| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Simultaneous equations with arc/area |
| Difficulty | Standard +0.3 This is a straightforward application of standard arc length and sector area formulas leading to simultaneous equations. Part (a) involves algebraic manipulation to reach a given quadratic, and part (b) requires solving the quadratic and back-substituting. While it requires multiple steps, the techniques are routine for P1 level with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct equations: \(\frac{1}{2}r^2\theta = 6\) and \(2r + r\theta = 10\) | B1 B1 | One correct equation (B1); both correct equations (B1); may use \(\varphi\) instead of \(\theta\); \(r+r+r\theta=10\) is acceptable; may be implied from later work |
| Eliminates \(r\): \(r = \frac{10}{2+\theta} \Rightarrow \frac{1}{2}\left(\frac{10}{2+\theta}\right)^2\theta = 6\) | M1 | Scored for eliminating \(r\) and reaching equation in \(\theta\) only; initial equations must be of similar form; condone errors rearranging but substitution must be correct |
| \(\Rightarrow 50\theta = 6(4 + 4\theta + \theta^2) \Rightarrow 3\theta^2 - 13\theta + 12 = 0\) * | A1* | Reaches given answer with no errors; at least one intermediate line of manipulation; \((\theta+2)^2\) must be multiplied out correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((3\theta - 4)(\theta - 3) = 0 \Rightarrow \theta = \frac{4}{3}, 3\) | B1 | Values may just appear from a calculator |
| \(\theta = \frac{4}{3}, r = 3\) and \(\theta = 3, r = 2\) | M1 A1 | M1: substitutes one value of \(\theta\) into an allowable equation and proceeds to \(r=\ldots\); A1: both correct pairs; withhold if \(r=-3/r=-2\) not ruled out; withhold if pairs incorrectly matched |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct equations: $\frac{1}{2}r^2\theta = 6$ and $2r + r\theta = 10$ | B1 B1 | One correct equation (B1); both correct equations (B1); may use $\varphi$ instead of $\theta$; $r+r+r\theta=10$ is acceptable; may be implied from later work |
| Eliminates $r$: $r = \frac{10}{2+\theta} \Rightarrow \frac{1}{2}\left(\frac{10}{2+\theta}\right)^2\theta = 6$ | M1 | Scored for eliminating $r$ and reaching equation in $\theta$ only; initial equations must be of similar form; condone errors rearranging but substitution must be correct |
| $\Rightarrow 50\theta = 6(4 + 4\theta + \theta^2) \Rightarrow 3\theta^2 - 13\theta + 12 = 0$ * | A1* | Reaches given answer with no errors; at least one intermediate line of manipulation; $(\theta+2)^2$ must be multiplied out correctly |
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## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3\theta - 4)(\theta - 3) = 0 \Rightarrow \theta = \frac{4}{3}, 3$ | B1 | Values may just appear from a calculator |
| $\theta = \frac{4}{3}, r = 3$ and $\theta = 3, r = 2$ | M1 A1 | M1: substitutes one value of $\theta$ into an allowable equation and proceeds to $r=\ldots$; A1: both correct pairs; withhold if $r=-3/r=-2$ not ruled out; withhold if pairs incorrectly matched |
---\begin{enumerate}
\item A sector $A O B$, of a circle centre $O$, has radius $r \mathrm {~cm}$ and angle $\theta$ radians.
\end{enumerate}
Given that the area of the sector is $6 \mathrm {~cm} ^ { 2 }$ and that the perimeter of the sector is 10 cm ,\\
(a) show that
$$3 \theta ^ { 2 } - 13 \theta + 12 = 0$$
(b) Hence find possible values of $r$ and $\theta$.\\
□\\
\includegraphics[max width=\textwidth, alt={}, center]{c8f8d35d-c2dd-4a1f-a4bb-a4fa06413d12-21_131_19_2627_1882}\\
\hfill \mbox{\textit{Edexcel P1 2019 Q10 [7]}}