| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Normal to circle at point |
| Difficulty | Moderate -0.8 This is a straightforward C1 circle question requiring completion of the square to find the centre, then using perpendicular gradients to find the normal equation. All steps are routine procedures with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple standard techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02q Use intersection points: of graphs to solve equations1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Set curve equal to line: \(x^3 - x^2 - 5x + 7 = x + 7\) | M1 | Must show this step |
| \(x^3 - x^2 - 6x = 0 \Rightarrow x(x^2 - x - 6) = 0\) | A1 | Factor out \(x\), leading to \(x^2 - x - 6 = 0\) shown |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)(x+2) = 0\) | M1 | Factorising \(x^2 - x - 6\) |
| \(x = 3, x = -2\) | A1 | Both \(x\) values |
| \(A = (-2, 5)\), \(C = (3, 10)\) | A1 | Both coordinates correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int(x^3 - x^2 - 5x + 7)\,dx = \frac{x^4}{4} - \frac{x^3}{3} - \frac{5x^2}{2} + 7x\) | M1 | Attempt at integration |
| Correct integration | A1 | All terms correct |
| +\(c\) not required | A1 | Fully correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Area under line from \(x=-2\) to \(x=0\): \(\int_{-2}^{0}(x+7)\,dx\) or area of trapezium | M1 | Correct method for line area |
| \(\int_{-2}^{0}(x^3 - x^2 - 5x + 7)\,dx\) evaluated | M1 | Substituting limits \(-2\) to \(0\) |
| Area \(= \left[\frac{x^4}{4} - \frac{x^3}{3} - \frac{5x^2}{2} + 7x\right]_{-2}^{0} - \left[\frac{x^2}{2} + 7x\right]_{-2}^{0}\) | DM1 | Subtracting line integral from curve integral |
| Area \(= \frac{8}{3}\) | A1 | Correct final answer |
# Question 6:
## Part (a)(i):
| Set curve equal to line: $x^3 - x^2 - 5x + 7 = x + 7$ | M1 | Must show this step |
|---|---|---|
| $x^3 - x^2 - 6x = 0 \Rightarrow x(x^2 - x - 6) = 0$ | A1 | Factor out $x$, leading to $x^2 - x - 6 = 0$ shown |
## Part (a)(ii):
| $(x-3)(x+2) = 0$ | M1 | Factorising $x^2 - x - 6$ |
|---|---|---|
| $x = 3, x = -2$ | A1 | Both $x$ values |
| $A = (-2, 5)$, $C = (3, 10)$ | A1 | Both coordinates correct |
## Part (b):
| $\int(x^3 - x^2 - 5x + 7)\,dx = \frac{x^4}{4} - \frac{x^3}{3} - \frac{5x^2}{2} + 7x$ | M1 | Attempt at integration |
|---|---|---|
| Correct integration | A1 | All terms correct |
| +$c$ not required | A1 | Fully correct expression |
## Part (c):
| Area under line from $x=-2$ to $x=0$: $\int_{-2}^{0}(x+7)\,dx$ or area of trapezium | M1 | Correct method for line area |
|---|---|---|
| $\int_{-2}^{0}(x^3 - x^2 - 5x + 7)\,dx$ evaluated | M1 | Substituting limits $-2$ to $0$ |
| Area $= \left[\frac{x^4}{4} - \frac{x^3}{3} - \frac{5x^2}{2} + 7x\right]_{-2}^{0} - \left[\frac{x^2}{2} + 7x\right]_{-2}^{0}$ | DM1 | Subtracting line integral from curve integral |
| Area $= \frac{8}{3}$ | A1 | Correct final answer |
---6 The diagram shows a curve and a line which intersect at the points $A , B$ and $C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f2124c89-79de-4758-b7b8-ff273345b9dd-7_574_844_349_609}
The curve has equation $y = x ^ { 3 } - x ^ { 2 } - 5 x + 7$ and the straight line has equation $y = x + 7$. The point $B$ has coordinates ( 0,7 ).
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinates of the points $A$ and $C$ satisfy the equation
$$x ^ { 2 } - x - 6 = 0$$
\item Find the coordinates of the points $A$ and $C$.
\end{enumerate}\item Find $\int \left( x ^ { 3 } - x ^ { 2 } - 5 x + 7 \right) \mathrm { d } x$.
\item Find the area of the shaded region $R$ bounded by the curve and the line segment $A B$.\\[0pt]
[4 marks]\\
$7 \quad$ A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } - 10 x + 12 y + 41 = 0$. The point $A ( 3 , - 2 )$ lies on the circle.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2014 Q6 [12]}}