| Exam Board | Edexcel |
|---|---|
| Module | AEA (Advanced Extension Award) |
| Year | 2017 |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with trigonometric functions |
| Difficulty | Challenging +1.8 This AEA question requires finding intersection points by solving tan²x = 4cos2x - 1, using the identity tan²x = sec²x - 1 to simplify to sec²x = 4cos2x, then integrating (4cos2x - sec²x) which needs the double angle formula cos2x = 2cos²x - 1. Multiple non-trivial steps including trigonometric manipulation, solving transcendental equations, and careful integration make this significantly harder than typical A-level questions, though the techniques are all within the syllabus. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
5.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{05b21c5d-5958-4267-b1e6-3d1ed20d5609-16_745_862_258_667}
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\caption{Figure 3}
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\end{figure}
Show that the area of the finite region between the curves $y = \tan ^ { 2 } x$ and $y = 4 \cos 2 x - 1$ in the interval $- \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$, shown shaded in Figure 3, is given by
$$2 \sqrt { 2 \sqrt { 3 } } - 2 \sqrt { 2 \sqrt { 3 } - 3 }$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{05b21c5d-5958-4267-b1e6-3d1ed20d5609-16_2255_51_315_1987}
\end{center}
\hfill \mbox{\textit{Edexcel AEA 2017 Q5 [13]}}