Paper 1
$a=2.73$
(Out of syllabus)
(i) $\overrightarrow{OP} = \begin{pmatrix}6 \\ 3 \\ -3 \end{pmatrix}$
(ii) $\theta = 87.8^{\circ}$
(iii) Area $=15\sqrt{3} \textrm{ units}^2$
(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
(i) $\frac{\pi}{9}$
(ii) $\frac{x^{n+1}}{n+1}\left( \ln x – \frac{1}{n+1} \right) +c$
$\frac{e^{n+1}}{n+1} + \frac{1-e^{n+1}}{(n+1)^2}$
(i) $a=2, b=\frac{3}{4}$
(ii) $1+4x+8x^2+\ldots$
$x=6.09, y=12.6$ gives a maximum area
(i) $-8$
(ii) $a=-3, b=6$
(iii) $z=1\pm\sqrt{3}i$ or $z=-\frac{1}{2}$
(ii) $y=f(x)$ is a horizontal line parallel to the $x$-axis
(ai) $n=34$. 1 October 2011
(aii) $\$6.08$
(b) $\$310$
(c) 81 months
$\left(-\frac{4}{11}, -\frac{4}{11}, \frac{7}{11} \right)$
$\mathbf{r} = \begin{pmatrix}-1\\-1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\1\end{pmatrix}, \lambda \in \mathbb{R}$
Paper 2
(ii) $x+x^2+\frac{x^3}{3}+\ldots$
(iv) $-1.96 < x < 1.56$
(i) 0.999
(ii) $\frac{4}{15} \pi \textrm{ units}^2$
(iii) $x=\frac{2}{3}$
(a)$|p|=1, \arg p = 2\theta$
$\theta=\frac{\pi}{5}, \frac{2\pi}{5}$
(ii)$f^{-1}(x)=4+\sqrt{x-1}$, $D_{f^{-1}}=(1, \infty)$
(iv)$y=x$ $x=\frac{9+\sqrt{13}}{2}$
(Out of syllabus)
(Out of syllabus)
(i) 0.5
(ii) 0.512
(iii) $\frac{7}{64}$
(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.
(Out of syllabus)
(i) 120
(ii) 9
(iii) 210
(iv) 485
(i) 0.0385
(ii) $E(Y)=110, Var(Y)=576$
$a=3, b=-40$
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