| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about x-axis, region between two curves |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring finding a line equation (routine coordinate geometry) and computing a volume using π∫(y₁² - y₂²)dx. The integration involves a simple power of (3x+1), which is standard C1/P1 material. The multi-part structure and 'show that' element add minimal difficulty since the algebra is routine. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=04.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = (0,\ 1)\), \(B = (5,\ \frac{1}{2})\) | B1 B1 | |
| \(y - 1 = -\dfrac{1}{10}(x - 0)\) | M1 | ft their \(A, B\) |
| \(y = -\dfrac{1}{10}x + 1\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Curve: \((\pi)\displaystyle\int_0^5 (3x+1)^{-1/2}\,dx\) | M1 | Attempt \(\displaystyle\int_0^5 y^2\,dx\) (\(\pi\) not vital) |
| \(\dfrac{2\pi}{3}[(3x+1)^{1/2}]_0^5\) | A1A1 | (\(\pi\) not vital); 2nd A mark is for \(\div 3\) |
| \(\dfrac{2\pi}{3}[4-1]\) | DM1 | Application of limits to their integral (in either integral); limits 0 to 5 only |
| \([2\pi]\) | ||
| Line: \((\pi)\displaystyle\int_0^5 \!\left(\frac{1}{100}x^2 - \frac{1}{5}x + 1\right)dx\) | M1 | Attempt \(\displaystyle\int_0^5 y^2\,dx\) (\(\pi\) not vital) |
| \((\pi)\!\left[\dfrac{1}{300}x^3 - \dfrac{1}{10}x^2 + x\right]_0^5\) | A2,1 | Also directly \(-\dfrac{10}{3}\!\left(-\dfrac{1}{10}x+1\right)^3\) or \(-\dfrac{10}{3}\!\left[\left(-\dfrac{1}{2}+1\right)^3 - 1^3\right]\) (\(\pi\) not vital) |
| \((\pi)\!\left[\dfrac{125}{300} - \dfrac{25}{10} + 5\right]\) | ||
| \(\left[\dfrac{35\pi}{12}\right]\) | ||
| Volume \(= \dfrac{35\pi}{12} - 2\pi = \dfrac{11\pi}{12}\) | DM1 A1 | Subtraction of their volumes; co |
## Question 11:
**Part (i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = (0,\ 1)$, $B = (5,\ \frac{1}{2})$ | B1 B1 | |
| $y - 1 = -\dfrac{1}{10}(x - 0)$ | M1 | ft their $A, B$ |
| $y = -\dfrac{1}{10}x + 1$ | A1 | **AG** |
**Part (ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Curve: $(\pi)\displaystyle\int_0^5 (3x+1)^{-1/2}\,dx$ | M1 | Attempt $\displaystyle\int_0^5 y^2\,dx$ ($\pi$ not vital) |
| $\dfrac{2\pi}{3}[(3x+1)^{1/2}]_0^5$ | A1A1 | ($\pi$ not vital); 2nd A mark is for $\div 3$ |
| $\dfrac{2\pi}{3}[4-1]$ | DM1 | Application of limits to their integral (in either integral); limits 0 to 5 only |
| $[2\pi]$ | | |
| Line: $(\pi)\displaystyle\int_0^5 \!\left(\frac{1}{100}x^2 - \frac{1}{5}x + 1\right)dx$ | M1 | Attempt $\displaystyle\int_0^5 y^2\,dx$ ($\pi$ not vital) |
| $(\pi)\!\left[\dfrac{1}{300}x^3 - \dfrac{1}{10}x^2 + x\right]_0^5$ | A2,1 | Also directly $-\dfrac{10}{3}\!\left(-\dfrac{1}{10}x+1\right)^3$ or $-\dfrac{10}{3}\!\left[\left(-\dfrac{1}{2}+1\right)^3 - 1^3\right]$ ($\pi$ not vital) |
| $(\pi)\!\left[\dfrac{125}{300} - \dfrac{25}{10} + 5\right]$ | | |
| $\left[\dfrac{35\pi}{12}\right]$ | | |
| Volume $= \dfrac{35\pi}{12} - 2\pi = \dfrac{11\pi}{12}$ | DM1 A1 | Subtraction of their volumes; co |11\\
\includegraphics[max width=\textwidth, alt={}, center]{ae57d8f1-5a0d-426c-952d-e8b99c6aeaba-5_609_897_255_625}
The diagram shows part of the curve $y = \frac { 1 } { ( 3 x + 1 ) ^ { \frac { 1 } { 4 } } }$. The curve cuts the $y$-axis at $A$ and the line $x = 5$ at $B$.\\
(i) Show that the equation of the line $A B$ is $y = - \frac { 1 } { 10 } x + 1$.\\
(ii) Find the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
\hfill \mbox{\textit{CAIE P1 2010 Q11 [13]}}