OCR C2 2015 June — Question 9 9 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Graphs & Exact Values
TypeCalculate intersection coordinates algebraically
DifficultyStandard +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.
Spec1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals

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 \(y = k\).
  1. The smallest positive solution of the equation \(2 \cos \frac { 1 } { 3 } x = k\) is denoted by \(\alpha\). State, in terms of \(\alpha\),
    1. the next smallest positive solution of the equation \(2 \cos \frac { 1 } { 3 } x = k\),
    2. the smallest positive solution of the equation \(2 \cos \frac { 1 } { 3 } x = - k\).
    3. 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 \(y = \sin \frac { 1 } { 3 } x\).
    4. 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}

Question 9:
Part (i)(a):
AnswerMarks Guidance
AnswerMark Guidance
\(6\pi - \alpha\)B1 Allow unsimplified equiv. Allow in degrees ie \(1080 - \alpha\), or equiv.
[1]
Part (i)(b):
AnswerMarks Guidance
AnswerMark Guidance
Use period of \(6\pi\) to make valid attempt at solutionM1 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):
AnswerMarks Guidance
AnswerMark 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 graphA1 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):
AnswerMarks Guidance
AnswerMark 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.
A1Obtain \(3.32\). 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]}}