| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and tangent/normal |
| Difficulty | Standard +0.3 Part (i) requires finding a derivative using the chain rule, evaluating at a point, finding the perpendicular gradient, and writing the normal equation—standard A-level calculus. Part (ii) involves setting up and evaluating a volume of revolution integral with a rational function, requiring substitution or partial fractions. Both parts are routine techniques with straightforward algebra, slightly easier than average due to the simple function form and clear structure. |
| 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 | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{-4}{(5-3x)^2} \times (-3)\) | B1 B1 | B1 without \(\times(-3)\), B1 for \(\times(-3)\) |
| Gradient of tangent \(= 3\), Gradient of normal \(= -\frac{1}{3}\) | \*M1 | Use of \(m_1m_2 = -1\) after calculus |
| \(\rightarrow\) eqn: \(y - 2 = -\frac{1}{3}(x-1)\) | DM1 | Correct form of equation, with \((1, \text{their } y)\), not \((1,0)\) |
| \(\rightarrow y = -\frac{1}{3}x + \frac{7}{3}\) | A1 | This mark needs to have come from \(y = 2\), \(y\) must be subject |
| Total: | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Vol} = \pi\int_{0}^{1} \frac{16}{(5-3x)^2}\, dx\) | M1 | Use of \(V = \pi\int y^2\, dx\) with an attempt at integration |
| \(\pi\left[\frac{-16}{(5-3x)} \div -3\right]\) | A1 A1 | A1 without \((\div -3)\), A1 for \((\div -3)\) |
| \(= \left(\pi\left(\frac{16}{6} - \frac{16}{15}\right)\right) = \frac{8\pi}{5}\) (if limits switched must show \(-\) to \(+\)) | M1 A1 | Use of both correct limits M1 |
| Total: | 5 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{-4}{(5-3x)^2} \times (-3)$ | **B1 B1** | **B1** without $\times(-3)$, **B1** for $\times(-3)$ |
| Gradient of tangent $= 3$, Gradient of normal $= -\frac{1}{3}$ | **\*M1** | Use of $m_1m_2 = -1$ after calculus |
| $\rightarrow$ eqn: $y - 2 = -\frac{1}{3}(x-1)$ | **DM1** | Correct form of equation, with $(1, \text{their } y)$, not $(1,0)$ |
| $\rightarrow y = -\frac{1}{3}x + \frac{7}{3}$ | **A1** | This mark needs to have come from $y = 2$, $y$ must be subject |
| **Total:** | **5** | |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Vol} = \pi\int_{0}^{1} \frac{16}{(5-3x)^2}\, dx$ | **M1** | Use of $V = \pi\int y^2\, dx$ with an attempt at integration |
| $\pi\left[\frac{-16}{(5-3x)} \div -3\right]$ | **A1 A1** | **A1** without $(\div -3)$, **A1** for $(\div -3)$ |
| $= \left(\pi\left(\frac{16}{6} - \frac{16}{15}\right)\right) = \frac{8\pi}{5}$ (if limits switched must show $-$ to $+$) | **M1 A1** | Use of both correct limits **M1** |
| **Total:** | **5** | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{028c7979-6b24-42d0-9857-c616a169b2b2-18_510_410_260_863}
The diagram shows part of the curve $y = \frac { 4 } { 5 - 3 x }$.\\
(i) Find the equation of the normal to the curve at the point where $x = 1$ in the form $y = m x + c$, where $m$ and $c$ are constants.\\
The shaded region is bounded by the curve, the coordinate axes and the line $x = 1$.\\
(ii) Find, showing all necessary working, the volume obtained when this shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
\hfill \mbox{\textit{CAIE P1 2017 Q10 [10]}}