| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove hyperbolic identity from exponentials |
| Difficulty | Standard +0.3 Part (i) is a straightforward derivation using the exponential definition of cosh x and basic algebra—standard for Further Maths students who've learned hyperbolic functions. Part (ii) requires solving for cosh x from the double angle formula then using the identity cosh²x - sinh²x = 1, which is routine manipulation. This is easier than average even for FM students, as it's a foundational exercise testing basic hyperbolic identities rather than problem-solving. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\text{RHS} = 2\left[\frac{1}{2}(e^x+e^{-x})\right]^2 - 1 = \frac{1}{2}(e^{2x}+e^{-2x}) = \text{LHS}\) | M1 | For correct squaring of \((e^x+e^{-x})\) |
| \(-\) | A1 | For completely correct proof |
| (ii) \(2\cosh^2 x - 1 = k \Rightarrow \cosh x = \pm\sqrt{\frac{1}{2}(1+k)}\) | M1 | For use of (i) and solving for \(\cosh x\) |
| \(-\) | A1 | For correct positive square root only |
| \(2\sinh^2 x + 1 = k \Rightarrow \sinh x = \pm\sqrt{\frac{1}{2}(k-1)}\) | M1 | For use of \(\cosh^2 x - \sinh^2 x = 1\), or equivalent |
| \(-\) | A1 | For both correct square roots |
| \(-\) | 6 |
(i) $\text{RHS} = 2\left[\frac{1}{2}(e^x+e^{-x})\right]^2 - 1 = \frac{1}{2}(e^{2x}+e^{-2x}) = \text{LHS}$ | M1 | For correct squaring of $(e^x+e^{-x})$ |
$-$ | A1 | For completely correct proof |
(ii) $2\cosh^2 x - 1 = k \Rightarrow \cosh x = \pm\sqrt{\frac{1}{2}(1+k)}$ | M1 | For use of (i) and solving for $\cosh x$ |
$-$ | A1 | For correct positive square root only |
$2\sinh^2 x + 1 = k \Rightarrow \sinh x = \pm\sqrt{\frac{1}{2}(k-1)}$ | M1 | For use of $\cosh^2 x - \sinh^2 x = 1$, or equivalent |
$-$ | A1 | For both correct square roots |
$-$ | **6** | |
---1 (i) Starting from the definition of $\cosh x$ in terms of $\mathrm { e } ^ { x }$, show that $\cosh 2 x = 2 \cosh ^ { 2 } x - 1$.\\
(ii) Given that $\cosh 2 x = k$, where $k > 1$, express each of $\cosh x$ and $\sinh x$ in terms of $k$.
\hfill \mbox{\textit{OCR FP2 Q1 [6]}}