Paper 1

1. $a=2.73$
2. (Out of syllabus)
3. 
   (i) $\overrightarrow{OP} = \begin{pmatrix}6 \\ 3 \\ -3 \end{pmatrix}$
   (ii) $\theta = 87.8^{\circ}$
   (iii) Area $=15\sqrt{3} \textrm{ units}^2$
4. 
   (i) $y=\frac{3}{2}\ln(x^2+1)+c$
   (ii) $y=\frac{3}{2}\ln(x^2+1)+2$
   (iii) Gradient of every solution curve approaches 0
5. 
   (i) $\frac{\pi}{9}$
   (ii) $\frac{x^{n+1}}{n+1}\left( \ln x – \frac{1}{n+1} \right) +c$
   $\frac{e^{n+1}}{n+1} + \frac{1-e^{n+1}}{(n+1)^2}$
6. 
   (i) $a=2, b=\frac{3}{4}$
   (ii) $1+4x+8x^2+\ldots$
7. $x=6.09, y=12.6$ gives a maximum area
8. 
   (i) $-8$
   (ii) $a=-3, b=6$
   (iii) $z=1\pm\sqrt{3}i$ or $z=-\frac{1}{2}$
9. 
   (ii) $y=f(x)$ is a horizontal line parallel to the $x$-axis
10. 
    (ai) $n=34$. 1 October 2011
    (aii) $\$6.08$
    (b) $\$310$
    (c) 81 months
11. 
    $\left(-\frac{4}{11}, -\frac{4}{11}, \frac{7}{11} \right)$
    $\mathbf{r} = \begin{pmatrix}-1\\-1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\1\end{pmatrix}, \lambda \in \mathbb{R}$

Paper 2

1. 
   (ii) $x+x^2+\frac{x^3}{3}+\ldots$
   (iv) $-1.96 < x < 1.56$
2. 
   (i) 0.999
   (ii) $\frac{4}{15} \pi \textrm{ units}^2$
   (iii) $x=\frac{2}{3}$
3. 
   (a)$|p|=1, \arg p = 2\theta$
   $\theta=\frac{\pi}{5}, \frac{2\pi}{5}$
4. 
   (ii)$f^{-1}(x)=4+\sqrt{x-1}$, $D_{f^{-1}}=(1, \infty)$
   (iv)$y=x$
   $x=\frac{9+\sqrt{13}}{2}$
5. (Out of syllabus)
6. (Out of syllabus)
7. 
   (i) 0.5
   (ii) 0.512
   (iii) $\frac{7}{64}$
8. 
   (i) 0.970
   (iii) $P(4.8,7.6)$ The remaining points are such that $t$ is increasing at a decreasing rate with respect to $x$, which is consistent with a logarithmic model.
   (iv) $t=1.42472 + 4.39656 \ln x$
   (v) 8.32
   (vi) The estimate is not reliable as $x=8.0$ lies outside the given date range from 1.2 to 6.9.
9. (Out of syllabus)
10. 
    (i) 120
    (ii) 9
    (iii) 210
    (iv) 485
11. 
    (i) 0.0385
    (ii) $E(Y)=110, Var(Y)=576$
    $a=3, b=-40$