| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Challenging +1.3 This is a multi-part Further Maths question involving hyperbolic functions, requiring quotient rule differentiation, algebraic manipulation to reach tanh x = 1/x, graphical analysis of intersections, and a second derivative calculation with substitution. While it tests multiple techniques and hyperbolic identities, each part follows standard procedures without requiring novel insight—the steps are guided and methodical rather than requiring creative problem-solving. |
| 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 |
| Answer | Marks |
|---|---|
| 10(a) | Uses quotient or product |
| Answer | Marks | Guidance |
|---|---|---|
| derivative | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| numerator equal to 0 | AO2.4 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | AO2.1 | R1 |
| (b)(i) | Sketches tanh x correctly | |
| including asymptotes | AO1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| correctly | AO1.2 | B1 |
| (ii) | Deduces correct number |
| Answer | Marks | Guidance |
|---|---|---|
| FT ‘their’ sketch in (b)(i) | AO2.2a | B1F |
| Q | Marking Instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 10(c) | Finds the second | |
| derivative | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| second derivative | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| part (a) | AO2.2a | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| and contains no slips | AO2.1 | R1 |
| Total | 10 | |
| Q | Marking Instructions | AO |
Question 10:
--- 10(a) ---
10(a) | Uses quotient or product
rule to obtain correct
derivative | AO1.1b | B1 | x dy coshx−xsinhx
y = ⇒ =
coshx dx cosh2x
dy
Stationary point ⇒ =0
dx
coshx−xsinhx
⇒ =0
cosh2x
⇒coshx−xsinhx=0
sinhx 1
⇒ =
cosh x x
1
⇒tanhx=
x
dy
Clearly sets ‘their’
dx
numerator equal to 0 | AO2.4 | R1
Rearranges to complete
a rigorous argument to
show the required result.
AG | AO2.1 | R1
(b)(i) | Sketches tanh x correctly
including asymptotes | AO1.2 | B1
1
Sketches
x
correctly | AO1.2 | B1
(ii) | Deduces correct number
of stationary points
FT ‘their’ sketch in (b)(i) | AO2.2a | B1F | 2 stationary points
Q | Marking Instructions | AO | Marks | Typical Solution
--- 10(c) ---
10(c) | Finds the second
derivative | AO1.1a | M1 | d2y cosh2x(sinhx−xcoshx−sinhx)
=
dx2 cosh4x
2coshxsinhx(coshx−xsinhx)
−
cosh4x
second term is zero at stationary points
d2y x
= − = −y
dx2 coshx
d2y
⇒ + y =0
dx2
Obtains a correct
expression for the
second derivative | AO1.1b | A1
Deduces that the
second term is zero
by using results from
part (a) | AO2.2a | R1
Completes a rigorous
argument to show the
required result. AG
Mark awarded if they
have a completely
correct solution, which
is clear, easy to follow
and contains no slips | AO2.1 | R1
Total | 10
Q | Marking Instructions | AO | Marks | Typical SolutionThe curve, $C$, has equation $y = \frac{x}{\cosh x}$
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinates of any stationary points of $C$ satisfy the equation $\tanh x = \frac{1}{x}$
[3 marks]
\item \begin{enumerate}[label=(\roman*)]
\item Sketch the graphs of $y = \tanh x$ and $y = \frac{1}{x}$ on the axes below.
[2 marks]
\item Hence determine the number of stationary points of the curve $C$.
[1 mark]
\end{enumerate}
\item Show that $\frac{d^2y}{dx^2} + y = 0$ at each of the stationary points of the curve $C$.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 Q10 [10]}}