| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| 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 parametric differentiation question requiring knowledge of derivatives of inverse trig and hyperbolic functions, followed by second derivative calculation. Part (a) involves straightforward application of dy/dx = (dy/dt)/(dx/dt) with standard derivatives, while part (b) requires the chain rule for d²y/dx². The algebra is moderately involved but follows standard techniques. Slightly above average due to Further Maths content and multi-step calculation, but remains a routine exercise for FM students. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07s Parametric and implicit differentiation4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dt} = 1 + \dfrac{1}{\sqrt{1+t^2}}\) | B1 | |
| \(\dfrac{dx}{dt} = -1 + \dfrac{1}{1+t^2}\) | B1 | |
| \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}} \times \dfrac{1+t^2}{-t^2}\) | M1 | Uses chain rule |
| \(-\sqrt{1+t^2}\left(\dfrac{\sqrt{1+t^2}+1}{t^2}\right) = \dfrac{-1+t^2+\sqrt{1+t^2}}{t^2}\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = -\dfrac{t^2\!\left(2t + t(1+t^2)^{-\frac{1}{2}}\right) - 2t\!\left(1+t^2+\sqrt{1+t^2}\right)}{t^4}\) | M1 A1 | Applies product or quotient rule |
| \(\dfrac{d^2y}{dx^2} = \dfrac{t^2(1+t^2)^{-\frac{1}{2}} - 2\!\left(1+\sqrt{1+t^2}\right)}{t^5} \times (1+t^2) = -\dfrac{\left(t^2+2+2\sqrt{t^2+1}\right)\sqrt{1+t^2}}{t^5}\) | M1 A1 | Uses chain rule. Simplification not required for A1 |
| \(= -\dfrac{80}{3}\) when \(t = \dfrac{3}{4}\) | A1 | |
| Total: 5 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dt} = 1 + \dfrac{1}{\sqrt{1+t^2}}$ | B1 | |
| $\dfrac{dx}{dt} = -1 + \dfrac{1}{1+t^2}$ | B1 | |
| $\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}} \times \dfrac{1+t^2}{-t^2}$ | M1 | Uses chain rule |
| $-\sqrt{1+t^2}\left(\dfrac{\sqrt{1+t^2}+1}{t^2}\right) = \dfrac{-1+t^2+\sqrt{1+t^2}}{t^2}$ | A1 | AG |
| **Total: 4** | | |
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## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = -\dfrac{t^2\!\left(2t + t(1+t^2)^{-\frac{1}{2}}\right) - 2t\!\left(1+t^2+\sqrt{1+t^2}\right)}{t^4}$ | M1 A1 | Applies product or quotient rule |
| $\dfrac{d^2y}{dx^2} = \dfrac{t^2(1+t^2)^{-\frac{1}{2}} - 2\!\left(1+\sqrt{1+t^2}\right)}{t^5} \times (1+t^2) = -\dfrac{\left(t^2+2+2\sqrt{t^2+1}\right)\sqrt{1+t^2}}{t^5}$ | M1 A1 | Uses chain rule. Simplification not required for A1 |
| $= -\dfrac{80}{3}$ when $t = \dfrac{3}{4}$ | A1 | |
| **Total: 5** | | |
---4 It is given that
$$x = - t + \tan ^ { - 1 } t \quad \text { and } \quad y = t + \sinh ^ { - 1 } t$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { d y } { d x } = - \frac { t ^ { 2 } + 1 + \sqrt { t ^ { 2 } + 1 } } { t ^ { 2 } }$.
\item Find the value of $\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }$ when $t = \frac { 3 } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q4 [9]}}