| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Area with Turning Points |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring finding a turning point by differentiation (routine), then calculating an area between a curve and a line. The setup is clearly defined with given coordinates, requiring straightforward polynomial integration and subtraction of areas. Slightly above average only due to the multi-step nature and need to set up the correct integral bounds. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks |
|---|---|
| (a) \(\frac{dy}{dx}=12x^{2}+18x-30\) | M1 |
| Either Substitute \(x=1\) to give \(\frac{dy}{dx}=12+18-30=0\) | A1 |
| Or Solve \(\frac{dy}{dx}=12x^{2}+18x-30=0\) to give \(x=\) | Al eso |
| So turning point (all correct work so far) | (3) |
| (b) Way 1: When \(x=1\), \(y=4+9-30-8=-25\) | B1 |
| Area of triangle \(ABP=\frac{1}{2}\times 1\times 25=12.5\) (Where \(P\) is at \((1,0)\)) | B1 |
| \(\int(4x^{3}+9x^{2}-30x-8)dx=x^{4}+3x^{3}-15x^{2}-8x+\{c\}\) | M1A1 |
| \([x^{4}+3x^{3}-15x^{2}-8x]_{-1}^{1}=(1+3-15-8)-((-\frac{1}{4})+3(-\frac{1}{4})-15(\frac{1}{4})-8(-\frac{1}{4}))\) | dM1 |
| \(=(-19)-\frac{261}{256}\) or \(-19-1.02\) | ddM1 |
| So Area = "\(\text{their } 12.5\)"+"\(\text{their } 20.02\)" or "\(12.5\)"+"\(\text{their } \frac{5125}{256}\)" | A1 |
| \(=32.52\) (NOT \(-32.52\)) | (7) |
| (a) $\frac{dy}{dx}=12x^{2}+18x-30$ | M1 |
|---|---|
| **Either** Substitute $x=1$ to give $\frac{dy}{dx}=12+18-30=0$ | A1 |
| **Or** Solve $\frac{dy}{dx}=12x^{2}+18x-30=0$ to give $x=$ | Al eso |
| So turning point (all correct work so far) | (3) |
| (b) **Way 1:** When $x=1$, $y=4+9-30-8=-25$ | B1 |
| Area of triangle $ABP=\frac{1}{2}\times 1\times 25=12.5$ (Where $P$ is at $(1,0)$) | B1 |
| $\int(4x^{3}+9x^{2}-30x-8)dx=x^{4}+3x^{3}-15x^{2}-8x+\{c\}$ | M1A1 |
| $[x^{4}+3x^{3}-15x^{2}-8x]_{-1}^{1}=(1+3-15-8)-((-\frac{1}{4})+3(-\frac{1}{4})-15(\frac{1}{4})-8(-\frac{1}{4}))$ | dM1 |
| $=(-19)-\frac{261}{256}$ or $-19-1.02$ | ddM1 |
| So Area = "$\text{their } 12.5$"+"$\text{their } 20.02$" or "$12.5$"+"$\text{their } \frac{5125}{256}$" | A1 |
| $=32.52$ (NOT $-32.52$) | (7) |
**Notes:**
- (a) M1: Attempt at differentiation - all powers reduced by 1 with $8\rightarrow 0$; A1: the derivative must be correct and uses substitutes $x=1$ to give 0. Ignore any reference to the other root ($-2.5$) for this mark; A1cso: obtains $x=1$ from correct work, or deduces turning point (if substitution used – may be implied by a preamble e.g. dy/10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{752efc6c-8d0e-46a6-b75d-5125956969d8-28_761_1120_258_411}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation
$$y = 4 x ^ { 3 } + 9 x ^ { 2 } - 30 x - 8 , \quad - 0.5 \leqslant x \leqslant 2.2$$
The curve has a turning point at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Using calculus, show that the $x$ coordinate of $A$ is 1
The curve crosses the $x$-axis at the points $B ( 2,0 )$ and $C \left( - \frac { 1 } { 4 } , 0 \right)$
The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the line $A B$, and the $x$-axis.
\item Use integration to find the area of the finite region $R$, giving your answer to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2017 Q10 [10]}}