| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Logo and design problems |
| Difficulty | Standard +0.3 This is a straightforward application of radians, arc length, and sector area formulas with some basic trigonometry. Part (a) uses the cosine rule in a standard way, parts (b) and (c) apply standard formulas for arc length and sector area plus kite area. The multi-step nature and combination of geometry with radians makes it slightly above average, but all techniques are routine for P1 students. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(OB^2 = 0.6^2 + 1.4^2 - 2 \times 0.6 \times 1.4\cos 2 \Rightarrow OB^2 = \ldots\) or \(OB = \ldots\) | M1 | Attempts cosine rule to get \(OB\) or \(OB^2\), seen in part (a) only |
| \(OB = 1.738\) | A1 | \(OB\) = awrt 1.74 or truncated as 1.7 or e.g. 1.73 |
| \(\frac{\sin AOB}{1.4} = \frac{\sin 2}{\text{"1.738"}} \Rightarrow AOB = 0.822\) or \(\frac{\sin ABO}{0.6} = \frac{\sin 2}{\text{"1.738"}} \Rightarrow ABO = 0.319\ldots \Rightarrow AOB = \pi - 2 - 0.319\ldots\) | dM1 | Attempts sine rule with sides and angles in correct positions to find \(AOB\) or \(\frac{1}{2}AOB\). Must include using inverse sine |
| \(\theta = 2 \times AOB = 2 \times 0.822 = 1.64\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(0.6 \times \alpha\) with \(\alpha = 2\pi - 1.64\) or \(\alpha = \pi - 1.64\), or attempts \(2 \times \pi \times 0.6 - 0.6 \times 1.64\) | M1 | |
| \(0.6 \times (2\pi - 1.64) + 2.8 = 5.6\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{1}{2} \times 0.6^2 \times \alpha\) with \(\alpha = 2\pi - 1.64\) or \(\alpha = \pi - 1.64\), or attempts \(\pi \times 0.6^2 - \frac{1}{2} \times 0.6^2 \times 1.64\) | M1 | |
| Attempts \(0.6 \times 1.4 \sin 2\) | M1 | |
| Full method: \(\frac{1}{2} \times 0.6^2 \times (2\pi - 1.64) + 0.6 \times 1.4\sin 2 = 1.6\) m\(^2\) | ddM1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(OB^2 = 0.6^2 + 1.4^2 - 2\times0.6\times1.4\cos\theta \Rightarrow OB^2 = ...\) or \(OB = ...\) | M1 | Cosine rule attempt |
| \(OB = 1.738\) | A1 | |
| \(\frac{\sin AOB}{1.4} = \frac{\sin\theta}{"1.738"}\) or sine rule equivalent | M1 | |
| \(\theta = 2 \times AOB = 1.64^*\) | A1* | Fully correct work, no obvious rounding errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(0.6\times\alpha\) with allowable \(\alpha\) | M1 | Accept \(\alpha = 2\pi-1.64\) (awrt 4.64) or \(\alpha = \pi-1.64\) (awrt 1.50) |
| \(0.6\times(2\pi-1.64)+2.8\) = awrt 5.6 m | A1 | Condone lack of units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{1}{2}\times0.6^2\times\alpha\) with \(\alpha = 2\pi-1.64\) or \(\alpha = \pi-1.64\) | M1 | Major sector area; alt: full circle minus minor sector |
| Attempts \(0.6\times1.4\sin 2\) or equivalent kite/triangle area | M1 | Kite \(ABCO\) or half of it |
| \(\frac{1}{2}\times0.6^2\times(2\pi-1.64)+0.6\times1.4\sin 2\) | ddM1 | Complete correct method; depends on both previous M marks |
| awrt \(1.6\text{ m}^2\) | A1 | Condone lack of units |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $OB^2 = 0.6^2 + 1.4^2 - 2 \times 0.6 \times 1.4\cos 2 \Rightarrow OB^2 = \ldots$ or $OB = \ldots$ | M1 | Attempts cosine rule to get $OB$ or $OB^2$, seen in part (a) only |
| $OB = 1.738$ | A1 | $OB$ = awrt 1.74 or truncated as 1.7 or e.g. 1.73 |
| $\frac{\sin AOB}{1.4} = \frac{\sin 2}{\text{"1.738"}} \Rightarrow AOB = 0.822$ or $\frac{\sin ABO}{0.6} = \frac{\sin 2}{\text{"1.738"}} \Rightarrow ABO = 0.319\ldots \Rightarrow AOB = \pi - 2 - 0.319\ldots$ | dM1 | Attempts sine rule with sides and angles in correct positions to find $AOB$ or $\frac{1}{2}AOB$. Must include using inverse sine |
| $\theta = 2 \times AOB = 2 \times 0.822 = 1.64$ | A1* | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $0.6 \times \alpha$ with $\alpha = 2\pi - 1.64$ or $\alpha = \pi - 1.64$, or attempts $2 \times \pi \times 0.6 - 0.6 \times 1.64$ | M1 | |
| $0.6 \times (2\pi - 1.64) + 2.8 = 5.6$ m | A1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{1}{2} \times 0.6^2 \times \alpha$ with $\alpha = 2\pi - 1.64$ or $\alpha = \pi - 1.64$, or attempts $\pi \times 0.6^2 - \frac{1}{2} \times 0.6^2 \times 1.64$ | M1 | |
| Attempts $0.6 \times 1.4 \sin 2$ | M1 | |
| Full method: $\frac{1}{2} \times 0.6^2 \times (2\pi - 1.64) + 0.6 \times 1.4\sin 2 = 1.6$ m$^2$ | ddM1 A1 | |
## Question 7 (Sector/Circle Problem):
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $OB^2 = 0.6^2 + 1.4^2 - 2\times0.6\times1.4\cos\theta \Rightarrow OB^2 = ...$ or $OB = ...$ | M1 | Cosine rule attempt |
| $OB = 1.738$ | A1 | |
| $\frac{\sin AOB}{1.4} = \frac{\sin\theta}{"1.738"}$ or sine rule equivalent | M1 | |
| $\theta = 2 \times AOB = 1.64^*$ | A1* | Fully correct work, no obvious rounding errors |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $0.6\times\alpha$ with allowable $\alpha$ | M1 | Accept $\alpha = 2\pi-1.64$ (awrt 4.64) or $\alpha = \pi-1.64$ (awrt 1.50) |
| $0.6\times(2\pi-1.64)+2.8$ = awrt 5.6 m | A1 | Condone lack of units |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{1}{2}\times0.6^2\times\alpha$ with $\alpha = 2\pi-1.64$ or $\alpha = \pi-1.64$ | M1 | Major sector area; alt: full circle minus minor sector |
| Attempts $0.6\times1.4\sin 2$ or equivalent kite/triangle area | M1 | Kite $ABCO$ or half of it |
| $\frac{1}{2}\times0.6^2\times(2\pi-1.64)+0.6\times1.4\sin 2$ | ddM1 | Complete correct method; depends on both previous M marks |
| awrt $1.6\text{ m}^2$ | A1 | Condone lack of units |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-18_428_894_210_525}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the design for a sign at a bird sanctuary.\\
The design consists of a kite $O A B C$ joined to a sector $O C X A$ of a circle centre $O$.\\
In the design
\begin{itemize}
\item $O A = O C = 0.6 \mathrm {~m}$
\item $A B = C B = 1.4 \mathrm {~m}$
\item Angle $O A B =$ Angle $O C B = 2$ radians
\item Angle $A O C = \theta$ radians, as shown in Figure 3
\end{itemize}
Making your method clear,
\begin{enumerate}[label=(\alph*)]
\item show that $\theta = 1.64$ radians to 3 significant figures,
\item find the perimeter of the sign, in metres to 2 significant figures,
\item find the area of the sign, in $\mathrm { m } ^ { 2 }$ to 2 significant figures.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q7 [10]}}