Edexcel AEA (Advanced Extension Award) 2019 June

1．（a）By writing \(u = \log _ { 4 } r\) ，where \(r > 0\) ，show that $$\log _ { 4 } r = \frac { 1 } { 2 } \log _ { 2 } r$$ （b）Solve the equation $$\log _ { 4 } \left( 5 x ^ { 2 } - 11 \right) = \log _ { 2 } ( 3 x - 5 )$$

2．The discrete random variable \(X\) follows the binomial distribution $$X \sim \mathrm {~B} ( n , p )$$ where \(0 < p < 1\) ．The mode of \(X\) is \(m\) ．
（a）Write down，in terms of \(m , n\) and \(p\) ，an expression for \(\mathrm { P } ( X = m )\)
（b）Determine，in terms of \(n\) and \(p\) ，an interval of width 1 ，in which \(m\) lies．
（c）Find a value of \(n\) where \(n > 100\) ，and a value of \(p\) where \(p < 0.2\) ，for which \(X\) has two modes． For your chosen values of \(n\) and \(p\) ，state these two modes．

3．Given that \(\phi = \frac { 1 } { 2 } ( \sqrt { 5 } + 1 )\) ，
（a）show that
（i）\(\phi ^ { 2 } = \phi + 1\)
（ii）\(\frac { 1 } { \phi } = \phi - 1\)
（b）The equations of two curves are $$\begin{array} { r l r l } y & = \frac { 1 } { x } & x > 0
\text { and } & y & = \ln x - x + k & x > 0 \end{array}$$ where \(k\) is a positive constant．
The curves touch at the point \(P\) ．
Find in terms of \(\phi\)
（i）the coordinates of \(P\) ，
（ii）the value of \(k\) ．

4．（a）Prove the identity $$( \sin x + \cos y ) \cos ( x - y ) \equiv ( 1 + \sin ( x - y ) ) ( \cos x + \sin y )$$ （b）Hence，or otherwise，show that $$\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } \equiv \frac { 1 + \tan \theta } { 1 - \tan \theta }$$ （c）Given that \(k > 1\) ，show that the equation \(\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } = k\) has a unique solution in the interval \(0 < \theta < \frac { \pi } { 4 }\)

1. Points \(A\) and \(B\) have position vectors \(\mathbf { a }\) and \(\mathbf { b }\), respectively, relative to an origin \(O\), and are such that \(O A B\) is a triangle with \(O A = a\) and \(O B = b\).

The point \(C\), with position vector \(\mathbf { c }\), lies on the line through \(O\) that bisects the angle \(A O B\).

1. Prove that the vector \(b \mathbf { a } - a \mathbf { b }\) is perpendicular to \(\mathbf { c }\). The point \(D\), with position vector \(\mathbf { d }\), lies on the line \(A B\) between \(A\) and \(B\).
2. Explain why \(\mathbf { d }\) can be expressed in the form \(\mathbf { d } = ( 1 - \lambda ) \mathbf { a } + \lambda \mathbf { b }\) for some scalar \(\lambda\) with \(0 < \lambda < 1\)
3. Given that \(D\) is also on the line \(O C\), find an expression for \(\lambda\) in terms of \(a\) and \(b\) only and hence show that $$D A : D B = O A : O B$$

6．Figure 1 shows a sketch of part of the curve with equation \(y = x \sin ( \ln x ) , x \geqslant 1\) \begin{figure}[h] \end{figure} For \(x > 1\) ，the curve first crosses the \(x\)－axis at the point \(A\) ．
（a）Find the \(x\) coordinate of \(A\) ．
（b）Differentiate \(x \sin ( \ln x )\) and \(x \cos ( \ln x )\) with respect to \(x\) and hence find $$\int \sin ( \ln x ) \mathrm { d } x \text { and } \int \cos ( \ln x ) \mathrm { d } x$$ （c）（i）Find \(\int x \sin ( \ln x ) \mathrm { d } x\) ．
（ii）Hence show that the area of the shaded region \(\boldsymbol { R }\) ，bounded by the curve and the \(x\)－axis between the points \(( 1,0 )\) and \(A\) ，is $$\frac { 1 } { 5 } \left( \mathrm { e } ^ { 2 \pi } + 1 \right)$$

7．Figure 2 shows a rectangular section of marshland，\(O A B C\) ，which is \(a\) metres long by \(b\) metres wide，where \(a > b\) ． \begin{figure}[h] \end{figure} Edgar intends to get from \(O\) to \(B\) in the shortest possible time．In order to do this，he runs along edge \(O A\) for a distance \(x\) metres \(( 0 \leqslant x < a )\) to the point \(D\) before wading through the marsh directly from \(D\) to \(B\) ． Edgar can wade through the marsh at a constant speed of \(1 \mathrm {~ms} ^ { - 1 }\) ，and he can run along the edge of the marsh at a constant speed of \(\lambda \mathrm { ms } ^ { - 1 }\) ，where \(\lambda > 1\)
（a）By finding an expression in terms of \(x\) for the time taken，\(t\) seconds，for Edgar to reach \(B\) from \(O\) ，show that $$\frac { \mathrm { d } t } { \mathrm {~d} x } = \frac { 1 } { \lambda } - \frac { a - x } { \sqrt { ( a - x ) ^ { 2 } + b ^ { 2 } } }$$ （b）（i）Find，in terms of \(a , b\) and \(\lambda\) ，the value of \(x\) for which \(\frac { \mathrm { d } t } { \mathrm {~d} x } = 0\)
（ii）Show that this value of \(x\) lies in the interval \(0 \leqslant x < a\) provided \(\lambda \geqslant \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\)
（iii）For \(\lambda\) in this range，show that the value of \(x\) found in（b）（i）gives a minimum value of \(t\) ．
（c）Find the minimum time taken for Edgar to get from \(O\) to \(B\) if
（i）\(\lambda \geqslant \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\)
（ii） \(1 < \lambda < \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\) Edgar＇s friend，Frankie，also runs at a constant speed of \(\lambda \mathrm { m } \mathrm { s } ^ { - 1 }\) ．Frankie runs along the edges \(O A\) and \(A B\) ．Given that \(\lambda \geqslant \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\)
（d）find the range of values of \(\lambda\) for which Frankie gets to \(B\) from \(O\) in a shorter time than Edgar＇s minimum time．