| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.8 This question requires deriving a stationary point equation using the quotient rule and trigonometric manipulation, then applying iterative methods and integration. Part (a) involves differentiating y = √x cos x and showing tan a = 1/(2a), which requires careful algebraic manipulation. The iteration in (b) is straightforward application. Part (c) requires volume of revolution integration with a product of functions. This is a substantial multi-part question requiring several A-level techniques, but each step follows standard methods without requiring exceptional insight—typical of harder P3 questions. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct product rule | M1 | |
| Obtain correct derivative in any form | A1 | e.g. \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\cos x - \sqrt{x}\sin x\). Accept in \(a\) or in \(x\) |
| Equate derivative to zero and obtain \(\tan a = \frac{1}{2a}\) | A1 | Obtain given answer from correct working. The question says 'show that ..' so there should be an intermediate step e.g. \(\cos x = 2x\sin x\). Allow \(\tan x = \frac{1}{2x}\) |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the iterative process correctly at least once (get one value and go on to use it in a second use of the formula) | M1 | Must be working in radians. Degrees gives 1, 12.6039, 5.4133, ... M0 |
| Obtain final answer 3.29 | A1 | Clear conclusion |
| Show sufficient iterations to at least 4 d.p. to justify 3.29, or show there is a sign change in the interval (3.285, 3.295) | A1 | 3, 3.3067, 3.2917, 3.2923. Allow more than 4 d.p. Condone truncation. |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply the indefinite integral for the volume is \(\pi\int\!\left(\sqrt{x}\cos x\right)^2 dx\) | B1 | If \(\pi\) omitted, or \(2\pi\) or \(\frac{1}{2}\pi\) used, give B0 and follow through. 4/6 available |
| Use correct \(\cos 2A\) formula, commence integration by parts and reach \(x(ax + b\sin 2x) \pm \int ax + b\sin 2x\, dx\) | *M1 | Alternative: \(\frac{x^2}{4} + \frac{x}{4}\sin 2x - \int\frac{1}{4}\sin 2x\, dx\) |
| Obtain \(x\!\left(\frac{1}{2}x + \frac{1}{4}\sin 2x\right) - \int\frac{1}{2}x + \frac{1}{4}\sin 2x\, dx\), or equivalent | A1 | |
| Complete integration and obtain \(\frac{1}{4}x^2 + \frac{1}{4}x\sin 2x + \frac{1}{8}\cos 2x\) | A1 | OE |
| Substitute limits \(x = 0\) and \(x = \frac{1}{2}\pi\), having integrated twice | DM1 | \(\frac{\pi}{2}\!\left[\frac{\pi^2}{8} + 0 - \frac{1}{4} - 0 - 0 - \frac{1}{4}\right]\) |
| Obtain answer \(\frac{1}{16}\pi\!\left(\pi^2 - 4\right)\), or exact equivalent | A1 | CAO |
| Total: 6 marks |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct product rule | M1 | |
| Obtain correct derivative in any form | A1 | e.g. $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\cos x - \sqrt{x}\sin x$. Accept in $a$ or in $x$ |
| Equate derivative to zero and obtain $\tan a = \frac{1}{2a}$ | A1 | Obtain given answer from correct working. The question says 'show that ..' so there should be an intermediate step e.g. $\cos x = 2x\sin x$. Allow $\tan x = \frac{1}{2x}$ |
| **Total: 3 marks** | | |
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the iterative process correctly at least once (get one value and go on to use it in a second use of the formula) | M1 | Must be working in radians. Degrees gives 1, 12.6039, 5.4133, ... M0 |
| Obtain final answer 3.29 | A1 | Clear conclusion |
| Show sufficient iterations to at least 4 d.p. to justify 3.29, or show there is a sign change in the interval (3.285, 3.295) | A1 | 3, 3.3067, 3.2917, 3.2923. Allow more than 4 d.p. Condone truncation. |
| **Total: 3 marks** | | |
---
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply the indefinite integral for the volume is $\pi\int\!\left(\sqrt{x}\cos x\right)^2 dx$ | B1 | If $\pi$ omitted, or $2\pi$ or $\frac{1}{2}\pi$ used, give B0 and follow through. 4/6 available |
| Use correct $\cos 2A$ formula, commence integration by parts and reach $x(ax + b\sin 2x) \pm \int ax + b\sin 2x\, dx$ | *M1 | Alternative: $\frac{x^2}{4} + \frac{x}{4}\sin 2x - \int\frac{1}{4}\sin 2x\, dx$ |
| Obtain $x\!\left(\frac{1}{2}x + \frac{1}{4}\sin 2x\right) - \int\frac{1}{2}x + \frac{1}{4}\sin 2x\, dx$, or equivalent | A1 | |
| Complete integration and obtain $\frac{1}{4}x^2 + \frac{1}{4}x\sin 2x + \frac{1}{8}\cos 2x$ | A1 | OE |
| Substitute limits $x = 0$ and $x = \frac{1}{2}\pi$, having integrated twice | DM1 | $\frac{\pi}{2}\!\left[\frac{\pi^2}{8} + 0 - \frac{1}{4} - 0 - 0 - \frac{1}{4}\right]$ |
| Obtain answer $\frac{1}{16}\pi\!\left(\pi^2 - 4\right)$, or exact equivalent | A1 | CAO |
| **Total: 6 marks** | | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{77a45360-8e1d-4f4f-9830-075d832a14cf-18_549_933_260_605}
The diagram shows the curve $y = \sqrt { x } \cos x$, for $0 \leqslant x \leqslant \frac { 3 } { 2 } \pi$, and its minimum point $M$, where $x = a$. The shaded region between the curve and the $x$-axis is denoted by $R$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a$ satisfies the equation $\tan a = \frac { 1 } { 2 a }$.
\item The sequence of values given by the iterative formula $a _ { n + 1 } = \pi + \tan ^ { - 1 } \left( \frac { 1 } { 2 a _ { n } } \right)$, with initial value $x _ { 1 } = 3$, converges to $a$.
Use this formula to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\item Find the volume of the solid obtained when the region $R$ is rotated completely 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.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q10 [12]}}