| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.8 This is a Further Maths implicit differentiation question requiring two differentiations. Part (a) is routine implicit differentiation to verify a given gradient. Part (b) requires finding the second derivative, which involves differentiating the first derivative expression (itself containing dy/dx terms) and substituting known values—a multi-step process requiring careful algebraic manipulation. While systematic, it's more demanding than standard single-variable calculus and above average difficulty for A-level, though still a recognizable technique for Further Maths students. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dx}(8y^3) = 24y^2 y'\) | B1 | Differentiates \(y^3\) correctly |
| \(\frac{d}{dx}(2xy) = 2(xy' + y)\) | B1 | Differentiates \(xy\) correctly |
| \(3(-2)^2 + 2(-2)y' + 2(-1) + 24(-1)^2 y' = 0 \Rightarrow 10 + 20y' = 0 \Rightarrow y' = -\frac{1}{2}\) | B1 | Substitutes \((-2,-1)\) into \(3x^2 + 2xy' + 2y + 24y^2 y' = 0\), AG. Must see intermediary step between substitution and final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6x + 2xy'' + 2y'\) | B1 | Differentiates \(3x^2 + 2xy'\) |
| \(+ 2y' + 24y^2 y'' + 24y'(2y)y' = 0\) | B1 B1 | Differentiates \(2y + 24y^2 y'\); B1 for \(2y' + 24y^2 y''\); B1 for \(48y(y')^2\) (allow alternative using quotient rule) |
| \(6(-2) + 2(-2)y'' + 4\left(-\frac{1}{2}\right) + 24(-1)^2 y'' + 48(-1)\left(-\frac{1}{2}\right)^2 = 0 \Rightarrow 20y'' - 26 = 0\) | M1 | Substitutes \((-2,-1)\) into their expression (must have derivatives \(y'\) and \(y''\)) |
| \(y'' = \frac{13}{10}\) | A1 | OE |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}(8y^3) = 24y^2 y'$ | B1 | Differentiates $y^3$ correctly |
| $\frac{d}{dx}(2xy) = 2(xy' + y)$ | B1 | Differentiates $xy$ correctly |
| $3(-2)^2 + 2(-2)y' + 2(-1) + 24(-1)^2 y' = 0 \Rightarrow 10 + 20y' = 0 \Rightarrow y' = -\frac{1}{2}$ | B1 | Substitutes $(-2,-1)$ into $3x^2 + 2xy' + 2y + 24y^2 y' = 0$, AG. Must see intermediary step between substitution and final answer |
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## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6x + 2xy'' + 2y'$ | B1 | Differentiates $3x^2 + 2xy'$ |
| $+ 2y' + 24y^2 y'' + 24y'(2y)y' = 0$ | B1 B1 | Differentiates $2y + 24y^2 y'$; B1 for $2y' + 24y^2 y''$; B1 for $48y(y')^2$ (allow alternative using quotient rule) |
| $6(-2) + 2(-2)y'' + 4\left(-\frac{1}{2}\right) + 24(-1)^2 y'' + 48(-1)\left(-\frac{1}{2}\right)^2 = 0 \Rightarrow 20y'' - 26 = 0$ | M1 | Substitutes $(-2,-1)$ into their expression (must have derivatives $y'$ and $y''$) |
| $y'' = \frac{13}{10}$ | A1 | OE |
---3 The curve $C$ has equation
$$x ^ { 3 } + 2 x y + 8 y ^ { 3 } = - 12$$
\begin{enumerate}[label=(\alph*)]
\item Show that, at the point $( - 2 , - 1 )$ on $C , \frac { \mathrm {~d} y } { \mathrm {~d} x } = - \frac { 1 } { 2 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-06_2714_37_143_2008}
\begin{center}
\end{center}
\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at the point $( - 2 , - 1 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q3 [8]}}