| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Moderate -0.3 This is a standard C2 differentiation question requiring routine techniques: differentiate a simple function with negative power, find gradient at a point, verify a normal equation, find stationary points, and solve simultaneous equations. All steps are straightforward applications of learned procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{8}{x^2} = 8x^{-2}\) | B1 | PI by derivative as \(16x^{-3}\) or \(-16x^{-3}\) |
| \(\frac{dy}{dx} = 2+16x^{-3}\) | M1, A1 | M1: differentiating either \(6+2x\) correctly or differentiating \(-8/x^2\) correctly; A1: \(2+16x^{-3}\) OE |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| At \(P(2,8)\): \(\frac{dy}{dx} = 2+16\times 2^{-3}\) \((= 4)\) | M1 | Attempt to find \(\frac{dy}{dx}\) when \(x=2\) |
| Gradient of normal at \(P = -\frac{1}{4}\) | m1 | \(m \times m' = -1\) used |
| \(y-8 = -\frac{1}{4}(x-2) \Rightarrow x+4y = 34\) | A1 | CSO AG |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| At stationary point: \(\frac{dy}{dx}=0\), \(2+16x^{-3}=0\) | M1 | Equating c's \(\frac{dy}{dx}\) to 0. Accept '\(\frac{dy}{dx}=0\) so \(x=-2\)' stated with no errors seen |
| \(16x^{-3} = -2 \quad x = -2\) | A1 | \(x=-2\) |
| When \(x=-2\), \(y=6-4-2=0\); \(M(-2,0)\) lies on \(x\)-axis | A1 | Need statement and correct coords |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Tangent at \(M\) has equation \(y=0\) | B1 | \(y=0\) OE |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Intersects normal at \(P\) when \(x+0=34\) | M1 | PI: solving tangent with normal (b), correctly eliminating one variable |
| \(T(34, 0)\) | A1 | Accept \(x=34\), \(y=0\) |
| Total: 2 marks |
# Question 5:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{8}{x^2} = 8x^{-2}$ | B1 | PI by derivative as $16x^{-3}$ or $-16x^{-3}$ |
| $\frac{dy}{dx} = 2+16x^{-3}$ | M1, A1 | M1: differentiating either $6+2x$ correctly or differentiating $-8/x^2$ correctly; A1: $2+16x^{-3}$ OE |
| **Total: 3 marks** | | |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| At $P(2,8)$: $\frac{dy}{dx} = 2+16\times 2^{-3}$ $(= 4)$ | M1 | Attempt to find $\frac{dy}{dx}$ when $x=2$ |
| Gradient of normal at $P = -\frac{1}{4}$ | m1 | $m \times m' = -1$ used |
| $y-8 = -\frac{1}{4}(x-2) \Rightarrow x+4y = 34$ | A1 | CSO AG |
| **Total: 3 marks** | | |
## Part (c)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| At stationary point: $\frac{dy}{dx}=0$, $2+16x^{-3}=0$ | M1 | Equating c's $\frac{dy}{dx}$ to 0. Accept '$\frac{dy}{dx}=0$ so $x=-2$' stated with no errors seen |
| $16x^{-3} = -2 \quad x = -2$ | A1 | $x=-2$ |
| When $x=-2$, $y=6-4-2=0$; $M(-2,0)$ lies on $x$-axis | A1 | Need statement and correct coords |
| **Total: 3 marks** | | |
## Part (c)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| Tangent at $M$ has equation $y=0$ | B1 | $y=0$ OE |
| **Total: 1 mark** | | |
## Part (d)
| Working | Marks | Guidance |
|---------|-------|----------|
| Intersects normal at $P$ when $x+0=34$ | M1 | PI: solving tangent with normal (b), correctly eliminating one variable |
| $T(34, 0)$ | A1 | Accept $x=34$, $y=0$ |
| **Total: 2 marks** | | |5 The point $P ( 2,8 )$ lies on a curve, and the point $M$ is the only stationary point of the curve.
The curve has equation $y = 6 + 2 x - \frac { 8 } { x ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that the normal to the curve at the point $P ( 2,8 )$ has equation $x + 4 y = 34$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the stationary point $M$ lies on the $x$-axis.
\item Hence write down the equation of the tangent to the curve at $M$.
\end{enumerate}\item The tangent to the curve at $M$ and the normal to the curve at $P$ intersect at the point $T$. Find the coordinates of $T$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q5 [12]}}