| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Region bounded by two curves |
| Difficulty | Moderate -0.3 This is a standard area-between-curves question requiring finding intersection points by solving x + 2 = 3√x (which reduces to a quadratic), then integrating the difference of functions. While it involves multiple steps (solving equation, setting up integral, integrating), all techniques are routine for P1 level 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 integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(x - 3\sqrt{x} + 2 \text{ or } k^2 - 3k + 2 \text{ or } (3\sqrt{x})^2 = (x + 2)^2\) | M1 | OR attempt to eliminate x eg sub \(x = \frac{y^2}{9}\), \(y^2 - 9y + 18 = 0\), \(y = 3\) or \(6\), \(x = 1\) or \(4\) |
| \(\sqrt{x} = 1\) or \(2\) or \(k = 1\) or \(2\) or \(x^2 - 5x + 4(= 0)\) | A1 | |
| \(x = 1\) or \(4\) | A1 | |
| \(y = 3\) or \(6\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[3x\frac{1}{2} - \text{dx} - \int[(x+2)]\text{ dx or attempt at trapezium}\right]\) | M1DM1 | Attempt to integrate. Subtract at some stage |
| \(2x\frac{3}{2} - \left[\frac{1}{2}x^2 + 2x\right]\) or \(\frac{1}{2}(y_2 + y_1)(x_2 - x_1)\) | A1A1 | Where \((x_1, y_1), (x_2, y_2)\) is their \((1, 3), (4, 6)\) |
| \((16 - 2) - \left[(8 + 8) - \left(\frac{1}{2} + 2\right)\right]\) or their \(1 \times 9 \times 3\) | DM1 | Apply their \(1→4\) limits correctly to curve |
| \(\frac{1}{2}\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\int(y - 2)\text{ dy or attempt at trap} - \int\frac{y^2}{9}\text{dy}\right]\) | M1DM1 | |
| \(\left[\frac{1}{2}y^2 - 2y \text{ or } \frac{1}{2}(x_1 + x_2)(y_2 - y_1) - \frac{y^3}{27}\right]\) | A1A1 | |
| \(\left[(18 - 12) - \left(4\frac{1}{2} - 6\right)\right]\) or \(\left[\frac{1}{2}1 - 5 \times 3\right] - [-8-1]\) | DM1 | Apply their \(3→6\) limits correctly to curve |
| \(\frac{1}{2}\) | A1 | [6] |
### (i)
$x - 3\sqrt{x} + 2 \text{ or } k^2 - 3k + 2 \text{ or } (3\sqrt{x})^2 = (x + 2)^2$ | **M1** | OR attempt to eliminate x eg sub $x = \frac{y^2}{9}$, $y^2 - 9y + 18 = 0$, $y = 3$ or $6$, $x = 1$ or $4$
$\sqrt{x} = 1$ or $2$ or $k = 1$ or $2$ or $x^2 - 5x + 4(= 0)$ | **A1** |
$x = 1$ or $4$ | **A1** |
$y = 3$ or $6$ | **A1** | [4]
### (ii)
$\left[3x\frac{1}{2} - \text{dx} - \int[(x+2)]\text{ dx or attempt at trapezium}\right]$ | **M1DM1** | Attempt to integrate. Subtract at some stage
$2x\frac{3}{2} - \left[\frac{1}{2}x^2 + 2x\right]$ or $\frac{1}{2}(y_2 + y_1)(x_2 - x_1)$ | **A1A1** | Where $(x_1, y_1), (x_2, y_2)$ is their $(1, 3), (4, 6)$
$(16 - 2) - \left[(8 + 8) - \left(\frac{1}{2} + 2\right)\right]$ or their $1 \times 9 \times 3$ | **DM1** | Apply their $1→4$ limits correctly to curve
$\frac{1}{2}$ | **A1** | [6]
OR
$\left[\int(y - 2)\text{ dy or attempt at trap} - \int\frac{y^2}{9}\text{dy}\right]$ | **M1DM1** |
$\left[\frac{1}{2}y^2 - 2y \text{ or } \frac{1}{2}(x_1 + x_2)(y_2 - y_1) - \frac{y^3}{27}\right]$ | **A1A1** |
$\left[(18 - 12) - \left(4\frac{1}{2} - 6\right)\right]$ or $\left[\frac{1}{2}1 - 5 \times 3\right] - [-8-1]$ | **DM1** | Apply their $3→6$ limits correctly to curve
$\frac{1}{2}$ | **A1** | [6]
---\includegraphics{figure_9}
The diagram shows parts of the graphs of $y = x + 2$ and $y = 3\sqrt{x}$ intersecting at points $A$ and $B$.
\begin{enumerate}[label=(\roman*)]
\item Write down an equation satisfied by the $x$-coordinates of $A$ and $B$. Solve this equation and hence find the coordinates of $A$ and $B$. [4]
\item Find by integration the area of the shaded region. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q9 [10]}}