| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.8 This is a straightforward C2 differentiation question requiring routine application of power rule (rewriting 1/x² as x^(-2)), finding gradient at a point, and using point-gradient form for normal/tangent lines. All techniques are standard with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = -\frac{2}{x^3} + 4\) | M1 A1 A1 | M1 for attempting differentiation of \(x^{-2}\); A1 for \(-2x^{-3}\); A1 for \(+4\) |
| Answer | Marks | Guidance |
|---|---|---|
| At \(P(-1,-3)\): \(\frac{dy}{dx} = -\frac{2}{(-1)^3} + 4 = 2 + 4 = 6\) | M1 | Substituting \(x = -1\) into their \(\frac{dy}{dx}\) |
| Gradient of normal \(= -\frac{1}{6}\) | M1 | Using \(m_1 m_2 = -1\) |
| \(y + 3 = -\frac{1}{6}(x + 1)\) or \(y = -\frac{1}{6}x - \frac{19}{6}\) | A1 | Accept any correct equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\frac{2}{x^3} + 4 = -12\) | M1 | Setting \(\frac{dy}{dx} = -12\) |
| \(-\frac{2}{x^3} = -16 \Rightarrow x^3 = \frac{1}{8} \Rightarrow x = \frac{1}{2}\) | M1 A1 | Solving for \(x\) |
| \(y = \frac{1}{(1/2)^2} + 4(\frac{1}{2}) = 4 + 2 = 6\) | M1 | Finding \(y\) coordinate |
| \(y - 6 = -12(x - \frac{1}{2})\) i.e. \(y = -12x + 12\) | A1 | Accept any correct equivalent form |
## Question 4:
### Part (a)
| $\frac{dy}{dx} = -\frac{2}{x^3} + 4$ | M1 A1 A1 | M1 for attempting differentiation of $x^{-2}$; A1 for $-2x^{-3}$; A1 for $+4$ |
### Part (b)
| At $P(-1,-3)$: $\frac{dy}{dx} = -\frac{2}{(-1)^3} + 4 = 2 + 4 = 6$ | M1 | Substituting $x = -1$ into their $\frac{dy}{dx}$ |
| Gradient of normal $= -\frac{1}{6}$ | M1 | Using $m_1 m_2 = -1$ |
| $y + 3 = -\frac{1}{6}(x + 1)$ or $y = -\frac{1}{6}x - \frac{19}{6}$ | A1 | Accept any correct equivalent form |
### Part (c)
| $-\frac{2}{x^3} + 4 = -12$ | M1 | Setting $\frac{dy}{dx} = -12$ |
| $-\frac{2}{x^3} = -16 \Rightarrow x^3 = \frac{1}{8} \Rightarrow x = \frac{1}{2}$ | M1 A1 | Solving for $x$ |
| $y = \frac{1}{(1/2)^2} + 4(\frac{1}{2}) = 4 + 2 = 6$ | M1 | Finding $y$ coordinate |
| $y - 6 = -12(x - \frac{1}{2})$ i.e. $y = -12x + 12$ | A1 | Accept any correct equivalent form |
---4 A curve has equation $y = \frac { 1 } { x ^ { 2 } } + 4 x$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item The point $P ( - 1 , - 3 )$ lies on the curve. Find an equation of the normal to the curve at the point $P$.
\item Find an equation of the tangent to the curve that is parallel to the line $y = - 12 x$.\\[0pt]
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2014 Q4 [11]}}