| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Curve from derivative information |
| Difficulty | Standard +0.3 This is a straightforward integration and coordinate geometry problem from C1. Part (a) requires basic integration with a constant found from the given point, and part (b) involves solving a quadratic to find roots and calculating distance. All techniques are standard with clear guidance ('show that' format reduces difficulty). The multi-step nature and 11 total marks place it slightly above average, but no novel insight is required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks |
|---|---|
| (a) \(y = \int (6 - 4x - 3x^2) \, dx\), \(y = 6x - 2x^2 - x^3 + c\) | M1 A2 |
| \((0,0) \therefore c = 0\) | M1 |
| \(y = 6x - 2x^2 - x^3\) | A1 |
| Answer | Marks |
|---|---|
| \(x = 0\) (at O) or \(6 - 2x - x^2 = 0\) | M1 |
| at A, B: \(x = \frac{2 \pm \sqrt{4 + 24}}{-2} = \frac{2 \pm 2\sqrt{7}}{-2} = -1 \pm \sqrt{7}\) | M2 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore AB = (-1 + \sqrt{7}) - (-1 - \sqrt{7}) = 2\sqrt{7}\) | M1 A1 | [\(k = 2\)] (11 marks) |
(a) $y = \int (6 - 4x - 3x^2) \, dx$, $y = 6x - 2x^2 - x^3 + c$ | M1 A2 |
$(0,0) \therefore c = 0$ | M1 |
$y = 6x - 2x^2 - x^3$ | A1 |
(b) $6x - 2x^2 - x^3 = 0$, $x(6 - 2x - x^2) = 0$
$x = 0$ (at O) or $6 - 2x - x^2 = 0$ | M1 |
at A, B: $x = \frac{2 \pm \sqrt{4 + 24}}{-2} = \frac{2 \pm 2\sqrt{7}}{-2} = -1 \pm \sqrt{7}$ | M2 A1 |
$A(-1 - \sqrt{7}, 0)$, $B(-1 + \sqrt{7}, 0)$
$\therefore AB = (-1 + \sqrt{7}) - (-1 - \sqrt{7}) = 2\sqrt{7}$ | M1 A1 | [$k = 2$] (11 marks)\includegraphics{figure_1}
Figure 1 shows the curve with equation $y = \text{f}(x)$ which crosses the $x$-axis at the origin and at the points $A$ and $B$.
Given that
$$\text{f}'(x) = 6 - 4x - 3x^2,$$
\begin{enumerate}[label=(\alph*)]
\item find an expression for $y$ in terms of $x$, [5]
\item show that $AB = k\sqrt{7}$, where $k$ is an integer to be found. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q9 [11]}}