| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Area with Turning Points |
| Difficulty | Standard +0.3 This is a straightforward C2 question requiring standard techniques: finding a turning point by differentiation (part a is a 'show that' making it easier), then calculating area between a curve and a line using integration. The algebra is manageable and the methods are routine textbook applications, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\frac{dy}{dx} = 8 + 2x - 3x^2\) | M1 A1 |
| \(3x^2 - 2x - 8 = 0\), \((3x+4)(x-2) = 0\), \(x = 2\) (ignore other solution) (*) | A1cso (3) | Final mark may also be scored by verifying \(\frac{dy}{dx} = 0\) at \(x = 2\) |
| (b) | Area of triangle \(= \frac{1}{2} \times 2 \times 22\) | M1 A1 |
| \(\int 10 + 8x + x^2 - x^3\, dx = 10x + \frac{8x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\) | M1 A1 A1 | M1: \(x^n \to x^{n+1}\) for one of the terms; integrating gradient function loses M mark; only 1 term correct: M1 A0 A0; 2 or 3 terms correct: M1 A1 A0 |
| \(\left[10x + \frac{8x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\right]_0^2 = 20 + 16 + \frac{8}{3} - 4\) | M1 | Substitute limit 2 into changed function; M can be awarded even if other limit wrong |
| Area of \(R = 34\frac{2}{3} - 22 = \frac{38}{3} \left(= 12\frac{2}{3}\right)\) (or \(12.\dot{6}\)) | M1 A1 (8) | Final M: dependent on use of calculus in (b) and correct overall strategy — subtract either way round; A: must be exact, not 12.67 or similar; negative area at end, even if subsequently made positive, loses A mark |
## Question 8:
**(a)** | $\frac{dy}{dx} = 8 + 2x - 3x^2$ | M1 A1 | M1: $x^n \to x^{n-1}$ for one of the terms, not just $10 \to 0$ |
| $3x^2 - 2x - 8 = 0$, $(3x+4)(x-2) = 0$, $x = 2$ (ignore other solution) (*) | A1cso (3) | Final mark may also be scored by verifying $\frac{dy}{dx} = 0$ at $x = 2$ |
**(b)** | Area of triangle $= \frac{1}{2} \times 2 \times 22$ | M1 A1 | M1: correct method to find area of triangle; area $= 22$ with no working acceptable |
| $\int 10 + 8x + x^2 - x^3\, dx = 10x + \frac{8x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$ | M1 A1 A1 | M1: $x^n \to x^{n+1}$ for one of the terms; integrating gradient function loses M mark; only 1 term correct: M1 A0 A0; 2 or 3 terms correct: M1 A1 A0 |
| $\left[10x + \frac{8x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\right]_0^2 = 20 + 16 + \frac{8}{3} - 4$ | M1 | Substitute limit 2 into changed function; M can be awarded even if other limit wrong |
| Area of $R = 34\frac{2}{3} - 22 = \frac{38}{3} \left(= 12\frac{2}{3}\right)$ (or $12.\dot{6}$) | M1 A1 (8) | Final M: dependent on use of calculus in (b) and correct overall strategy — subtract either way round; A: must be exact, not 12.67 or similar; negative area at end, even if subsequently made positive, loses A mark |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0454f5f6-b5ee-40b1-bc6a-ff8aeb06a455-11_668_1267_292_367}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = 10 + 8 x + x ^ { 2 } - x ^ { 3 }$.\\
The curve has a maximum turning point $A$.
\begin{enumerate}[label=(\alph*)]
\item Using calculus, show that the $x$-coordinate of $A$ is 2 .
The region $R$, shown shaded in Figure 2, is bounded by the curve, the $y$-axis and the line from $O$ to $A$, where $O$ is the origin.
\item Using calculus, find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q8 [11]}}