| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question with straightforward application of the product rule and algebraic manipulation. Part (a) requires routine differentiation technique, while part (b) involves solving simultaneous equations by substitution—both are typical C4 exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(x^2 - 3xy - 4y^2 + 64 = 0\) | |
| \(\frac{dx}{dy} = 2x\left(\frac{3y}{dx}\right) - 3y - 3x\frac{dy}{dx} - 8y\frac{dy}{dx} = 0\) | M1 A1 |
| M1 | |
| \(2x - 3y + (-3x - 8y)\frac{dy}{dx} = 0\) | dM1 |
| \(\frac{dy}{dx} = \frac{2x-3y}{3x+8y}\) or \(\frac{3y-2x}{-3x-8y}\) | o.e. A1 cso |
| [5] |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = 0 \Rightarrow 2x - 3y = 0\) | M1 |
| \(y = \frac{2}{3}x\) | A1 in |
| \(x^2 - 3x\left(\frac{2}{3}x\right) - 4\left(\frac{2}{3}x\right)^2 + 64 = 0\) | dM1 |
| \(x^2 - 2x^2 - \frac{16}{9}x^2 + 64 = 0 \Rightarrow -\frac{25}{9}x^2 + 64 = 0\) | |
| \(\Rightarrow x^2 = \frac{256}{25} \Rightarrow x = \frac{24}{5}\) or \(-\frac{24}{5}\) | A1 cso |
| When \(x = \frac{24}{5}\), \(y = \frac{2}{3}\left(\frac{24}{5}\right)\) and \(-\frac{3}{5}\left(\frac{24}{5}\right)\) | |
| \(\left(\frac{24}{5}, \frac{16}{5}\right)\) and \(\left(-\frac{24}{5}, -\frac{16}{5}\right)\) or \(x = \frac{24}{5}, y = \frac{16}{5}\) and \(x = -\frac{24}{5}, y = -\frac{16}{5}\) | ddM1 |
| \(\left(\frac{24}{5}, -\frac{16}{5}\right)\) | cso A1 |
$x^2 - 3xy - 4y^2 + 64 = 0$ | |
$\frac{dx}{dy} = 2x\left(\frac{3y}{dx}\right) - 3y - 3x\frac{dy}{dx} - 8y\frac{dy}{dx} = 0$ | M1 A1 |
| M1 |
$2x - 3y + (-3x - 8y)\frac{dy}{dx} = 0$ | dM1 |
$\frac{dy}{dx} = \frac{2x-3y}{3x+8y}$ or $\frac{3y-2x}{-3x-8y}$ | o.e. A1 cso |
| [5] |
## Part (b)
$\frac{dy}{dx} = 0 \Rightarrow 2x - 3y = 0$ | M1 |
$y = \frac{2}{3}x$ | A1 in |
$x^2 - 3x\left(\frac{2}{3}x\right) - 4\left(\frac{2}{3}x\right)^2 + 64 = 0$ | dM1 |
$x^2 - 2x^2 - \frac{16}{9}x^2 + 64 = 0 \Rightarrow -\frac{25}{9}x^2 + 64 = 0$ | |
$\Rightarrow x^2 = \frac{256}{25} \Rightarrow x = \frac{24}{5}$ or $-\frac{24}{5}$ | A1 cso |
When $x = \frac{24}{5}$, $y = \frac{2}{3}\left(\frac{24}{5}\right)$ and $-\frac{3}{5}\left(\frac{24}{5}\right)$ | |
$\left(\frac{24}{5}, \frac{16}{5}\right)$ and $\left(-\frac{24}{5}, -\frac{16}{5}\right)$ or $x = \frac{24}{5}, y = \frac{16}{5}$ and $x = -\frac{24}{5}, y = -\frac{16}{5}$ | ddM1 |
$\left(\frac{24}{5}, -\frac{16}{5}\right)$ | cso A1 |
---The curve $C$ has equation
$$x^2 - 3xy - 4y^2 + 64 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$ in terms of $x$ and $y$.
[5]
\item Find the coordinates of the points on $C$ where $\frac{dy}{dx} = 0$
(Solutions based entirely on graphical or numerical methods are not acceptable.)
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2015 Q2 [11]}}