Math Repository

We have digitized the answers and solutions from 2007 to 2022 at
Math Repo.

In particular, you can start navigating the 2007 answers and full worked solutions from
2007P1Q1.

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Math Pro

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We are still under construction, but 6 pure math topics have been completed so far, with more to come!

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) $\frac{2}{5} \textrm{ units}^2$.

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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.