| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using substitution u = cosh x or u = sinh x |
| Difficulty | Standard +0.3 This is a standard Further Maths hyperbolic functions question with routine techniques: proving a fundamental identity from definitions, solving via substitution using cosh²x - sinh²x = 1 to get a quadratic, finding stationary points by differentiation, and integrating standard hyperbolic forms. All parts follow textbook methods with no novel insight required, though it's slightly above average difficulty due to being Further Maths content. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions |
4 (i) Starting from the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials, prove that
$$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
(ii) Solve the equation $4 \cosh ^ { 2 } x + 9 \sinh x = 13$, giving the answers in exact logarithmic form.\\
(iii) Show that there is only one stationary point on the curve
$$y = 4 \cosh ^ { 2 } x + 9 \sinh x$$
and find the $y$-coordinate of the stationary point.\\
(iv) Show that $\int _ { 0 } ^ { \ln 2 } \left( 4 \cosh ^ { 2 } x + 9 \sinh x \right) \mathrm { d } x = 2 \ln 2 + \frac { 33 } { 8 }$.
\hfill \mbox{\textit{OCR MEI FP2 2008 Q4 [18]}}