Aussie Maths & Science Teachers: Save your time with SmarterEd
When Stuart stops drinking alcohol at 11:30 pm, he has a blood alcohol content (BAC) of 0.08625.
The number of hours required for a person to reach zero BAC after they stop consuming alcohol is given by the formula:
\(\text{Time}=\dfrac{BAC}{0.015}\).
At what time on the next day should Stuart expect his BAC to be 0.05?
This shape is made up of two right-angled triangle and a regular hexagon.
The area of a regular hexagon can be estimated using the formula \(A=2.598S^2\) where \(S\) is the hexagon's side-length.
Calculate the total area of the shape using this formula. (3 marks)
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Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`
Part of the graph of `y=f(x)` is shown below.
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A chemist uses spectroscopy to identify an unknown organic molecule, Molecule \(\text{J}\), that contains chlorine. The \({}^{13}\text{C NMR}\) spectrum of Molecule \(\text{J}\) is shown below. The infra-red (IR) spectrum of Molecule \(\text{J}\) is shown below. --- 2 WORK AREA LINES (style=lined) --- The mass spectrum of Molecule \(\text{J}\) is shown below --- 2 WORK AREA LINES (style=lined) --- The \({ }^1 \text{H NMR}\) spectrum of Molecule \(\text{J}\) is shown below. --- 3 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=blank) --- a. Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. b. The peak at 110 m/z: c. The two singlet peaks indicate: d. Either of the two molecules shown below are correct: a. Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. b. The peak at 110 m/z: c. The two singlet peaks indicate: d. Either of the two molecules shown below are correct:
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♦ Mean mark (a) 41%.
♦♦♦ Mean mark (b) 15%.
COMMENT: Know the masses of common isotopes.
♦ Mean mark (d) 40%.
Which one of the following statements about IR spectroscopy is correct?
The infrared (IR) spectrum of the molecule 3-methyl-2-butanone is shown below. Explain why different frequencies of infrared radiation can be absorbed by the same molecule as shown in the spectrum above. (3 marks) --- 7 WORK AREA LINES (style=lined) ---
The strength of the eggshell of birds is determined by the calcium carbonate, \(\ce{CaCO3}\), content of the eggshell. The percentage of calcium carbonate in the eggshell can be determined by gravimetric analysis. 0.412 g of clean, dry eggshell was completely dissolved in a minimum volume of dilute hydrochloric acid. \(\ce{CaCO3(s) + 2H+(aq)\rightarrow Ca^2+(aq) + CO2(g) + H2O(l)}\) An excess of a basic solution of ammonium oxalate, \(\ce{(NH4)2C2O4}\), was then added to form crystals of calcium oxalate monohydrate, \(\ce{CaC2O4.H2O}\). The suspension was filtered and the crystals were then dried to constant mass. 0.523 g of \(\ce{CaC2O4.H2O}\) was collected. --- 1 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) ---
Organic compounds are numerous and diverse due to the nature of the carbon atom. There are international conventions for the naming and representation of organic compounds. --- 5 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- i. ii. \(\ce{C4H6}\) iii. • Longest carbon chain is 6, hence hexane i. ii. \(\ce{C4H6}\) iii. Longest carbon chain is 6, hence hexane
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♦♦♦ Mean mark (c) 27%.
Brass is an alloy of copper and zinc.
To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of standard solutions of copper\(\text{(II)}\) ions and measured their absorbance using an atomic absorption spectrometer (AAS).
The calibration curve obtained is shown below.
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Students are comparing the diffraction patterns produced by electrons and X-rays, in which the same spacing of bands is observed in the patterns, as shown schematically in the diagram. Note that both patterns shown are to the same scale.
The electron diffraction pattern is produced by 3.0 × 10\(^3\) eV electrons.
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A 250 g toy car performs a loop in the apparatus shown in the diagram below. The car starts from rest at point \(\text{A}\) and travels along the track without any air resistance or retarding frictional forces. The radius of the car's path in the loop is 0.20 m. When the car reaches point \(\text{B}\) it is travelling at a speed of 3.0 m s\(^{-1}\). --- 6 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
A home owner on a large property creates a backyard entertainment area. The entertainment area has a low-voltage lighting system. To operate correctly, the lighting system requires a voltage of 12 V. The lighting system has a resistance of 12 \(\Omega\). --- 3 WORK AREA LINES (style=lined) --- To operate the lighting system, the home owner installs an ideal transformer at the house to reduce the voltage from 240 V to 12 V. The home owner then runs a 200 m long heavy-duty outdoor extension lead, which has a total resistance of 3 \( \Omega\), from the transformer to the entertainment area. --- 7 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
A particle of mass \(m\) and charge \(q\) travelling at velocity \(v\) enters a uniform magnetic field \(\text{B}\), as shown in the diagram. --- 2 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) ---
An astronaut has left Earth and is travelling on a spaceship at 0.800\(c\) directly towards the star known as Sirius, which is located 8.61 light-years away from Earth, as measured by observers on Earth. --- 4 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) ---
Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.
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The Ionospheric Connection Explorer (ICON) space weather satellite, constructed to study Earth's ionosphere, was launched in October 2019. ICON will study the link between space weather and Earth's weather at its orbital altitude of 600 km above Earth's surface. Assume that ICON's orbit is a circular orbit.
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Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).
Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
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A student stirs 2.80 g of silver \(\text{(I)}\) nitrate powder into 250.0 mL of 1.00 mol L\(^{-1}\) sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.
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The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
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A student stirs 2.45 g of copper \(\text{(II)}\) nitrate powder into 200.0 mL of 1.25 mol L\(^{-1}\) sodium carbonate solution until it is fully dissolved. A reaction occurs and a precipitate appears.
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The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
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The function \(g\) is defined as follows.
\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)
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Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).
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| a. |  |
| b.i | \(y=2x-3\) |
| b.ii |  |
| c. | \(\dfrac{3}{2}=1.5\) |
| d. | \(0.0036\) |
| e.i | \(k=2.397\) |
| e.ii |  |
a.
| b.i | \(g(x)\) | \(=3\log_{e}x-x\) |
| \(g(1)\) | \(=3\log_{e}1-1=-1\) | |
| \(g^{\prime}(x)\) | \(=\dfrac{3}{x}-1\) | |
| \(g^{\prime}(1)\) | \(=\dfrac{3}{1}-1=2\) |
\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)
\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)
b.ii
c. \(\text{Newton’s Method}\)
| \(x_1\) | \(=x_0-\dfrac{g(x)}{g'(x)}\) |
| \(=1-\left(\dfrac{-1}{2}\right)\) | |
| \(=\dfrac{3}{2}=1.5\) |
\(\text{Using CAS:}\)
d. \(\text{Using CAS}\)
\(x\text{-intercept}:\ x=1.85718\)
\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)
e.i. \(\text{Using CAS}\)
| \(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\) | \(=1.5\) |
| \(k>1\ \therefore\ \ k\) | \(=2.397\) |
e.ii
During a laboratory experiment, a gas is collected in a sealed syringe. Initially, the gas has a volume of 5.0 litres and a pressure of 1.0 atmosphere.
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Explain why the boiling points of hydrogen halides \(\ce{(HF, HCl, HBr}\), and \(\ce{HI)}\) increase from \(\ce{HF}\) to \(\ce{HI}\).
Include in your answer the types of intermolecular forces involved and how molecular mass affects these boiling points. (3 marks)
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A particle moves in a straight line so that its distance, \(x\) metres, from a fixed origin \(O\) after time \(t\) seconds is given by the differential equation \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\), where \(x=0\) when \(t=0\).
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Two seconds after the first particle passed through \(O\), a second particle passes through \(O\).
Its distance \(x\) metres from \(O, t\) seconds after the first particle passed through \(O\), is given by \(x=\log _e\left(\tan ^{-1}(3 t-6)+1\right).\)
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a.i. \(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)
a.ii. \(x=\log _e\left(\tan ^{-1}(2 t)+1\right) \)
b.i. \(\text{Asymptote: }\ x=\log _e\left(\dfrac{\pi}{2}+1\right)\)
b.ii.
c. \(0.02\)
d. \(\text {Particle positions at}\ \ t=6:\)
\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)
\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)
e. \(\dfrac{v_1}{v_2}=\dfrac{2}{3}\)
a.i. \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\)
\(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)
a.ii. \(e^x=\tan ^{-1}(2 t)+c\)
\(\text {When}\ \ t=0, \ x=0 \ \Rightarrow \ c=1\)
\begin{aligned}
e^x& =\tan ^{-1}(2 t)+1 \\
x&=\log _e\left(\tan ^{-1}(2 t)+1\right)
\end{aligned}
b.i. \(\text {As}\ \ t \rightarrow \infty, \ \tan ^{-1}(2 t) \rightarrow \dfrac{\pi}{2}\)
\(x \rightarrow \log _e\left(\dfrac{\pi}{2}+1\right)\)
\(\therefore \text{Asymptote: }\ x=\log _e\left(\dfrac{\pi}{2}+1\right)\)
b.ii. \(\log _e\left(\dfrac{\pi}{2}+1\right) \approx 0.944\)
\(\text{When}\ \ t=10 :\)
\(x=\log _e\left( \tan ^{-1}(20)+1\right)=0.92\ \text{(2 d.p.)}\)
c. \(\text {At}\ \ t=3, x=\log _e\left(\tan ^{-1}(6)+1\right)\)
\(\text {Substitute } t \text { and } x \text { into } \dfrac{d x}{d t}:\)
\(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}=0.02\, \text{ms} ^{-1}\ \text{(2 d.p.)}\)
d. \(\text {Particle positions at}\ \ t=6:\)
\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)
\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)
e. \(e^x=e^{\log _e\left(\tan ^{-1}(12)+1\right)}=\tan ^{-1}(12)+1\)
\(\text {At}\ \ t=6:\)
\(\dfrac{d x_1}{d t}=\dfrac{2}{1+4 t^2} \times \dfrac{1}{\tan ^{-1}(2 t)+1}=\dfrac{2}{145\left(\tan ^{-1}(12)+1\right)}\)
\(\dfrac{d x^2}{d t}=\dfrac{3}{1+(3 t-6)^2} \times \dfrac{1}{\tan ^{-1}(3 t-6)-1}=\dfrac{3}{145\left(\tan ^{-1}(12)+1\right)}\)
\(\therefore \dfrac{v_1}{v_2}=\dfrac{2}{3}\)
Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.
Use the following table of values for `f` and `f^{\prime}`.
| `\quad x \quad` | `\quad\quad 1/2\quad\quad` | `\quad\quad(sqrt{2})/2\quad\quad` | `\quad\quad(sqrt{3})/2\quad\quad` |
| `f(x)` | `-2` | `5` | `3` |
| `\quad\quad f^{prime}(x)\quad\quad` | `7` | `0` | `1/9` |
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The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.
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Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`
Part of the graph of `y=f(x)` is shown below.
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Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.
Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.
Mika flips the coin five times.
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The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
\(f(h)=\begin{cases} ah^2+bh+c &\ \ 1.5\leq h\leq 3 \\ \\ 0 &\ \ \text{elsewhere} \\ \end{cases}\)
where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.
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A particle moving in a straight line with constant acceleration has a velocity of 7 ms\(^{-1}\) at point \(A\) and 17 ms\(^{-1}\) at point \(B\).
The velocity of the particle, in metres per second, at the midpoint of \(AB\) is
An ideal transformer has an input DC voltage of 240 V, 2000 turns in the primary coil and 80 turns in the secondary coil.
The output voltage is closest to
On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.
The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.
One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.
The graph has been drawn to scale.
The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.
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The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.
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The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
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Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.
The population of rabbits over a longer period of time can be modelled by the rule
`s(t)=1700cdote^(-0.003t)cdot sin((pit)/80)+2500,\qquad text(for all)\ t>=0`
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The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.
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The tangent to `f` at point `M` has gradient `-2` .
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The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.
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A soccer player kicks a ball with an angle of elevation of `theta` °, where `theta` is a normally distributed random variable with a mean of 42° and a standard deviation of 8°.
The horizontal distance that the ball travels before landing is given by the function `d=50 \ sin (2\theta)`.
The probability that the ball travels more than 40 m horizontally before landing is closest to
Consider the curve given by `5 x^2 y-3 x y+y^2=10`.
The equation of the tangent to this curve at the point `(1, m)`, where `m` is a real constant, will have a negative gradient when
Consider the function \(g:R \to R, g(x)=2^x+5\).
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ii. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places. (2 marks)
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Let \(h:R\to R, h(x)=2^x-x^2\).
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a. \(\text{As }x\to -\infty,\ \ 2^x\to 0\)
\(\therefore\ 2^x+5\to 5\)
| b. | \(g(x)\) | \(=2^x+5\) |
| \(=\Big(e^{\log_{e}{2}}\Big)^x\) | ||
| \(=e\ ^{x\log_{e}{2}}+5\) | ||
| \(g^{\prime}(x)\) | \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\) | |
| \(=\log_{e}{2}\times 2^x\) | ||
| \(\therefore\ k\) | \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\) |
| \(y-(2^a+5)\) | \(=\log_{e}{2}\times2^a(x-a)\) |
| \(\therefore\ y\) | \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
cii. \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)
| \( y\) | \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
| \(0\) | \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
| \(0\) | \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)
\(\text{Equation of tangent when }\ a\approx 2.6178\)
| \( y\) | \(=2^{2.6178..}\ \log_{e}{(2)x}+0\) |
| \(\therefore\ y\) | \(=4.255x\) |
| d. | \(h(x)\) | \(=2^x-x^2\) |
| \(h^{\prime}(x)\) | \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\) | |
| \(h^{”}(x)\) | \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\) |
\(\text{Solving }h^{”}(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)
\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)
\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)
e. \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)
\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)
f. \(\text{Newton’s Method }\Rightarrow\ x_a-\dfrac{h(x_a)}{h'(x_a)}\)
\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)
\(h(x)=2x-x^2\ \text{and }h^{\prime}(x)=\ln{2}\times 2^x-2x\)
\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} & 0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} & -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} & -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}
g. \(\text{The denominator in Newton’s Method is}\ h^{\prime}(x)=\log_{e}{(2)}\cdot 2^x-2x\)
\(\text{and the calculation will be undefined if }h^{\prime}(x)=0\ \text{as the tangent lines are horizontal}.\)
\(\therefore\ \text{The solution to }h^{\prime}(x)=0\ \text{cannot be used for }x_0.\)
h. \(\text{For a local minimum }f(x)=0\)
\(\rightarrow\ n^x-x^n=0\)
\(\rightarrow\ n^x=x^n\ \ \ (1)\)
\(\text{Also for a local minimum }f^{\prime}(x)=0\)
\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)
\(\text{Substitute (1) into (2)}\)
\(\ln(n)\cdot x^n-nx^{n-1}=0\)
\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)
\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)
\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)
\(\therefore\ n=e\)
The digram shows shows a stationary electron (e\(^{-}\)) in a uniform magnetic field between two parallel plates. The plates are separated by a distance of 6.0 × 10\(^{-3}\) m, and they are connected to a 200 V power supply and a switch. Initially, the plates are uncharged. Assume that gravitational effects on the electron are negligible. --- 3 WORK AREA LINES (style=lined) --- The switch is now closed. --- 5 WORK AREA LINES (style=lined) --- Ravi and Mia discuss what they think will happen regarding the size and the direction of the magnetic force on the electron after the switch is closed. Ravi says that there will be a magnetic force of constant magnitude, but it will be continually changing direction. Mia says that there will be a constantly increasing magnetic force, but it will always be acting in the same direction. Evaluate these two statements, giving clear reasons for your answer. (4 marks) --- 8 WORK AREA LINES (style=lined) ---
A schematic side view of one design of an audio loudspeaker is shown in Diagram 1 below. It uses a current carrying coil that interacts with permanent magnets to create sound by moving a cone in and out. Diagram 2 shows a schematic view of the loudspeaker from the position of the eye shown in Diagram 1. The direction of the current is clockwise, as shown. --- 0 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- a. b. \(E\) c. \(2.51\ \text{N}\) a. Magnetic fields run from the north pole to the south pole. b. Apply the right-hand rule: \(\Rightarrow E\) c. \(l= 2\pi \times 5 \times 10^{-2} = 0.1\pi\) \(\therefore F=nlIB=20 \times 0.1\pi \times 2 \times 0.2=2.51\ \text{N}\)
A.
left
B.
right
C.
up the page
D.
down the page
E.
into the page
F.
out of the page
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Show Worked Solution
♦♦♦ Mean mark 24%.
The following diagram represents an observation wheel, with its centre at point \(P\). Passengers are seated in pods, which are carried around as the wheel turns. The wheel moves anticlockwise with constant speed and completes one full rotation every 30 minutes.When a pod is at the lowest point of the wheel (point \(A\)), it is 15 metres above the ground. The wheel has a radius of 60 metres.
Consider the function \(h(t)=-60\ \cos(bt)+c\) for some \(b, c \in R\), which models the height above the ground of a pod originally situated at point \(A\), after time \(t\) minutes.
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After 15 minutes, the wheel stops moving and remains stationary for 5 minutes. After this, it continues moving at double its previous speed for another 7.5 minutes.
The height above the ground of a pod that was initially at point \(A\), after \(t\) minutes, can be modelled by the piecewise function \(w\):
\(w(t) = \begin {cases}
h(t) &\ \ 0 \leq t < 15 \\
k &\ \ 15 \leq t < 20 \\
h(mt+n) &\ \ 20\leq t\leq 27.5
\end{cases}\)
where \(k\geq 0, m\geq 0\) and \(n \in R\).
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iii. Sketch the graph of the piecewise function \(w\) on the axes below, showing the coordinates of the endpoints. (3 marks)
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a. \(\text{See worked solution}\)
b. \(\approx 36.80\ \text{m}\)
c. \(8\)
d.i. \(k=135, m=2\)
d.ii. \(n=30p+5,\ p\in Z\)
d.iii.
| a. | \(\text{Period:}\) | \(\dfrac{2\pi}{b}\) | \(=30\) |
| \(\therefore\ b\) | \(=\dfrac{\pi}{15}\) |
\(\text{Given }h(t)=-60\cos(bt)+c,\ \text{evaluate when }t=0, h=15\ \text{to find }c.\)
| \(-60\ \cos(0)+c\) | \(=15\) |
| \(c\) | \(=75\) |
\(\therefore\ h(t)=-60\cos(\dfrac{\pi t}{15})+75\)
| b. | \(\text{Average height}\) | \(=\dfrac{1}{\frac{30}{4}-0}\displaystyle\int_0^{\frac{30}{4}}\Bigg(-60\cos\Bigg(\dfrac{\pi}{15}t\Bigg)+75\Bigg)\,dt\) |
| \(=\dfrac{2}{15}\left[75t-\dfrac{900\ \sin(\frac{\pi}{15}t)}{\pi}\right]_{0}^{7.5}\) | ||
| \(=\dfrac{2}{15}\Bigg[75\times 7.5-\dfrac{900\ \sin(\frac{\pi}{15}\times 7.5)}{\pi}\Bigg]-\Bigg[0\Bigg]\) | ||
| \(=\dfrac{75\pi-120}{\pi}=\dfrac{15(5\pi-8)}{\pi}\) | ||
| \(=36.802\dots\approx 36.80\ \text{m}\) |
c. \(\text{Av rate of change of height}\)
| \(=\dfrac{h(7.5)-h(0)}{7.5}\) | |
| \(=\dfrac{\Bigg(75-60\cos(\dfrac{\pi \times 7.5}{15})\Bigg)-\Bigg(75-60\cos(\dfrac{\pi\times 0}{15})\Bigg)}{7.5}\) | |
| \(=\dfrac{75-15}{7.5}=8\) |
| di. | \(\text{Period is 30 minutes, so after 15 minutes pod}\) \(\text{is at the top of the wheel.}\) |
| \(\therefore\ k=75-60\ \cos(\frac{\pi}{15}\times 15)=135\) |
\(\text{Pod is travelling at twice its previous speed}\)
\(\text{so one revolution takes 15 minutes}\)
| \(\therefore\ \text{Period}\rightarrow\) | \(\dfrac{2\pi}{bm}\) | \(=15\) |
| \(bm\) | \(=\dfrac{2\pi}{15}\) | |
| \(\dfrac{\pi}{15}\cdot m\) | \(=\dfrac{2\pi}{15}\) | |
| \(m\) | \(=\dfrac{2\pi}{15}\times \dfrac{15}{\pi}\) | |
| \(=2\) |
| dii. | \(\text{When }t=20\ \text{the pod is at the top of the wheel and height is 135.}\) \(\text{When }t=27.5\ \text{the pod is back at the start and height is 15.}\) |
\(\text{Using CAS solve for }t=20:\rightarrow\ h(2(20)+n)=135\ \rightarrow\ n=30p+5\)
\(\text{Using CAS solve for }t=27.5:\rightarrow\ h(2(27.5)+n)=15\ \rightarrow\ n=30p+5\)
\(\therefore\ n=30p+5,\ p\in Z\)
diii.
Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.
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There are over 400 geostationary satellites above Earth in circular orbits. The period of orbit is one day (86 400 seconds). Each geostationary satellite remains stationary in relation to a fixed point on the equator. The diagram shows an example of a geostationary satellite that is in orbit relative to a fixed point, \(\text{X}\), on the equator. --- 4 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) ---
Let \(w=\text{cis}\left(\dfrac{2 \pi}{7}\right)\). --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=blank) --- --- 0 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- use De Moivre's theorem to show that --- 8 WORK AREA LINES (style=lined) --- a. \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\) \(\therefore w\ \text{is a root of}\ \ z^7-1=0 \) b. \(\text{Roots of}\ \ z^7-1=0:\) \(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \) c. d.i. d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \) \(\arg(z-1)=\dfrac{9\pi}{14} \) f.i. \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)} \) f.ii. \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \) \((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \) \(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\) \(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\) \(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\) a. \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\) \(\therefore w\ \text{is a root of}\ \ z^7-1=0 \) b. \(\text{Roots of}\ \ z^7-1=0:\) \(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \) c. d.i. d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \) \(\arg(z-1)=\dfrac{9\pi}{14} \) f.i. \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)} \) f.ii. \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \) \((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \) \(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\) \(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\) \(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)
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e.
\((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=z^7-1\)
Show Worked Solution
e.
\((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=z^7-1\)
A manufacturer produces tennis balls.
The diameter of the tennis balls is a normally distributed random variable \(D\), which has a mean of 6.7 cm and a standard deviation of 0.1 cm.
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Tennis balls are packed and sold in cylindrical containers. A tennis ball can fit through the opening at the top of the container if its diameter is smaller than 6.95 cm.
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A tennis ball is classed as grade A if its diameter is between 6.54 cm and 6.86 cm, otherwise it is classed as grade B.
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A tennis coach uses both grade A and grade B balls. The serving speed, in metres per second, of a grade A ball is a continuous random variable, \(V\), with the probability density function
\(f(v) = \begin {cases}
\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg) &\ \ 30 \leq v \leq 3\pi^2+30 \\
0 &\ \ \text{elsewhere}
\end{cases}\)
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The serving speed of a grade B ball is given by a continuous random variable, \(W\), with the probability density function \(g(w)\).
A transformation maps the graph of \(f\) to the graph of \(g\), where \(g(w)=af\Bigg(\dfrac{w}{b}\Bigg)\).
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Let \(f:R \to R, f(x)=e^x+e^{-x}\) and \(g:R \to R, g(x)=\dfrac{1}{2}f(2-x)\).
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Two functions \(g_1\) and \(g_2\) are created, both with the same rule as \(g\) but with distinct domains, such that \(g_1\) is strictly increasing and \(g_2\) is strictly decreasing.
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Shown below is the graph of \(g\), the inverse of \(g_1\) and \(g_2\), and the line \(y=x\).
The intersection points between the graphs of \(y=x, y=g(x)\) and the inverses of \(g_1\) and \(g_2\), are labelled \(P\) and \(Q\).
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Let \(h:R\to R, h(x)=\dfrac{1}{k}f(k-x)\), where \(k\in (o, \infty)\).
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Let \(h_1:[k, \infty)\to R, h_1(x)=h(x)\).
The rule for the inverse of \(h_1\) is \(y=\log_{e}\Bigg(\dfrac{1}{k}x+\dfrac{1}{2}\sqrt{k^2x^2-4}\Bigg)+k\)
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It is possible for the graphs of \(h\) and the inverse of \(h_1\) to intersect twice. This occurs when \(k=5\).
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Two functions, \(f\) and \(g\), are continuous and differentiable for all \(x\in R\). It is given that \(f(-2)=-7,\ g(-2)=8\) and \(f^{′}(-2)=3,\ g^{′}(-2)=2\).
The gradient of the graph \(y=f(x)\times g(x)\) at the point where \(x=-2\) is
Part of the graph of `y=f(x)` is shown below. The rule `A(k)=k \ sin(k)` gives the area bounded by the graph of `f`, the horizontal axis and the line `x=k`.
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A tilemaker wants to make square tiles of size 20 cm × 20 cm.
The front surface of the tiles is to be painted with two different colours that meet the following conditions:
An example is shown below.
There are two types of tiles: Type A and Type B.
For Type A, the colours on the tiles are divided using the rule `f(x)=4 \sin \left(\frac{\pi x}{10}\right)+a`, where `a \in R`.
The corners of each tile have the coordinates (0,0), (20,0), (20,20) and (0,20), as shown below.
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ii. Find the value of `a` so that a Type A tile meets Condition 1. (1 mark)
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Type B tiles, an example of which is shown below, are divided using the rule `g(x)=-\frac{1}{100} x^3+\frac{3}{10} x^2-2 x+10`.
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The graph of `y=f(x)`, where `f:[0,2 \pi] \rightarrow R, f(x)=2 \sin(2x)-1`, is shown below.
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a. Graph `y=g(x)`
b. `\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`
c.i. `b=2`
cii. `=\frac{\pi}{2}`
ciii. `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`
a.
| b. `2 \sin (2 k)-1` | `=0` `0<=k<=2\pi` | |
| `sin (2 k)` | `=1/2` | |
| `2k` | `=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}` | |
| `k` | `=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}` |
| c.i ` 2 \sin 2(x-a)-1+b` | `=-(2 \sin 2 x-1)` | |
| `:.\ -1+b` | `=1` | |
| `b` | `=2` |
| c.ii `2 \sin 2(x-a)` | `=-(2 \sin 2 x-1)` | |
| `\sin (2 x-2 a)` | `=-\sin 2 x` | |
| `\therefore 2 a` | `=\pi` | |
| `a` | `=\frac{\pi}{2}` |
c.iii The domain for `f(x)` is `[0,2pi]`
`:. \ D` is `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`
The shapes of two walking tracks are shown below.
Track 1 is described by the function \(f(x)=a-x(x-2)^2\).
Track 2 is defined by the function \(g(x)=12x-bx^2\).
The unit of length is kilometres.
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Let \(\hat{P}\) be the random variable that represents the sample proportion of households in a given suburb that have solar panels installed.
From a sample of randomly selected households in a given suburb, an approximate 95% confidence interval for the proportion \(p\) of households having solar panels installed was determined to be (0.04, 0.16).
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Use \(z=2\) to approximate the 95% confidence interval.
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Suppose that the queuing time, \(T\) (in minutes), at a customer service desk has a probability density function given by
\(f(t) = \begin {cases}
kt(16-t^2) &\ \ 0 \leq t \leq 4 \\
\\
0 &\ \ \text{elsewhere}
\end{cases}\)
for some \(K \in R\).
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One of the landmarks in state \(A\) requires a renovation project.
This project involves 12 activities, \(A\) to \(L\). The directed network below shows these activities and their completion times, in days.
The table below shows the 12 activities that need to be completed for the renovation project.
It also shows the earliest start time (EST), the duration, and the immediate predecessors for the activities.
The immediate predecessor(s) for activity \(I\) and the EST for activity \(J\) are missing.
\begin{array} {|c|c|c|}
\hline
\quad \textbf{Activity} \quad & \quad\quad\textbf{EST} \quad\quad& \quad\textbf{Duration}\quad & \textbf{Immediate} \\
& & & \textbf{predecessor(s)} \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 0 & 6 & - \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 0 & 4 & - \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 6 & 7 & A \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 & 5 & B \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 4 & 10 & B \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 13 & 4 & C \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 9 & 3 & D \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 9 & 7 & D \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 13 & 6 & - \\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & - & 6 & E, H \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 19 & 4 & F, I \\
\hline
\rule{0pt}{2.5ex} L \rule[-1ex]{0pt}{0pt} & 23 & 1 & J, K \\
\hline
\end{array}
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The managers of the project are able to reduce the time, in days, of six activities.
These reductions will result in an increase in the cost of completing the activity.
The maximum decrease in time of any activity is two days.
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \quad A \quad & \quad B \quad& \quad F \quad & \quad H \quad & \quad I \quad & \quad K \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{Daily cost (\$)} \rule[-1ex]{0pt}{0pt} & 1500 & 2000 & 2500 & 1000 & 1500 & 3000 \\
\hline
\end{array}
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\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}
a. \(\text{Immediate predecessors of}\ I:\ C, G\)
\(\text{Dummy activity before activity}\ C\ \text{does not effect this.}\)
b. \(\text{Scan network:}\)
\(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)
c. \(\text{Critical Path:}\ A\ C\ I\ K\ L\)
\(\text{Activities on the critical path have a float time of zero.}\)
\(\Rightarrow \ \text{5 activities have a float time of zero.}\)
d. \(\text{If activities}\ A\ \text{and}\ B\ \text{are reduced by 2 days,}\)
\(\text{the critical path remains:}\ A\ C\ I\ K\ L\ \text{(22 days)}\)
\(\text{Maximum reduction time = 2 days}\)
e.
\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 1\\
\hline
\end{array}
The state \(A\) has nine landmarks, \(G, H, I, J, K, L, M, N\) and \(O\).
The edges on the graph represent the roads between the landmarks.
The numbers on each edge represent the length, in kilometres, along each road.
Three friends, Eden, Reynold and Shyla, meet at landmark \(G\).
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Arthur takes out a new loan of $60 000 to pay for an overseas holiday. Interest on this loan compounds weekly. The balance of the loan, in dollars, after \(n\) weeks, \(V_n\), can be determined using a recurrence relation of the form \(V_0=60\ 000, \quad V_{n+1}=1.0015\,V_n-d\) --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) ---
Arthur fully repays the loan in five years. Round your answer to the nearest cent. (1 mark)
Consider the graph below.
The number of edges that need to be removed for this graph to be planar is
\(\text{The graph can be drawn as follows with no crossing edges.}\)
\(\Rightarrow A\)
Two types of computers - laptops `(L)` and desktops `(D)` - can be serviced by Henry `(H)`, Irvine `(I)` or Jean `(J)`.
Matrix `N` shows the time, in minutes, it takes each person to service a laptop and a desktop.
`{:(qquadqquadquad\ LquadqquadD),(N = [(18,8),(10,17),(12,9)]{:(H),(I),(J):}):}`
Matrix `Q` shows the number of laptops and desktops in four different departments: marketing `(M)`, advertising `(A)`, publishing `(P)` and editing `(E)`.
`{:(qquadqquadquad\ LquadqquadD),(Q = [(6,8),(4,7),(5,5),(10,12)]{:(M),(A),(P),(E):}):}`
A calculation that determines the total time that it would take each of Henry, Irvine or Jean, working alone, to service all the laptops and desktops in all four departments is
The following table provides information about three organic compounds, \(\text{X, Y}\) and \(\text{Z}\).
Which one of the following is the best estimate for the boiling point of Compound \(\text{Z}\)?
The complex numbers \(w\) and \(z\) both have modulus 1, and \(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{z}{w}\Big{)} \lt \pi\), where \(\text{Arg}\) denotes the principal argument.
For real numbers \(x\) and \(y\), consider the complex number \(\dfrac{xz+yw}{z}\).
On an \(xy\)-plane, clearly sketch the region that contains all points \((x,y)\) for which \(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{xz+yw}{z}\Big{)} \lt \pi\).
\(\text{One strategy:}\)
\(\text{Let}\ \ u=\dfrac{z}{w},\ \text{where}\ \abs{z} = \abs{w} = 1\ \ \Rightarrow \ \ \abs{u}=1. \)
\(\dfrac{\pi}{2} \lt \text{Arg} (u) \lt \pi\ \text{(given)}\ \ \Rightarrow \text{Re}(u) \lt 0\ \ \text{and}\ \ \text{Im}(u) \gt 0 \)
\(\text{Using}\ \ \dfrac{z}{w} = \dfrac{1}{u} = \dfrac{\bar u}{\abs{u}} = \bar u, \)
\(\text{find all real numbers}\ x\ \text{and}\ y\ \text{such that:} \)
\(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{xz+yw}{z}\Big{)} \lt \pi\)
\(\dfrac{\pi}{2} \lt \text{Arg} (x+ \bar y u) \lt \pi\)
\(\text{i.e.}\ \ \text{Re}(x+y \bar u) = x + y \times \text{Re}(u) \lt 0\ \ \text{and}\ \ \text{Im}(y \bar u) \gt 0 \)
\(\therefore y \lt 0\ \ \text{and} \)
\(x \lt -y \times \text{Re}(u) = -y \times \cos (\text{Arg}(u)) = -y \times \cos \Bigg{(}\text{Arg}\Big{(} \dfrac{z}{w} \Big{)} \Bigg{)} \)
\(\text{Region is all real}\ (x,y)\ \text{that lie below the}\ x\text{-axis and to the} \)
\(\text{left of the line}\ \ y=-\sec \Bigg{(}\text{Arg}\Big{(} \dfrac{z}{w} \Big{)} \Bigg{)}x \)
Consider any three-dimensional vectors \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\) and \(\underset{\sim}{c}=\overrightarrow{O C}\) that satisfy these three conditions
\(\underset{\sim}{a} \cdot \underset{\sim}{b}=1\)
\(\underset{\sim}{b} \cdot \underset{\sim}{c}=2\)
\(\underset{\sim}{c} \cdot \underset{\sim}{a}=3\).
Which of the following statements about the vectors is true?
A particular building project has ten activities that must be completed.
These activities and their immediate predecessor(s) are shown in the table below.
\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \textbf{Immediate predecessor(s)} \\
\hline
A & - \\
\hline
B & - \\
\hline
C & A \\
\hline
D & A \\
\hline
E & B \\
\hline
F & D, E \\
\hline
G & C, F \\
\hline
H & F \\
\hline
I & D, E \\
\hline
J & H, I \\
\hline
\end{array}
A directed graph that could represent this project is
Let \(w\) be the complex number \(z=e^{\small{\dfrac{2i \pi}{3}}} \).
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The vertices of a triangle can be labelled \(A, B\) and \(C\) in anticlockwise or clockwise direction, as shown.
Three complex numbers \(a, b\) and \(c\) are represented in the complex plane by points \(A, B\) and \(C\) respectively.
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\(a^2+b^2+c^2=a b+b c+c a .\) (2 marks)
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Tavi took out a loan of $20 000, with interest compounding quarterly. She makes quarterly repayments of $653.65.
The graph below represents the balance in dollars of Tavi's loan at the end of each quarter of the first year of the loan.
The effective interest rate for the first year of Tavi's loan is closest to
Maia is at a skatepark. She stands on her skateboard as it rolls in a straight line down a gentle slope at a constant speed of 3.0 m s\(^{-1}\), as shown in the figure below. The slope is 5° to the horizontal. The combined mass of Maia and the skateboard is 65 kg. --- 0 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- Near the bottom of the ramp, Maia takes hold of a large pole and comes to a complete rest while still standing on the skateboard. Maia and the skateboard now have no momentum or kinetic energy. --- 3 WORK AREA LINES (style=lined) --- a. b. 55.5 N c. Momentum: Kinetic energy: a. b. Total frictional forces: c. Momentum: Kinetic energy:
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\(F_{\text{down slope}}\)
\(=F_f\)
\(=mg\, \sin\, \theta\)
\(=65 \times 9.8 \times \sin\,5^{\circ}\)
\(=55.5\ \text{N}\)
♦ Mean mark (b) 50%.
♦♦♦ Mean mark (c) 26%.
