| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Second derivative relations with hyperbolics |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring differentiation of inverse hyperbolic functions and algebraic manipulation to verify given identities. While it involves multiple steps and careful algebra with the chain rule applied twice, the techniques are standard for F3 level—students follow a clear path using known derivatives of arcosh and systematic substitution. The 'show that' format provides the target, reducing problem-solving demand compared to open-ended questions. |
| Spec | 4.07d Differentiate/integrate: hyperbolic functions4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{dy}{dx} = 2\text{arcosh}(3x) \times \dfrac{3}{\sqrt{9x^2-1}}\) | M1 A1 A1 | Chain rule and derivative of \(\text{arcosh}\) |
| \(\sqrt{9x^2-1}\,\dfrac{dy}{dx} = 6\,\text{arcosh}(3x)\) | ||
| \((9x^2-1)\left(\dfrac{dy}{dx}\right)^2 = 36(\text{arcosh}(3x))^2\) | dM1 | Squaring both sides |
| \((9x^2-1)\left(\dfrac{dy}{dx}\right)^2 = 36y\) \(\ast\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left\{18x\left(\dfrac{dy}{dx}\right)^2 + (9x^2-1)\times 2\dfrac{dy}{dx}\times\dfrac{d^2y}{dx^2}\right\} = 36\dfrac{dy}{dx}\) | M1 {A1} A1 | Differentiating implicitly |
| \((9x^2-1)\dfrac{d^2y}{dx^2} + 9x\dfrac{dy}{dx} = 18\) \(\ast\) | A1 | (4) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = 2\text{arcosh}(3x) \times \dfrac{3}{\sqrt{9x^2-1}}$ | M1 A1 A1 | Chain rule and derivative of $\text{arcosh}$ |
| $\sqrt{9x^2-1}\,\dfrac{dy}{dx} = 6\,\text{arcosh}(3x)$ | | |
| $(9x^2-1)\left(\dfrac{dy}{dx}\right)^2 = 36(\text{arcosh}(3x))^2$ | dM1 | Squaring both sides |
| $(9x^2-1)\left(\dfrac{dy}{dx}\right)^2 = 36y$ $\ast$ | A1 | (5) | Using $y = (\text{arcosh}(3x))^2$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left\{18x\left(\dfrac{dy}{dx}\right)^2 + (9x^2-1)\times 2\dfrac{dy}{dx}\times\dfrac{d^2y}{dx^2}\right\} = 36\dfrac{dy}{dx}$ | M1 {A1} A1 | Differentiating implicitly |
| $(9x^2-1)\dfrac{d^2y}{dx^2} + 9x\dfrac{dy}{dx} = 18$ $\ast$ | A1 | (4) | Dividing through by $2\frac{dy}{dx}$ |
---5. Given that $y = ( \operatorname { arcosh } 3 x ) ^ { 2 }$, where $3 x > 1$, show that
\begin{enumerate}[label=(\alph*)]
\item $\left( 9 x ^ { 2 } - 1 \right) \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 36 y$,
\item $\left( 9 x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 9 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 18$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 Q5 [9]}}