| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing routine differentiation to find a tangent, substitution to verify a point lies on both curve and tangent, and basic integration with area calculation. All techniques are standard with clear signposting and no problem-solving insight required—easier than average A-level questions. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 5x^4 - 4x\) | M1 | one of these terms correct |
| A1 | all correct (no +c etc) | |
| \(\left(= 5(-1)^4 - 4(-1)\right) = 9\) | A1 | 9 marks total |
| Tangent has equation \(y = \text{'their'9'} \cdot x + c\) and \(6 = \text{'their'9'}(-1) + c \Rightarrow c = ...\) | m1 | tangent using 'their' gradient, and attempt to find \(c\) using \(x = -1\) and \(y = 6\) |
| \(\Rightarrow y = 9x + 15\) | A1 | equation must be seen in this form; 5 marks total |
| (b)(i) When \(x = 2\), \(y = 2^5 - 2 \times 2^2 + 9 = 32 - 8 + 9 = 33\) | B1 | NMS \(k = 33\) scores B0; 1 mark total |
| \((k = 33)\) | be convinced that they are using curve equation | |
| (ii) When \(x = 2\), \(y = 9 \times 2 + 15 = 33\) so lies on tangent | B1 | be convinced that they are using tangent equation and have statement; 1 mark total |
| (c)(i) \(\frac{x^6}{6} - \frac{2x^3}{3} + 9x\) | M1 | one of these terms correct |
| A1 | another term correct | |
| A1 | all correct (may have +c) | |
| \(\left[\frac{2^6}{6} - \frac{2 \times 2^3}{3} + 9 \times 2\right] - \left[\frac{(-1)^6}{6} - \frac{2 \times (-1)^3}{3} + 9 \times (-1)\right]\) | m1 | F(2) − F(−1) unsimplified FT 'their terms' from integration |
| \(\left[\frac{64}{6} - \frac{16}{3} + 18\right] - \left[\frac{1}{6} + \frac{2}{3} - 9\right]\) | \(= \frac{70}{3} - \left(-\frac{49}{6}\right)\) | |
| \(= 31.5\) | A1 | condone single fraction \(\left(\text{or } \frac{189}{6} \text{ etc}\right)\); 5 marks total |
| (ii) Area of trapezium \(= \frac{1}{2} \times 3 \times (6 + \text{'their'k})\) | B1 | = 58.5 when \(k = 33\) |
| Shaded area = Trapezium − 'their' (c)(i) value | M1 | |
| \(= 27\) | A1 | OE eg \(\frac{162}{6}\); 3 marks total |
**(a)** $\frac{dy}{dx} = 5x^4 - 4x$ | M1 | one of these terms correct
| | | A1 | all correct (no +c etc)
$\left(= 5(-1)^4 - 4(-1)\right) = 9$ | A1 | 9 marks total
Tangent has equation $y = \text{'their'9'} \cdot x + c$ and $6 = \text{'their'9'}(-1) + c \Rightarrow c = ...$ | m1 | tangent using 'their' gradient, and attempt to find $c$ using $x = -1$ and $y = 6$
$\Rightarrow y = 9x + 15$ | A1 | equation must be seen in this form; 5 marks total
**(b)(i)** When $x = 2$, $y = 2^5 - 2 \times 2^2 + 9 = 32 - 8 + 9 = 33$ | B1 | NMS $k = 33$ scores B0; 1 mark total
$(k = 33)$ | | be convinced that they are using curve equation
**(ii)** When $x = 2$, $y = 9 \times 2 + 15 = 33$ so lies on tangent | B1 | be convinced that they are using tangent equation and have statement; 1 mark total
**(c)(i)** $\frac{x^6}{6} - \frac{2x^3}{3} + 9x$ | M1 | one of these terms correct
| | | A1 | another term correct
| | | A1 | all correct (may have +c)
$\left[\frac{2^6}{6} - \frac{2 \times 2^3}{3} + 9 \times 2\right] - \left[\frac{(-1)^6}{6} - \frac{2 \times (-1)^3}{3} + 9 \times (-1)\right]$ | m1 | F(2) − F(−1) unsimplified FT 'their terms' from integration
$\left[\frac{64}{6} - \frac{16}{3} + 18\right] - \left[\frac{1}{6} + \frac{2}{3} - 9\right]$ | | $= \frac{70}{3} - \left(-\frac{49}{6}\right)$
$= 31.5$ | A1 | condone single fraction $\left(\text{or } \frac{189}{6} \text{ etc}\right)$; 5 marks total
**(ii)** Area of trapezium $= \frac{1}{2} \times 3 \times (6 + \text{'their'k})$ | B1 | = 58.5 when $k = 33$
Shaded area = Trapezium − 'their' (c)(i) value | M1 |
$= 27$ | A1 | OE eg $\frac{162}{6}$; 3 marks total
**Total: 15 marks**
---6 A curve has equation $y = x ^ { 5 } - 2 x ^ { 2 } + 9$. The point $P$ with coordinates $( - 1,6 )$ lies on the curve.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve at the point $P$, giving your answer in the form $y = m x + c$.
\item The point $Q$ with coordinates $( 2 , k )$ lies on the curve.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $k$.
\item Verify that $Q$ also lies on the tangent to the curve at the point $P$.
\end{enumerate}\item The curve and the tangent to the curve at $P$ are sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{aa42b4fd-1e37-48b8-90ee-269916c4db2c-4_721_887_936_589}
\begin{enumerate}[label=(\roman*)]
\item Find $\int _ { - 1 } ^ { 2 } \left( x ^ { 5 } - 2 x ^ { 2 } + 9 \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve and the tangent to the curve at $P$.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q6 [15]}}