| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line Intersection Area |
| Difficulty | Standard +0.3 This is a straightforward multi-part integration question requiring: (i) substituting a point to find k, (ii) verifying an intersection by substitution, and (iii) computing area between curves using standard integration. All steps are routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2 = k(8 - 28 + 24) \rightarrow k = 1/2\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x=5\), \(y = [\frac{1}{2}](125 - 175 + 60) = 5\) | M1 | Or solve \([\frac{1}{2}](x^3 - 7x^2 + 12x) = x \Rightarrow x = 5\ [x=0,2]\) |
| Which lies on \(y = x\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int[\frac{1}{2}(x^3 - 7x^2 + 12x) - x]dx\) | M1 | Expect \(\int\frac{1}{2}x^3 - \frac{7}{2}x^2 + 5x\) |
| \(\frac{1}{8}x^4 - \frac{7}{6}x^3 + \frac{5}{2}x^2\) | B2,1,0FT | Ft on their \(k\) |
| \(2 - 28/3 + 10\) | DM1 | Apply limits \(0 \rightarrow 2\) |
| \(8/3\) | A1 | |
| OR \(\frac{1}{8}x^4 - \frac{7}{6}x^3 + 3x^2\) | B2,1,0FT | Integrate to find area under curve, Ft on their \(k\) |
| \(2 - 28/3 + 12\) | M1 | Apply limits \(0 \rightarrow 2\). Dep on integration attempted |
| Area \(\Delta = \frac{1}{2} \times 2 \times 2\) or \(\int_0^2 x\,dx = [\frac{1}{2}x^2] = 2\) | M1 | |
| \(8/3\) | A1 | |
| Total: 5 |
## Question 7:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2 = k(8 - 28 + 24) \rightarrow k = 1/2$ | B1 | |
| | **Total: 1** | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=5$, $y = [\frac{1}{2}](125 - 175 + 60) = 5$ | M1 | Or solve $[\frac{1}{2}](x^3 - 7x^2 + 12x) = x \Rightarrow x = 5\ [x=0,2]$ |
| Which lies on $y = x$ | A1 | |
| | **Total: 2** | |
**Part (iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int[\frac{1}{2}(x^3 - 7x^2 + 12x) - x]dx$ | M1 | Expect $\int\frac{1}{2}x^3 - \frac{7}{2}x^2 + 5x$ |
| $\frac{1}{8}x^4 - \frac{7}{6}x^3 + \frac{5}{2}x^2$ | B2,1,0FT | Ft on their $k$ |
| $2 - 28/3 + 10$ | DM1 | Apply limits $0 \rightarrow 2$ |
| $8/3$ | A1 | |
| OR $\frac{1}{8}x^4 - \frac{7}{6}x^3 + 3x^2$ | B2,1,0FT | Integrate to find area under curve, Ft on their $k$ |
| $2 - 28/3 + 12$ | M1 | Apply limits $0 \rightarrow 2$. Dep on integration attempted |
| Area $\Delta = \frac{1}{2} \times 2 \times 2$ or $\int_0^2 x\,dx = [\frac{1}{2}x^2] = 2$ | M1 | |
| $8/3$ | A1 | |
| | **Total: 5** | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{d178603a-f59a-4986-b5ab-b47eceedb2fc-10_503_853_260_641}
The diagram shows part of the curve with equation $y = k \left( x ^ { 3 } - 7 x ^ { 2 } + 12 x \right)$ for some constant $k$. The curve intersects the line $y = x$ at the origin $O$ and at the point $A ( 2,2 )$.\\
(i) Find the value of $k$.\\
(ii) Verify that the curve meets the line $y = x$ again when $x = 5$.\\
(iii) Find, showing all necessary working, the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2018 Q7 [8]}}