Edexcel AEA 2007 June — Question 6 17 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2007
SessionJune
Marks17
PaperDownload PDF ↗
TopicStationary points and optimisation
TypeOptimise perimeter or area of 2D region
DifficultyHard +2.3 This AEA question requires multiple sophisticated techniques: deriving an area expression from a geometric setup, verifying a complex derivative involving trigonometric and composite functions, proving bounds on a critical point using inequality analysis, establishing a non-trivial trigonometric identity, and finally combining results to prove an inequality. The multi-part structure with interdependent steps, the need for trigonometric manipulation skills, and the requirement to prove rather than just calculate elevate this significantly above standard A-level fare.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08d Evaluate definite integrals: between limits

  1. Find an expression, in terms of \(x\), for the area \(A\) of \(R\).
  2. Show that \(\frac { \mathrm { d } A } { \mathrm {~d} x } = \frac { 1 } { 4 } ( \pi - 2 x - 2 \sin x ) \sec ^ { 2 } \frac { x } { 2 }\).
  3. Prove that the maximum value of \(A\) occurs when \(\frac { \pi } { 4 } < x < \frac { \pi } { 3 }\).
  4. Prove that \(\tan \frac { \pi } { 8 } = \sqrt { } 2 - 1\).
  5. Show that the maximum value of \(A > \frac { \pi } { 4 } ( \sqrt { } 2 - 1 )\).

\begin{enumerate}[label=(\alph*)]
\item Find an expression, in terms of $x$, for the area $A$ of $R$.
\item Show that $\frac { \mathrm { d } A } { \mathrm {~d} x } = \frac { 1 } { 4 } ( \pi - 2 x - 2 \sin x ) \sec ^ { 2 } \frac { x } { 2 }$.
\item Prove that the maximum value of $A$ occurs when $\frac { \pi } { 4 } < x < \frac { \pi } { 3 }$.
\item Prove that $\tan \frac { \pi } { 8 } = \sqrt { } 2 - 1$.
\item Show that the maximum value of $A > \frac { \pi } { 4 } ( \sqrt { } 2 - 1 )$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2007 Q6 [17]}}