| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.8 This is a Further Maths implicit differentiation question requiring students to find vertical tangent points by setting the denominator to zero. While the differentiation in part (a) is routine, part (b) requires the insight that furthest west/east corresponds to vertical tangents (dy/dx undefined), then solving the resulting simultaneous equations. This conceptual leap and multi-step algebraic manipulation elevates it above standard A-level questions. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either \(3y^2 \to Ay\frac{dy}{dx}\) or \(2xy \to 2x\frac{dy}{dx}+2y\) | M1 | Selecting appropriate implicit differentiation method; condone \(-2xy\to -2x\frac{dy}{dx}+2y\) unless evidence of incorrect product rule \(vu'-uv'\) |
| \(2x - 2x\frac{dy}{dx} - 2y + 6y\frac{dy}{dx} = 0\) | A1 | Fully correct derivative; allow \(2x\,dx - 2x\,dy - 2y\,dx + 6y\,dy = 0\) |
| \((6y-2x)\frac{dy}{dx} = 2y-2x\) | M1 | Collecting \(\frac{dy}{dx}\) terms |
| \(\frac{dy}{dx} = \frac{2y-2x}{6y-2x} = \frac{y-x}{3y-x}\) | A1* | Completes proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(P\) and \(Q\), \(\frac{dy}{dx}\to\infty \Rightarrow\) deduces \(3y-x=0\) | M1 | |
| Solves \(y=\frac{1}{3}x\) and \(x^2-2xy+3y^2=50\) simultaneously | M1 | |
| \(x=(\pm)5\sqrt{3}\) OR \(y=(\pm)\frac{5}{3}\sqrt{3}\) | A1 | |
| Using \(y=\frac{1}{3}x\) to find both \(x=\ldots\) AND \(y=\ldots\) | dM1 | |
| \(P=\left(-5\sqrt{3},\,-\frac{5}{3}\sqrt{3}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Explains that you need to solve \(y=x\) and \(x^2-2xy+3y^2=50\) simultaneously and choose the positive solution | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{2y-2x}{6y-2x}\) | M1 | Valid attempt making \(\frac{dy}{dx}\) subject with two terms in \(\frac{dy}{dx}\) coming from \(3y^2\) and \(2xy\) |
| \(\frac{dy}{dx} = \frac{y-x}{3y-x}\) | A1* | No errors or omissions; previous line \(\frac{dy}{dx} = \frac{2y-2x}{6y-2x}\) must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(3y - x = 0\) | M1 | oe |
| Solves \(y = \frac{1}{3}x\) with \(x^2 - 2xy + 3y^2 = 50\) simultaneously | M1 | Allow for finding quadratic in \(x\) or \(y\); may be awarded using \(y=x\) |
| \(x = (\pm)5\sqrt{3}\) OR \(y = (\pm)\frac{5}{3}\sqrt{3}\) | A1 | |
| Finds \(y\) coordinate from \(x\) (or vice versa) | dM1 | Dependent on previous M; also may follow from numerator \(= 0\) using \(y=x\) |
| \(P = \left(-5\sqrt{3}, -\frac{5}{3}\sqrt{3}\right)\) | A1 | OE; allow \(x=\ldots\), \(y=\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Explains \(\frac{dy}{dx}=0\) where \(y=x\), solve \(y=x\) and \(x^2-2xy+3y^2=50\) simultaneously, choose positive solution | B1ft | Alternatively: complete the square \((x-y)^2+2y^2=50\), state max \(y\) when \(x=y\), choose positive solution from \(2y^2=50\) |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either $3y^2 \to Ay\frac{dy}{dx}$ or $2xy \to 2x\frac{dy}{dx}+2y$ | M1 | Selecting appropriate implicit differentiation method; condone $-2xy\to -2x\frac{dy}{dx}+2y$ unless evidence of incorrect product rule $vu'-uv'$ |
| $2x - 2x\frac{dy}{dx} - 2y + 6y\frac{dy}{dx} = 0$ | A1 | Fully correct derivative; allow $2x\,dx - 2x\,dy - 2y\,dx + 6y\,dy = 0$ |
| $(6y-2x)\frac{dy}{dx} = 2y-2x$ | M1 | Collecting $\frac{dy}{dx}$ terms |
| $\frac{dy}{dx} = \frac{2y-2x}{6y-2x} = \frac{y-x}{3y-x}$ | A1* | Completes proof with no errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $P$ and $Q$, $\frac{dy}{dx}\to\infty \Rightarrow$ deduces $3y-x=0$ | M1 | |
| Solves $y=\frac{1}{3}x$ and $x^2-2xy+3y^2=50$ simultaneously | M1 | |
| $x=(\pm)5\sqrt{3}$ OR $y=(\pm)\frac{5}{3}\sqrt{3}$ | A1 | |
| Using $y=\frac{1}{3}x$ to find both $x=\ldots$ AND $y=\ldots$ | dM1 | |
| $P=\left(-5\sqrt{3},\,-\frac{5}{3}\sqrt{3}\right)$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Explains that you need to solve $y=x$ and $x^2-2xy+3y^2=50$ simultaneously and choose the positive solution | B1ft | |
# Question (Implicit - Differentiation/Curve):
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{2y-2x}{6y-2x}$ | M1 | Valid attempt making $\frac{dy}{dx}$ subject with two terms in $\frac{dy}{dx}$ coming from $3y^2$ and $2xy$ |
| $\frac{dy}{dx} = \frac{y-x}{3y-x}$ | A1* | No errors or omissions; previous line $\frac{dy}{dx} = \frac{2y-2x}{6y-2x}$ must be seen |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $3y - x = 0$ | M1 | oe |
| Solves $y = \frac{1}{3}x$ with $x^2 - 2xy + 3y^2 = 50$ simultaneously | M1 | Allow for finding quadratic in $x$ or $y$; may be awarded using $y=x$ |
| $x = (\pm)5\sqrt{3}$ OR $y = (\pm)\frac{5}{3}\sqrt{3}$ | A1 | |
| Finds $y$ coordinate from $x$ (or vice versa) | dM1 | Dependent on previous M; also may follow from numerator $= 0$ using $y=x$ |
| $P = \left(-5\sqrt{3}, -\frac{5}{3}\sqrt{3}\right)$ | A1 | OE; allow $x=\ldots$, $y=\ldots$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Explains $\frac{dy}{dx}=0$ where $y=x$, solve $y=x$ and $x^2-2xy+3y^2=50$ simultaneously, choose positive solution | B1ft | Alternatively: complete the square $(x-y)^2+2y^2=50$, state max $y$ when $x=y$, choose positive solution from $2y^2=50$ |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b5f50f17-9f1b-4b4c-baf3-e50de5f2ea9c-22_537_748_242_662}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve with equation $x ^ { 2 } - 2 x y + 3 y ^ { 2 } = 50$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x } { 3 y - x }$
The curve is used to model the shape of a cycle track with both $x$ and $y$ measured in km .\\
The points $P$ and $Q$ represent points that are furthest west and furthest east of the origin $O$, as shown in Figure 4.
Using part (a),
\item find the exact coordinates of the point $P$.
\item Explain briefly how to find the coordinates of the point that is furthest north of the origin $O$. (You do not need to carry out this calculation).
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2018 Q9 [10]}}