Edexcel AEA (Advanced Extension Award) 2015 June

1．（a）Sketch the graph of the curve with equation $$y = | \ln ( 2 x + 5 ) | \quad x > - \frac { 5 } { 2 }$$ On your sketch you should clearly state the equations of any asymptotes and mark the coordinates of points where the curve meets the coordinate axes．
（b）Solve the equation \(| \ln ( 2 x + 5 ) | = \ln 9\)

2．（a）Show that \(( x + 1 )\) is a factor of \(2 x ^ { 3 } + 3 x ^ { 2 } - 1\)
（b）Solve the equation
（b）Solve the equation $$\sqrt { x ^ { 2 } + 2 x + 5 } = x + \sqrt { 2 x + 3 }$$

3．Solve for \(0 < x < 360 ^ { \circ }\) $$\cot 2 x - \tan 78 ^ { \circ } = \frac { ( \sec x ) \left( \sec 78 ^ { \circ } \right) } { 2 }$$ where \(x\) is not an integer multiple of \(90 ^ { \circ }\)

4．（a）Find the binomial series expansion for \(( 4 + y ) ^ { \frac { 1 } { 2 } }\) in ascending powers of \(y\) up to and including the term in \(y ^ { 3 }\) ．Simplify the coefficient of each term．
（3）
（b）Hence show that the binomial series expansion for \(\left( 4 + 5 x + x ^ { 2 } \right) ^ { \frac { 1 } { 2 } }\) in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\) is $$2 + \frac { 5 x } { 4 } - \frac { 9 x ^ { 2 } } { 64 } + \frac { 45 x ^ { 3 } } { 512 }$$ （c）Show that the binomial series expansion of \(\left( 4 + 5 x + x ^ { 2 } \right) ^ { \frac { 1 } { 2 } }\) will converge for \(- \frac { 1 } { 2 } \leqslant x \leqslant \frac { 1 } { 2 }\)
（d）Use the result in part（b）to estimate $$\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \sqrt { 4 + 5 x + x ^ { 2 } } d x$$ Give your answer as a single fraction．

5.
\includegraphics[max width=\textwidth, alt={}, center]{3e18cb7c-1a67-4152-8628-76847e368882-4_639_1177_264_445} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = \frac { x ^ { 2 } + 16 } { 3 x } \quad x \neq 0$$ The curve has a maximum at the point \(A\) with coordinates \(( a , b )\).

1. Find the value of \(a\) and the value of \(b\). The function g is defined as $$\mathrm { g } : x \mapsto \frac { x ^ { 2 } + 16 } { 3 x } \quad a \leqslant x < 0$$ where \(a\) is the value found in part (a).
2. Write down the range of g .
3. On the same axes sketch \(y = \mathrm { g } ( x )\) and \(y = \mathrm { g } ^ { - 1 } ( x )\).
4. Find an expression for \(\mathrm { g } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { g } ^ { - 1 }\)
5. Solve the equation \(\mathrm { g } ( x ) = \mathrm { g } ^ { - 1 } ( x )\).

6．The lines \(L _ { 1 }\) and \(L _ { 2 }\) have vector equations $$\begin{aligned} & L _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 1
10
- 3 \end{array} \right) + \lambda \left( \begin{array} { r } 2
- 5
4 \end{array} \right)
& L _ { 2 } : \mathbf { r } = \left( \begin{array} { r } - 1
2
3 \end{array} \right) + \mu \left( \begin{array} { l } 1
2
2 \end{array} \right) \end{aligned}$$ （a）Show that \(L _ { 1 }\) and \(L _ { 2 }\) are perpendicular．
（b）Show that \(L _ { 1 }\) and \(L _ { 2 }\) are skew lines． The point \(A\) with position vector \(- \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }\) lies on \(L _ { 2 }\) and the point \(X\) lies on \(L _ { 1 }\) such that \(\overrightarrow { A X }\) is perpendicular to \(L _ { 1 }\)
（c）Find the position vector of \(X\) ．
（d）Find \(| \overrightarrow { A X } |\) The point \(B\)（distinct from \(A\) ）also lies on \(L _ { 2 }\) and \(| \overrightarrow { B X } | = | \overrightarrow { A X } |\)
（e）Find the position vector of \(B\) ．
（f）Find the cosine of angle \(A X B\) ．

10
- 3 \end{array} \right) + \lambda \left( \begin{array} { r } 2
- 5
4 \end{array} \right)
& L _ { 2 } : \mathbf { r } = \left( \begin{array} { r } - 1
2
3 \end{array} \right) + \mu \left( \begin{array} { l } 1
2
2 \end{array} \right) \end{aligned}$$ （a）Show that \(L _ { 1 }\) and \(L _ { 2 }\) are perpendicular．
（b）Show that \(L _ { 1 }\) and \(L _ { 2 }\) are skew lines． The point \(A\) with position vector \(- \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }\) lies on \(L _ { 2 }\) and the point \(X\) lies on \(L _ { 1 }\) such that \(\overrightarrow { A X }\) is perpendicular to \(L _ { 1 }\)
（c）Find the position vector of \(X\) ．
（d）Find \(| \overrightarrow { A X } |\) The point \(B\)（distinct from \(A\) ）also lies on \(L _ { 2 }\) and \(| \overrightarrow { B X } | = | \overrightarrow { A X } |\)
（e）Find the position vector of \(B\) ．
（f）Find the cosine of angle \(A X B\) ． 7．（a）Use the substitution \(x = \sec \theta\) to show that $$\int _ { \sqrt { 2 } } ^ { 2 } \frac { 1 } { \left( x ^ { 2 } - 1 \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x = \frac { \sqrt { 6 } - 2 } { \sqrt { 3 } }$$ （b）Use integration by parts to show that $$\int \operatorname { cosec } \theta \cot ^ { 2 } \theta \mathrm {~d} \theta = \frac { 1 } { 2 } [ \ln | \operatorname { cosec } \theta + \cot \theta | - \operatorname { cosec } \theta \cot \theta ] + c$$ （6） \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \frac { 1 } { \left( x ^ { 2 } - 1 \right) ^ { \frac { 3 } { 2 } } }\) for \(x > 1\)
The region \(R\) ，shown shaded in Figure 2，is bounded by the curve，the \(x\)－axis and the lines \(x = \sqrt { 2 }\) and \(x = 2\)
The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)－axis．
（c）Show that the volume of the solid formed is $$\pi \left[ \frac { 3 } { 8 } \ln \left( \frac { 1 + \sqrt { 2 } } { \sqrt { 3 } } \right) + \frac { 7 } { 36 } - \frac { \sqrt { 2 } } { 8 } \right]$$