| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| 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 area-between-curves problem requiring finding intersection points by solving a quadratic (after substituting u=√x), then integrating the difference of two simple functions. All techniques are standard P1 content with no novel insight 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 |
| EITHER: \(4 - 3\sqrt{x} = 3 - 2x \rightarrow 2x - 3\sqrt{x} + 1 (=0)\) or e.g. \(2k^2 - 3k + 1 (=0)\) | (M1) | Form 3-term quad & attempt to solve for \(\sqrt{x}\) |
| \(\sqrt{x} = \frac{1}{2}, 1\) | A1 | Or \(k = \frac{1}{2}\) or \(1\) (where \(k = \sqrt{x}\)) |
| \(x = \frac{1}{4}, 1\) | A1) | |
| OR1: \((3\sqrt{x})^2 = (1+2x)^2\) | (M1) | |
| \(4x^2 - 5x + 1\ (=0)\) | A1 | |
| \(x = \frac{1}{4}, 1\) | A1) | |
| OR2: \(\frac{3-y}{2} = \left(\frac{4-y}{3}\right)^2 \left(\rightarrow 2y^2 - 7y + 5 (=0)\right)\) | (M1) | Eliminate \(x\) |
| \(y = \frac{5}{2}, 1\) | A1 | |
| \(x = \frac{1}{4}, 1\) | A1) | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: Area under line \(= \int(3-2x)\,dx = 3x - x^2\) | (B1) | |
| \(= \left[(3-1) - \left(\frac{3}{4} - \frac{1}{16}\right)\right]\) | M1 | Apply *their* limits (e.g. \(\frac{1}{4} \to 1\)) after integn. |
| Area under curve \(= \int\left(4 - 3x^{1/2}\right)dx = 4x - 2x^{3/2}\) | B1 | |
| \(\left[(4-2) - \left(1 - \frac{1}{4}\right)\right]\) | M1 | Apply *their* limits (e.g. \(\frac{1}{4} \to 1\)) after integration |
| Required area \(= \frac{21}{16} - \frac{5}{4} = \frac{1}{16}\) (or 0.0625) | A1) | |
| OR: \(\pm\int(3-2x) - \left(4 - 3x^{\frac{1}{2}}\right) = +/-\int(-1 - 2x + 3x^{\frac{1}{2}})\) | (*M1) | Subtract functions and then attempt integration |
| \(+/-\left[-x - x^2 + \frac{3x^{3/2}}{3/2}\right]\) | A2, 1, 0 FT | FT on *their* subtraction. Deduct 1 mark for each term incorrect |
| \(+/-\left[-1-1+2-\left(-\frac{1}{4}+\frac{1}{16}+\frac{1}{8}\right)\right] = \frac{1}{16}\) (or 0.0625) | DM1 A1) | Apply *their* limits \(\frac{1}{4} \to 1\) |
| Total | 5 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** $4 - 3\sqrt{x} = 3 - 2x \rightarrow 2x - 3\sqrt{x} + 1 (=0)$ or e.g. $2k^2 - 3k + 1 (=0)$ | (M1) | Form 3-term quad & attempt to solve for $\sqrt{x}$ |
| $\sqrt{x} = \frac{1}{2}, 1$ | A1 | Or $k = \frac{1}{2}$ or $1$ (where $k = \sqrt{x}$) |
| $x = \frac{1}{4}, 1$ | A1) | |
| **OR1:** $(3\sqrt{x})^2 = (1+2x)^2$ | (M1) | |
| $4x^2 - 5x + 1\ (=0)$ | A1 | |
| $x = \frac{1}{4}, 1$ | A1) | |
| **OR2:** $\frac{3-y}{2} = \left(\frac{4-y}{3}\right)^2 \left(\rightarrow 2y^2 - 7y + 5 (=0)\right)$ | (M1) | Eliminate $x$ |
| $y = \frac{5}{2}, 1$ | A1 | |
| $x = \frac{1}{4}, 1$ | A1) | |
| **Total** | **3** | |
---
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** Area under line $= \int(3-2x)\,dx = 3x - x^2$ | (B1) | |
| $= \left[(3-1) - \left(\frac{3}{4} - \frac{1}{16}\right)\right]$ | M1 | Apply *their* limits (e.g. $\frac{1}{4} \to 1$) after integn. |
| Area under curve $= \int\left(4 - 3x^{1/2}\right)dx = 4x - 2x^{3/2}$ | B1 | |
| $\left[(4-2) - \left(1 - \frac{1}{4}\right)\right]$ | M1 | Apply *their* limits (e.g. $\frac{1}{4} \to 1$) after integration |
| Required area $= \frac{21}{16} - \frac{5}{4} = \frac{1}{16}$ (or 0.0625) | A1) | |
| **OR:** $\pm\int(3-2x) - \left(4 - 3x^{\frac{1}{2}}\right) = +/-\int(-1 - 2x + 3x^{\frac{1}{2}})$ | (*M1) | Subtract functions and then attempt integration |
| $+/-\left[-x - x^2 + \frac{3x^{3/2}}{3/2}\right]$ | A2, 1, 0 FT | FT on *their* subtraction. Deduct 1 mark for each term incorrect |
| $+/-\left[-1-1+2-\left(-\frac{1}{4}+\frac{1}{16}+\frac{1}{8}\right)\right] = \frac{1}{16}$ (or 0.0625) | DM1 A1) | Apply *their* limits $\frac{1}{4} \to 1$ |
| **Total** | **5** | |
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{17ca6dd2-271b-4b06-8433-354493feaf06-12_485_570_262_790}
The diagram shows parts of the graphs of $y = 3 - 2 x$ and $y = 4 - 3 \sqrt { } x$ intersecting at points $A$ and $B$.\\
(i) Find by calculation the $x$-coordinates of $A$ and $B$.\\
(ii) Find, showing all necessary working, the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2017 Q8 [8]}}