CAIE
FP1
2009
November
Q2
6 marks
Standard +0.8
2 Relative to an origin \(O\), the points \(A , B , C\) have position vectors
$$\mathbf { i } , \quad \mathbf { j } + \mathbf { k } , \quad \mathbf { i } + \mathbf { j } + \theta \mathbf { k }$$
respectively. The shortest distance between the lines \(A B\) and \(O C\) is \(\frac { 1 } { \sqrt { 2 } }\). Find the value of \(\theta\).
CAIE
FP1
2009
November
Q5
9 marks
Challenging +1.2
5 The equation
$$x ^ { 3 } + 5 x + 3 = 0$$
has roots \(\alpha , \beta , \gamma\). Use the substitution \(x = - \frac { 3 } { y }\) to find a cubic equation in \(y\) and show that the roots of this equation are \(\beta \gamma , \gamma \alpha , \alpha \beta\).
Find the exact values of \(\beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 } + \alpha ^ { 2 } \beta ^ { 2 }\) and \(\beta ^ { 3 } \gamma ^ { 3 } + \gamma ^ { 3 } \alpha ^ { 3 } + \alpha ^ { 3 } \beta ^ { 3 }\).
CAIE
FP1
2009
November
Q7
9 marks
Challenging +1.2
7 Use de Moivre's theorem to express \(\sin ^ { 6 } \theta\) in the form
$$a + b \cos 2 \theta + c \cos 4 \theta + d \cos 6 \theta$$
where \(a , b , c , d\) are constants to be found.
Hence evaluate
$$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 6 } 2 x d x$$
leaving your answer in terms of \(\pi\).
CAIE
FP1
2009
November
Q11 EITHER
Challenging +1.2
Prove by induction that
$$\sum _ { n = 1 } ^ { N } n ^ { 3 } = \frac { 1 } { 4 } N ^ { 2 } ( N + 1 ) ^ { 2 }$$
Use this result, together with the formula for \(\sum _ { n = 1 } ^ { N } n ^ { 2 }\), to show that
$$\sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } \right) = N ( N + 1 ) ( N + 3 ) ( 5 N + 2 ) .$$
Let
$$S _ { N } = \sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } + \mu n \right)$$
Find the value of the constant \(\mu\) such that \(S _ { N }\) is of the form \(N ^ { 2 } ( N + 1 ) ( a N + b )\), where the constants \(a\) and \(b\) are to be determined.
Show that, for this value of \(\mu\),
$$5 + \frac { 22 } { N } < N ^ { - 4 } S _ { N } < 5 + \frac { 23 } { N }$$
for all \(N \geqslant 18\).
CAIE
FP1
2010
November
Q6
8 marks
Challenging +1.2
6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { A }\), where
$$\mathbf { A } = \left( \begin{array} { r r r r }
1 & 2 & - 1 & \alpha \\
2 & 3 & - 1 & 0 \\
2 & 1 & 2 & - 2 \\
0 & 1 & - 3 & - 2
\end{array} \right)$$
Given that the dimension of the range space of T is 4 , show that \(\alpha \neq 1\).
It is now given that \(\alpha = 1\). Show that the vectors
$$\left( \begin{array} { l }
1 \\
2 \\
2 \\
0
\end{array} \right) , \quad \left( \begin{array} { l }
2 \\
3 \\
1 \\
1
\end{array} \right) \quad \text { and } \quad \left( \begin{array} { r }
- 1 \\
- 1 \\
2 \\
- 3
\end{array} \right)$$
form a basis for the range space of T .
Given also that the vector \(\left( \begin{array} { c } p \\ 1 \\ 1 \\ q \end{array} \right)\) is in the range space of T , find a condition satisfied by \(p\) and \(q\).
CAIE
FP1
2010
November
Q7
9 marks
Challenging +1.3
7 The roots of the equation \(x ^ { 3 } + 4 x - 1 = 0\) are \(\alpha , \beta\) and \(\gamma\). Use the substitution \(y = \frac { 1 } { 1 + x }\) to show that the equation \(6 y ^ { 3 } - 7 y ^ { 2 } + 3 y - 1 = 0\) has roots \(\frac { 1 } { \alpha + 1 } , \frac { 1 } { \beta + 1 }\) and \(\frac { 1 } { \gamma + 1 }\).
For the cases \(n = 1\) and \(n = 2\), find the value of
$$\frac { 1 } { ( \alpha + 1 ) ^ { n } } + \frac { 1 } { ( \beta + 1 ) ^ { n } } + \frac { 1 } { ( \gamma + 1 ) ^ { n } }$$
Deduce the value of \(\frac { 1 } { ( \alpha + 1 ) ^ { 3 } } + \frac { 1 } { ( \beta + 1 ) ^ { 3 } } + \frac { 1 } { ( \gamma + 1 ) ^ { 3 } }\).
Hence show that \(\frac { ( \beta + 1 ) ( \gamma + 1 ) } { ( \alpha + 1 ) ^ { 2 } } + \frac { ( \gamma + 1 ) ( \alpha + 1 ) } { ( \beta + 1 ) ^ { 2 } } + \frac { ( \alpha + 1 ) ( \beta + 1 ) } { ( \gamma + 1 ) ^ { 2 } } = \frac { 73 } { 36 }\).
CAIE
FP1
2010
November
Q10
10 marks
Challenging +1.3
10 By using de Moivre's theorem to express \(\sin 5 \theta\) and \(\cos 5 \theta\) in terms of \(\sin \theta\) and \(\cos \theta\), show that
$$\tan 5 \theta = \frac { 5 t - 10 t ^ { 3 } + t ^ { 5 } } { 1 - 10 t ^ { 2 } + 5 t ^ { 4 } }$$
where \(t = \tan \theta\).
Show that the roots of the equation \(x ^ { 4 } - 10 x ^ { 2 } + 5 = 0\) are \(\tan \left( \frac { 1 } { 5 } n \pi \right)\) for \(n = 1,2,3,4\).
By considering the product of the roots of this equation, find the exact value of \(\tan \left( \frac { 1 } { 5 } \pi \right) \tan \left( \frac { 2 } { 5 } \pi \right)\).