| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Find period or state transformations |
| Difficulty | Easy -1.2 This is a straightforward question on basic trigonometric transformations. Part (a) is routine graph sketching, part (b) requires recognizing a horizontal stretch (scale factor 1/5), and part (c) requires identifying a horizontal translation of 2°. These are standard textbook exercises requiring only recall of transformation rules with no problem-solving or novel insight needed. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05f Trigonometric function graphs: symmetries and periodicities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct sine curve shape with amplitude 1 | B1 | Must pass through origin, reach max of 1 and min of -1 |
| Correct period of 360°, passing through (180°, 0) and (360°, 0) | B1 | Both zeroes at 180° and 360° marked correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Stretch | B1 | Must use word "stretch" |
| Scale factor \(\frac{1}{5}\) in the \(x\)-direction (horizontal stretch) | B1 | Must state direction and scale factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Translation | B1 | Must use word "translation" |
| By \(2°\) to the left / \(\begin{pmatrix}-2\\0\end{pmatrix}\) | B1 | \(y = \sin(5x+10°) = \sin 5(x+2°)\), so translation of \(-2°\) in \(x\)-direction |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct sine curve shape with amplitude 1 | B1 | Must pass through origin, reach max of 1 and min of -1 |
| Correct period of 360°, passing through (180°, 0) and (360°, 0) | B1 | Both zeroes at 180° and 360° marked correctly |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Stretch | B1 | Must use word "stretch" |
| Scale factor $\frac{1}{5}$ in the $x$-direction (horizontal stretch) | B1 | Must state direction and scale factor |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Translation | B1 | Must use word "translation" |
| By $2°$ to the left / $\begin{pmatrix}-2\\0\end{pmatrix}$ | B1 | $y = \sin(5x+10°) = \sin 5(x+2°)$, so translation of $-2°$ in $x$-direction |
---6
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the axes given below, the graph of $y = \sin x$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\item Describe the geometrical transformation that maps the graph of $y = \sin x$ onto the graph of $y = \sin 5 x$.
\item Describe the single geometrical transformation that maps the graph of $y = \sin 5 x$ onto the graph of $y = \sin \left( 5 x + 10 ^ { \circ } \right)$.\\[0pt]
[2 marks]
(a)\\
\includegraphics[max width=\textwidth, alt={}, center]{30ccdbe9-0c91-4011-a3f9-3ce01862215d-12_675_1417_906_370}
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2014 Q6 [6]}}