| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| 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 that explicitly guides students to use the cone formula rather than integration. Part (i) requires finding a derivative and checking perpendicularity (routine calculus), while part (ii) involves recognizing the shaded region forms a cone when rotated and applying the given formula—a direct application with minimal problem-solving required. Slightly easier than average due to the helpful hint and geometric simplification. |
| 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 | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \left[\frac{1}{2}(1+4x)^{-1/2}\right]\times[4]\) | B1B1 | |
| At \(x=6\), \(\frac{dy}{dx} = \frac{2}{5}\) | B1 | |
| Gradient of normal at \(P = -\frac{1}{2}\) | B1↑ | OR eqn of norm: \(y - 5 = their\,-\frac{5}{2}(x-6)\); When \(y=0\), \(x=8\) hence result |
| Gradient of \(PQ = -\frac{5}{2}\) hence \(PQ\) is a normal, or \(m_1 m_2 = -1\) | B1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vol for curve \(= (\pi)\int(1+4x)\) and attempt to integrate \(y^2\) | M1 | |
| \(= (\pi)\left[x + 2x^2\right]\) ignore '\(+c\)' | A1 | |
| \(= (\pi)[6 + 72 - 0]\) | DM1 | Apply limits \(0 \to 6\) (allow reversed if corrected later) |
| \(= 78(\pi)\) | A1 | |
| Vol for line \(= \frac{1}{3}\times(\pi)\times 5^2 \times 2\) | M1 | OR \((\pi)\left[\frac{\left(-\frac{5}{2}x+20\right)^3}{3\times-\frac{5}{2}}\right]_6^8\) |
| \(= \frac{50}{3}(\pi)\) | A1 | |
| Total Vol \(= 78\pi + 50\pi/3 = 94\frac{2}{3}\pi\) (or \(284\pi/3\)) | A1 [7] |
## Question 11:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \left[\frac{1}{2}(1+4x)^{-1/2}\right]\times[4]$ | B1B1 | |
| At $x=6$, $\frac{dy}{dx} = \frac{2}{5}$ | B1 | |
| Gradient of normal at $P = -\frac{1}{2}$ | B1↑ | **OR** eqn of norm: $y - 5 = their\,-\frac{5}{2}(x-6)$; When $y=0$, $x=8$ hence result |
| Gradient of $PQ = -\frac{5}{2}$ hence $PQ$ is a normal, or $m_1 m_2 = -1$ | B1 [5] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vol for curve $= (\pi)\int(1+4x)$ and attempt to integrate $y^2$ | M1 | |
| $= (\pi)\left[x + 2x^2\right]$ ignore '$+c$' | A1 | |
| $= (\pi)[6 + 72 - 0]$ | DM1 | Apply limits $0 \to 6$ (allow reversed if corrected later) |
| $= 78(\pi)$ | A1 | |
| Vol for line $= \frac{1}{3}\times(\pi)\times 5^2 \times 2$ | M1 | **OR** $(\pi)\left[\frac{\left(-\frac{5}{2}x+20\right)^3}{3\times-\frac{5}{2}}\right]_6^8$ |
| $= \frac{50}{3}(\pi)$ | A1 | |
| Total Vol $= 78\pi + 50\pi/3 = 94\frac{2}{3}\pi$ (or $284\pi/3$) | A1 [7] | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{a9e04003-1e43-40c4-991a-36aa3a93654b-4_517_857_1594_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 2015 Q11 [12]}}