| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Challenging +1.2 This is a standard Further Maths parametric differentiation question requiring the chain rule for first derivative and quotient rule for second derivative. While it involves inverse trig functions and algebraic manipulation, the techniques are routine for Further Maths students and the 'show that' format provides the target answer, making it moderately above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -\frac{1}{\sqrt{1-t^2}} \times (-t^2) = \frac{t^2}{\sqrt{1-t^2}}\) | M1A1 | Uses chain rule, AG |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dt}\!\left(\frac{dy}{dx}\right) = \frac{2t\sqrt{1-t^2} + t^3\left(1-t^2\right)^{-\frac{1}{2}}}{1-t^2}\) | M1A1 | Applies Quotient rule |
| \(\frac{d^2y}{dx^2} = \frac{2t\sqrt{1-t^2} + t^3\left(1-t^2\right)^{-\frac{1}{2}}}{1-t^2} \cdot (-t^2) = -2t^3\left(1-t^2\right)^{-\frac{1}{2}} - t^5\left(1-t^2\right)^{-\frac{3}{2}}\) \(= -t^3\left(1-t^2\right)^{-\frac{3}{2}}\!\left(2\left(1-t^2\right)+t^2\right) = -t^3\left(1-t^2\right)^{-\frac{3}{2}}\!\left(2-t^2\right)\) | M1A1 | Uses chain rule |
| Total: 4 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -\frac{1}{\sqrt{1-t^2}} \times (-t^2) = \frac{t^2}{\sqrt{1-t^2}}$ | M1A1 | Uses chain rule, AG |
| **Total: 2** | | |
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## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dt}\!\left(\frac{dy}{dx}\right) = \frac{2t\sqrt{1-t^2} + t^3\left(1-t^2\right)^{-\frac{1}{2}}}{1-t^2}$ | M1A1 | Applies Quotient rule |
| $\frac{d^2y}{dx^2} = \frac{2t\sqrt{1-t^2} + t^3\left(1-t^2\right)^{-\frac{1}{2}}}{1-t^2} \cdot (-t^2) = -2t^3\left(1-t^2\right)^{-\frac{1}{2}} - t^5\left(1-t^2\right)^{-\frac{3}{2}}$ $= -t^3\left(1-t^2\right)^{-\frac{3}{2}}\!\left(2\left(1-t^2\right)+t^2\right) = -t^3\left(1-t^2\right)^{-\frac{3}{2}}\!\left(2-t^2\right)$ | M1A1 | Uses chain rule |
| **Total: 4** | | |2 It is given that
$$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = \cos ^ { - 1 } t \quad \text { for } 0 < t < 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } } { \sqrt { 1 - t ^ { 2 } } }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-05_2723_33_99_22}
\item Show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - t ^ { a } \left( 1 - t ^ { 2 } \right) ^ { b } \left( 2 - t ^ { 2 } \right)$, where $a$ and $b$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q2 [6]}}