-6e^x+1)=0\) | M2 | or \((e^x-e^{-x})(e^x+e^{-x}-6)=0\) |
| \(x=0\) | B1 | |
| \(x=\ln(3\pm\sqrt{8})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx}=6\cosh x-2\cosh 2x\) | B1 | |
| If \(\frac{dy}{dx}=5\) then \(6\cosh x-2(2\cosh^2 x-1)=5\) | M1 | Using \(\cosh 2x=2\cosh^2 x-1\) |
| \(4\cosh^2 x-6\cosh x+3=0\) | M1 | |
| Discriminant \(D=6^2-4\times 4\times 3=-12\) | M1 | Considering \(D\), or completing square, or considering turning point |
| Since \(D<0\) there are no solutions | A1 | |
| 4 | ||
| OR Gradient \(g=6\cosh x-2\cosh 2x\) | B1 | |
| \(g'=6\sinh x-4\sinh 2x=2\sinh x(3-4\cosh x) = 0\) when \(x=0\) (only) | M1 | |
| \(g''=6\cosh x-8\cosh 2x=-2\) when \(x=0\) | M1 | |
| Max value \(g=4\) when \(x=0\) | A1 | |
| So \(g\) is never equal to 5 | A1 | Final A1 requires a complete proof showing this is the only turning point |
## 4(i)
| $\frac{1}{2}(e^x+e^{-x})=k$, $e^{2x}-2ke^x+1=0$ | M1, M1 | or $\cosh x+\sinh x=e^x$; or $k \pm \sqrt{k^2-1}=e^x$ |
|---|---|---|
| $e^x=\frac{2k\pm\sqrt{4k^2-4}}{2}=k\pm\sqrt{k^2-1}$ | A1 | One value sufficient |
|---|---|---|
| $x=\ln(k+\sqrt{k^2-1})$ or $\ln(k-\sqrt{k^2-1})$ | A1 | |
|---|---|---|
| $(k+\sqrt{k^2-1})(k-\sqrt{k^2-1})=k^2-(k^2-1)=1$ | M1 | or $\cosh x$ is an even function (or equivalent) |
|---|---|---|
| $\ln(k-\sqrt{k^2-1})=\ln\left(\frac{1}{k+\sqrt{k^2-1}}\right)=-\ln(k+\sqrt{k^2-1})$ | A1 ag | |
|---|---|---|
| $x=\pm\ln(k+\sqrt{k^2-1})$ | | |
|---|---|---|
| | 5 | |
## 4(ii)
| $\int_1^2 \frac{1}{\sqrt{4x^2-1}}dx = \left[\frac{1}{2}\text{arcosh}2x\right]_1^2$ | M1, A1 | For arcosh or $\ln(2x+\sqrt{2x^2-\ldots})$ or any cosh substitution; For arcosh $2x$ or $2x=\cosh u$ or $\ln(2x+\sqrt{4x^2-1})$ or $\ln(x+\sqrt{x^2-\frac{1}{4}})$; For $\frac{1}{2}$ or $\frac{1}{4}du$ |
|---|---|---|
| $= \frac{1}{2}(\text{arcosh}4-\text{arcosh}2) = \frac{1}{2}\left(\ln(4+\sqrt{15})-\ln(2+\sqrt{3})\right)$ | M1, A1 | Exact numerical logarithmic form |
|---|---|---|
| | 5 | |
## 4(iii)
| $6\sinh x-2\sinh x\cosh x=0$, $\cosh x=3$ (or $\sinh x=0$) | M1, M1 | Obtaining a value for $\cosh x$ |
|---|---|---|
| $x=0$ | B1 | |
|---|---|---|
| $x=\pm\ln(3+\sqrt{8})$ | A1 | or $x=\ln(3\pm\sqrt{8})$ |
|---|---|---|
| | 4 | |
| OR $e^{4x}-6e^{3x}+6e^x-1=0$, $(e^{2x}-1)(e^{2x}-6e^x+1)=0$ | M2 | or $(e^x-e^{-x})(e^x+e^{-x}-6)=0$ |
|---|---|---|
| $x=0$ | B1 | |
|---|---|---|
| $x=\ln(3\pm\sqrt{8})$ | A1 | |
|---|---|---|
## 4(iv)
| $\frac{dy}{dx}=6\cosh x-2\cosh 2x$ | B1 | |
|---|---|---|
| If $\frac{dy}{dx}=5$ then $6\cosh x-2(2\cosh^2 x-1)=5$ | M1 | Using $\cosh 2x=2\cosh^2 x-1$ |
|---|---|---|
| $4\cosh^2 x-6\cosh x+3=0$ | M1 | |
|---|---|---|
| Discriminant $D=6^2-4\times 4\times 3=-12$ | M1 | Considering $D$, or completing square, or considering turning point |
|---|---|---|
| Since $D<0$ there are no solutions | A1 | |
|---|---|---|
| | 4 | |
| OR Gradient $g=6\cosh x-2\cosh 2x$ | B1 | |
|---|---|---|
| $g'=6\sinh x-4\sinh 2x=2\sinh x(3-4\cosh x) = 0$ when $x=0$ (only) | M1 | |
|---|---|---|
| $g''=6\cosh x-8\cosh 2x=-2$ when $x=0$ | M1 | |
|---|---|---|
| Max value $g=4$ when $x=0$ | A1 | |
|---|---|---|
| So $g$ is never equal to 5 | A1 | Final A1 requires a complete proof showing this is the only turning point |
|---|---|---|4 (i) Given that $k \geqslant 1$ and $\cosh x = k$, show that $x = \pm \ln \left( k + \sqrt { k ^ { 2 } - 1 } \right)$.\\
(ii) Find $\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 1 } } \mathrm {~d} x$, giving the answer in an exact logarithmic form.\\
(iii) Solve the equation $6 \sinh x - \sinh 2 x = 0$, giving the answers in an exact form, using logarithms where appropriate.\\
(iv) Show that there is no point on the curve $y = 6 \sinh x - \sinh 2 x$ at which the gradient is 5 .
\hfill \mbox{\textit{OCR MEI FP2 2008 Q4 [18]}}