| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line-Axis Bounded Region |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring finding intersection points by solving a quadratic equation, then calculating area between curves using definite integration. The algebra is straightforward (factorizable quadratic), and the integration involves only polynomial terms with clear bounds. Slightly easier than average due to routine methods and no conceptual challenges. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Sets \(5 - 3x = 20x - 12x^2\) → 3TQ | M1 | Attempts to set equations equal and proceeds to a 3TQ |
| \(12x^2 - 23x + 5 = 0\) → \((4x-1)(3x-5) = 0\) → \(x = \frac{1}{4}, \frac{5}{3}\) | M1 A1 | Uses algebraic method to solve; both \(x\) values correct |
| \(P = \left(\frac{1}{4}, \frac{17}{4}\right)\) and \(Q = \left(\frac{5}{3}, 0\right)\) | A1, B1 | A1: correct \(P\); B1: correct \(Q\) (independent, can be found from \(5-3x=0\) or \(20x-12x^2=0\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int 20x - 12x^2\,dx = \left[10x^2 - 4x^3\right]\) | M1 A1 | M1: at least one correct power; A1: correct integral |
| Area under curve \(= \left[10x^2 - 4x^3\right]_0^{\frac{1}{4}} = \frac{9}{16}\) | dM1 | Attempts to use correct limits \(0\) to \(\frac{1}{4}\) |
| Area of triangle \(= \frac{1}{2} \times \left(\frac{5}{3} - \frac{1}{4}\right) \times \frac{17}{4} = \frac{289}{96}\) | ddM1 | For finding complementary area; dependent on both previous M marks |
| Total area \(= \frac{9}{16} + \frac{289}{96} = \frac{343}{96}\) | dddM1, A1 | dddM1: correct method combining areas; dependent on all three M marks; A1: \(\frac{343}{96}\) |
## Question 15:
### Part 15(a):
| Working | Mark | Guidance |
|---------|------|----------|
| Sets $5 - 3x = 20x - 12x^2$ → 3TQ | M1 | Attempts to set equations equal and proceeds to a 3TQ |
| $12x^2 - 23x + 5 = 0$ → $(4x-1)(3x-5) = 0$ → $x = \frac{1}{4}, \frac{5}{3}$ | M1 A1 | Uses algebraic method to solve; both $x$ values correct |
| $P = \left(\frac{1}{4}, \frac{17}{4}\right)$ and $Q = \left(\frac{5}{3}, 0\right)$ | A1, B1 | A1: correct $P$; B1: correct $Q$ (independent, can be found from $5-3x=0$ or $20x-12x^2=0$) |
**Total: 5 marks**
---
### Part 15(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int 20x - 12x^2\,dx = \left[10x^2 - 4x^3\right]$ | M1 A1 | M1: at least one correct power; A1: correct integral |
| Area under curve $= \left[10x^2 - 4x^3\right]_0^{\frac{1}{4}} = \frac{9}{16}$ | dM1 | Attempts to use correct limits $0$ to $\frac{1}{4}$ |
| Area of triangle $= \frac{1}{2} \times \left(\frac{5}{3} - \frac{1}{4}\right) \times \frac{17}{4} = \frac{289}{96}$ | ddM1 | For finding complementary area; dependent on both previous M marks |
| Total area $= \frac{9}{16} + \frac{289}{96} = \frac{343}{96}$ | dddM1, A1 | dddM1: correct method combining areas; dependent on all three M marks; A1: $\frac{343}{96}$ |
**Total: 6 marks**
**Note:** No marks available to candidates who don't show intent to integrate either Curve or $\pm$(Curve $-$ Line)15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{75d68987-2314-4c8f-8160-24977c5c4e34-40_545_794_294_584}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The straight line $l$ with equation $y = 5 - 3 x$ cuts the curve $C$, with equation $y = 20 x - 12 x ^ { 2 }$, at the points $P$ and $Q$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Use algebra to find the exact coordinates of the points $P$ and $Q$.
The finite region $R$, shown shaded in Figure 3, is bounded by the line $l$, the $x$-axis and the curve $C$.
\item Use calculus to find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q15 [11]}}