| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with trigonometric functions |
| Difficulty | Standard +0.8 This is a multi-step volumes of revolution problem requiring: (i) integration of cos²x + sec²x using double angle formula and standard result, (ii) recognizing that V = π∫y² dx and connecting part (i) to part (ii). While the individual techniques are standard A-level (double angle formula, sec²x integration, volume formula), the problem requires careful algebraic manipulation and the insight to use part (i) in part (ii), making it moderately challenging but within typical P2 scope. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Express \(\cos^2 x\) in form \(k_1 + k_2 \cos 2x\) | M1 | |
| Obtain correct \(\frac{1}{2} + \frac{1}{2}\cos 2x\) | A1 | |
| Rewrite second term as \(\sec^2 x\) | B1 | |
| Integrate to obtain at least terms \(k_3 \sin 2x\) and \(k_4 \tan x\) | M1 | |
| Obtain \(\frac{1}{2}x + \frac{1}{4}\sin 2x + \tan x\) | A1 | |
| Confirm given result \(\frac{1}{6}\pi + \frac{8}{9}\sqrt{3}\) | A1 | [6] |
| (ii) State volume is \(\pi \int (\cos x + \frac{1}{\cos x})^2\) (\(\pi\) maybe implied by later appearance) | B1 | |
| Expand to obtain \(\pi \int [\cos^2 x + \frac{1}{\cos^2 x} + 2]dx\) or \(\pi \int [\cos^2 x + \frac{1}{\cos^2 x} + 2]dx\) | B1 | |
| Integrate integrand involving three terms (in part using part (i)) or otherwise i.e. \(k_1 \sin 2x + k_4 \tan x + k_5 x\) | M1 | |
| Obtain \(\frac{7}{6}\pi^2 + \frac{8}{9}\sqrt{3}\) or exact equivalent | A1 | [4] |
(i) Express $\cos^2 x$ in form $k_1 + k_2 \cos 2x$ | M1 |
Obtain correct $\frac{1}{2} + \frac{1}{2}\cos 2x$ | A1 |
Rewrite second term as $\sec^2 x$ | B1 |
Integrate to obtain at least terms $k_3 \sin 2x$ and $k_4 \tan x$ | M1 |
Obtain $\frac{1}{2}x + \frac{1}{4}\sin 2x + \tan x$ | A1 |
Confirm given result $\frac{1}{6}\pi + \frac{8}{9}\sqrt{3}$ | A1 | [6]
(ii) State volume is $\pi \int (\cos x + \frac{1}{\cos x})^2$ ($\pi$ maybe implied by later appearance) | B1 |
Expand to obtain $\pi \int [\cos^2 x + \frac{1}{\cos^2 x} + 2]dx$ or $\pi \int [\cos^2 x + \frac{1}{\cos^2 x} + 2]dx$ | B1 |
Integrate integrand involving three terms (in part using part (i)) or otherwise i.e. $k_1 \sin 2x + k_4 \tan x + k_5 x$ | M1 |
Obtain $\frac{7}{6}\pi^2 + \frac{8}{9}\sqrt{3}$ or exact equivalent | A1 | [4]7 (i) Show that the exact value of $\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \left( \cos ^ { 2 } x + \frac { 1 } { \cos ^ { 2 } x } \right) \mathrm { d } x$ is $\frac { 1 } { 6 } \pi + \frac { 9 } { 8 } \sqrt { } 3$.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{7e100be2-9768-4fcd-b516-c714e53b0665-3_444_495_1523_865}
The diagram shows the curve $y = \cos x + \frac { 1 } { \cos x }$ for $0 \leqslant x \leqslant \frac { 1 } { 3 } \pi$. The shaded region is bounded by the curve and the lines $x = 0 , x = \frac { 1 } { 3 } \pi$ and $y = 0$. Find the exact volume of the solid obtained when the shaded region is rotated completely about the $x$-axis.
\hfill \mbox{\textit{CAIE P2 2015 Q7 [10]}}