(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) $f^{-1}(x)=\frac{1}{x}+3$, $D_{f^{-1}} = R_f = (-\infty,0) \cup (0,\infty)$.
(b) $w=-1+2i$.
$I = \frac{2}{3} (1+2\mathrm{e}^{-\frac{3}{4}t})$.
For large values of $t$, the current approaches $\frac{2}{3}$.
$A=2, B=3$.
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.
(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) $\sqrt{35} \textrm{ units}^2$.
(ii) Since $r=\frac{2}{3}, |r|<1$. Hence the geometric series is convergent and $S_\infty = 3a$.
(iii) $\{ n: 6 \leq n \leq 13, n \in \mathbb{Z} \}$.
(i)
(iii) $\frac{2}{5} \textrm{ units}^2$.
Paper 2
$7.65.
(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) $1-\frac{1}{N^2}$.
(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 $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.
(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) $5.391 \textrm{ units}^3$.
(Out of syllabus)
0.933.
(i) *0.717.
(ii) *0.616
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.
(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).
$x=-0.260t+66.2$.
$x=-11.8.$ 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.
You must be logged in to post a comment.