| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question with standard steps: substituting to find a coordinate, differentiating implicitly to find dy/dx, and writing a tangent equation. All techniques are routine for C4 level with no conceptual challenges or 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 | Guidance |
|---|---|---|
| \(x = 1, 5a^2 - a - 4 = 0\) | M1 | |
| \((5a+4)(a-1) = 0, a = 1\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} + 4\) | B1B1 | |
| \(= 10xy^2 + 10x^2y\frac{dy}{dx}\) | M1 | |
| M1 | condone 5 × 2 for 10 | |
| A1 | ||
| \(x = 1, y = 1\) \(\frac{dy}{dx} + 4 = 10 + 10\frac{dy}{dx}\) | M1 | |
| \(\frac{dy}{dx} = -\frac{6}{9} = -\left(-\frac{2}{3}\right)\) | A1 | 7 marks |
| Alt (for last two marks) | ||
| \(\frac{dy}{dx} = \frac{10xy^2-4}{1-10x^2y}\) | M1 | |
| \((1,1) \Rightarrow \frac{10-4}{1-10} = \frac{6}{-9}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{y-1}{x-1} = \frac{2}{3}\) (OE) | B1F | 1 mark |
**5(a)**
| $x = 1, 5a^2 - a - 4 = 0$ | M1 | | condone $y$ for $a$ |
| $(5a+4)(a-1) = 0, a = 1$ | A1 | 2 marks | AG – be convinced, both factors seen or $1 \Rightarrow a = 1$ or $1 \Rightarrow a = -\frac{4}{5}$ or $a = -\frac{4}{5}$ or $1 \Rightarrow a = 1$ A0 for 2 positive roots (substitute $(1,1) \Rightarrow 5 = 5$ no marks) |
**5(b)**
| $\frac{dy}{dx} + 4$ | B1B1 | | (Ignore $\frac{dy}{dt} =$ if not used, otherwise loses final A1) |
| $= 10xy^2 + 10x^2y\frac{dy}{dx}$ | M1 | | attempt product rule, see two terms added; chain rule, $\frac{dy}{dx}$ attached to one term only |
| | M1 | | condone 5 × 2 for 10 |
| | A1 | | |
| $x = 1, y = 1$ $\frac{dy}{dx} + 4 = 10 + 10\frac{dy}{dx}$ | M1 | | two terms, or more, in $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = -\frac{6}{9} = -\left(-\frac{2}{3}\right)$ | A1 | 7 marks | CSO |
| **Alt (for last two marks)** | | | |
| $\frac{dy}{dx} = \frac{10xy^2-4}{1-10x^2y}$ | M1 | | find $\frac{dy}{dx}$ in terms of $x, y$ and substitute $x = 1, y = 1$ must be from expression with two terms or more in $\frac{dy}{dx}$ |
| $(1,1) \Rightarrow \frac{10-4}{1-10} = \frac{6}{-9}$ | A1 | | |
**5(c)**
| $\frac{y-1}{x-1} = \frac{2}{3}$ (OE) | B1F | 1 mark | ft on gradient; ISW after any correct form |5 The point $P ( 1 , a )$, where $a > 0$, lies on the curve $y + 4 x = 5 x ^ { 2 } y ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = 1$.
\item Find the gradient of the curve at $P$.
\item Find an equation of the tangent to the curve at $P$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q5 [10]}}