| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Optimization with sectors |
| Difficulty | Standard +0.8 This is a multi-step optimization problem requiring geometric insight to set up equal areas, trigonometric manipulation to derive the given equation, and careful calculation of perimeters. Part (i) requires recognizing that area of triangle OAC equals area of segment, leading to non-trivial algebraic manipulation. The problem is above average difficulty but uses standard A-level techniques without requiring exceptional insight. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05g Exact trigonometric values: for standard angles1.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(OC = r\cos\alpha\) or \(AC = r\sin\alpha\) or oe soi | M1 | |
| Area \(\Delta OAC = \frac{1}{2}r^2\sin\alpha\cos\alpha\) | A1 | |
| \(\frac{1}{2}r^2\sin\alpha\cos\alpha = \frac{1}{2} \times \frac{1}{2}r^2\alpha\) oe | M1 | Or e.g. \(\frac{1}{2}r^2\alpha - \frac{1}{2}r^2\cos\alpha\sin\alpha = \frac{1}{4}r^2\alpha\) or \(\frac{1}{2}r^2\alpha - \frac{1}{2}r^2\cos\alpha\sin\alpha = \frac{1}{2}r^2\cos\alpha\sin\alpha\) |
| \(\sin\alpha\cos\alpha = \frac{1}{2}\alpha\) | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| Perimeter \(\Delta OAC = r + r\sin\alpha + r\cos\alpha = 2.4(0)r\) | M1A1 | Allow with \(r\) a number. 2.0164 gets M1A0 |
| Perim. \(ACB = r\alpha + r\sin\alpha + r - r\cos\alpha = 2.18r\) or \(2.17r\) | M1A1 | Allow with \(r\) a number. 0.9644 gets M1A0. Allow 2.2 www. |
| Ratio \(= \dfrac{2.4(0)}{2.18 \text{ or } 2.17} : 1 = 1.1 : 1\) | A1 [5] | Use of \(\cos = 0.6\), \(\sin = 0.8\), \(\alpha = 0.9\) is PA |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(54.3°\) cao | B1 [1] |
## Question 11:
**Part (i):**
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $OC = r\cos\alpha$ or $AC = r\sin\alpha$ or oe soi | M1 | |
| Area $\Delta OAC = \frac{1}{2}r^2\sin\alpha\cos\alpha$ | A1 | |
| $\frac{1}{2}r^2\sin\alpha\cos\alpha = \frac{1}{2} \times \frac{1}{2}r^2\alpha$ oe | M1 | Or e.g. $\frac{1}{2}r^2\alpha - \frac{1}{2}r^2\cos\alpha\sin\alpha = \frac{1}{4}r^2\alpha$ or $\frac{1}{2}r^2\alpha - \frac{1}{2}r^2\cos\alpha\sin\alpha = \frac{1}{2}r^2\cos\alpha\sin\alpha$ |
| $\sin\alpha\cos\alpha = \frac{1}{2}\alpha$ | A1 [4] | AG |
**Part (ii):**
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Perimeter $\Delta OAC = r + r\sin\alpha + r\cos\alpha = 2.4(0)r$ | M1A1 | Allow with $r$ a number. 2.0164 gets M1A0 |
| Perim. $ACB = r\alpha + r\sin\alpha + r - r\cos\alpha = 2.18r$ or $2.17r$ | M1A1 | Allow with $r$ a number. 0.9644 gets M1A0. Allow 2.2 www. |
| Ratio $= \dfrac{2.4(0)}{2.18 \text{ or } 2.17} : 1 = 1.1 : 1$ | A1 [5] | Use of $\cos = 0.6$, $\sin = 0.8$, $\alpha = 0.9$ is PA |
**Part (iii):**
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $54.3°$ cao | B1 [1] | |11\\
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In the diagram, $O A B$ is a sector of a circle with centre $O$ and radius $r$. The point $C$ on $O B$ is such that angle $A C O$ is a right angle. Angle $A O B$ is $\alpha$ radians and is such that $A C$ divides the sector into two regions of equal area.\\
(i) Show that $\sin \alpha \cos \alpha = \frac { 1 } { 2 } \alpha$.
It is given that the solution of the equation in part (i) is $\alpha = 0.9477$, correct to 4 decimal places.\\
(ii) Find the ratio perimeter of region $O A C$ : perimeter of region $A C B$, giving your answer in the form $k : 1$, where $k$ is given correct to 1 decimal place.\\
(iii) Find angle $A O B$ in degrees.
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\hfill \mbox{\textit{CAIE P1 2015 Q11 [10]}}