| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2006 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 Part (i) is a standard differentiation exercise using the product rule to find stationary points. Part (ii) applies a given iterative formula with straightforward calculator work. Part (iii) is routine integration by parts with a trigonometric function. All parts follow standard A-level techniques with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks |
|---|---|
| (i) Use product rule | M1 |
| Obtain correct derivative \(\cos 2x - 2x\sin 2x\) | A1 |
| Equate derivative to zero and obtain given answer correctly | A1 |
| (ii) Use the iterative formula correctly at least once | M1 |
| Obtain final answer 0.43 | A1 |
| Show sufficient iterations to at least 3.d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.425, 0.435) | A1 |
| (iii) Attempt integration by parts and obtain \(\frac{1}{2}x\sin 2x \pm \int \sin 2x dx\), where \(k,l = \frac{1}{2}, l,\) or \(2\) | M1* |
| Obtain \(\frac{1}{2}x\sin 2x - [\frac{1}{4}\sin 2x dx\) | A1 |
| Obtain indefinite integral \(\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x\) | A1 |
| Use limits \(x = 0\) and \(x = \frac{\pi}{4}\) having integrated twice | M1(dep)* |
| Obtain answer \(\frac{\pi}{8} - \frac{1}{4}\), or exact equivalent | A1 |
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| 3 | |
| 5 |
(i) Use product rule | M1 |
Obtain correct derivative $\cos 2x - 2x\sin 2x$ | A1 |
Equate derivative to zero and obtain given answer correctly | A1 |
(ii) Use the iterative formula correctly at least once | M1 |
Obtain final answer 0.43 | A1 |
Show sufficient iterations to at least 3.d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.425, 0.435) | A1 |
(iii) Attempt integration by parts and obtain $\frac{1}{2}x\sin 2x \pm \int \sin 2x dx$, where $k,l = \frac{1}{2}, l,$ or $2$ | M1* |
Obtain $\frac{1}{2}x\sin 2x - [\frac{1}{4}\sin 2x dx$ | A1 |
Obtain indefinite integral $\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x$ | A1 |
Use limits $x = 0$ and $x = \frac{\pi}{4}$ having integrated twice | M1(dep)* |
Obtain answer $\frac{\pi}{8} - \frac{1}{4}$, or exact equivalent | A1 |
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| | 3 |
| | 5 |10\\
\includegraphics[max width=\textwidth, alt={}, center]{9e3b4c96-0989-4ffb-bd74-0e73b79ca45a-3_430_807_1375_667}
The diagram shows the curve $y = x \cos 2 x$ for $0 \leqslant x \leqslant \frac { 1 } { 4 } \pi$. The point $M$ is a maximum point.\\
(i) Show that the $x$-coordinate of $M$ satisfies the equation $1 = 2 x \tan 2 x$.\\
(ii) The equation in part (i) can be rearranged in the form $x = \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 x } \right)$. Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 x _ { n } } \right) ,$$
with initial value $x _ { 1 } = 0.4$, to calculate the $x$-coordinate of $M$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.\\
(iii) Use integration by parts to find the exact area of the region enclosed between the curve and the $x$-axis from 0 to $\frac { 1 } { 4 } \pi$.
\hfill \mbox{\textit{CAIE P3 2006 Q10 [11]}}