Paper 1

1. $x<-3$ or $-2< x <-1$ or $x>7$.
2. (i) $fg$ does not exist since $R_g = [0,\infty) \not \subseteq (-\infty,3)\cup (3,\infty) = D_f$.
   $gf: x \mapsto \frac{1}{(x-3)^2}, \quad \textrm{for } x \in \mathbb{R},x\neq 3$.
   (ii) $f^{-1}(x)=\frac{1}{x}+3$, $D_{f^{-1}} = R_f = (-\infty,0) \cup (0,\infty)$.
3. (b) $w=-1+2i$.
4. $I = \frac{2}{3} (1+2\mathrm{e}^{-\frac{3}{4}t})$.
   For large values of $t$, the current approaches $\frac{2}{3}$.
5. $A=2, B=3$.
   Transformation 1: Translate the graph of $y=\frac{1}{x}$ along the negative $x$-axis direction by 2 units.
   Transformation 2: Scale the resulting graph by a scale factor of 3 parallel to the $y$-axis.
6. (i) Since $\overrightarrow{OA}\cdot\overrightarrow{OB}=0, OA$ is perpendicular to $OB$.
   (ii) $\overrightarrow{OM} = \frac{1}{3} \begin{pmatrix} 4 \\ 2 \\ 5 \end{pmatrix}$.
   (iii) $\sqrt{35} \textrm{ units}^2$.
7. (i) $r\mathrm{e}^{-i\theta}$.
8. (i) $\left ( \frac{5}{2}, \frac{3}{2}, \frac{11}{2} \right )$.
   (ii) $78.8^\circ$.
   (iii) $\frac{4\sqrt{14}}{7}$.
9. (i) $\alpha = 0.619, \beta = 1.512$.
10. (ii) Since $r=\frac{2}{3}, |r|<1$. Hence the geometric series is convergent and $S_\infty = 3a$.
    (iii) $\{ n: 6 \leq n \leq 13, n \in \mathbb{Z} \}$.
11. (i)
    (iii) $\frac{2}{5} \textrm{ units}^2$.

Paper 2

1. $7.65.
2. (ii) $1-\frac{1}{(N+1)^2}$.
   (iii) As $N\to \infty$, $\frac{1}{(N+1)^2}\to 0$ so $1-\frac{1}{(N+1)^2}\to 1$. Hence the series is convergent and the sum to infinity is $1$.
   (iv) $1-\frac{1}{N^2}$.
3. (i)$(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3+\ldots$
   (ii) $8 – 3x + \frac{387}{16}x^2 – \frac{1151}{128}x^3+\ldots$
   (iii) Expansion is valid for $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.
4. (i) $\int^{\frac{5}{3}}_0 \sin^2 x \mathrm{ d}x = \frac{5\pi}{6}+\frac{\sqrt{3}}{8}$.
   $\int^{\frac{5}{3}}_0 \cos^2 x \mathrm{ d}x = \frac{5\pi}{6}-\frac{\sqrt{3}}{8}$.
   (ii) (a) $(\pi – 2) \textrm{ units}^2$.
   (b) $5.391 \textrm{ units}^3$.
5. (Out of syllabus)
6. 0.933.
   (i) *0.717.
   (ii) *0.616
7. Unbiased estimate of population mean = 30.84.
   Unbiased estimate of population variance = 33.7.
   $p$-value = 0.0382. Reject $H_0$.
   We have sufficient evidence at 5% level of significance to conclude that the mean time for a student to complete the project exceeds 30 hours.
   No assumptions are needed since the samplae size is large. By Central Limit Theorem, the distribution of $\overline{X}$ is normally distributed approximately.
8. (i) 0.395.
   (ii) 0.160.
   (iii) 0.392.
   (iv) It is because the event in (ii) is a subset of the event in (iii).
9. (i) (a) 479 001 600.
   (b) 46 080.
   (ii)(a) 39 916 800.
   (b) 86 400.
   (c) 240.
10. (i) $\frac{1}{64}$.
    $\frac{21}{256}.$
    $\frac{13}{17}$.
11. $x=-0.260t+66.2$.
    $x=-11.8.$ The linear model is not suitable as the value of $x$, the concentration, should not be negative.
    (i) $r=-0.994$. This indicates a strong negative linear correlation between $\ln x$ and $t$.
    (ii) $\ln x = -0.0123t+4.62.$
    $t=155$ minutes.