| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Challenging +1.2 Part (a) requires identifying a common algebraic error (dividing by an expression that could be zero) in a hyperbolic function context—straightforward for Further Maths students. Part (b) involves standard manipulation using hyperbolic identities (cosh²-sinh²=1) to form a quadratic in cosh(2x), then solving and applying inverse hyperbolic functions. While multi-step, these are routine techniques for Further Maths students with no novel insight required. The 7-mark total and straightforward structure place this slightly above average difficulty. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks |
|---|---|
| 9(a) | Identifies the 2nd line or as the error. |
| Answer | Marks | Guidance |
|---|---|---|
| 1−cosh 𝑥𝑥 | 2.3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| sinh 𝑥𝑥 1−cosh 𝑥𝑥 | 2.4 | E1 |
| Answer | Marks |
|---|---|
| 9(b) | 2 2 |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. uses to write the equation in terms of only | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Possibly implied by a correct value for their function. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| cosh2𝑥𝑥 < 1 | 2.4 | E1F |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| ±2cosh �1+√3� Total | 7 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Q | Marking instructions | AO |
Question 9:
--- 9(a) ---
9(a) | Identifies the 2nd line or as the error.
Possibly implied by the explanatio2n.
1−cosh 𝑥𝑥 | 2.3 | B1 | In the 2nd line, should not have
been replaced with2
sinh 𝑥𝑥
2
1−cosh 𝑥𝑥
Explains that should not be replaced with
Accept any correct2 arrangement of the identity 2
sinh 𝑥𝑥 1−cosh 𝑥𝑥 | 2.4 | E1
--- 9(b) ---
9(b) | 2 2
Uses a correct identity to express the equation cino stherm𝑥𝑥s− osfi nohne𝑥𝑥 fu=n1ction.
e.g. uses to write the equation in terms of only | 1.1b | B1 | 2
cosh (2𝑥𝑥)−1−2cosh(2𝑥𝑥)= 1
2
cosh (2𝑥𝑥)−2cosh(2𝑥𝑥)−2 = 0
cosh(2𝑥𝑥)= 1±√3
but
cosh(2𝑥𝑥)≥ 1 ∴ cosh(2𝑥𝑥)= 1+√3
1 −1
𝑥𝑥 = ± cosh �1+√3�
2 2
Rearrangecso sthhe(ir2 e𝑥𝑥q)u−astiionnh in(2to𝑥𝑥 )a =qu1adratic=0 or quartic=0 cosh(2𝑥𝑥)
Must be in terms of just one function.
Possibly implied by a correct value for their function. | 1.1a | M1
Finds a correct exact value for their function,
e.g.
√3
ISW cionschor𝑥𝑥re=ct± w�o1rk+ fo llowing a correct unsimplified answer.
2 | 1.1b | A1
Correctly explains why they reject at least one of their solutions,
e.g. cannot be a solution.
Follow through B1M1 only.
cosh2𝑥𝑥 < 1 | 2.4 | E1F
Finds both correct sets of values for p, q and r and no others.
Accept for full marks.
1 | 1.1b | A1
−1
±2cosh �1+√3� Total | 7 | 2
𝑝𝑝 = ±2 , 𝑞𝑞 = 1 , 𝑟𝑟 = 3
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item Saul is solving the equation
$$2\cosh x + \sinh^2 x = 1$$
He writes his steps as follows:
$$2\cosh x + \sinh^2 x = 1$$
$$2\cosh x + 1 - \cosh^2 x = 1$$
$$2\cosh x - \cosh^2 x = 0$$
$$\cosh x \neq 0 \therefore 2 - \cosh x = 0$$
$$\cosh x = 2$$
$$x = \pm \cosh^{-1}(2)$$
Identify and explain the error in Saul's method. [2 marks]
\item Anna is solving the different equation
$$\sinh^2(2x) - 2\cosh(2x) = 1$$
and finds the correct answers in the form $x = \frac{1}{p}\cosh^{-1}(q + \sqrt{r})$, where $p$, $q$ and $r$ are integers.
Find the possible values of $p$, $q$ and $r$.
Fully justify your answer. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2019 Q9 [7]}}