| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| 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 involving inverse hyperbolic and trigonometric functions with chain rule differentiation. Part (a) is guided standard work (differentiating cos y = t implicitly). Part (b) requires computing d²y/dx² = (dy'/dt)/(dx/dt) using quotient rule and knowing d(sinh⁻¹t)/dt = 1/√(1+t²), then simplifying. While it involves Further Maths content and multi-step calculus, the techniques are standard applications without requiring novel insight—moderately above average difficulty. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\sin y \frac{dy}{dt} = 1\) | M1 A1 | Differentiates both sides with respect to \(t\) |
| \(0 < y < \pi \Rightarrow \sin y > 0 \Rightarrow -\sqrt{1 - \cos^2 y}\frac{dy}{dt} = 1\) | M1 | Applies \(\sin^2 y + \cos^2 y = 1\) |
| \(\frac{dy}{dt} = -\frac{1}{\sqrt{1-t^2}}\) | A1 | AG, justifies taking positive square root |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{dt} = \frac{1}{\sqrt{1+t^2}}\) | B1 | |
| \(\frac{dy}{dx} = -\sqrt{\frac{1+t^2}{1-t^2}}\) | B1 | Finds first derivative |
| \(\frac{d}{dt}\left(-\sqrt{\frac{1+t^2}{1-t^2}}\right) = -\frac{t(1-t^2)^{\frac{1}{2}}(1+t^2)^{-\frac{1}{2}} + t(1+t^2)^{\frac{1}{2}}(1-t^2)^{-\frac{1}{2}}}{1-t^2}\) | M1 | Differentiates \(\frac{dy}{dx}\) with respect to \(t\) |
| \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(-\sqrt{\frac{1+t^2}{1-t^2}}\right) \times \frac{dt}{dx}\) | M1 | Applies chain rule |
| \(= -\frac{t\left((1-t^2)^{\frac{1}{2}} + (1+t^2)(1-t^2)^{-\frac{1}{2}}\right)}{1-t^2} \left(= -\frac{2t}{(1-t^2)^{\frac{3}{2}}}\right)\) | A1 | OE (simplified) |
| Total: 5 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\sin y \frac{dy}{dt} = 1$ | M1 A1 | Differentiates both sides with respect to $t$ |
| $0 < y < \pi \Rightarrow \sin y > 0 \Rightarrow -\sqrt{1 - \cos^2 y}\frac{dy}{dt} = 1$ | M1 | Applies $\sin^2 y + \cos^2 y = 1$ |
| $\frac{dy}{dt} = -\frac{1}{\sqrt{1-t^2}}$ | A1 | AG, justifies taking positive square root |
| **Total: 4** | | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = \frac{1}{\sqrt{1+t^2}}$ | B1 | |
| $\frac{dy}{dx} = -\sqrt{\frac{1+t^2}{1-t^2}}$ | B1 | Finds first derivative |
| $\frac{d}{dt}\left(-\sqrt{\frac{1+t^2}{1-t^2}}\right) = -\frac{t(1-t^2)^{\frac{1}{2}}(1+t^2)^{-\frac{1}{2}} + t(1+t^2)^{\frac{1}{2}}(1-t^2)^{-\frac{1}{2}}}{1-t^2}$ | M1 | Differentiates $\frac{dy}{dx}$ with respect to $t$ |
| $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(-\sqrt{\frac{1+t^2}{1-t^2}}\right) \times \frac{dt}{dx}$ | M1 | Applies chain rule |
| $= -\frac{t\left((1-t^2)^{\frac{1}{2}} + (1+t^2)(1-t^2)^{-\frac{1}{2}}\right)}{1-t^2} \left(= -\frac{2t}{(1-t^2)^{\frac{3}{2}}}\right)$ | A1 | OE (simplified) |
| **Total: 5** | | |5 It is given that
$$x = \sinh ^ { - 1 } t , \quad y = \cos ^ { - 1 } t$$
where $- 1 < t < 1$.
\begin{enumerate}[label=(\alph*)]
\item By differentiating $\cos y$ with respect to $t$, show that $\frac { d y } { d t } = - \frac { 1 } { \sqrt { 1 - t ^ { 2 } } }$.
\item Find $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ in terms of $t$, simplifying your answer.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q5 [9]}}