| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and tangent/normal |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question with standard techniques. Part (i) requires finding tangent equations at two points (routine differentiation and algebra), while part (ii) involves a standard volume integral with simple algebraic manipulation. The 'show that' in part (i) provides the target, making it easier than an open-ended problem. Slightly above average due to the two-part nature and algebraic manipulation required, but well within typical A-level scope. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \dfrac{x}{2} + \dfrac{6}{x} = 4 \rightarrow x = 2\) or \(6\) | B1 B1 | Inspection or guesswork OK |
| \(\dfrac{dy}{dx} = \dfrac{1}{2} - \dfrac{6}{x^2}\) | B1 | Unsimplified OK |
| When \(x = 2\), \(m = -1 \rightarrow x + y = 6\) | *M1 | Correct method for either tangent |
| When \(x = 6\), \(m = \dfrac{1}{3} \rightarrow y = \dfrac{1}{3}x + 2\) | ||
| Attempt to solve simultaneous equations | DM1 | Could solve BOTH equations separately with \(y = x\) and get \(x = 3\) both times |
| \((3, 3)\) | A1 | Statement about \(y = x\) not required |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = (\pi) \int \left(\frac{x^2}{4} + 6 + \frac{36}{x^2}\right) dx\) | \*M1 | Integrate using \(\pi \int y^2 dx\) (doesn't need \(\pi\) or \(dx\)). Allow incorrect squaring. Not awarded for \(\pi \int \left\{4 - \left(\frac{x}{2} + \frac{6}{x}\right)\right\}^2 dx\). Integration indicated by increase in any power by 1. |
| Integration \(\rightarrow \frac{x^3}{12} + 6x - \frac{36}{x}\) | A2,1 | 3 things wanted — 1 each error, allow \(+ C\). (Doesn't need \(\pi\)) |
| Using limits 'their 2' to 'their 6' \(\left(53\frac{1}{3}\pi,\ \frac{160}{3}\pi,\ 168 \text{ awrt}\right)\) | DM1 | Evidence of their values 6 and 2 from (i) substituted into their integrand and then subtracted. \(48 - \left(-\frac{16}{3}\right)\) is enough. |
| Vol for line: integration or cylinder \((\rightarrow 64\pi)\) | M1 | Use of \(\pi r^2 h\) or integration of \(4^2\) (could be from \(\left\{4 - \left(\frac{x}{2} + \frac{6}{x}\right)\right\}^2\)) |
| Subtracts \(\rightarrow 10\frac{2}{3}\pi\) oe \(\left(\text{e.g. } \frac{32}{3}\pi,\ 33.5 \text{ awrt}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = (\pi) \int 4^2 - \left(\frac{x}{2} + \frac{6}{x}\right)^2 dx\) | M1 \*M1 | Integrate using \(\pi \int y^2 dx\) (doesn't need \(\pi\) or \(dx\)). Integration indicated by increase in any power by 1. |
| \(= (\pi) \int 16 - \left(\frac{x^2}{4} + 6 + \frac{36}{x^2}\right) dx\) | ||
| \(= (\pi) \left[16x - \left(\frac{x^3}{12} + 6x - \frac{36}{x}\right)\right]\) | A2,1 | Or \(\left[10x - \frac{x^3}{12} + \frac{36}{x}\right]\) |
| \(= (\pi)(48 - 37\frac{1}{3})\) | DM1 | Evidence of their values 6 and 2 from (i) substituted |
| \(= 10\frac{2}{3}\pi\) oe \(\left(\text{e.g. } \frac{32}{3}\pi,\ 33.5 \text{ awrt}\right)\) | A1 | |
| 6 |
## Question 11:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \dfrac{x}{2} + \dfrac{6}{x} = 4 \rightarrow x = 2$ or $6$ | B1 B1 | Inspection or guesswork OK |
| $\dfrac{dy}{dx} = \dfrac{1}{2} - \dfrac{6}{x^2}$ | B1 | Unsimplified OK |
| When $x = 2$, $m = -1 \rightarrow x + y = 6$ | *M1 | Correct method for either tangent |
| When $x = 6$, $m = \dfrac{1}{3} \rightarrow y = \dfrac{1}{3}x + 2$ | | |
| Attempt to solve simultaneous equations | DM1 | Could solve BOTH equations separately with $y = x$ and get $x = 3$ both times |
| $(3, 3)$ | A1 | Statement about $y = x$ not required |
| **Total** | **6** | |
## Question 11(ii):
**Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = (\pi) \int \left(\frac{x^2}{4} + 6 + \frac{36}{x^2}\right) dx$ | \*M1 | Integrate using $\pi \int y^2 dx$ (doesn't need $\pi$ or $dx$). Allow incorrect squaring. Not awarded for $\pi \int \left\{4 - \left(\frac{x}{2} + \frac{6}{x}\right)\right\}^2 dx$. Integration indicated by increase in any power by 1. |
| Integration $\rightarrow \frac{x^3}{12} + 6x - \frac{36}{x}$ | A2,1 | 3 things wanted — 1 each error, allow $+ C$. (Doesn't need $\pi$) |
| Using limits 'their 2' to 'their 6' $\left(53\frac{1}{3}\pi,\ \frac{160}{3}\pi,\ 168 \text{ awrt}\right)$ | DM1 | Evidence of their values 6 and 2 from (i) substituted into their integrand and then subtracted. $48 - \left(-\frac{16}{3}\right)$ is enough. |
| Vol for line: integration or cylinder $(\rightarrow 64\pi)$ | M1 | Use of $\pi r^2 h$ or integration of $4^2$ (could be from $\left\{4 - \left(\frac{x}{2} + \frac{6}{x}\right)\right\}^2$) |
| Subtracts $\rightarrow 10\frac{2}{3}\pi$ oe $\left(\text{e.g. } \frac{32}{3}\pi,\ 33.5 \text{ awrt}\right)$ | A1 | |
---
**Method 2 (OR):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = (\pi) \int 4^2 - \left(\frac{x}{2} + \frac{6}{x}\right)^2 dx$ | M1 \*M1 | Integrate using $\pi \int y^2 dx$ (doesn't need $\pi$ or $dx$). Integration indicated by increase in any power by 1. |
| $= (\pi) \int 16 - \left(\frac{x^2}{4} + 6 + \frac{36}{x^2}\right) dx$ | | |
| $= (\pi) \left[16x - \left(\frac{x^3}{12} + 6x - \frac{36}{x}\right)\right]$ | A2,1 | Or $\left[10x - \frac{x^3}{12} + \frac{36}{x}\right]$ |
| $= (\pi)(48 - 37\frac{1}{3})$ | DM1 | Evidence of their values 6 and 2 from (i) substituted |
| $= 10\frac{2}{3}\pi$ oe $\left(\text{e.g. } \frac{32}{3}\pi,\ 33.5 \text{ awrt}\right)$ | A1 | |
| | **6** | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{58d65166-2b1a-4b58-9859-afe919c0a3a9-18_643_969_258_587}
The diagram shows part of the curve $y = \frac { x } { 2 } + \frac { 6 } { x }$. The line $y = 4$ intersects the curve at the points $P$ and $Q$.\\
(i) Show that the tangents to the curve at $P$ and $Q$ meet at a point on the line $y = x$.\\
(ii) Find, showing all necessary working, the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis. Give your answer in terms of $\pi$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE P1 2018 Q11 [12]}}