Pre-U Pre-U 9794/2 2012 June — Question 9 9 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2012
SessionJune
Marks9
TopicTangents, normals and gradients
TypeOptimization with constraints
DifficultyModerate -0.3 This is a straightforward optimization problem using standard circle sector formulas. Part (i) requires basic recall of sector area and arc length formulas. Part (ii) involves simple algebraic manipulation to eliminate r using the constraint P=20. Part (iii) is routine differentiation and finding a maximum. While it requires multiple steps, each step uses standard A-level techniques with no novel insight needed, making it slightly easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives

\includegraphics{figure_9} The diagram shows a sector of a circle, \(OMN\). The angle \(MON\) is \(2x\) 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 perimeter, \(P\), of the sector. [2]
  2. Given that \(P = 20\), show that \(A = \frac{100x}{(1 + x)^2}\). [2]
  3. Find \(\frac{dA}{dx}\), and hence find the value of \(x\) for which the area of the sector is a maximum. [5]

AnswerMarks Guidance
(i) \(P = 2r + 2rx\); \(A = r^2x\)B1, B1 [2]
(ii) \(P = 20\) implies \(r = \frac{10}{1+x}\)M1 \(r = f(x)\)
so \(A = \left(\frac{10}{1+x}\right)^2 x = \frac{100x}{(1+x)^2}\) AGA1 [2]
(iii) Use quotient rule: \(\frac{dA}{dx} = \frac{100(1+x)^2 - 200x(1+x)}{(1+x)^4} = \left[\frac{100(1-x)}{(1+x)^3}\right]\)M1, A1, A1
Set equal to zero and find \(x = 1\); Attempt to show with first differential test that it is max; Completely correct solutionM1, A1 [5] [9]
**(i)** $P = 2r + 2rx$; $A = r^2x$ | B1, B1 [2] |

**(ii)** $P = 20$ implies $r = \frac{10}{1+x}$ | M1 | $r = f(x)$

so $A = \left(\frac{10}{1+x}\right)^2 x = \frac{100x}{(1+x)^2}$ AG | A1 [2] |

**(iii)** Use quotient rule: $\frac{dA}{dx} = \frac{100(1+x)^2 - 200x(1+x)}{(1+x)^4} = \left[\frac{100(1-x)}{(1+x)^3}\right]$ | M1, A1, A1 |

Set equal to zero and find $x = 1$; Attempt to show with first differential test that it is max; Completely correct solution | M1, A1 [5] | [9] | Allow $\pm 1$; Or equivalent
\includegraphics{figure_9}

The diagram shows a sector of a circle, $OMN$. The angle $MON$ is $2x$ radians, the radius of the circle is $r$ and $O$ is the centre.

\begin{enumerate}[label=(\roman*)]
\item Find expressions, in terms of $r$ and $x$, for the area, $A$, and perimeter, $P$, of the sector. [2]

\item Given that $P = 20$, show that $A = \frac{100x}{(1 + x)^2}$. [2]

\item Find $\frac{dA}{dx}$, and hence find the value of $x$ for which the area of the sector is a maximum. [5]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2012 Q9 [9]}}