| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and tangent/normal |
| Difficulty | Standard +0.3 This question combines standard differentiation to find a normal, coordinate geometry, and volume of revolution using the standard formula. All steps are routine A-level techniques with no novel insight required. The 'show that' format provides checkpoints, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations4.08d Volumes of revolution: about x and y axes |
| Answer | Marks |
|---|---|
| M1 A1 DM1 DM1 [5] | Any attempt at differentiation; For \(-\frac{1}{2}\); Use of \(m_1m_2 = -1\); Correct method for eqn of line |
| Answer | Marks |
|---|---|
| A1 | Ans given – beware fortuitous answers. |
| Answer | Marks |
|---|---|
| M1 A1 | Use of \(\int y^2 dx\) for M, correct(needs \(\pi\)) for A |
| Answer | Marks |
|---|---|
| DM1 | Use of 6 and 4.5 |
| Answer | Marks |
|---|---|
| A1 [4] | Beware fortuitous answers (ans given) |
**(i)** Differentiate $y = 18/x \to -18x^{-2}$ Gradient of tangent = $-\frac{1}{2}$ Gradient of normal = $2$ Eqn of normal: $y-3 = 2(x-6)$ $(y=2x-9)$
| M1 A1 DM1 DM1 [5] | Any attempt at differentiation; For $-\frac{1}{2}$; Use of $m_1m_2 = -1$; Correct method for eqn of line |
If $y = 0$, $x = 4\frac{1}{2}$
| A1 | Ans given – beware fortuitous answers. |
**(ii)** Vol $= \pi \int y^2 dx = \pi \int \frac{324}{x^2} dx = \pi[-324x^{-1}]$
| M1 A1 | Use of $\int y^2 dx$ for M, correct(needs $\pi$) for A |
Uses value at $x=6$ – value at $x = 4.5$
| DM1 | Use of 6 and 4.5 |
$-54\pi - (-72\pi) = 18\pi$
| A1 [4] | Beware fortuitous answers (ans given) |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{22a31966-4433-4d7d-8a75-bcd536acfa24-3_646_841_593_651}
The diagram shows part of the graph of $y = \frac { 18 } { x }$ and the normal to the curve at $P ( 6,3 )$. This normal meets the $x$-axis at $R$. The point $Q$ on the $x$-axis and the point $S$ on the curve are such that $P Q$ and $S R$ are parallel to the $y$-axis.\\
(i) Find the equation of the normal at $P$ and show that $R$ is the point ( $4 \frac { 1 } { 2 } , 0$ ).\\
(ii) Show that the volume of the solid obtained when the shaded region $P Q R S$ is rotated through $360 ^ { \circ }$ about the $x$-axis is $18 \pi$.
\hfill \mbox{\textit{CAIE P1 2004 Q7 [9]}}