| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.8 This question requires implicit differentiation of a cubic curve (non-trivial algebra), setting dy/dx = 0 to find stationary points, then solving the resulting system of equations x³ + y³ = 3xy and y = x². While the techniques are standard C3 content, the algebraic manipulation and solving the coupled equations requires more problem-solving than typical textbook exercises. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3 + y^3 = 3xy \Rightarrow 3x^2 + 3y^2\frac{dy}{dx} = 3x\frac{dy}{dx} + 3y\) | B1B1 | LHS, RHS; condone \(3xdy/dx+y\) (missing bracket) if recovered thereafter; or equivalent if re-arranged |
| \((3y^2 - 3x)\frac{dy}{dx} = 3y - 3x^2\) | M1 | Collecting terms in \(dy/dx\) and factorising; ft correct algebra on incorrect expressions with two \(dy/dx\) terms; ignore starting with '\(dy/dx = \ldots\)' unless pursued |
| \(\frac{dy}{dx} = \frac{3y-3x^2}{3y^2-3x} = \frac{y-x^2}{y^2-x}\) | A1cao | NB AG |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| TP when \(y - x^2 = 0 \Rightarrow y = x^2\) | M1 | or \(x = \sqrt{y}\) |
| \(x^3 + x^6 = 3x \cdot x^2\) | M1 | Substituting for \(y\) in implicit equation (allow one slip, e.g. \(x^5\)); or \(x\) for \(y\) (i.e. \(y^{3/2} + y^3 = 3y^{1/2}y\) o.e.) |
| \(x^6 = 2x^3\) | A1 | o.e. (so must be exact); or \(y^{3/2} = 2\) |
| \(x^3 = 2\) (or \(x = 0\)) | ||
| \(x = \sqrt[3]{2}\) | A1cao | \(x = 1.2599\ldots\) is A0 (but can isw \(x = \sqrt[3]{2}\)) |
| [4] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 + y^3 = 3xy \Rightarrow 3x^2 + 3y^2\frac{dy}{dx} = 3x\frac{dy}{dx} + 3y$ | B1B1 | LHS, RHS; condone $3xdy/dx+y$ (missing bracket) if recovered thereafter; or equivalent if re-arranged |
| $(3y^2 - 3x)\frac{dy}{dx} = 3y - 3x^2$ | M1 | Collecting terms in $dy/dx$ and factorising; ft correct algebra on incorrect expressions with two $dy/dx$ terms; ignore starting with '$dy/dx = \ldots$' unless pursued |
| $\frac{dy}{dx} = \frac{3y-3x^2}{3y^2-3x} = \frac{y-x^2}{y^2-x}$ | A1cao | **NB AG** |
| **[4]** | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| TP when $y - x^2 = 0 \Rightarrow y = x^2$ | M1 | or $x = \sqrt{y}$ |
| $x^3 + x^6 = 3x \cdot x^2$ | M1 | Substituting for $y$ in implicit equation (allow one slip, e.g. $x^5$); or $x$ for $y$ (i.e. $y^{3/2} + y^3 = 3y^{1/2}y$ o.e.) |
| $x^6 = 2x^3$ | A1 | o.e. (so must be exact); or $y^{3/2} = 2$ |
| $x^3 = 2$ (or $x = 0$) | | |
| $x = \sqrt[3]{2}$ | A1cao | $x = 1.2599\ldots$ is A0 (but can isw $x = \sqrt[3]{2}$) |
| **[4]** | | |
---4 Fig. 7 shows the curve $x ^ { 3 } + y ^ { 3 } = 3 x y$. The point P is a turning point of the curve.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{09d318c7-27b9-43aa-b4a0-e32ea8bd53c5-1_593_531_1573_805}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { y ^ { 2 } - x }$.\\
(ii) Hence find the exact $x$-coordinate of P .
\hfill \mbox{\textit{OCR MEI C3 Q4 [8]}}