Edexcel AEA 2009 June — Question 6 17 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2009
SessionJune
Marks17
PaperDownload PDF ↗
TopicParametric integration
TypeParametric area under curve
DifficultyChallenging +1.8 This AEA question requires finding a tangent equation to a parametric curve involving ln(sec t), then computing an area bounded by the curve, tangent, and x-axis. Part (a) involves standard parametric differentiation and tangent equations. Part (b) requires setting up and evaluating a parametric integral with non-trivial trigonometric and logarithmic expressions, demanding careful algebraic manipulation and integration techniques beyond standard A-level, but the structure is guided by the 'show that' format.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals

6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{dfb57dc0-5831-4bbb-b1e5-58c4798215cb-5_700_684_246_694} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of the curve \(C\) with parametric equations $$x = 2 \sin t , \quad y = \ln ( \sec t ) , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ The tangent to \(C\) at the point \(P\) ,where \(t = \frac { \pi } { 3 }\) ,cuts the \(x\)-axis at \(A\) .
  1. Show that the \(x\)-coordinate of \(A\) is \(\frac { \sqrt { } 3 } { 3 } ( 3 - \ln 2 )\) . The shaded region \(R\) lies between \(C\) ,the positive \(x\)-axis and the tangent \(A P\) as shown in Figure 2 .
  2. Show that the area of \(R\) is \(\sqrt { 3 } ( 1 + \ln 2 ) - 2 \ln ( 2 + \sqrt { 3 } ) - \frac { \sqrt { 3 } } { 6 } ( \ln 2 ) ^ { 2 }\) .

6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{dfb57dc0-5831-4bbb-b1e5-58c4798215cb-5_700_684_246_694}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

Figure 2 shows a sketch of the curve $C$ with parametric equations

$$x = 2 \sin t , \quad y = \ln ( \sec t ) , \quad 0 \leqslant t < \frac { \pi } { 2 }$$

The tangent to $C$ at the point $P$ ,where $t = \frac { \pi } { 3 }$ ,cuts the $x$-axis at $A$ .
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of $A$ is $\frac { \sqrt { } 3 } { 3 } ( 3 - \ln 2 )$ .

The shaded region $R$ lies between $C$ ,the positive $x$-axis and the tangent $A P$ as shown in Figure 2 .
\item Show that the area of $R$ is $\sqrt { 3 } ( 1 + \ln 2 ) - 2 \ln ( 2 + \sqrt { 3 } ) - \frac { \sqrt { 3 } } { 6 } ( \ln 2 ) ^ { 2 }$ .
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2009 Q6 [17]}}