| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Standard +0.8 This is a Further Maths question requiring manipulation of hyperbolic functions. Part (a) involves algebraic manipulation with a common denominator and applying the identity cosh²θ - sinh²θ = 1, which is standard but requires careful execution. Part (b) is straightforward once part (a) is established. The question tests technique rather than deep insight, making it moderately above average difficulty but not exceptional 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 |
|---|---|
| 11(a) | Commences proof by considering |
| Answer | Marks | Guidance |
|---|---|---|
| divided by RHS | AO2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| eliminate (or introduce) sinh2 θ | AO2.4 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| factorised expressions) | AO1.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| is clear and contains no slips. | AO2.1 | R1 |
| (b) | Uses result from part (a) to deduce |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | AO2.2a | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| correct exact form | AO1.1b | A1 |
| Total | 6 | |
| Q | Marking Instructions | AO |
Question 11:
--- 11(a) ---
11(a) | Commences proof by considering
one side of the identity only: if
considering LHS combines terms as
a single fraction with a common
denominator.
If considering RHS writes coth θ as
a fraction and introduces factor of
(1 + cosh θ to both numerator and
denominator)
Note alternative valid approaches
include commencing proof by
considering LHS minus RHS or LHS
divided by RHS | AO2.1 | R1 | sinhθ 1+coshθ
+
1+coshθ sinhθ
sinh2θ+1+cosh2θ+2coshθ
≡
(1+coshθ)sinhθ
cosh2θ+cosh2θ+2coshθ
≡ ,
(1+coshθ)sinhθ
1+sinh2θ≡cosh2θ
2coshθ(1+coshθ)
≡
(1+coshθ)sinhθ
2coshθ
≡
sinhθ
≡2cothθ
Explicitly states identity
cosh2 θ − sinh2 θ ≡ 1 and uses it to
eliminate (or introduce) sinh2 θ | AO2.4 | R1
Factorises numerator and cancels
correctly for ‘their’ fraction
(if considering RHS rearranges
‘their’ numerator correctly into two
factorised expressions) | AO1.1b | B1F
Completes rigorous proof to obtain
result AG
Only award if they have a
completely correct argument, which
is clear and contains no slips. | AO2.1 | R1
(b) | Uses result from part (a) to deduce
1
that tanh θ =
2 | AO2.2a | R1 | 2coth θ = 4
1
tanh θ =
2
1 1
θ = tanh-1 = ln 3
2 2
Uses natural log form and
substitutes correct value to obtain
correct exact form | AO1.1b | A1
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution\begin{enumerate}[label=(\alph*)]
\item Prove that $\frac{\sinh \theta}{1 + \cosh \theta} + \frac{1 + \cosh \theta}{\sinh \theta} \equiv 2\coth \theta$
Explicitly state any hyperbolic identities that you use within your proof.
[4 marks]
\item Solve $\frac{\sinh \theta}{1 + \cosh \theta} + \frac{1 + \cosh \theta}{\sinh \theta} = 4$ giving your answer in an exact form.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 Q11 [6]}}