| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Calculate intersection coordinates algebraically |
| Difficulty | Standard +0.3 This is a straightforward C2 trigonometry question requiring understanding of cosine graph symmetry and solving a basic trig equation. Part (i) tests pattern recognition in solutions, part (ii) is a routine sketch, and part (iii) involves solving 2cos(x/3) = sin(x/3) using tan = 1/2, which is standard technique. Slightly easier than average due to clear structure and standard methods throughout. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6\pi - \alpha\) | B1 | Allow unsimplified equiv. Allow in degrees ie \(1080 - \alpha\), or equiv. |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use period of \(6\pi\) to make valid attempt at solution | M1 | Allow any unsimplified equiv. Allow in degrees ie \(540 - \alpha\), or equiv. |
| \(3\pi - \alpha\) | A1 | Must be simplified, and in radians. Allow \(a\) or alpha for \(\alpha\). |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct graph shape for \(y = k\sin\frac{1}{3}x\) | M1 | Must be one complete (positive) sin cycle, starting at \((0,0)\) and clearly intended to have a final root at the same \(x\)-value as the end point of the given curve. Allow the curve to extend beyond this final root. Allow any amplitude. Condone a slightly inaccurate \(x\)-intercept for the middle root. Condone poor curvature, including overly straight sections and stationary values that are pointed rather than curved. |
| Fully correct graph | A1 | Curve should clearly be intended to have an amplitude that is half of the given curve, but explicit labels of 1 and -1 are not required. A0 if an incorrect scale is given. A smooth, symmetrical curve is now required, with correct \(x\)-intercepts clearly intended. Ignore any scale, correct or incorrect, on the \(x\)-axis. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan\frac{1}{3}x = 2\) | B1 | Allow B1 for correct equation even if no, or an incorrect, attempt to solve. Give BOD on notation e.g. \(\frac{\sin}{\cos}(\frac{1}{3}x) = 2\). If \(\tan\frac{1}{3}x = 2\) is obtained fortuitously from incorrect algebra then mark as B0M1A0A0. |
| \(\frac{1}{3}x = 1.107,\ 4.249\) | M1 | Attempt to solve \(\tan\frac{1}{3}x = k\). Attempt \(3\tan^{-1}(k)\), any non-zero numerical \(k\). M0 for \(\tan^{-1}(3k)\). Allow if attempted in degrees not radians. M1 could be implied rather than explicit. |
| A1 | Obtain \(3.32\). Must be radians and not degrees. Allow answers in range \([3.32, 3.33]\). A0 for answer given as a multiple of \(\pi\). | |
| \(x = 3.32,\ 12.7\) | A1 | Obtain \(12.7\). Must be radians and not degrees. Allow answers in range \([12.7, 12.8]\). A0 for answer given as a multiple of \(\pi\). Max of 3/4 if additional solutions given in range \([0, 6\pi]\) but ignore any solutions outside of this range. Answer only with no method shown is 0/4. |
| Alt method: B1 Obtain \(5\sin^2\frac{1}{3}x = 4\) or \(5\cos^2\frac{1}{3}x = 1\). M1 Attempt to solve \(\sin^2\frac{1}{3}x = k\) or \(\cos^2\frac{1}{3}x = k\) (allow M1 if just the positive square root used). A1 Obtain \(3.32\). A1 Obtain \(12.7\) (max 3/4 if additional solutions in range). | ||
| [4] |
# Question 9:
## Part (i)(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $6\pi - \alpha$ | B1 | Allow unsimplified equiv. Allow in degrees ie $1080 - \alpha$, or equiv. |
| | **[1]** | |
## Part (i)(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use period of $6\pi$ to make valid attempt at solution | M1 | Allow any unsimplified equiv. Allow in degrees ie $540 - \alpha$, or equiv. |
| $3\pi - \alpha$ | A1 | Must be simplified, and in radians. Allow $a$ or alpha for $\alpha$. |
| | **[2]** | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct graph shape for $y = k\sin\frac{1}{3}x$ | M1 | Must be one complete (positive) sin cycle, starting at $(0,0)$ and clearly intended to have a final root at the same $x$-value as the end point of the given curve. Allow the curve to extend beyond this final root. Allow any amplitude. Condone a slightly inaccurate $x$-intercept for the middle root. Condone poor curvature, including overly straight sections and stationary values that are pointed rather than curved. |
| Fully correct graph | A1 | Curve should clearly be intended to have an amplitude that is half of the given curve, but explicit labels of 1 and -1 are not required. A0 if an incorrect scale is given. A smooth, symmetrical curve is now required, with correct $x$-intercepts clearly intended. Ignore any scale, correct or incorrect, on the $x$-axis. |
| | **[2]** | |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan\frac{1}{3}x = 2$ | B1 | Allow B1 for correct equation even if no, or an incorrect, attempt to solve. Give BOD on notation e.g. $\frac{\sin}{\cos}(\frac{1}{3}x) = 2$. If $\tan\frac{1}{3}x = 2$ is obtained fortuitously from incorrect algebra then mark as B0M1A0A0. |
| $\frac{1}{3}x = 1.107,\ 4.249$ | M1 | Attempt to solve $\tan\frac{1}{3}x = k$. Attempt $3\tan^{-1}(k)$, any non-zero numerical $k$. M0 for $\tan^{-1}(3k)$. Allow if attempted in degrees not radians. M1 could be implied rather than explicit. |
| | A1 | Obtain $3.32$. Must be radians and not degrees. Allow answers in range $[3.32, 3.33]$. A0 for answer given as a multiple of $\pi$. |
| $x = 3.32,\ 12.7$ | A1 | Obtain $12.7$. Must be radians and not degrees. Allow answers in range $[12.7, 12.8]$. A0 for answer given as a multiple of $\pi$. Max of 3/4 if additional solutions given in range $[0, 6\pi]$ but ignore any solutions outside of this range. Answer only with no method shown is 0/4. |
| | | **Alt method:** **B1** Obtain $5\sin^2\frac{1}{3}x = 4$ or $5\cos^2\frac{1}{3}x = 1$. **M1** Attempt to solve $\sin^2\frac{1}{3}x = k$ or $\cos^2\frac{1}{3}x = k$ (allow M1 if just the positive square root used). **A1** Obtain $3.32$. **A1** Obtain $12.7$ (max 3/4 if additional solutions in range). |
| | **[4]** | |
9\\
\includegraphics[max width=\textwidth, alt={}, center]{6dd10d03-5fe2-4a70-b5a2-03347dff0360-4_406_625_248_721}
The diagram shows part of the curve $y = 2 \cos \frac { 1 } { 3 } x$, where $x$ is in radians, and the line $y = k$.\\
(i) The smallest positive solution of the equation $2 \cos \frac { 1 } { 3 } x = k$ is denoted by $\alpha$. State, in terms of $\alpha$,
\begin{enumerate}[label=(\alph*)]
\item the next smallest positive solution of the equation $2 \cos \frac { 1 } { 3 } x = k$,
\item the smallest positive solution of the equation $2 \cos \frac { 1 } { 3 } x = - k$.\\
(ii) The curve $y = 2 \cos \frac { 1 } { 3 } x$ is shown in the Printed Answer Book. On the diagram, and for the same values of $x$, sketch the curve of $y = \sin \frac { 1 } { 3 } x$.\\
(iii) Calculate the $x$-coordinates of the points of intersection of the curves in part (ii). Give your answers in radians correct to 3 significant figures.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2015 Q9 [9]}}