| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve inequality involving trig |
| Difficulty | Moderate -0.8 This is a straightforward C2 trigonometry question testing standard skills: stating period and coordinates from a graph, solving a basic inequality graphically, and solving a trigonometric equation using tan substitution. All parts follow routine procedures with no problem-solving insight required, making it easier than average but not trivial due to the exact value requirement. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(\pi\) radians | B1 | 1 |
| (ii) \((\frac{7}{2}, -1)\) | B1 | State \(x = {}^s\!/\!_2\). Allow 1.57 radians, or better. Allow \(A = {}^7\!/\!_2\). B0 for 90°. |
| State \(y = -1\) | B1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) \(\cos 2x = 0.5\), \(2x = {}^{\pi}\!/\!_3, {}^{5\pi}\!/\!_{6}\), \(x = {}^{\pi}\!/\!_6, {}^{5\pi}\!/\!_{12}\), hence \({}^{\pi}\!/\!_6 \le x \le {}^{5\pi}\!/\!_{12}\) | M1 | Attempt correct solution method. Inverse cos and then divide by 2, to find at least one angle. |
| Obtain \({}^{\pi}\!/\!_6\) (allow 0.524 or 30°) | A1 | Just mark angle, ignore any (in)equality signs. |
| Obtain \({}^{5\pi}\!/\!_{12}\) (allow 2.62 or 150°) | A1 | Needs to be single term so \(\pi - {}^{\pi}\!/\!_6\) is A0. Just mark angle, ignore any (in)equality signs. A0 if any other angles in range \(0 \le x \le \pi\). |
| Obtain \({}^{\pi}\!/\!_6 \le x \le {}^{5\pi}\!/\!_{12}\) (exact radians only) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(\tan 2x = {}^1\!/\!_{\sqrt{3}}\) with \(2x = {}^{\pi}\!/\!_6, {}^{7\pi}\!/\!_6\), \(x = {}^{\pi}\!/\!_{12}, {}^{7\pi}\!/\!_{12}\) | B1 | Obtain \(\tan 2x = {}^1\!/\!_{\sqrt{3}}\). Allow for decimal equiv ie \(\tan 2x = 0.577\). Allow for \(\sqrt{3} \tan 2x = 1\). |
| Attempt correct solution method of \(\tan 2x = k\) | M1 | Inverse tan and then divide by 2, to find at least one angle. Could follow error eg tan \(2x = \sqrt{3}\), even if \(\tan 2x = {}^{\cos 2x}/\!_{\sin 2x}\) clearly used. |
| Obtain one correct angle | A1 | Could be exact (\({}^{\pi}\!/\!_{12}\) or \({}^{7\pi}\!/\!_{12}\) |
**(a)(i)** $\pi$ radians | B1 | 1 | State $\pi$. Allow 3.14 radians or 180°. B0 for $0 \le x \le \pi$.
**(ii)** $(\frac{7}{2}, -1)$ | B1 | State $x = {}^s\!/\!_2$. Allow 1.57 radians, or better. Allow $A = {}^7\!/\!_2$. B0 for 90°.
**State $y = -1$** | B1 | 2 | Allow $\cos 2A = -1$.
SR Award B1 for $(-1, {}^7\!/\!_2)$
**(iii)** $\cos 2x = 0.5$, $2x = {}^{\pi}\!/\!_3, {}^{5\pi}\!/\!_{6}$, $x = {}^{\pi}\!/\!_6, {}^{5\pi}\!/\!_{12}$, hence ${}^{\pi}\!/\!_6 \le x \le {}^{5\pi}\!/\!_{12}$ | M1 | Attempt correct solution method. Inverse cos and then divide by 2, to find at least one angle.
**Obtain ${}^{\pi}\!/\!_6$ (allow 0.524 or 30°)** | A1 | Just mark angle, ignore any (in)equality signs.
**Obtain ${}^{5\pi}\!/\!_{12}$ (allow 2.62 or 150°)** | A1 | Needs to be single term so $\pi - {}^{\pi}\!/\!_6$ is A0. Just mark angle, ignore any (in)equality signs. A0 if any other angles in range $0 \le x \le \pi$.
**Obtain ${}^{\pi}\!/\!_6 \le x \le {}^{5\pi}\!/\!_{12}$ (exact radians only)** | A1 | 4 | Allow two separate inequalities as long as both correct and linked by 'and' (not 'or', a comma or no link). Mark final answer and condone incorrect inequality signs elsewhere in solution.
SR If alternative methods (eg double angle formulae) or inspection are used (or no method shown at all) then mark as:
- B2 Obtain one correct angle (degrees or radians) with no errors seen
- A1 Obtain second correct angle – and no others
- A1 Obtain correct inequality (exact radians only)
**(b)** $\tan 2x = {}^1\!/\!_{\sqrt{3}}$ with $2x = {}^{\pi}\!/\!_6, {}^{7\pi}\!/\!_6$, $x = {}^{\pi}\!/\!_{12}, {}^{7\pi}\!/\!_{12}$ | B1 | Obtain $\tan 2x = {}^1\!/\!_{\sqrt{3}}$. Allow for decimal equiv ie $\tan 2x = 0.577$. Allow for $\sqrt{3} \tan 2x = 1$.
**Attempt correct solution method of $\tan 2x = k$** | M1 | Inverse tan and then divide by 2, to find at least one angle. Could follow error eg tan $2x = \sqrt{3}$, even if $\tan 2x = {}^{\cos 2x}/\!_{\sin 2x}$ clearly used.
**Obtain one correct angle** | A1 | Could be exact (${}^{\pi}\!/\!_{12}$ or ${}^{7\pi}\!/\!_{12}$9
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{4d03f4e3-ae6c-4a0d-ae4d-d89258d2919a-4_362_979_1505_625}
The diagram shows part of the curve $y = \cos 2 x$, where $x$ is in radians. The point $A$ is the minimum point of this part of the curve.
\begin{enumerate}[label=(\roman*)]
\item State the period of $y = \cos 2 x$.
\item State the coordinates of $A$.
\item Solve the inequality $\cos 2 x \leqslant 0.5$ for $0 \leqslant x \leqslant \pi$, giving your answers exactly.
\end{enumerate}\item Solve the equation $\cos 2 x = \sqrt { 3 } \sin 2 x$ for $0 \leqslant x \leqslant \pi$, giving your answers exactly.
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2011 Q9 [11]}}