| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Moderate -0.8 This is a structured, multi-part question that guides students through standard procedures: substitution, algebraic expansion, differentiation (either by chain rule or after expansion), and finding a normal line equation. Each part is routine with clear scaffolding, making it easier than average despite covering multiple techniques. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | \(y_P = 4\) | B1 |
| 5(b) | \(y = 1 + \frac{2}{x} + \frac{2}{x} + \frac{4}{x^2}\) or \(y = 1 + 4x^{-1} + 4x^{-2}\) | B2, 1, 0 |
| 5(c) | \(\frac{dy}{dx} = -4x^{-2} - 8x^{-3}\) | M1, A1ft, A1 |
| 5(d) | When \(x = 2, \quad \frac{dy}{dx} = -4 \times 2^{-2} - 8 \times 2^{-3}\) | M1 |
| Gradient \(= -1 - 1 = -2\) | A1 | 2 marks |
| 5(e) | \(-2 \times m' = -1\) | M1 |
| \(y - 4 = m(x - 2)\) | M1 | |
| \(y - 4 = \frac{1}{2}(x - 2)\) | A1ft | |
| \(x - 2y + 6 = 0\) | A1 | 4 marks |
**5(a)** | $y_P = 4$ | B1 | 1 mark |
**5(b)** | $y = 1 + \frac{2}{x} + \frac{2}{x} + \frac{4}{x^2}$ or $y = 1 + 4x^{-1} + 4x^{-2}$ | B2, 1, 0 | 2 marks | (B1 if only one error in the expansion) For B2 the last line of the candidate's solution must be correct |
**5(c)** | $\frac{dy}{dx} = -4x^{-2} - 8x^{-3}$ | M1, A1ft, A1 | 3 marks | Index reduced by 1 after differentiating $x$ to a negative power; At least 1 term in $x$ correct ft on expn; CSO Full correct solution. ACF |
**5(d)** | When $x = 2, \quad \frac{dy}{dx} = -4 \times 2^{-2} - 8 \times 2^{-3}$ | M1 | | Attempt to find $y'(2)$ |
| | Gradient $= -1 - 1 = -2$ | A1 | 2 marks | AG (be convinced–no errors seen) |
**5(e)** | $-2 \times m' = -1$ | M1 | | $m_1 \times m_2 = -1$ OE stated or used. PI |
| | $y - 4 = m(x - 2)$ | M1 | | C's $y_P$ from part (a) if not recovered; $m$ must be numerical |
| | $y - 4 = \frac{1}{2}(x - 2)$ | A1ft | | Ft on candidate's $y_P$ from part (a) if not recovered |
| | $x - 2y + 6 = 0$ | A1 | 4 marks | CAO Must be this or $0 = x - 2y + 6$ |
---5 A curve is defined for $x > 0$ by the equation
$$y = \left( 1 + \frac { 2 } { x } \right) ^ { 2 }$$
The point $P$ lies on the curve where $x = 2$.
\begin{enumerate}[label=(\alph*)]
\item Find the $y$-coordinate of $P$.
\item Expand $\left( 1 + \frac { 2 } { x } \right) ^ { 2 }$.
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence show that the gradient of the curve at $P$ is - 2 .
\item Find the equation of the normal to the curve at $P$, giving your answer in the form $x + b y + c = 0$, where $b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q5 [12]}}