| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard two-part harmonic form question requiring routine application of the R-formula (finding R=2, α=π/6) followed by solving a trigonometric equation. While it requires multiple steps and exact form answers, the techniques are well-practiced at A-level with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \([\cos x + \sqrt{3}\sin x = R\sin(x+\alpha)]\) \(\Rightarrow R\cos\alpha = \sqrt{3}\), \(R\sin\alpha = 1\) | M1 | 1.1a |
| \(\tan\alpha = \frac{1}{\sqrt{3}}\) | M1 | 1.1, 1.2 |
| \([\alpha] = \frac{\pi}{6}\) | A1 | |
| \(\Rightarrow R^2=4, R=2\) | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos x+\sqrt{3}\sin x = \sqrt{3} \Rightarrow 2\sin\left(x+\frac{\pi}{6}\right)=\sqrt{3}\) | M1 | 3.1a |
| \(\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\) | M1 | 1.1 |
| \(x+\frac{\pi}{6}=\frac{\pi}{3}\) or \(\frac{2\pi}{3}\) | M1 | 1.1 |
| \(x=\frac{\pi}{6}, \frac{\pi}{2}\) | A1 | 1.1 |
## Question 8(a):
$[\cos x + \sqrt{3}\sin x = R\sin(x+\alpha)]$ $\Rightarrow R\cos\alpha = \sqrt{3}$, $R\sin\alpha = 1$ | M1 | 1.1a | M0 if $R$ missing but next M mark available; must be using $\alpha$ and not $x$
$\tan\alpha = \frac{1}{\sqrt{3}}$ | M1 | 1.1, 1.2 | Allow other equivalent methods using their values of sin and cos; allow equivalent surds; dependent on M2
$[\alpha] = \frac{\pi}{6}$ | A1 | | Dependent on M2
$\Rightarrow R^2=4, R=2$ | B1 | 1.1 | soi anywhere
**[4 marks]**
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## Question 8(b):
$\cos x+\sqrt{3}\sin x = \sqrt{3} \Rightarrow 2\sin\left(x+\frac{\pi}{6}\right)=\sqrt{3}$ | M1 | 3.1a | Use of their result from (a)
$\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$ | M1 | 1.1 | Value for a trig function following on from first step
$x+\frac{\pi}{6}=\frac{\pi}{3}$ or $\frac{2\pi}{3}$ | M1 | 1.1 | At least one value for their $x+\frac{\pi}{6}$; condone use of inequality symbols if recovered; condone working consistently in degrees up to this point
$x=\frac{\pi}{6}, \frac{\pi}{2}$ | A1 | 1.1 | Both roots correct in radians only; if extra roots then A0; candidates who square both sides and solve should check solutions and discard erroneous ones — maximum M3A0 if no check
**[4 marks]**8 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Express $\cos x + \sqrt { 3 } \sin x$ in the form $\mathrm { R } \sin ( \mathrm { x } + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$. Give the values of $R$ and $\alpha$ in exact form.
\item Hence solve the equation $\cos x = \sqrt { 3 } ( 1 - \sin x )$ for values of $x$ in the interval $- \pi \leqslant x \leqslant \pi$. Give the roots of this equation in exact form.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2024 Q8 [8]}}