Edexcel P1 2022 June — Question 9 8 marks

Exam BoardEdexcel
ModuleP1 (Pure Mathematics 1)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Graphs & Exact Values
TypeSketch two trig curves and count intersections/solutions
DifficultyModerate -0.3 This is a slightly below-average A-level question. Part (a) tests routine application of sine symmetry properties with straightforward angle transformations. Parts (b) and (c) require sketching sin 2x and using the graph to find intersections with y=p, which is a standard textbook exercise in graph transformations and solving trigonometric equations graphically. The multi-part structure adds some length but each component is procedural with no novel insight required.
Spec1.05a Sine, cosine, tangent: definitions for all arguments1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals
9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-26_428_1354_251_287} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows part of the graph of the curve with equation \(y = \sin x\) Given that \(\sin \alpha = p\), where \(0 < \alpha < 90 ^ { \circ }\)
  1. state, in terms of \(p\), the value of
    1. \(2 \sin \left( 180 ^ { \circ } - \alpha \right)\)
    2. \(\sin \left( \alpha - 180 ^ { \circ } \right)\)
    3. \(3 + \sin \left( 180 ^ { \circ } + \alpha \right)\) A copy of Figure 4, labelled Diagram 1, is shown on page 27. On Diagram 1,
  2. sketch the graph of \(y = \sin 2 x\)
  3. Hence find, in terms of \(\alpha\), the \(x\) coordinates of any points in the interval \(0 < x < 180 ^ { \circ }\) where $$\sin 2 x = p$$
    \includegraphics[max width=\textwidth, alt={}]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-27_433_1331_296_310}
    \section*{Diagram 1}
Question 9:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
(i) \(2p\)B1 Condone \(p+p\); award when \(2p\) is the \(y\)-coordinate of a pair e.g. \((180-\alpha, 2p)\)
(ii) \(-p\)B1 Award when \(-p\) is the \(y\)-coordinate of a pair e.g. \((\alpha-180, -p)\)
(iii) \(3-p\)B1 Award when \(3-p\) is the \(y\)-coordinate of a pair e.g. \((180+\alpha, 3-p)\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Correct shape, same height starting at \(O\), scaling may be incorrectM1 Same shape, starting at \(O\), same height as original; one complete sine curve on just one side of \(y\)-axis sufficient for M mark
Two repeats of the \(\sin x\) graph each sideA1 Look for intersections at \((\pm360°, 0)\) and \((\pm180°, 0)\) in addition
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = \frac{\alpha}{2}\)B1 Given as one solution
Attempt at second root e.g. \(x = \frac{180° - \alpha}{2}\)M1 A second solution which may not be in given range; any of \(\frac{180°-\alpha}{2}\), \(90°-\frac{\alpha}{2}\), \(180°+\frac{\alpha}{2}\), \(-90°-\frac{\alpha}{2}\), \(-180°+\frac{\alpha}{2}\), \(360°+\frac{\alpha}{2}\) are examples
\(x = 90° - \frac{\alpha}{2}\)A1 Or \(\frac{180°-\alpha}{2}\) as second solution with no additional solutions within given range 
## Question 9:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $2p$ | B1 | Condone $p+p$; award when $2p$ is the $y$-coordinate of a pair e.g. $(180-\alpha, 2p)$ |
| (ii) $-p$ | B1 | Award when $-p$ is the $y$-coordinate of a pair e.g. $(\alpha-180, -p)$ |
| (iii) $3-p$ | B1 | Award when $3-p$ is the $y$-coordinate of a pair e.g. $(180+\alpha, 3-p)$ |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape, same height starting at $O$, scaling may be incorrect | M1 | Same shape, starting at $O$, same height as original; one complete sine curve on just one side of $y$-axis sufficient for M mark |
| Two repeats of the $\sin x$ graph each side | A1 | Look for intersections at $(\pm360°, 0)$ and $(\pm180°, 0)$ in addition |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{\alpha}{2}$ | B1 | Given as one solution |
| Attempt at second root e.g. $x = \frac{180° - \alpha}{2}$ | M1 | A second solution which may not be in given range; any of $\frac{180°-\alpha}{2}$, $90°-\frac{\alpha}{2}$, $180°+\frac{\alpha}{2}$, $-90°-\frac{\alpha}{2}$, $-180°+\frac{\alpha}{2}$, $360°+\frac{\alpha}{2}$ are examples |
| $x = 90° - \frac{\alpha}{2}$ | A1 | Or $\frac{180°-\alpha}{2}$ as second solution with no additional solutions within given range |

---
9.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-26_428_1354_251_287}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

Figure 4 shows part of the graph of the curve with equation $y = \sin x$

Given that $\sin \alpha = p$, where $0 < \alpha < 90 ^ { \circ }$
\begin{enumerate}[label=(\alph*)]
\item state, in terms of $p$, the value of
\begin{enumerate}[label=(\roman*)]
\item $2 \sin \left( 180 ^ { \circ } - \alpha \right)$
\item $\sin \left( \alpha - 180 ^ { \circ } \right)$
\item $3 + \sin \left( 180 ^ { \circ } + \alpha \right)$

A copy of Figure 4, labelled Diagram 1, is shown on page 27.

On Diagram 1,
\end{enumerate}\item sketch the graph of $y = \sin 2 x$
\item Hence find, in terms of $\alpha$, the $x$ coordinates of any points in the interval $0 < x < 180 ^ { \circ }$ where

$$\sin 2 x = p$$

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-27_433_1331_296_310}
\end{center}

\section*{Diagram 1}
\end{enumerate}

\hfill \mbox{\textit{Edexcel P1 2022 Q9 [8]}}
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