| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 5 |
| Topic | Radians, Arc Length and Sector Area |
| Type | Optimization with sectors |
| Difficulty | Standard +0.3 This is a straightforward optimization problem involving standard sector formulas (area = ½r²θ, arc length = rθ) with a constraint. Part (a) is direct recall, part (b) is algebraic manipulation using the constraint P=20, and part (c) is routine differentiation and finding a maximum. The problem requires multiple steps but uses only standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
**(a)**
$P = 2r + 2rx$ — **B1**
$A = r^2x$ — **B1** [2]
**(b)**
$P = 20$ implies $r = \frac{10}{1+x}$ — **M1**
so $A = \left(\frac{10}{1+x}\right)^2 x = \frac{100x}{(1+x)^2}$ **AG** — **A1** [2]
**(c)**
Use quotient rule — **M1**
$\frac{dA}{dx} = \frac{100(1+x)^2 - 200x(1+x)}{(1+x)^4} = \frac{100(1-x)}{(1+x)^3}$ — **A1**
Set equal to zero and find $x = 1$ — **A1**
Show with first differential test that it is maximum, o.e. — **M1 A1** [5]5\\
\includegraphics[max width=\textwidth, alt={}, center]{48b63de9-f022-4881-a187-f08e3c7d9f1a-3_570_734_219_667}
The diagram shows a sector of a circle, $O M N$. The angle $M O N$ is $2 x$ radians, the radius of the circle is $r$ and $O$ is the centre.
\begin{enumerate}[label=(\alph*)]
\item Find expressions, in terms of $r$ and $x$, for the area, $A$, and the perimeter, $P$, of the sector.
\item Given that $P = 20$, show that $A = \left( \frac { 10 } { 1 + x } \right) ^ { 2 }$.
\item Find $\frac { \mathrm { d } A } { \mathrm {~d} x }$, and hence find the value of $x$ for which the area of the sector is a maximum.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2019 Q5 [5]}}