Edexcel AEA 2011 June — Question 13

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2011
SessionJune
TopicVectors: Cross Product & Distances

13
- 3
- 8 \end{array} \right) + t \left( \begin{array} { r } - 5
3
4 \end{array} \right)$$ The point \(P\) has position vector \(\left( \begin{array} { r } - 7
2
7 \end{array} \right)\) .
The point \(P ^ { \prime }\) is the reflection of \(P\) in \(L\) .
(a)Find the position vector of \(P ^ { \prime }\) .
(b)Show that the point \(A\) with position vector \(\left( \begin{array} { r } - 7
9
8 \end{array} \right)\) lies on \(L\) .
(c)Show that angle \(P A P ^ { \prime } = 120 ^ { \circ }\) . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{12d3f92f-8464-4ba1-93a2-c7b841e3d3de-5_483_1367_1263_347} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The point \(B\) lies on \(L\) and \(A P B P ^ { \prime }\) forms a kite as shown in Figure 3.
The area of the kite is \(50 \sqrt { } 3\)
(d)Find the position vector of the point \(B\) .
(e)Show that angle \(B P A = 90 ^ { \circ }\) . The circle \(C\) passes through the points \(A , P , P ^ { \prime }\) and \(B\) .
(f)Find the position vector of the centre of \(C\) .
7.
\includegraphics[max width=\textwidth, alt={}, center]{12d3f92f-8464-4ba1-93a2-c7b841e3d3de-6_675_1145_237_459} \section*{Figure 4} (a)Figure 4 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) ,where $$\mathrm { f } ( x ) = \frac { x ^ { 2 } - 5 } { 3 - x } , \quad x \in \mathbb { R } , x \neq 3$$ The curve has a minimum at the point \(A\) ,with \(x\)-coordinate \(\alpha\) ,and a maximum at the point \(B\) , with \(x\)-coordinate \(\beta\) . Find the value of \(\alpha\) ,the value of \(\beta\) and the \(y\)-coordinates of the points \(A\) and \(B\) .
(b) The functions g and h are defined as follows $$\begin{array} { l l } \mathrm { g } : x \rightarrow x + p & x \in \mathbb { R }
\mathrm {~h} : x \rightarrow | x | & x \in \mathbb { R } \end{array}$$ where \(p\) is a constant. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{12d3f92f-8464-4ba1-93a2-c7b841e3d3de-7_673_1338_591_367} \captionsetup{labelformat=empty} \caption{Figure 5}
\end{figure} Figure 5 shows a sketch of the curve with equation \(y = \mathrm { h } ( \mathrm { fg } ( x ) + q ) , x \in \mathbb { R } , x \neq 0\), where \(q\) is a constant. The curve is symmetric about the \(y\)-axis and has minimum points at \(C\) and \(D\).
  1. Find the value of \(p\) and the value of \(q\).
  2. Write down the coordinates of \(D\).
    (c) The function \(m\) is given by $$\mathrm { m } ( x ) = \frac { x ^ { 2 } - 5 } { 3 - x } , \quad x \in \mathbb { R } , x \leqslant \alpha$$ where \(\alpha\) is the \(x\)-coordinate of \(A\) as found in part (a).
  3. Find \(\mathrm { m } ^ { - 1 }\)
  4. Write down the domain of \(\mathrm { m } ^ { - 1 }\)
  5. Find the value of \(t\) such that \(\mathrm { m } ( t ) = \mathrm { m } ^ { - 1 } ( t )\)