10
\(\sqrt { 10 }\)
\(10 - 2 \mathrm { i }\)
\(10 + 2 i\)
2 Three matrices \(\mathbf { A } , \mathbf { B }\) and \(\mathbf { C }\) are given by
\(\mathbf { A } = \left[ \begin{array} { c c c } 5 & 2 & - 3
0 & 7 & 6
4 & 1 & 0 \end{array} \right]\),
\(\mathbf { B } = \left[ \begin{array} { c c } 1 & 0
3 & - 5
- 2 & 6 \end{array} \right]\)
and \(\mathbf { C } = \left[ \begin{array} { l l l } 6 & 4 & 3
1 & 2 & 0 \end{array} \right]\)
Which of the following cannot be calculated?
Circle your answer.
[0pt]
[1 mark]
AB
AC
BC
\(\mathrm { A } ^ { \mathbf { 2 } }\)
3 Which of the following functions has the fourth term \(- \frac { 1 } { 720 } x ^ { 6 }\) in its Maclaurin series expansion?
Circle your answer.
[0pt]
[1 mark]
\(\sin x\)
\(\cos x\)
\(\mathrm { e } ^ { x }\)
\(\ln ( 1 + x )\)
4 Sketch the graph given by the polar equation
$$r = \frac { a } { \cos \theta }$$
where \(a\) is a positive constant.
\includegraphics[max width=\textwidth, alt={}, center]{1d017497-11b1-4096-b83a-63314188307e-03_74_960_1018_541}
5 Describe fully the transformation given by the matrix \(\left[ \begin{array} { c c c } - \frac { 1 } { 2 } & - \frac { \sqrt { 3 } } { 2 } & 0
\frac { \sqrt { 3 } } { 2 } & - \frac { 1 } { 2 } & 0
0 & 0 & 1 \end{array} \right]\)
6
- Matthew is finding a formula for the inverse function \(\operatorname { arsinh } x\).
He writes his steps as follows:
$$\begin{gathered}
\text { Let } y = \sinh x
y = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } \right)
2 y = \mathrm { e } ^ { x } - \mathrm { e } ^ { - x }
0 = \mathrm { e } ^ { x } - 2 y - \mathrm { e } ^ { - x }
0 = \left( \mathrm { e } ^ { x } \right) ^ { 2 } - 2 y \mathrm { e } ^ { x } - 1
0 = \left( \mathrm { e } ^ { x } - y \right) ^ { 2 } - y ^ { 2 } - 1
y ^ { 2 } + 1 = \left( \mathrm { e } ^ { x } - y \right) ^ { 2 }
\pm \sqrt { y ^ { 2 } + 1 } = \mathrm { e } ^ { x } - y
y \pm \sqrt { y ^ { 2 } + 1 } = \mathrm { e } ^ { x }
\end{gathered}$$
To find the inverse function, swap \(x\) and \(y : x \pm \sqrt { x ^ { 2 } + 1 } = \mathrm { e } ^ { y }\)
$$\begin{gathered}
\ln \left( x \pm \sqrt { x ^ { 2 } + 1 } \right) = y
\operatorname { arsinh } x = \ln \left( x \pm \sqrt { x ^ { 2 } + 1 } \right)
\end{gathered}$$
Identify, and explain, the error in Matthew's proof.
6 - Solve \(\ln \left( x + \sqrt { x ^ { 2 } + 1 } \right) = 3\)
7 Find two invariant points under the transformation given by \(\left[ \begin{array} { l l } 2 & 3
1 & 4 \end{array} \right]\)
\(82 - 3 \mathrm { i }\) is one root of the equation
$$z ^ { 3 } + m z + 52 = 0$$
where \(m\) is real.
8 - Find the other roots.
- Determine the value of \(m\).
9
- Sketch the graph of \(y ^ { 2 } = 4 x\)
\includegraphics[max width=\textwidth, alt={}, center]{1d017497-11b1-4096-b83a-63314188307e-08_871_1052_413_493}
9 - Ben is using a 3D printer to make a plastic bowl which holds exactly \(1000 \mathrm {~cm} ^ { 3 }\) of water.
Ben models the bowl as a region which is rotated through \(2 \pi\) radians about the \(x\)-axis.
He uses the finite region enclosed by the lines \(x = d\) and \(y = 0\) and the curve with equation \(y ^ { 2 } = 4 x\) for \(y \geq 0\)
9
- Find the depth of the bowl to the nearest millimetre.
9
- (ii) What assumption has Ben made about the bowl?
| 10 | Prove by induction that, for all integers \(n \geq 1\), \(\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }\) |
- Hence show that
$$\sum _ { r = 1 } ^ { 2 n } r ( r - 1 ) ( r + 1 ) = n ( n + 1 ) ( 2 n - 1 ) ( 2 n + 1 )$$