| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| 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 integration question requiring finding k from a turning point condition (standard differentiation), then calculating an area using definite integration. The steps are routine: differentiate to find k, substitute to get y-coordinate, integrate the cubic, and evaluate. Slightly above average difficulty only because it requires careful setup of the area calculation with correct limits. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 3x^2 - 20x + k\) | M1 A1 | Differentiation is required |
| At \(x=2\), \(\frac{dy}{dx}=0\), so \(12-40+k=0\), \(k=28\) | A1 cso | The \('=0'\) must be seen at some stage to score the final mark |
| Alternatively using \(k=28\): \(\frac{dy}{dx}=3x^2-20x+28\) | (M1 A1) | 'Assuming' \(k=28\) only scores final cso mark if there is justification that \(\frac{dy}{dx}=0\) at \(x=2\) represents the maximum turning point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int(x^3-10x^2+28x)\,dx = \frac{x^4}{4}-\frac{10x^3}{3}+\frac{28x^2}{2}\) | M1 A1 | Allow \(\frac{kx^2}{2}\) for \(\frac{28x^2}{2}\) |
| \(\left[\frac{x^4}{4}-\frac{10x^3}{3}+14x^2\right]_0^2 = \ldots \quad \left(=4-\frac{80}{3}+56=\frac{100}{3}\right)\) | M1 | With limits 0 to 2, substitute limit 2 into a 'changed function' |
| \(y\)-coordinate of \(P = 8-40+56=24\) | B1 | Allow if seen in part (a); B1 for 24 may be scored by implication from later working |
| Area of rectangle \(= 2 \times\) (their \(y\)-coordinate of \(P\)) | ||
| Area of \(R = (\text{their } 48) - \left(\text{their }\frac{100}{3}\right) = \frac{44}{3}\left(14\frac{2}{3} \text{ or } 14.\dot{6}\right)\) | M1 A1 | If subtraction is 'wrong way round', final A mark is lost |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 3x^2 - 20x + k$ | M1 A1 | Differentiation is required |
| At $x=2$, $\frac{dy}{dx}=0$, so $12-40+k=0$, $k=28$ | A1 cso | The $'=0'$ must be seen at some stage to score the final mark |
| Alternatively using $k=28$: $\frac{dy}{dx}=3x^2-20x+28$ | (M1 A1) | 'Assuming' $k=28$ only scores final cso mark if there is justification that $\frac{dy}{dx}=0$ at $x=2$ represents the maximum turning point |
**(3 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int(x^3-10x^2+28x)\,dx = \frac{x^4}{4}-\frac{10x^3}{3}+\frac{28x^2}{2}$ | M1 A1 | Allow $\frac{kx^2}{2}$ for $\frac{28x^2}{2}$ |
| $\left[\frac{x^4}{4}-\frac{10x^3}{3}+14x^2\right]_0^2 = \ldots \quad \left(=4-\frac{80}{3}+56=\frac{100}{3}\right)$ | M1 | With limits 0 to 2, substitute limit 2 into a 'changed function' |
| $y$-coordinate of $P = 8-40+56=24$ | B1 | Allow if seen in part (a); B1 for 24 may be scored by implication from later working |
| Area of rectangle $= 2 \times$ (their $y$-coordinate of $P$) | | |
| Area of $R = (\text{their } 48) - \left(\text{their }\frac{100}{3}\right) = \frac{44}{3}\left(14\frac{2}{3} \text{ or } 14.\dot{6}\right)$ | M1 A1 | If subtraction is 'wrong way round', final A mark is lost |
**(6 marks, 9 total)**
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{571780c2-945b-4636-b7c3-0bd558d28710-10_611_831_210_575}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with equation
$$y = x ^ { 3 } - 10 x ^ { 2 } + k x$$
where $k$ is a constant.
The point $P$ on $C$ is the maximum turning point.\\
Given that the $x$-coordinate of $P$ is 2 ,
\begin{enumerate}[label=(\alph*)]
\item show that $k = 28$.
The line through $P$ parallel to the $x$-axis cuts the $y$-axis at the point $N$. The region $R$ is bounded by $C$, the $y$-axis and $P N$, as shown shaded in Figure 2.
\item Use calculus to find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2010 Q8 [9]}}