Edexcel AEA 2008 June — Question 4 13 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2008
SessionJune
Marks13
PaperDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeFind stationary points - logarithmic functions
DifficultyHard +2.3 This AEA question requires finding maxima of a composite transcendental function involving ln(sec x), then computing a complex area involving integration of cos x ln(sec x). Part (a) demands differentiation using product and chain rules with trigonometric identities to solve a transcendental equation. Part (b) requires setting up and evaluating a challenging integral with substitution techniques, leading to an expression involving inverse trigonometric functions and surds. The multi-step reasoning, non-standard function, and algebraically intensive integration place this significantly above average difficulty.
Spec1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals

\includegraphics{figure_1} Figure 1 shows a sketch of the curve \(C\) with equation $$y = \cos x \ln(\sec x), \quad -\frac{\pi}{2} < x < \frac{\pi}{2}$$ The points \(A\) and \(B\) are maximum points on \(C\).
  1. Find the coordinates of \(B\) in terms of e. [5]
The finite region \(R\) lies between \(C\) and the line \(AB\).
  1. Show that the area of \(R\) is $$\frac{2}{e} \arccos \left(\frac{1}{e}\right) + 2\ln \left(e + \sqrt{(e^2 - 1)}\right) - \frac{4}{e} \sqrt{(e^2 - 1)}.$$ [arccos \(x\) is an alternative notation for \(\cos^{-1} x\)] [8]

\includegraphics{figure_1}

Figure 1 shows a sketch of the curve $C$ with equation
$$y = \cos x \ln(\sec x), \quad -\frac{\pi}{2} < x < \frac{\pi}{2}$$

The points $A$ and $B$ are maximum points on $C$.

\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $B$ in terms of e. [5]
\end{enumerate}

The finite region $R$ lies between $C$ and the line $AB$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the area of $R$ is
$$\frac{2}{e} \arccos \left(\frac{1}{e}\right) + 2\ln \left(e + \sqrt{(e^2 - 1)}\right) - \frac{4}{e} \sqrt{(e^2 - 1)}.$$
[arccos $x$ is an alternative notation for $\cos^{-1} x$] [8]
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2008 Q4 [13]}}