Pre-U Pre-U 9794/2 2020 Specimen — Question 5 5 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2020
SessionSpecimen
Marks5
TopicRadians, Arc Length and Sector Area
TypeOptimization with sectors
DifficultyStandard +0.3 This is a straightforward optimization problem involving standard sector formulas (area = ½r²θ, arc length = rθ) and basic calculus. Part (a) requires recall of formulas, part (b) involves algebraic manipulation to eliminate r using the constraint P=20, and part (c) requires routine differentiation and finding a maximum. The problem is slightly easier than average because it's highly structured with clear steps and uses standard techniques without requiring geometric insight or novel problem-solving approaches.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07a Derivative as gradient: of tangent to curve1.07n Stationary points: find maxima, minima using derivatives

5 \includegraphics[max width=\textwidth, alt={}, center]{8a0a6e46-99cf-4217-93ad-5ed6e9d7c4ef-3_565_730_219_669} 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.
  1. Find expressions, in terms of \(r\) and \(x\), for the area, \(A\), and the perimeter, \(P\), of the sector.
  2. Given that \(P = 20\), show that \(A = \left( \frac { 10 } { 1 + x } \right) ^ { 2 }\).
  3. 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.

(a)
\(P = 2r + 2rx\) — B1
\(A = r^2x\) — B1 [2]
(b)
\(P = 20\) implies \(r = \frac{10}{1+x}\) — M1
so \(A = \left(\frac{10}{1+x}\right)^2 x = \frac{100x}{(1+x)^2}\) AGA1 [2]
(c)
Use quotient rule — M1
\(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{100(1+x)^2 - 200x(1+x)}{(1+x)^4} = \frac{100(1-x)}{(1+x)^3}\) — A1
Set equal to zero and find \(x = 1\) — A1
Show with first differential test that it is maximum, o.e. — M1 A1 [5]
**(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**

$\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{100(1+x)^2 - 200x(1+x)}{(1+x)^4} = \frac{100(1-x)}{(1+x)^3}$ — **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]{8a0a6e46-99cf-4217-93ad-5ed6e9d7c4ef-3_565_730_219_669}

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 2020 Q5 [5]}}