| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Two Curves Intersection Area |
| Difficulty | Standard +0.2 This is a standard P2/C2 area between curves question with straightforward algebra. Part (a) requires solving a cubic equation where one root is given, making factorization routine. Part (b) is a direct application of the standard integration formula for area between curves with simple polynomial integration. The question follows a very familiar textbook pattern with no novel insights required. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3 - 6x + 9 = -2x^2 + 7x - 1 \Rightarrow \ldots\) | M1 | Sets \(C_1 = C_2\) and collects terms |
| \(\pm(x^3 + 2x^2 - 13x + 10) = 0\) | A1 | Correct cubic equation; "\(= 0\)" may be implied by attempt to solve |
| Attempts to factorise using \((x-1)\) as factor or long division to obtain quadratic, then solves | M1 | e.g. \(x^3 + 2x^2 - 13x + 10 = (x-1)(x^2+3x-10)\); or attempts 3 factors directly; or solves via calculator |
| \(x = 2,\ y = 5\) or \((2, 5)\) | A1 | Correct values from correct cubic. Allow as coordinate pair or written separately |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^n \rightarrow x^{n+1}\) | M1 | For increasing any power of \(x\) by 1, for \(C_1\) or \(C_2\) or \(\pm(C_1 - C_2)\) |
| \(\pm\int\{-2x^2+7x-1-(x^3-6x+9)\}dx = \pm\int(-x^3-2x^2+13x-10)dx\) \(= \pm\left(-\frac{x^4}{4} - \frac{2x^3}{3} + \frac{13x^2}{2} - 10x\right)\) | dM1, A1 | dM1: correct integration of 1 term for \(C_1\) and one term for \(C_2\), or 2 terms of \(\pm(C_1-C_2)\). A1: fully correct integration of both; award as soon as seen, ignore subsequent work |
| \(= -\frac{2^4}{4} - \frac{2(2)^3}{3} + \frac{13(2)^2}{2} - 10(2) - \left(-\frac{1^4}{4} - \frac{2(1)^3}{3} + \frac{13(1)^2}{2} - 10(1)\right)\) | ddM1 | Fully correct strategy for area. Depends on both previous M marks. Uses limits 2 and 1, subtracts either way round |
| \(= \dfrac{13}{12}\) | A1 | If \(-\frac{13}{12}\) obtained, allow recovery if answer made positive |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 - 6x + 9 = -2x^2 + 7x - 1 \Rightarrow \ldots$ | M1 | Sets $C_1 = C_2$ and collects terms |
| $\pm(x^3 + 2x^2 - 13x + 10) = 0$ | A1 | Correct cubic equation; "$= 0$" may be implied by attempt to solve |
| Attempts to factorise using $(x-1)$ as factor or long division to obtain quadratic, then solves | M1 | e.g. $x^3 + 2x^2 - 13x + 10 = (x-1)(x^2+3x-10)$; or attempts 3 factors directly; or solves via calculator |
| $x = 2,\ y = 5$ or $(2, 5)$ | A1 | Correct values from correct cubic. Allow as coordinate pair or written separately |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^n \rightarrow x^{n+1}$ | M1 | For increasing any power of $x$ by 1, for $C_1$ or $C_2$ or $\pm(C_1 - C_2)$ |
| $\pm\int\{-2x^2+7x-1-(x^3-6x+9)\}dx = \pm\int(-x^3-2x^2+13x-10)dx$ $= \pm\left(-\frac{x^4}{4} - \frac{2x^3}{3} + \frac{13x^2}{2} - 10x\right)$ | dM1, A1 | dM1: correct integration of 1 term for $C_1$ **and** one term for $C_2$, or 2 terms of $\pm(C_1-C_2)$. A1: fully correct integration of both; **award as soon as seen, ignore subsequent work** |
| $= -\frac{2^4}{4} - \frac{2(2)^3}{3} + \frac{13(2)^2}{2} - 10(2) - \left(-\frac{1^4}{4} - \frac{2(1)^3}{3} + \frac{13(1)^2}{2} - 10(1)\right)$ | ddM1 | Fully correct strategy for area. Depends on both previous M marks. Uses limits 2 and 1, subtracts either way round |
| $= \dfrac{13}{12}$ | A1 | If $-\frac{13}{12}$ obtained, allow recovery if answer made positive |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0e107b51-2fb3-4ad7-8542-5aa0da13b127-20_978_1292_267_328}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curves $C _ { 1 }$ and $C _ { 2 }$ with equations
$$\begin{array} { l l }
C _ { 1 } : y = x ^ { 3 } - 6 x + 9 & x \geqslant 0 \\
C _ { 2 } : y = - 2 x ^ { 2 } + 7 x - 1 & x \geqslant 0
\end{array}$$
The curves $C _ { 1 }$ and $C _ { 2 }$ intersect at the points $A$ and $B$ as shown in Figure 1 .\\
The point $A$ has coordinates (1,4).
Using algebra and showing all steps of your working,
\begin{enumerate}[label=(\alph*)]
\item find the coordinates of the point $B$.
The finite region $R$, shown shaded in Figure 1, is bounded by $C _ { 1 }$ and $C _ { 2 }$
\item Use algebraic integration to find the exact area of $R$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2020 Q6 [9]}}