| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and stationary points |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring standard techniques: solving a quadratic to find intersection points, differentiation to find a stationary point, and applying the standard volume formula π∫(y²)dx. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.07n Stationary points: find maxima, minima using derivatives4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x + \frac{4}{x} = 5 \to A(1,5), B(4,5)\) | B1 B1 | co. co. |
| \(\frac{dy}{dx} = 1 - \frac{4}{x^2}\) | M1 | Differentiates. |
| \(= 0\) when \(x = 2, M(2, 4).\) | DM1 A1 | Setting to 0. co. |
| [5] | ||
| (ii) Vol of cylinder \(= \pi 5^2 \cdot 3\) | B1 | Any valid method. |
| Vol under curve \(= \pi \int y^2 \, dx\) | M1 | Attempt at integrating \(y^2\) |
| Integral \(= \frac{x^3}{3} - \frac{16}{x} + 8x\) | A2, 1, 0 | Allow if no \(\pi\) present. |
| Uses his limits "1 to 4" | DM1 | Using his limits. |
| \(\to 75\pi - 57\pi = 18\pi\) | A1 | co. |
| [6] |
$y = x + \frac{4}{x}$
(i) $x + \frac{4}{x} = 5 \to A(1,5), B(4,5)$ | B1 B1 | co. co.
$\frac{dy}{dx} = 1 - \frac{4}{x^2}$ | M1 | Differentiates.
$= 0$ when $x = 2, M(2, 4).$ | DM1 A1 | Setting to 0. co.
| [5] |
(ii) Vol of cylinder $= \pi 5^2 \cdot 3$ | B1 | Any valid method.
Vol under curve $= \pi \int y^2 \, dx$ | M1 | Attempt at integrating $y^2$
Integral $= \frac{x^3}{3} - \frac{16}{x} + 8x$ | A2, 1, 0 | Allow if no $\pi$ present.
Uses his limits "1 to 4" | DM1 | Using his limits.
$\to 75\pi - 57\pi = 18\pi$ | A1 | co.
| [6] |
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{71fe6352-e0dc-4c3a-8b54-99709a1782ca-4_602_899_248_625}
The diagram shows part of the curve $y = x + \frac { 4 } { x }$ which has a minimum point at $M$. The line $y = 5$ intersects the curve at the points $A$ and $B$.\\
(i) Find the coordinates of $A , B$ and $M$.\\
(ii) Find the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
\hfill \mbox{\textit{CAIE P1 2010 Q9 [11]}}