| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Find coordinates of turning points |
| Difficulty | Moderate -0.8 This is a straightforward question on cosine graph properties requiring recall of standard values and basic graph reading. Part (a) asks for coordinates of a maximum (Q = (0,1)) and where the graph crosses the x-axis (k = 90), both standard features. Part (b) requires understanding that for exactly two solutions, p must be between the minimum and the value at the endpoint, giving -1 < p < 0. All steps are routine with no problem-solving or novel insight required. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((90, -1)\) | B1 | One coordinate correct in correct position. Award for \(x = 90\) or \(y = -1\). Condone \(x = 90°\) and \(\frac{180}{2}\) for 90 |
| B1 | Fully correct \((90, -1)\). Special case: \(\left(-1, 90\right)\) or \(\left(-1, \frac{\pi}{2}\right)\) allow B1B0. \((-1, 180)\) scores B0B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(225\) | B1 | Condone \(225°\). Also allow \((225, 0)\) or \((225°, 0)\) or \(225\) marked at \(R\) on diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| One of \(-1 < p < 0\), \(p = 1\) | M1 | One of the correct pair, ft on "\(-1\)" from (a). Condone use of \(y\) rather than \(p\) |
| Both \(-1 < p < 0\), \(p = 1\) | A1 | Fully correct in terms of \(p\) |
## Question 4(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(90, -1)$ | B1 | One coordinate correct in correct position. Award for $x = 90$ or $y = -1$. Condone $x = 90°$ and $\frac{180}{2}$ for 90 |
| | B1 | Fully correct $(90, -1)$. Special case: $\left(-1, 90\right)$ or $\left(-1, \frac{\pi}{2}\right)$ allow B1B0. $(-1, 180)$ scores B0B0 |
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## Question 4(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $225$ | B1 | Condone $225°$. Also allow $(225, 0)$ or $(225°, 0)$ or $225$ marked at $R$ on diagram |
---
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| One of $-1 < p < 0$, $p = 1$ | M1 | One of the correct pair, ft on "$-1$" from (a). Condone use of $y$ rather than $p$ |
| Both $-1 < p < 0$, $p = 1$ | A1 | Fully correct in terms of $p$ |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-08_721_855_214_550}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = \cos 2 x ^ { \circ } \quad 0 \leqslant x \leqslant k$$
The point $Q$ and the point $R ( k , 0 )$ lie on the curve and are shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item State
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $Q$,
\item the value of $k$.
\end{enumerate}\item Given that there are exactly two solutions to the equation
$$\cos 2 x ^ { \circ } = p \quad \text { in the region } 0 \leqslant x \leqslant k$$
find the range of possible values for $p$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q4 [5]}}