| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - single tangent line |
| Difficulty | Standard +0.3 This is a straightforward application of standard arc length and sector area formulas with a tangent segment. Part (i) requires recognizing that RT = r tan θ and applying basic formulas. Part (ii) is a simple algebraic verification showing A ≠ rP by substitution. The question is slightly easier than average as it involves direct formula application with minimal problem-solving insight required. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
**Question 11:**
**(i)**
$RS = r\theta$
$RT = r\tan\theta$
$OT = r\sec\theta$
$ST = r\sec\theta - r$
$P = r\sec\theta - r + r\theta + r\tan\theta$
$A = \frac{1}{2}RT \times OR - \frac{1}{2}r^2\theta = \frac{1}{2}r^2(\tan\theta - \theta)$
- **B1**: Correct $RS$
- **B1**: Correct $RT$
- **M1**: Attempt $ST$
- **A1**: Fully correct expression for $P$ (could be $\frac{1}{\cos\theta}$ for $\sec\theta$, but not $\sqrt{1+\tan^2\theta}$)
- **M1**: Attempt area of triangle – must be valid attempt at $RT$
- **B1**: State correct area of sector
- **A1**: Correct expression for $A$
**Total for (i): [7]**
**(ii)**
Let $A=rP$:
$$r^2\sec\theta - r^2 + r^2\theta + r^2\tan\theta = \frac{1}{2}r^2(\tan\theta-\theta)$$
so $2\sec\theta - 2 + \tan\theta + 3\theta = 0$
For $0<\theta<\frac{1}{2}\pi$, $\sec\theta>1$, so $2\sec\theta-2>0$.
Since $\tan\theta>0$ and $\theta>0$, equality is impossible, so we have a contradiction.
- **M1**: Equate $A$ with $rP$ (allow use of $\neq$)
- **M1**: Attempt to justify why no solutions – correct equation only
- **B1**: State $\sec\theta>1$, or equivalent
- **A1**: Fully correct and convincing argument
**Total for (ii): [4]**11\\
\includegraphics[max width=\textwidth, alt={}, center]{2f48a6ee-e8ce-47e4-a07f-2c55a6904e7d-3_661_953_767_596}
The diagram shows a circle, centre $O$, radius $r$. The points $R$ and $S$ lie on the circumference of the circle, and the line $R T$ is a tangent to the circle at $R$. The angle $R O S$ is $\theta$ radians where $0 < \theta < \frac { 1 } { 2 } \pi$.\\
(i) Find expressions for the perimeter, $P$, and the area, $A$, of the shaded region in terms of $r$ and $\theta$.\\
(ii) Hence show that $A \neq r P$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2015 Q11 [11]}}