| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Standard +0.3 Part (i) is a straightforward proof using the exponential definition of cosh x and algebraic manipulation. Part (ii) requires substituting the double angle formula to get a quadratic in cosh x, then solving using the inverse hyperbolic function. This is a standard Further Maths exercise with clear steps and no novel insight required, making it slightly easier than average overall. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Correct definition of cosh \(x\) or cosh \(2x\) | B1 | |
| Attempt to sub. in RHS and simplify | M1 | or LHS if used |
| Clearly produce A.G. | A1 | |
| (ii) Write as quadratic in cosh \(x\) | M1 | \((2\cosh^2 x - 7\cosh x - 4 = 0)\) |
| Solve their quadratic accurately | A1 | Factorise/quadratic formula |
| Justify one cosh \(x \geq 1\)/graph; allow \(\geq 0\) | B1 | State cosh \(x \geq 1\)/graph; allow \(\geq 0\) |
| Give \(\ln(4 + \sqrt{15})\) | A1 | cao; any one of \(\pm \ln(4 \pm \sqrt{15})\) or decimal equivalent of \(\ln()\) |
**(i)** Correct definition of cosh $x$ or cosh $2x$ | B1 |
Attempt to sub. in RHS and simplify | M1 | or LHS if used
Clearly produce A.G. | A1 |
**(ii)** Write as quadratic in cosh $x$ | M1 | $(2\cosh^2 x - 7\cosh x - 4 = 0)$
Solve their quadratic accurately | A1 | Factorise/quadratic formula
Justify one cosh $x \geq 1$/graph; allow $\geq 0$ | B1 | State cosh $x \geq 1$/graph; allow $\geq 0$
Give $\ln(4 + \sqrt{15})$ | A1 | cao; any one of $\pm \ln(4 \pm \sqrt{15})$ or decimal equivalent of $\ln()$4 (i) Using the definition of $\cosh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, prove that
$$\cosh 2 x = 2 \cosh ^ { 2 } x - 1$$
(ii) Hence solve the equation
$$\cosh 2 x - 7 \cosh x = 3$$
giving your answer in logarithmic form.
\hfill \mbox{\textit{OCR FP2 2006 Q4 [7]}}