| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove hyperbolic identity from exponentials |
| Difficulty | Standard +0.3 Part (a) is a straightforward proof using standard exponential definitions and algebraic manipulation—routine for Further Maths students. Part (b) requires substituting the identity, forming a quadratic in sinh x, and applying discriminant conditions, which is a standard technique. Both parts follow predictable patterns with no novel insight required, making this slightly easier than average even for Further Maths. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cosh x = \frac{1}{2}(e^x + e^{-x})\), \(\sinh x = \frac{1}{2}(e^x - e^{-x})\) | B1 | |
| \(\frac{1}{2}(e^x - e^{-x})^2 + 1 = \frac{1}{2}(e^{2x} - e^{-2x}) = \cosh 2x\) | M1 A1 | Expands, AG |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\sinh^2 x - k\sinh x + 1 = 0\) | M1 A1 | Applies identity |
| \(k^2 - 8 > 0\) | M1 A1 | Sets discriminant positive |
| \(k < -\sqrt{8},\ k > \sqrt{8}\) | A1 | |
| Total | 5 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh x = \frac{1}{2}(e^x + e^{-x})$, $\sinh x = \frac{1}{2}(e^x - e^{-x})$ | B1 | |
| $\frac{1}{2}(e^x - e^{-x})^2 + 1 = \frac{1}{2}(e^{2x} - e^{-2x}) = \cosh 2x$ | M1 A1 | Expands, AG |
| **Total** | **3** | |
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## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\sinh^2 x - k\sinh x + 1 = 0$ | M1 A1 | Applies identity |
| $k^2 - 8 > 0$ | M1 A1 | Sets discriminant positive |
| $k < -\sqrt{8},\ k > \sqrt{8}$ | A1 | |
| **Total** | **5** | |2
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of cosh and sinh in terms of exponentials, prove that
$$\cosh 2 x = 2 \sinh ^ { 2 } x + 1$$
\item Find the set of values of $k$ for which $\cosh 2 \mathrm { x } = \mathrm { ksinh } \mathrm { x }$ has two distinct real roots.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q2 [8]}}