Edexcel AEA (Advanced Extension Award) 2018 June

1．（a）Show that \(\sqrt { \frac { 1 + x } { 1 - x } }\) can be written in the form \(\frac { 1 + x } { \sqrt { 1 - x ^ { 2 } } }\) for \(| x | < 1\)
（b）Hence，or otherwise，find the expansion，in ascending powers of \(x\) up to and including the term in \(x ^ { 5 }\) ，of \(\sqrt { \frac { 1 + x } { 1 - x } }\)

2．Solve，for \(0 \leqslant x \leqslant 360 ^ { \circ }\) $$\sin 47 ^ { \circ } \cos ^ { 3 } x + \cos 47 ^ { \circ } \sin x \cos ^ { 2 } x = \frac { 1 } { 2 } \cos ^ { 2 } x$$

3．The lines \(L _ { 1 }\) and \(L _ { 2 }\) have the equations $$L _ { 1 } : \mathbf { r } = \left( \begin{array} { l } 1
0
9 \end{array} \right) + s \left( \begin{array} { l } 2
p
6 \end{array} \right) \quad \text { and } \quad L _ { 2 } : \mathbf { r } = \left( \begin{array} { r } - 15
12
- 9 \end{array} \right) + t \left( \begin{array} { r } 4
- 5
2 \end{array} \right)$$ where \(p\) is a constant．
The acute angle between \(L _ { 1 }\) and \(L _ { 2 }\) is \(\theta\) where \(\cos \theta = \frac { \sqrt { 5 } } { 3 }\)
（a）Find the value of \(p\) ． The line \(L _ { 3 }\) has equation \(\mathbf { r } = \left( \begin{array} { r } - 15
12
- 9 \end{array} \right) + u \left( \begin{array} { r } 8
- 6
- 5 \end{array} \right)\) and the lines \(L _ { 3 }\) and \(L _ { 2 }\) intersect at the point \(A\) ．
The point \(B\) on \(L _ { 2 }\) has position vector \(\left( \begin{array} { r } 5
- 13
1 \end{array} \right)\) and point \(C\) lies on \(L _ { 3 }\) such that \(A B D C\) is a rhombus．
（b）Find the two possible position vectors of \(D\) ．

4．A curve \(C\) has equation \(y = \mathrm { f } ( x )\) where \(x \in \mathbb { R }\) and f is a one－one function．
（a）Describe a single transformation that transforms \(C\) to the curve with equation \(y = - \mathrm { f } ( - x )\) ． The curve \(C\) is reflected in the line with equation \(y = - x\) to give the curve \(V\) ． The equation of \(V\) is \(y = \mathrm { g } ( x )\) ．
（b）Explain why \(\mathrm { g } ^ { - 1 } ( x ) = - \mathrm { f } ( - x )\) ． \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \frac { 3 ( x - 1 ) } { x - 2 } \quad x \in \mathbb { R } , x \neq 2$$ The curve has asymptotes with equations \(x = p\) and \(y = q\) and \(C\) crosses the \(x\)－axis at the point \(A\) and the \(y\)－axis at the point \(B\) ．
（c）Write down the value of \(p\) and the value of \(q\) ．
（d）Write down the coordinates of the point \(A\) and the coordinates of the point \(B\) ． Given the definition of \(\mathrm { g } ( x )\) in part（b），
（e）find the function g ．
（f）Solve \(\mathrm { g } ^ { - 1 } \mathrm { f } ( x ) = x\)

5. \begin{figure}[h] \end{figure} Figure 2 shows part of the curve \(T\) with equation \(y = \cos 2 x\) and the circle \(C _ { 1 }\) that touches \(T\) at \(\left( \frac { \pi } { 4 } , 0 \right)\) and \(\left( \frac { 3 \pi } { 4 } , 0 \right)\) ．
（a）Find the radius of \(C _ { 1 }\) \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of \(T\) and part of a circle \(C _ { 2 }\) that touches \(T\) at the point \(P\) with coordinates \(\left( \frac { \pi } { 2 } , - 1 \right)\) ．For values of \(x\) close to \(\frac { \pi } { 2 }\) the curve \(T\) lies inside \(C _ { 2 }\) as shown in Figure 3.
（b）Without doing any calculation，explain why the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) for \(C _ { 2 }\) at \(P\) is less than the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) for \(T\) at \(P\) ． The radius of \(C _ { 2 }\) is \(r\) ．
（c）Use the result from part（b）to find a value of \(k\) such that \(r > k\) ． Given that \(C _ { 2 }\) cuts \(T\) at the point \(( 0,1 )\) ，
（d）find the value of \(r\) ．

6. (a) Use the substitution \(u = \sqrt { t }\) to show that $$\int _ { 1 } ^ { x } \frac { \ln t } { \sqrt { t } } \mathrm {~d} t = 4 - 4 \sqrt { x } + 2 \sqrt { x } \ln x \quad x \geqslant 1$$ (b) The function g is such that $$\int _ { 1 } ^ { x } \mathrm {~g} ( t ) \mathrm { d } t = x - \sqrt { x } \ln x - 1 \quad x \geqslant 1$$

1. Use differentiation to find the function g .
2. Evaluate \(\int _ { 4 } ^ { 16 } \mathrm {~g} ( t ) \mathrm { d } t\) and simplify your answer.
   (c) Find the value of \(x\) (where \(x > 1\) ) that gives the maximum value of $$\int _ { x } ^ { x + 1 } \frac { \ln t } { 2 ^ { t } } \mathrm {~d} t$$

7. \begin{figure}[h] \end{figure} Figure 4 shows a shape \(S ( \theta )\) made up of five line segments \(A B , B C , C D , D E\) and \(E A\) ． The lengths of the sides are \(A B = B C = 5 \mathrm {~cm} , C D = E A = 3 \mathrm {~cm}\) and \(D E = 7 \mathrm {~cm}\) ． Angle \(B A E =\) angle \(B C D = \theta\) radians． The length of each line segment always remains the same but the value of \(\theta\) can be varied so that different symmetrical shapes can be formed，with the added restriction that none of the line segments cross．
（a）Sketch \(S ( \pi )\) ，labelling the vertices clearly． The shape \(S ( \phi )\) is a trapezium．
（b）Sketch \(S ( \phi )\) and calculate the value of \(\phi\) ． The smallest possible value for \(\theta\) is \(\alpha\) ，where \(\alpha > 0\) ，and the largest possible value for \(\theta\) is \(\beta\) ， where \(\beta > \pi\) ．
（c）Show that \(\alpha = \arccos \left( \frac { 29 } { 40 } \right) \cdot \left[ \arccos ( x ) \right.\) is an alternative notation for \(\left. \cos ^ { - 1 } ( x ) \right]\)
（d）Find the value of \(\beta\) ． The area，in \(\mathrm { cm } ^ { 2 }\) ，of shape \(S ( \theta )\) is \(R ( \theta )\) ．
（e）Show that for \(\alpha \leqslant \theta < \pi\) $$R ( \theta ) = 15 \sin \theta + \frac { 7 } { 4 } \sqrt { 87 - 120 \cos \theta }$$ Given that this formula for \(R ( \theta )\) holds for \(\alpha \leqslant \theta \leqslant \beta\)
(f) show that \(R ( \theta )\) has only one stationary point and that this occurs when \(\theta = \frac { 2 \pi } { 3 }\)
(g) find the maximum and minimum values of \(R ( \theta )\). FOR STYLE, CLARITY AND PRESENTATION: 7 MARKS TOTAL FOR PAPER: 100 MARKS
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