| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and stationary points |
| Difficulty | Challenging +1.2 This question requires finding a stationary point using the product rule (with chain rule for the exponential and square root terms), then computing a volume of revolution. The differentiation is moderately involved but follows standard techniques. The integration for volume of revolution simplifies nicely after squaring the function, likely yielding a straightforward integral. This is a typical P3/C4 level question requiring multiple standard techniques but no novel insight—slightly above average difficulty due to the algebraic complexity of the product/chain rule application. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Either use correct product or quotient rule, or square both sides, use correct product rule and make a reasonable attempt at applying the chain rule | M1 | |
| Obtain correct result of differentiation in any form | A1 | |
| Set derivative equal to zero and solve for \(x\) | M1 | |
| Obtain \(x = \frac{3}{2}\) only, correctly | A1 | [4 marks] |
| (ii) State or imply the indefinite integral for the volume is \(\pi\int e^{-x}(1 + 2x)dx\) | B1 | |
| Integrate by parts and reach \(\pm e^{-x}(1 + 2x) \pm \int 2e^{-x}dx\) | M1 | |
| Obtain \(-e^{-x}(1 + 2x) + \int 2e^{-x}dx\), or equivalent | A1 | |
| Complete integration correctly, obtaining \(-e^{-x}(1 + 2x) - 2e^{-x}\), or equivalent | A1 | |
| Use limits \(x = -\frac{1}{2}\) and \(x = 0\) correctly, having integrated twice | M1 | |
| Obtain exact answer \(\pi(2\sqrt{e} - 3)\), or equivalent | A1 | [If \(\pi\) omitted initially or \(2\pi\) or \(\pi 2\) used, give B0 and then follow through.] |
**(i)** Either use correct product or quotient rule, or square both sides, use correct product rule and make a reasonable attempt at applying the chain rule | M1 |
Obtain correct result of differentiation in any form | A1 |
Set derivative equal to zero and solve for $x$ | M1 |
Obtain $x = \frac{3}{2}$ only, correctly | A1 | [4 marks]
**(ii)** State or imply the indefinite integral for the volume is $\pi\int e^{-x}(1 + 2x)dx$ | B1 |
Integrate by parts and reach $\pm e^{-x}(1 + 2x) \pm \int 2e^{-x}dx$ | M1 |
Obtain $-e^{-x}(1 + 2x) + \int 2e^{-x}dx$, or equivalent | A1 |
Complete integration correctly, obtaining $-e^{-x}(1 + 2x) - 2e^{-x}$, or equivalent | A1 |
Use limits $x = -\frac{1}{2}$ and $x = 0$ correctly, having integrated twice | M1 |
Obtain exact answer $\pi(2\sqrt{e} - 3)$, or equivalent | A1 | [If $\pi$ omitted initially or $2\pi$ or $\pi 2$ used, give B0 and then follow through.] | [6 marks]
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\includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-4_547_1401_264_370}
The diagram shows the curve $y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \sqrt { } ( 1 + 2 x )$ and its maximum point $M$. The shaded region between the curve and the axes is denoted by $R$.\\
(i) Find the $x$-coordinate of $M$.\\
(ii) Find by integration the volume of the solid obtained when $R$ is rotated completely about the $x$-axis. Give your answer in terms of $\pi$ and e.
\hfill \mbox{\textit{CAIE P3 2008 Q9 [10]}}