| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume using cone or cylinder formula |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question with helpful scaffolding. Part (i) is routine verification (finding gradient and checking perpendicularity). Part (ii) explicitly hints to use the cone formula, making it a direct application: subtract cone PQ from the volume under the curve. The integration of (1+4x) is standard AS-level work. Slightly above average difficulty only due to the geometric decomposition step. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \left[\frac{1}{2}(1+4x)^{-1/2}\right] \times [4]\) | 2 | B1B1 |
| At \(x=6\), \(\frac{dy}{dx} = \frac{2}{5}\) | 1 | B1 |
| Gradient of normal at \(P = -\frac{1}{2}\) | 1 | B1^ |
| Gradient of \(PQ = -\frac{5}{2}\) hence \(PQ\) is a normal, or \(m_1 m_2 = -1\) | 1 | B1 |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vol for curve \(= (\pi)\int(1+4x)\) and attempt to integrate \(y^2\) | 1 | M1\* |
| \(= (\pi)\left[x + 2x^2\right]\) ignore '\(+c\)' | 1 | A1 |
| \(= (\pi)[6 + 72 - 0]\) | 1 | DM1 |
| \(= 78(\pi)\) | 1 | A1 |
| Vol for line \(= \frac{1}{3} \times (\pi) \times 5^2 \times 2\) | 1 | M1 |
| \(= \frac{50}{3}(\pi)\) | 1 | A1 |
| Total Vol \(= 78\pi + 50\pi/3 = 94\frac{2}{3}\pi\) (or \(284\pi/3\)) | 1 | A1 |
| Total | 7 |
## Question 11(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \left[\frac{1}{2}(1+4x)^{-1/2}\right] \times [4]$ | 2 | **B1B1** | |
| At $x=6$, $\frac{dy}{dx} = \frac{2}{5}$ | 1 | **B1** | |
| Gradient of normal at $P = -\frac{1}{2}$ | 1 | **B1^** | **OR** eqn of norm $y - 5 = their\left(-\frac{5}{2}\right)(x-6)$; When $y=0$, $x=8$ hence result |
| Gradient of $PQ = -\frac{5}{2}$ hence $PQ$ is a normal, or $m_1 m_2 = -1$ | 1 | **B1** | |
| **Total** | **5** | |
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## Question 11(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vol for curve $= (\pi)\int(1+4x)$ and attempt to integrate $y^2$ | 1 | **M1\*** | |
| $= (\pi)\left[x + 2x^2\right]$ ignore '$+c$' | 1 | **A1** | |
| $= (\pi)[6 + 72 - 0]$ | 1 | **DM1** | Apply limits $0 \to 6$ (allow reversed if corrected later) |
| $= 78(\pi)$ | 1 | **A1** | |
| Vol for line $= \frac{1}{3} \times (\pi) \times 5^2 \times 2$ | 1 | **M1** | **OR** $(\pi)\left[\dfrac{\left(-\frac{5}{2}x+20\right)^3}{3 \times -\frac{5}{2}}\right]_6^8$ |
| $= \frac{50}{3}(\pi)$ | 1 | **A1** | |
| Total Vol $= 78\pi + 50\pi/3 = 94\frac{2}{3}\pi$ (or $284\pi/3$) | 1 | **A1** | |
| **Total** | **7** | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{097c5d00-9f92-4c3e-8056-7de09347fbb6-18_515_853_260_644}
The diagram shows part of the curve $y = ( 1 + 4 x ) ^ { \frac { 1 } { 2 } }$ and a point $P ( 6,5 )$ lying on the curve. The line $P Q$ intersects the $x$-axis at $Q ( 8,0 )$.\\
(i) Show that $P Q$ is a normal to the curve.\\
(ii) Find, showing all necessary working, the exact volume of revolution obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.\\[0pt]
[In part (ii) you may find it useful to apply the fact that the volume, $V$, of a cone of base radius $r$ and vertical height $h$, is given by $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$.]\\
\hfill \mbox{\textit{CAIE P1 Q11 [12]}}