| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area of sector/segment problems |
| Difficulty | Moderate -0.3 This is a standard sector/segment geometry problem requiring routine application of circle theorems (tangent perpendicular to radius), sector area formula, and triangle area. Part (i) involves algebraic manipulation of standard formulas, while part (ii) is straightforward numerical substitution. The question is slightly easier than average as it follows a well-practiced template with clear geometric setup and no novel insight required. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly and integrate at least one side | B1 | |
| Obtain term \(\ln x\) | B1 | |
| Obtain term \(-\frac{2}{3}kt\sqrt{t}\), or equivalent | B1 | |
| Evaluate a constant, or use limits \(x = 100\) and \(t = 0\), in a solution containing terms \(a\ln x\) and \(bt\sqrt{t}\) | M1 | |
| Obtain correct solution in any form, e.g. \(\ln x = -\frac{2}{3}kt\sqrt{t} + \ln 100\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(x = 80\) and \(t = 25\) to form equation in \(k\) | M1 | |
| Substitute \(x = 40\) and eliminate \(k\) | M1 | |
| Obtain answer \(t = 64.1\) | A1 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly and integrate at least one side | B1 | |
| Obtain term $\ln x$ | B1 | |
| Obtain term $-\frac{2}{3}kt\sqrt{t}$, or equivalent | B1 | |
| Evaluate a constant, or use limits $x = 100$ and $t = 0$, in a solution containing terms $a\ln x$ and $bt\sqrt{t}$ | M1 | |
| Obtain correct solution in any form, e.g. $\ln x = -\frac{2}{3}kt\sqrt{t} + \ln 100$ | A1 | |
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = 80$ and $t = 25$ to form equation in $k$ | M1 | |
| Substitute $x = 40$ and eliminate $k$ | M1 | |
| Obtain answer $t = 64.1$ | A1 | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{e835a60b-fbeb-49fb-ba6b-ac12c702d487-10_499_922_262_607}
The diagram shows a circle with centre $O$ and radius $r \mathrm {~cm}$. The points $A$ and $B$ lie on the circle and $A T$ is a tangent to the circle. Angle $A O B = \theta$ radians and $O B T$ is a straight line.\\
(i) Express the area of the shaded region in terms of $r$ and $\theta$.\\
(ii) In the case where $r = 3$ and $\theta = 1.2$, find the perimeter of the shaded region.\\
\hfill \mbox{\textit{CAIE P3 2018 Q6 [7]}}