| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question requiring routine application of the product rule and chain rule, followed by straightforward algebraic manipulation. Part (ii) is verification only, and part (iii) requires solving simultaneous equations with the derivative equal to a known value. All techniques are core C4 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(2x - 4y = 4x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0\) | M1 A1 |
| \(\frac{dy}{dx} = \frac{2x - 4y}{4x - 4y} = \frac{x - 2y}{2x - 2y}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\text{grad} = \frac{3}{2}\) | M1 |
| \(\therefore y - 2 = \frac{3}{2}(x - 1)\) | M1 |
| \(2y - 4 = 3x - 3\) | |
| \(3x - 2y + 1 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x - 2y}{2x - 2y} = \frac{3}{2}\) | M1 | |
| \(2(x - 2y) = 3(2x - 2y), \quad y = 2x\) | A1 | |
| sub. \(\Rightarrow x^2 - 8x + 8x^2 = 1\) | M1 | |
| \(x^2 = 1, \quad x = 1 \text{ (at } P \text{) or } -1\) | A1 | |
| \(\therefore Q(-1, -2)\) | A1 | (11) |
## (i)
$2x - 4y = 4x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0$ | M1 A1 |
$\frac{dy}{dx} = \frac{2x - 4y}{4x - 4y} = \frac{x - 2y}{2x - 2y}$ | M1 A1 |
## (ii)
$\text{grad} = \frac{3}{2}$ | M1 |
$\therefore y - 2 = \frac{3}{2}(x - 1)$ | M1 |
$2y - 4 = 3x - 3$ | |
$3x - 2y + 1 = 0$ | A1 |
## (iii)
$\frac{x - 2y}{2x - 2y} = \frac{3}{2}$ | M1 |
$2(x - 2y) = 3(2x - 2y), \quad y = 2x$ | A1 |
sub. $\Rightarrow x^2 - 8x + 8x^2 = 1$ | M1 |
$x^2 = 1, \quad x = 1 \text{ (at } P \text{) or } -1$ | A1 |
$\therefore Q(-1, -2)$ | A1 | (11) |
---A curve has the equation
$$x^2 - 4xy + 2y^2 = 1.$$
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $\frac{dy}{dx}$ in its simplest form in terms of $x$ and $y$. [4]
\item Show that the tangent to the curve at the point $P(1, 2)$ has the equation
$$3x - 2y + 1 = 0.$$ [3]
\end{enumerate}
The tangent to the curve at the point $Q$ is parallel to the tangent at $P$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the coordinates of $Q$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q8 [11]}}