| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with trigonometric functions |
| Difficulty | Standard +0.3 This question requires expanding a squared trigonometric expression using the double angle formula (cos²2x = (1+cos4x)/2), then integrating term-by-term. While it involves multiple steps and careful algebraic manipulation, it follows a standard pattern for P3/C3 integration questions. The identity work in part (a) is routine, and part (b) requires finding where the curve touches the x-axis (standard equation solving) followed by straightforward integration. This is slightly easier than average because it's a well-practiced technique with clear signposting. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+2\cos 2x)^2 = 1 + 4\cos 2x + 4\cos^2 2x\); uses \(\cos 4x = 2\cos^2 2x - 1\) | M1 | Must attempt to multiply out \((1+2\cos 2x)^2\) and use \(\cos 4x = 2\cos^2 2x - 1\) to obtain form \(p + q\cos 2x + r\cos 4x\). Beware: \((1+2\cos 2x)^2 = 1+4\cos 2x + 4\cos^2 4x\) is M0A0 |
| \(= 3 + 4\cos 2x + 2\cos 4x\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = \frac{2\pi}{3}\) | B1 | Deduces \(a = \frac{2\pi}{3}\) (allow \(120°\)). If more than one angle found, look for which is substituted into integrated expression |
| \(\int 3 + 4\cos 2x + 2\cos 4x\, dx = 3x + 2\sin 2x + \frac{1}{2}\sin 4x\) | M1 A1ft | M1: Integrates \(q\cos 2x + r\cos 4x \rightarrow \pm\ldots\sin 2x \pm\ldots\sin 4x\). A1ft: \(px + \frac{q}{2}\sin 2x + \frac{r}{4}\sin 4x\) unsimplified, \(p,q\) and \(r \neq 0\) |
| \(\text{Area} = \left[3x + 2\sin 2x + \frac{1}{2}\sin 4x\right]_0^{\frac{2\pi}{3}} = 2\pi - \frac{3}{4}\sqrt{3}\) | dM1 A1 | dM1: Substitutes \(0\) and \(a = \frac{2\pi}{3}\) (or awrt 2.09) into valid function and subtracts. Note \(q,r \neq 0\) but no \(px\) term needed. Also allow \(a = \frac{\pi}{3}\) or \(a = \frac{4\pi}{3}\); \(a\) must be in radians. A1: \(2\pi - \frac{3}{4}\sqrt{3}\) or simplified equivalent |
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+2\cos 2x)^2 = 1 + 4\cos 2x + 4\cos^2 2x$; uses $\cos 4x = 2\cos^2 2x - 1$ | M1 | Must attempt to multiply out $(1+2\cos 2x)^2$ and use $\cos 4x = 2\cos^2 2x - 1$ to obtain form $p + q\cos 2x + r\cos 4x$. Beware: $(1+2\cos 2x)^2 = 1+4\cos 2x + 4\cos^2 4x$ is M0A0 |
| $= 3 + 4\cos 2x + 2\cos 4x$ | A1 | Correct answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \frac{2\pi}{3}$ | B1 | Deduces $a = \frac{2\pi}{3}$ (allow $120°$). If more than one angle found, look for which is substituted into integrated expression |
| $\int 3 + 4\cos 2x + 2\cos 4x\, dx = 3x + 2\sin 2x + \frac{1}{2}\sin 4x$ | M1 A1ft | M1: Integrates $q\cos 2x + r\cos 4x \rightarrow \pm\ldots\sin 2x \pm\ldots\sin 4x$. A1ft: $px + \frac{q}{2}\sin 2x + \frac{r}{4}\sin 4x$ unsimplified, $p,q$ and $r \neq 0$ |
| $\text{Area} = \left[3x + 2\sin 2x + \frac{1}{2}\sin 4x\right]_0^{\frac{2\pi}{3}} = 2\pi - \frac{3}{4}\sqrt{3}$ | dM1 A1 | dM1: Substitutes $0$ and $a = \frac{2\pi}{3}$ (or awrt 2.09) into valid function and subtracts. Note $q,r \neq 0$ but no $px$ term needed. Also allow $a = \frac{\pi}{3}$ or $a = \frac{4\pi}{3}$; $a$ must be in radians. A1: $2\pi - \frac{3}{4}\sqrt{3}$ or simplified equivalent |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b0b8db0-79fd-4ad5-88c9-737447d9f894-30_515_673_255_639}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve with equation
$$y = ( 1 + 2 \cos 2 x ) ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$( 1 + 2 \cos 2 x ) ^ { 2 } \equiv p + q \cos 2 x + r \cos 4 x$$
where $p , q$ and $r$ are constants to be found.
The curve touches the positive $x$-axis for the second time when $x = a$, as shown in Figure 4. The regions bounded by the curve, the $y$-axis and the $x$-axis up to $x = a$ are shown shaded in Figure 4.
\item Find, using algebraic integration and making your method clear, the exact total area of the shaded regions. Write your answer in simplest form.
\includegraphics[max width=\textwidth, alt={}, center]{9b0b8db0-79fd-4ad5-88c9-737447d9f894-32_2255_51_313_1980}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q10 [7]}}