| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then evaluate integral |
| Difficulty | Standard +0.3 This is a standard P3/C3 question testing routine application of addition formulae and double angle identities. Part (i) is a guided proof requiring straightforward expansion of compound angles. Part (ii) involves direct integration of the simplified form. Part (iii) requires differentiation and solving a trigonometric equation. All steps are procedural with clear guidance, making this slightly easier than average for A-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State correct expansion of \(\sin(3x+x)\) or \(\sin(3x-x)\) | B1 | B0 if their formula retains \(\pm\) in the middle |
| Substitute expansions in \(\frac{1}{2}(\sin 4x + \sin 2x)\) | M1 | |
| Obtain \(\sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)\) correctly | A1 | Must see the \(\sin 4x\) and \(\sin 2x\) or reference to LHS and RHS for A1. AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate and obtain \(-\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x\) | B1 B1 | |
| Substitute limits \(x=0\) and \(x=\frac{1}{3}\pi\) correctly | M1 | In their expression |
| Obtain answer \(\frac{9}{16}\) | A1 | From correct working seen |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State correct derivative \(2\cos 4x + \cos 2x\) | B1 | |
| Using correct double angle formula, express derivative in terms of \(\cos 2x\) and equate the result to zero | M1 | |
| Obtain \(4\cos^2 2x + \cos 2x - 2 = 0\) | A1 | |
| Solve for \(x\) or \(2x\): \(\cos 2x = \dfrac{-1 \pm \sqrt{33}}{8}\) | M1 | Must see working if solving an incorrect quadratic. Roots of correct quadratic are \(-0.843\) and \(0.593\). Need to get as far as \(x = \ldots\) The wrong value of \(x\) is \(0.468\) and can imply M1 if correct quadratic seen. Could be working from a quartic in \(\cos x\): \(16\cos^4 x - 14\cos^2 x + 1 = 0\) |
| Obtain answer \(x = 1.29\) only | A1 | |
| [5] |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State correct expansion of $\sin(3x+x)$ or $\sin(3x-x)$ | **B1** | B0 if their formula retains $\pm$ in the middle |
| Substitute expansions in $\frac{1}{2}(\sin 4x + \sin 2x)$ | **M1** | |
| Obtain $\sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)$ correctly | **A1** | Must see the $\sin 4x$ and $\sin 2x$ or reference to LHS and RHS for A1. **AG** |
| | **[3]** | |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate and obtain $-\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x$ | **B1 B1** | |
| Substitute limits $x=0$ and $x=\frac{1}{3}\pi$ correctly | **M1** | In their expression |
| Obtain answer $\frac{9}{16}$ | **A1** | From correct working seen |
| | **[4]** | |
---
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State correct derivative $2\cos 4x + \cos 2x$ | **B1** | |
| Using correct double angle formula, express derivative in terms of $\cos 2x$ and equate the result to zero | **M1** | |
| Obtain $4\cos^2 2x + \cos 2x - 2 = 0$ | **A1** | |
| Solve for $x$ or $2x$: $\cos 2x = \dfrac{-1 \pm \sqrt{33}}{8}$ | **M1** | Must see working if solving an incorrect quadratic. Roots of correct quadratic are $-0.843$ and $0.593$. Need to get as far as $x = \ldots$ The wrong value of $x$ is $0.468$ and can imply M1 if correct quadratic seen. Could be working from a quartic in $\cos x$: $16\cos^4 x - 14\cos^2 x + 1 = 0$ |
| Obtain answer $x = 1.29$ only | **A1** | |
| | **[5]** | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{772393d7-6e81-4b99-913a-63c9f87d1af2-16_524_689_260_726}
The diagram shows the curve $y = \sin 3 x \cos x$ for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$ and its minimum point $M$. The shaded region $R$ is bounded by the curve and the $x$-axis.\\
(i) By expanding $\sin ( 3 x + x )$ and $\sin ( 3 x - x )$ show that
$$\sin 3 x \cos x = \frac { 1 } { 2 } ( \sin 4 x + \sin 2 x ) .$$
(ii) Using the result of part (i) and showing all necessary working, find the exact area of the region $R$.\\
(iii) Using the result of part (i), express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\cos 2 x$ and hence find the $x$-coordinate of $M$, giving your answer correct to 2 decimal places.\\
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 P3 2019 Q10 [12]}}