Edexcel AEA (Advanced Extension Award) 2017 June

1．The function f is given by $$\mathrm { f } ( x ) = \sqrt { x + 2 } \quad \text { for } \quad x \in \mathbb { R } , x \geqslant 0$$ （a）Find \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\) The function g is given by $$\mathrm { g } ( x ) = x ^ { 2 } - 4 x + 5 \text { for } x \in \mathbb { R } , x \geqslant 0$$ （b）Find the range of g ．
（c）Solve the equation \(\operatorname { fg } ( x ) = x\) ．

2．（a）Show that the equation $$\tan x = \frac { \sqrt { 3 } } { 1 + 4 \cos x }$$ can be written in the form $$\sin 2 x = \sin \left( 60 ^ { \circ } - x \right)$$ （b）Solve，for \(0 < x < 180 ^ { \circ }\) $$\tan x = \frac { \sqrt { 3 } } { 1 + 4 \cos x }$$

1. The line \(L _ { 1 }\) has equation \(\mathbf { r } = \left( \begin{array} { c } - 13
   7
   - 1 \end{array} \right) + t \left( \begin{array} { c } 6
   - 2
   3 \end{array} \right)\). The line \(L _ { 2 }\) passes through the point \(A\) with position vector \(\left( \begin{array} { c } 1
   p
   10 \end{array} \right)\) and is parallel to \(\left( \begin{array} { c } - 2
   11
   - 5 \end{array} \right)\), where \(p\)
   is a constant. The lines \(L _ { 1 }\) and \(L _ { 2 }\) intersect at the point \(B\).
   1. Find
      1. the value of \(p\),
      2. the position vector of \(B\).
   The point \(C\) lies on \(L _ { 1 }\) and angle \(A C B\) is \(90 ^ { \circ }\)
2. Find the position vector of \(C\). The point \(D\) also lies on \(L _ { 1 }\) and triangle \(A B D\) is isosceles with \(A B = A D\).
3. Find the area of triangle \(A B D\).

4. \begin{figure}[h] \end{figure} Figure 1 shows the equilateral triangle \(L M N\) of side 2 cm ．The point \(P\) lies on \(L M\) such that \(L P = x \mathrm {~cm}\) and the point \(Q\) lies on \(L N\) such that \(L Q = y \mathrm {~cm}\) ．The points \(P\) and \(Q\) are chosen so that the area of triangle \(L P Q\) is half the area of triangle \(L M N\) ．
（a）Show that \(x y = 2\)
（b）Find the shortest possible length of \(P Q\) ，justifying your answer． Mathematicians know that for any closed curve or polygon enclosing a fixed area，the ratio \(\frac { \text { area enclosed } } { \text { perimeter } }\) is a maximum when the closed curve is a circle． By considering 6 copies of triangle \(L M N\) suitably arranged，
（c）find the length of the shortest line or curve that can be drawn from a point on \(L M\) to a point on \(L N\) to divide the area of triangle \(L M N\) in half．Justify your answer．
（6）

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = \frac { 4 ( x - 1 ) } { x ( x - 3 ) }$$ The curve cuts the \(x\)-axis at \(( a , 0 )\). The lines \(y = 0 , x = 0\) and \(x = b\) are asymptotes to the curve.

1. Write down the value of \(a\) and the value of \(b\).
   (2)
2. On separate axes, sketch the curves with the following equations. On your sketches, you should mark the coordinates of any intersections with the coordinate axes and state the equations of any asymptotes.
   1. \(y = \mathrm { f } ( x + 2 ) - 4\)
   2. \(y = \mathrm { f } ( | x | ) - 3\)

6．（a）Show that $$\frac { \mathrm { d } } { \mathrm {~d} u } \ln \left( u + \sqrt { u ^ { 2 } - 1 } \right) = \frac { 1 } { \sqrt { u ^ { 2 } - 1 } }$$ （b）Use the result from part（a）and the substitution \(x + 3 = \frac { 1 } { t }\) to find $$\int \frac { 1 } { ( x + 3 ) \sqrt { 2 x + 7 } } \mathrm {~d} x$$ （6）
（c）Express \(\frac { 1 } { 2 x ^ { 2 } + 13 x + 21 }\) in partial fractions．
（d）Find $$\int _ { 1 } ^ { 9 } \frac { 1 } { \left( 2 x ^ { 2 } + 13 x + 21 \right) \sqrt { 2 x + 7 } } \mathrm {~d} x$$ giving your answer in the form \(\ln r - s\) where \(r\) and \(s\) are rational numbers．

7. \begin{figure}[h] \end{figure} Figure 3 shows part of the curve \(C\) with equation \(y = x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - 34 x\) and the line \(L\) with equation \(y = m x + c\) ． The line \(L\) touches \(C\) at the points \(P\) and \(Q\) with \(x\) coordinates \(p\) and \(q\) respectively．
（a）Explain why $$x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - ( 34 + m ) x - c = ( x - p ) ^ { 2 } ( x - q ) ^ { 2 }$$ The finite region \(R\) ，shown shaded in Figure 3，is bounded by \(C\) and \(L\) ．
（b）Use integration by parts to show that the area of \(R\) is \(\frac { ( q - p ) ^ { 5 } } { 30 }\)
（c）Show that $$( x - p ) ^ { 2 } ( x - q ) ^ { 2 } = x ^ { 4 } - 2 ( p + q ) x ^ { 3 } + S x ^ { 2 } - T x + U$$ where \(S , T\) and \(U\) are expressions to be found in terms of \(p\) and \(q\) ．
（d）Using part（a）and part（c）find the value of \(p\) ，the value of \(q\) and the equation of \(L\) ．