Which energy system produces ATP at the fastest rate?
- Glycolytic
- Aerobic
- ATP-PCr
- All systems produce ATP at the same rate
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Which energy system produces ATP at the fastest rate?
\(C\)
Consider Option C: ATP-PCr
Other Options:
\(\Rightarrow C\)
Outline how inefficient jumping technique can affect the skeletal system and require first aid intervention. (3 marks)
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Sample Answer
Sample Answer
During a cross-country run, an athlete experiences severe abdominal cramping. Which first aid response would be most appropriate?
\(D\)
Consider Option D: Stop activity and lie in a comfortable position
Other Options:
\(\Rightarrow D\)
Outline how the muscular and skeletal systems work together during movement and identify when first aid intervention is required. (3 marks)
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Sample Answer
Sample Answer
Outline how the skeletal and muscular systems work together during a squat movement. (3 marks)
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Sample Answer
Descent:
Bottom Position:
Rising:
Sample Answer
Descent:
Bottom Position:
Rising:
During a netball game, a player performs a layup shot. Which body systems are working together to execute this movement?
\(C\)
Consider Option C: Skeletal, muscular and nervous
Other Options:
\(\Rightarrow C\)
The muscle group indicated in the image below is primarily responsible for:
\(B\)
Consider Option B:
Other Options:
\(\Rightarrow B\)
The function \(h:[0, \infty) \rightarrow R, \ h(t)=\dfrac{3000}{t+1}\) models the population of a town after \(t\) years. --- 2 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. \(600\) b. \(\text{Transformations:}\) \(\text{1- Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\) \(\text{2- Translation of 1500 units upwards}\) ci. \(\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\ \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\) cii. \(300\) a. \(h(4)=\dfrac{3000}{4+1}=600\)
\(\text{Given}\ \ h(0)=h_1(0):\) \(\dfrac{1500}{0+1} +C= 3000\ \ \Rightarrow\ \ C=1500\) \(h_1(t)=\dfrac{1500}{t+1}+1500\) \(\text{Transformations:}\) \(\text{1- Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\) \(\text{2- Translation of 1500 units upwards}\) ci. \(\hat{p}=\dfrac{60}{100}=\dfrac{3}{5},\quad 1-\hat{p}=\dfrac{2}{5},\quad z=2\)
b.
\(h(t)\)
\(=3000(t+1)^{-1}\)
\(h^{\prime}(t)\)
\(=-\dfrac{3000}{(t+1)^2}\)
\(h_1^{\prime}(t)\)
\(=\dfrac{1}{2}h^{\prime}(t)=-\dfrac{1500}{(t+1)^2}\)
\(h_1(t)\)
\(=\dfrac{1500}{t+1}+C\)
\(\text{Approx CI}\)
\(=\left(\dfrac{3}{5}-2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}},\ \dfrac{3}{5}+2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}}\right)\)
\(=\left(\dfrac{3}{5}-\dfrac{2\sqrt{6}}{50},\quad \dfrac{3}{5}+\dfrac{2\sqrt{6}}{50}\right)\)
\(=\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\quad \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)
cii.
\(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\)
\(=\dfrac{\sqrt{2}}{50}\)
\(\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}\)
\(=\dfrac{\sqrt{2}}{50}\)
\(\dfrac{6}{25n}\)
\(=\dfrac{2}{2500}\)
\(\dfrac{25n}{6}\)
\(=1250\)
\(n\)
\(=\dfrac{6}{25}\times 1250\)
\(=300\)
A study monitored the changes in the body temperature of a kookaburra (an Australian bird) and a human over a 24-hour period. The results of the study are shown in the graph. --- 2 WORK AREA LINES (style=lined) --- Some endothermic organisms can display torpor (a significant decrease in physiological activity). With reference to the graph, explain whether the human or the kookaburra was displaying torpor and if so, state the time this occurred. (3 marks) --- 7 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. 4 am b. Signs of torpor: → The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity → In contrast, the kookaburra exhibited classic torpor behaviour. → Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period. c. Kookaburra adaptation: → Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them. → This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier → The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently. a. 4 am b. Signs of torpor: → The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity → In contrast, the kookaburra exhibited classic torpor behaviour. → Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period. c. Kookaburra adaptation: → Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them. → This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier → The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.
Table 1 lists the Olympic year, \(\textit{year}\), and the gold medal-winning height for the men's high jump, \(\textit{Mgold}\), in metres, for each Olympic Games held from 1928 to 2020. No Olympic Games were held in 1940 or 1944, and the 2020 Olympic Games were held in 2021. Table 1 \begin{array}{|c|c|} --- 1 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
a.i. \(2.39\) a.ii. \(\dfrac{11}{22}=50\%\) \(Q_1=2.12, \ Q_3=2.36\) \(\text{Min}\ =1.94, \ \text{Max}\ =2.39\) e. \(\text{A coefficient of determination of 85.7% shows the variation in}\) \(\text{the}\ Mgold\ \text{that is explained by the variation in the }year.\)
\hline \quad \textit{year} \quad & \textit{Mgold}\,\text{(m)} \\
\hline 1928 & 1.94 \\
\hline 1932 & 1.97 \\
\hline 1936 & 2.03 \\
\hline 1948 & 1.98 \\
\hline 1952 & 2.04 \\
\hline 1956 & 2.12 \\
\hline 1960 & 2.16 \\
\hline 1964 & 2.18 \\
\hline 1968 & 2.24 \\
\hline 1972 & 2.23 \\
\hline 1976 & 2.25 \\
\hline 1980 & 2.36 \\
\hline 1984 & 2.35 \\
\hline 1988 & 2.38 \\
\hline 1992 & 2.34 \\
\hline 1996 & 2.39 \\
\hline 2000 & 2.35 \\
\hline 2004 & 2.36 \\
\hline 2008 & 2.36 \\
\hline 2012 & 2.33 \\
\hline 2016 & 2.38 \\
\hline 2020 & 2.37 \\
\hline
\end{array}
b.
\(z\)
\(=\dfrac{x-\overline x}{s_x}\)
\(=\dfrac{2.35-2.23}{0.15}\)
\(=0.8\)
c. \(Q_2=\dfrac{2.33+2.25}{2}=2.29\)
The vector \(\underset{\sim}{a}\) is \(\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) and the vector \(\underset{\sim}{b}\) is \(\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\). --- 4 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- 1. \(\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\) ii. \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\) \( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\) \(\therefore\ \text {Vectors are perpendicular.}\) i. \(\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\) \(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\) \( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)
ii. \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)
\(\therefore\ \text{Vectors are perpendicular.}\)
--- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- i. \(2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\) ii. \(-64 \sqrt{3}-64 i\) i. \(z=\sqrt{3}+i\) \(|z|=\sqrt{3+1}=2\) \(\arg (z)=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\dfrac{\pi}{6}\) \(z=2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)
ii.
\((\sqrt{3}+i)^7\)
\(=2^7\left(\cos \left(\dfrac{7 \pi}{6}\right)+i \sin \left(\dfrac{7 \pi}{6}\right)\right)\)
\(=128\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} i\right)\)
\(=-64 \sqrt{3}-64 i\)
Let \(z=2+3 i\) and \(w=1-5 i\). --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- i. \(z+\bar{w}=3+8 i\) ii. \(z^2=-5+12 i\) i. \(z=2+3 i\) \(w=1-5 i \ \Rightarrow \ \bar{w}=1+5 i\) \(z+\bar{w}=2+3 i+1+5 i=3+8 i\)
ii.
\(z^2\)
\(=(2+3 i)^{2}\)
\(=4+12 i+9 i^2\)
\(=-5+12 i\)
Find \(\displaystyle \int x e^x\, d x\) (2 marks) --- 5 WORK AREA LINES (style=lined) --- \(x e^x-e^x+c\) \(u=x \quad \ \ u^{\prime}=1\) \(v^{\prime}=e^x \quad v=e^x\)
\(\displaystyle\int x e^x \,d x\)
\(=u v^{\prime}-\displaystyle \int v u^{\prime}\, d x\)
\(=x e^x- \displaystyle \int e^x \cdot 1\, d x\)
\(=x e^x-e^x+c\)
Consider the vectors \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}\) and \(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\). --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- i. \(\displaystyle \binom{7}{0}\) ii. \(5\) i. \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}, \ \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\) \(2 \underset{\sim}{a}-\underset{\sim}{b}=2 \displaystyle \binom{3}{2}-\binom{-1}{4}=\binom{6}{4}-\binom{-1}{4}=\binom{7}{0}\) ii. \(\underset{\sim}{a} \cdot \underset{\sim}{b}=\displaystyle\binom{3}{2}\binom{-1}{4}=3 \times(-1)+2 \times 4=5\).
Four people completed the same fitness activity.
The graph shows the heart rate for each person before and after completing the activity.
Which person had the LEAST difference in heart rate?
\(B\)
\(\text{Option A: }\) | \(\text{Jo}\) | \(=120-80=40\) |
\(\text{Option B: }\) | \(\text{Kim}\) | \(=120-100=20\ \checkmark\) |
\(\text{Option C: }\) | \(\text{Lee}\) | \(=120-90=30\) |
\(\text{Option D: }\) | \(\text{Mal}\) | \(=150-100=50\) |
\(\Rightarrow B\)
Consider the statement:
'If a polygon is a square, then it is a rectangle.'
Which of the following is the converse of the statement above?
\(A\)
\(\text{Statement:}\ P \Rightarrow \ Q\)
\(\text{Converse of statement:}\ Q \Rightarrow \ P\)
\(\Rightarrow A\)
Consider the following statement written in the formal language of proof
\(\forall \theta \in\biggl(\dfrac{\pi}{2}, \pi\biggr) \exists\ \phi \in\biggl(\pi, \dfrac{3 \pi}{2}\biggr) ; \ \sin \theta=-\cos \phi\).
Which of the following best represents this statement?
\(C\)
\(\Rightarrow C\)
A vertical tower \(T C\) is 40 metres high. The point \(A\) is due east of the base of the tower \(C\). The angle of elevation to the top \(T\) of the tower from \(A\) is 35°. A second point \(B\) is on a different bearing from the tower as shown. The angle of elevation to the top of the tower from \(B\) is 30°. The points \(A\) and \(B\) are 100 metres apart.
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\(194^{\circ}\)
a. \(\text{In}\ \Delta TCA:\)
\(\tan 35°\) | \( =\dfrac{40}{AC}\) | |
\(AC\) | \( =\dfrac{40}{\tan 35°}\) | |
\(=57.125…\) | ||
\(=57.13\ \text{m (2 d.p.)}\) |
b. \(\text{In}\ \Delta TCB:\)
\(\tan 30°\) | \( =\dfrac{40}{BC}\) | |
\(BC\) | \( =\dfrac{40}{\tan 30°}\) | |
\(=69.28\ \text{m}\) |
\( \text{Find} \ \angle BCA \ \text{using cosine rule:}\)
\(\cos \angle B CA\) | \( = \dfrac{57.13^2+69.28^2-100^2}{2 \times 57.13 \times 69.28}\) | |
\(= -0.2446 \)… | ||
\(\angle BCA\) | \( = 104.2° \) |
\(\therefore\ \text{Bearing of}\ B\ \text{from}\ C= 90+104=194^{\circ} \text{(nearest degree)} \)
A network of towns and the distances between them in kilometres is shown. --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. \(TYWH\) b. \(\text{Length of shortest path}\ (YWHMG) = 89 \text{km}\) a. \(TYH=30+38=68, \quad TYWH=30+15+20=65\) \(\therefore \text{ Shortest Path is}\ TYWH.\) b. \(Y W C M G=15+25+25+25=90\) \(YWHMG=15+20+29+25=89\) \(\Rightarrow \ \text{All other paths are longer.}\) \(\therefore\text{ Length of shortest path = 89 km}\)
Which of the following structures is present in both prokaryotic and eukaryotic cells?
\(C\)
→ Both prokaryotic and eukaryotic cells contain ribosomes, which are responsible for protein synthesis.
→ Prokaryotes lack membrane-bound organelles like mitochondria or the Golgi apparatus.
\(\Rightarrow C\)
Farmers and supermarkets agree that green beans are bought more frequently than yellow beans. A supermarket has asked a farmer to produce only green beans.
One way this could be achieved is by
\(C\)
→ Selective breeding involves choosing parents with desirable traits and breeding them to produce offspring with those traits.
→ In this case, the farmer would selectively breed green bean plants to produce more green beans, as that is the preferred variety by consumers and supermarkets.
→ Selective breeding does not involve genetic engineering techniques like DNA hybridization (option B) and is a common agricultural practice used to enhance desirable traits in crops, unlike options A and D which are not relevant to this scenario.
\(\Rightarrow C\)
Which one of the following statements about proteins is correct?
\(A\)
Consider option A:
→ Changes in temperature or pH can disrupt the interactions that stabilise a protein’s higher-order structure.
→ This causes it to lose its three-dimensional shape and become denatured which usually results in a non-functional protein (Option A is Correct)
\(\Rightarrow A\)
A study assessed the effectiveness and safety of a drug called doxycycline. One hundred and fifty adults hospitalised with malaria were involved. These adults were randomly placed into two groups of equal size. One group received doxycycline in addition to standard care. The other group received standard care only.
The group receiving standard care only was the
\(A\)
→ A control group does not receive the experimental treatment or any intervention being tested.
\(\Rightarrow A\)
Conversion graphs can be used to convert from one currency to another.
Abbie converted 70 New Zealand dollars into Euros. She then converted all of these Euros into Australian dollars.
How much money, in Australian dollars, should Abbie have?
\(C\)
\(\text{Using the graphs:}\)
\($70\ \text{New Zealand}\) | \(=40\ \text{Euro}\) |
\(40\ \text{Euro}\) | \(=$55\ \text{Australian}\) |
\(\Rightarrow C\)
Nicole is 38 years old, and likes to keep fit by doing cross-fit classes.
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a. \(182\ \text{bpm}\)
b. \(91-155\ \text{bpm}\)
a. | \(\text{Max heart rate}\) | \(=220-38\) |
\(=182\ \text{bpm}\) |
b. \(\text{50% max heart rate}\ = 0.5\times 182 = 91\ \text{bpm}\)
\(\text{85% max heart rate}\ = 0.85\times 182 = 154.7\ \text{bpm}\)
\(\therefore\ \text{Nicole should aim for between 91 and 155 bpm during exercise.}\)
The formula \(C=80n+b\) is used to calculate the cost of producing desktop computers, where \(C\) is the cost in dollars, \(n\) is the number of desktop computers produced and \(b\) is the fixed cost in dollars.
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a. \(\text{Find}\ C,\ \text{given}\ n=2458\ \text{and}\ b=18\ 230\)
\(C\) | \(=80\times 2458+18\ 230\) | |
\(=$214\ 870\) |
b. \(\text{Find}\ n,\ \text{given}\ C=103\ 330\ \text{and}\ a=35\)
\(C\) | \(=80n+an+18\ 230\) |
\(103\ 330\) | \(=80n+35n+18\ 230\) |
\(115n\) | \(=85\ 100\) |
\(n\) | \(=\dfrac{85\ 100}{115}\) |
\(=740\ \text{desktop computers}\) |
What is the value of \(\dfrac{x-y}{6}\), if \(x=184\) and \(y=46\)?
\(B\)
\(\dfrac{x-y}{6}\) | \(=\dfrac{184-46}{6}\) |
\(=23\) |
\(\Rightarrow B\)
If \(V=\dfrac{4}{3}\pi r^3\), what is the value of \(V\) when \(r = 5\), correct to two decimal places?
\(D\)
\(V =\dfrac{4}{3}\pi r^3\)
\(\text{When}\ r = 2,\)
\(V\) | \(=\dfrac{4}{3}\pi\times 5^3\) |
\(=523.598\dots\) |
\(\Rightarrow D\)
Which of the following equations has \(x=7\) as the solution?
\(C\)
\(2x\) | \(=14\) |
\(x\) | \(=\dfrac{14}{2}\) |
\(\therefore\ x\) | \(=7\) |
\(\Rightarrow C\)
If \(A=P(1 + r)^n\), find \(A\) given \(P=$500\), \(r=0.09\) and \(n=5\) (give your answer to the nearest cent). (2 marks)
\($769.31\ \text{(nearest cent)}\)
\(A\) | \(=P(1 + r)^n\) |
\(=500(1 + 0.09)^5\) | |
\(=500(1.09)^5\) | |
\(=769.311\dots\) | |
\(=$769.31\ \text{(nearest cent)}\) |
Find the value of \(b\) given \(\dfrac{b}{9}-5=3\). (1 mark)
\(72\)
\(\dfrac{b}{9}-5\) | \(=3\) |
\(\dfrac{b}{9}\) | \(=8\) |
\(\therefore\ b\) | \(=72\) |
If \(\dfrac{x-8}{9}=2\), find \(x\). (1 mark)
\(26\)
\(\dfrac{x-8}{9}\) | \(=2\) |
\(x-8\) | \(=18\) |
\(x\) | \(=26\) |
It is given that \(I=\dfrac{3}{2}MR^2\).
What is the value of \(I\) when \(M =19.12\) and \(R = 1.02\), correct to two decimal places?
\(B\)
\(I\) | \(=\dfrac{3}{2}\times 19.12\times 1.02^2\) |
\(=29.84\) |
\(\Rightarrow B\)
Magnetic and gravitational forces have a variety of properties.
Which of the following best describes the attraction/repulsion properties of magnetic and gravitational forces?
\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \textbf{Magnetic forces}\rule[-1ex]{0pt}{0pt}& \ \textbf{Gravitational forces} \\
\hline
\rule{0pt}{2.5ex}\text{either attract or repel}\rule[-1ex]{0pt}{0pt}&\text{only attract}\\
\hline
\rule{0pt}{2.5ex}\text{only repel}\rule[-1ex]{0pt}{0pt}& \text{neither attract nor repel}\\
\hline
\rule{0pt}{2.5ex}\text{only attract}\rule[-1ex]{0pt}{0pt}& \text{only attract} \\
\hline
\rule{0pt}{2.5ex}\text{either attract or repel}\rule[-1ex]{0pt}{0pt}& \text{either attract or repel} \\
\hline
\end{array}
\end{align*}
\(A\)
→ In magnets, like poles repel and opposite poles attract.
→ Gravitational forces are only attractive.
\(\Rightarrow A\)
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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ii. Find \(\text{Pr}(X \geq 2).\) (1 mark)
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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. i. `frac{1}{32}` ii. `frac{13}{16}` iii. `0.806` (3 d.p.)
a. iv `text{E}(X)=5/2, text{sd}(X)=\frac{\sqrt{5}}{2}`
b. i. `1` ii. `a=-frac{4}{5}, b=frac{17}{5}, c=-frac{167}{60}`
b. iii. `r=-1, s=3`
c. i. `text{Discrete}` ii. `(0.208, 0.592)` iii. `n=100`
a.i `X ~ text{Bi}(5 , frac{1}{2})`
`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
a.ii By CAS: `text{binomCdf}(5,0.5,2,5)` `0.8125`
`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`
a.iii `\text{Pr}(X \geq 2 | X<5)`
`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
`= 0.806452 ~~ 0.806` (3 decimal places)
a.iv `X ~ text{Bi}(5 , frac{1}{2})`
`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`
`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`
b.i `\int_{1.5}^3 f(h) d h = 1`
b.ii By CAS:
`f(h):= a\·\h^2 + b\·\h +c`
`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`
`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\ c = =-2.78 \dot{3} = frac{-167}{60}`
b.iii `h + d = 3`
`:.\ f(h) = f(3 – d) = f(- d + 3)`
`:.\ r = – 1 ` and ` s = 3`
c.i `\hat{p}` is discrete.
The number of coin flips must be zero or a positive integer so `\hat{p}` is countable and therefore discrete.
c.ii `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`
`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`
`\approx(0.208\ ,0.592)`
c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.
`:.\ \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`
`:.\ n = 100`
She would need to flip the coin 100 times
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
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ai. `r(0)=2500`
aii. Minimum population of rabbits `= 800`
Maximum population of rabbits `= 4200`
aiii. `160` weeks
b. See worked solution.
c. `~~ 5339` (nearest whole number)
d. Weeks between the periods is 160
e. `~~ 4142` (nearest whole number)
f. Average rate of change `= – 3.6` rabbits/week (1 d.p.)
g. `t = 156` weeks (nearest whole number)
h. ` s → 2500`
ai. Initial population of rabbits
From graph when `t=0, \ r(0) = 2500`
Using formula when `t=0`
`r(t)` | `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500` | |
`r(0)` | `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500` rabbits |
aii. From graph,
Minimum population of rabbits `= 800`
Maximum population of rabbits `= 4200`
OR
Using formula
Minimum is when `t = 120`
`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`
Maximum is when `t = 40`
`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`
aiii. Number of weeks between maximum populations of rabbits `= 200 – 40 = 160` weeks
b. Period of foxes = period of rabbits = 160:
`frac{\2pi}{b} = 160`
`:.\ b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.
Using the point `(100 , 2500)`
Amplitude when `b = frac{\pi}{80}`:
`f(t)` | `=a \ sin (pi/80(t-60))+1600` | |
`f(100)` | `= 2500` | |
`2500` | `= a \ sin (pi/80(100-60))+1600` | |
`2500` | `= a \ sin (pi/2)+1600` | |
`a` | `= 2500 – 1600 = 900` |
`:.\ f(t)= 900 \ sin (pi/80)(t – 60) + 1600`
c. Using CAS find `h(t) = f(t) + r(t)`:
`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`
`text{fMax}(h(t),t)|0 <= t <= 160` `t = 53.7306….`
`h(53.7306…)=5339.46`
Maximum combined population `~~ 5339` (nearest whole number)
d. Using CAS, check by changing domain to 0 to 320.
`text{fMax}(h(t),t)|0 <= t <= 320` `t = 213.7305…`
`h(213.7305…)=5339.4568….`
Therefore, the number of weeks between the periods is 160.
e. Fox population:
`t^{\prime} = frac{90}{pi}t + 60` → `t = frac{pi}{90}(t^{\prime} – 60)`
`y^{\prime} = 900y+1600` → `y = frac{1}{900}(y^{\prime} – 1600)`
`frac{y^{\prime} – 1600}{900} = sin(frac{pi(t^{\prime} – 60)}{90})`
`:.\ f(t) = 900\ sin\frac{pi}{90}(t – 60) + 1600`
Average combined population [Using CAS]
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
`= 4142.2646….. ~~ 4142` (nearest whole number)
f. Using CAS
`s(t):= 1700e^(- 0.003t) dot\sin\frac{pit}{80} + 2500`
`text{fMax}(s(t),t)|0<=t<=320` `x = 38.0584….`
`s(38.0584….)=4012.1666….`
`text{fMax}(s(t),t)|160<=t<=320` `x = 198.0584….`
`s(198.0584….)=3435.7035….`
Av rate of change between the points
`(38.058 , 4012.167)` and `(198.058 , 3435.704)`
`= frac{4012.1666…. – 3435.7035….}{38.0584…. – 198.0584….} = – 3.60289….`
`:.` Average rate of change `= – 3.6` rabbits/week (1 d.p.)
g. Using CAS
`s^(”)(t) = 0` , `t = 80(n – 0.049) \ \forall n \in Z`
After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2`
`t` | `= 80(n – 0.049)` | |
`= 80(2 – 0.049)` | ||
`= 156.08` |
`:. \ t = 156` weeks (nearest whole number)
h. As `t → ∞`, `e^(- 0.003t) → 0`
`:.\ s → 2500`
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. `x=0`
b. ` f^{\prime}(x)=1/6x`
c. `x=-12`
di. `y=1/2x + 18`
dii. Area`= 375` units²
e. `b = 2a^2`
a. Axis of symmetry: `x=0`
b. | `f(x)` | `=\frac{x^2}{12}` | |
` f^{\prime}(x)` | `= 1/6x` |
c. At `M` gradient `= -2`
`1/6x` | `= -2` | |
`x` | `= -12` |
When `x = -12`
`f(x) = (-12)^2/12 = 12`
Equation of tangent at `(-12 , 12)`:
`y – y_1` | `=m(x – x_1)` | |
`y – 12` | `= -2(x + 12)` | |
`y` | `= -2x -12` |
d.i Gradient of tangent `= -2`
`:.` gradient of normal `= 1/2`
Equation at `M(- 12 , 12)`
`y – y_1` | `=m(x – x_1)` | |
`y – 12` | `= 1/2(x + 12)` | |
`y` | `=1/2x + 18` |
d.ii Points of intersection of `f(x)` and normal are at `M` and `N`.
So equate ` y = x^2/12` and `y = 1/2x + 18` to find `N`
`x^2/12` | `=1/2x + 18` | |
`x^2 – 6x – 216` | `=0` | |
`(x + 12)(x – 18)` | `=0` |
`:.\ x = -12` or `x = 18`
Area | `= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x` | |
`= [x^2/4 + 18x – x^3/36]_(-12)^18` | ||
`= [18^2/4 +18^2 – 18^3/36] – [12^2/4 + 18 xx (-12) – (-12)^3/36]` | ||
`= 375` units² |
e. `g(x) = x^2/(4a^2)` `a > 0`
At `x = – b` `y = (-b)^2/(4a^2) = b^2/4a^2`
`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`
Gradient of tangent `= (-b)/(2a^2)`
Gradient of normal `= (2a^2)/b`
Equation of normal at `(- b , b^2/(4a^2))`
`y – y_1` | `= m(x – x_1)` | |
`y – b^2/(4a^2)` | `= (2a^2)/b(x – (-b))` | |
`y` | `= (2a^2x)/b + 2a^2 + b^2/(4a^2)` | |
`y` | `= (2a^2x)/b +(8a^4 + b^2)/(4a^2)` |
Points of intersection of normal and parabola (Using CAS)
solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`
`x = – b` or `x = (8a^4+b^2)/b`
Calculate area using CAS
`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2} – frac{x^2}{4a^2} \right) dx`
`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`
Using CAS Solve derivative of `A = 0` with respect to `b` to find `b`
solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`
`b = – 2a^2` and `b = 2a^2`
Given `b > 0`
`b = 2a^2`
The period of the function `f(x)=3 \ cos (2 x+\pi)` is
`B`
Period | `= (2pi)/n` | `(n = 2)` |
`= (2pi)/2` | ||
`= pi` |
`=>B`
Consider the function \(g:R \to R, g(x)=2^x+5\).
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Let \(h:R\to R, h(x)=2^x-x^2\).
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a. \(5\)
b. \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
ci. \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)
\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
cii. \(y=4.255x\)
d. \((2.06 , -0.07)\)
e. \([0.49, 3.21]\)
f.
\begin{array} {|c|c|}
\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} & -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} & -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} & -0.773 \\
\hline
\end{array}
g. \(\text{See worked solution.}\)
h. \(n=e\)
a. \(\text{As }x\to -\infty,\ \ 2^x\to 0\)
\(\therefore\ 2^x+5\to 5\)
b. | \(g(x)\) | \(=2^x+5\) |
\(=\Big(e^{\log_{e}{2}}\Big)^x\) | ||
\(=e\ ^{x\log_{e}{2}}+5\) | ||
\(g'(x)\) | \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\) | |
\(=\log_{e}{2}\times 2^x\) | ||
\(\therefore\ k\) | \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\) |
\(y-(2^a+5)\) | \(=\log_{e}{2}\times2^a(x-a)\) |
\(\therefore\ y\) | \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\) |
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. | \(h(x)\) | \(=2^x-x^2\) |
\(h'(x)\) | \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\) | |
\(h”(x)\) | \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\) |
\(\text{Solving }h”(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)
\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)
\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)
f. \(\text{Newton’s Method }\Rightarrow\ x_a-\dfrac{h(x_a)}{h'(x_a)}\)
\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)
\(h(x)=2x-x^2\ \text{and }h'(x)=\ln{2}\times 2^x-2x\)
\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. \(\text{The denominator in Newton’s Method is}\ h'(x)=\log_{e}{(2)}\cdot 2^x-2x\)
\(\text{and the calculation will be undefined if }h'(x)=0\ \text{as the tangent lines are horizontal}.\)
\(\therefore\ \text{The solution to }h'(x)=0\ \text{cannot be used for }x_0.\)
h. \(\text{For a local minimum }f(x)=0\)
\(\rightarrow\ n^x-x^n=0\)
\(\rightarrow\ n^x=x^n\ \ \ (1)\)
\(\text{Also for a local minimum }f'(x)=0\)
\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)
\(\text{Substitute (1) into (2)}\)
\(\ln(n)\cdot x^n-nx^{n-1}=0\)
\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)
\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)
\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)
\(\therefore\ n=e\)
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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a. \((-1, 0), (0, 0), (2, 0)\)
b. \(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
c.i. \(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)
c.ii. \(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)
\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
c.iii. \(5.95\)
d. \(\text{1st case }\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)
\(\text{2nd case }\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)
a. \((-1, 0), (0, 0), (2, 0)\)
b. \(\text{Using CAS solve for}\ x:\)
\(\dfrac{d}{dx}(x(x-2)(x+1))=0\)
\(\therefore\ x=\dfrac{1-\sqrt{7}}{3}\ \text{and }x=\dfrac{1+\sqrt{7}}{3}\)
\(\text{Substitute }x\ \text{values into }f(x)\ \text{using CAS to get}\ y\ \text{values}\)
\(\text{The stationary points of }f\ \text{are}:\)
\(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
ci \(\text{Given }f(x)=g(x)\)
\(x(x-2)(x+1)\) | \(=x-2\) |
\(x(x-2)(x+1)(x-2)\) | \(=0\) |
\((x-2)(x(x+1)-1)\) | \(=0\) |
\((x-2)(x^2+x-1)\) | \(=0\) |
\(\therefore\ \text{Using CAS: } \)
\(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)
cii \(\text{Area of bounded region:}\)
\(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)
\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
ciii | \(\text{Solve the integral in c.ii above using CAS:}\) |
\(\text{Total area}=5.946045..\approx 5.95\) |
d. \(\text{Method 1 – Equating coefficients}\)
\((x-a)(x-b)^2=x(x-2)(x+1)+k\)
\(x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k\)
\((x^3-(a+2b)x^2+(2ab+b^2)x-ab^2=x^3-x^2-2x+k\)
\(\therefore\ -(a+2b)=-1\ \to\ a=1-2b …(1)\)
\(2ab+b^2=-2\ \ …(2)\)
\(\text{Substitute (1) into (2) and solve for }b.\)
\(2b(1-2b)+b^2\) | \(=-2\) |
\(3b^2-2b-2\) | \(=0\) |
\(b\) | \(=\dfrac{1\pm \sqrt{7}}{3}\) |
\(\text{When }b\) | \(=\dfrac{1+\sqrt{7}}{3}\) |
\(a\) | \(=1-2\Bigg(\dfrac{1+\sqrt{7}}{3}\Bigg)=\dfrac{-2\sqrt{7}+1}{3}\) |
\(\text{When }b\) | \(=\dfrac{1-\sqrt{7}}{3}\) |
\(a\) | \(=1-2\Bigg(\dfrac{1-\sqrt{7}}{3}\Bigg)=\dfrac{2\sqrt{7}+1}{3}\) |
\(\text{Method 2 – Using transformations}\)
\(\text{The squared factor in }(x-a)(x-b)^2=x(x-2)(x+1)+k,\)
\(\text{shows that the turning point is on the }x\ \text{axis}.\)
\(\therefore\ \text{Lowering }f(x)\ \text{by }\dfrac{2(7\sqrt{7}-10)}{27}\ \text{and raising }f(x)\ \text{by }\dfrac{2(7\sqrt{7}+10)}{27}\)
\(\text{will give the 2 possible sets of values for }a\ \text{and}\ b.\)
\(\text{1st case – lowering using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)-\dfrac{2(7\sqrt{7}-10)}{27}\)
\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)
\(\text{2nd case – raising using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)+\dfrac{2(7\sqrt{7}+10)}{27}\)
\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)
The following data shows the sizes of a sample of 20 oysters rated as small, medium or large.
\begin{array} {ccccc}
\text{small} & \text{small} & \text{large} & \text{medium} & \text{medium} \\
\text{medium} & \text{large} & \text{small} & \text{medium} & \text{medium}\\
\text{small} & \text{medium} & \text{small} & \text{small} & \text{medium}\\
\text{medium} & \text{medium} & \text{medium} & \text{small} & \text{large}
\end{array}
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i.
\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Size} \rule[-1ex]{0pt}{0pt} & \textbf{Number} & \textbf{Percentage (%) } \\
\hline
\rule{0pt}{2.5ex} \text{small} \rule[-1ex]{0pt}{0pt} & 7 & 35 \\
\hline
\rule{0pt}{2.5ex} \text{medium} \rule[-1ex]{0pt}{0pt} & 10 & 50 \\
\hline
\rule{0pt}{2.5ex} \text{large} \rule[-1ex]{0pt}{0pt} & 3 & 15 \\
\hline
\rule{0pt}{2.5ex} \textbf{Total} \rule[-1ex]{0pt}{0pt} & 20 & 100 \\
\hline
\end{array}
ii.
i.
\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Size} \rule[-1ex]{0pt}{0pt} & \textbf{Number} & \textbf{Percentage (%) } \\
\hline
\rule{0pt}{2.5ex} \text{small} \rule[-1ex]{0pt}{0pt} & 7 & 35 \\
\hline
\rule{0pt}{2.5ex} \text{medium} \rule[-1ex]{0pt}{0pt} & 10 & 50 \\
\hline
\rule{0pt}{2.5ex} \text{large} \rule[-1ex]{0pt}{0pt} & 3 & 15 \\
\hline
\rule{0pt}{2.5ex} \textbf{Total} \rule[-1ex]{0pt}{0pt} & 20 & 100 \\
\hline
\end{array}
ii.
The dot plot below shows the times, in seconds, of 40 runners in the qualifying heats of their 800 m club championship.
Question 1
The median time, in seconds, of these runners is
Question 2
The shape of this distribution is best described as
\(\text{Question 1:}\ B\)
\(\text{Question 2:}\ A\)
\(\text{Question 1}\)
\(\text{40 data points}\ \Rightarrow \ \text{Median = average of 20th and 21st data points}\)
\(\text{Median}\ = \dfrac{136 + 136}{2} = 136\)
\(\Rightarrow B\)
\(\text{Question 2}\)
\(\text{Distribution is positive skewed (tail stretches to the right)} \)
\(\text{Q}_1 = \dfrac{135+135}{2} = 135\)
\(\text{Q}_3 = \dfrac{138+138}{2} = 138\)
\(\text{IQR} = 138-135=3 \)
\(\text{Outlier (upper fence)}\ = 138+ 1.5 \times 3 = 142.5\)
\(\Rightarrow A\)
Why is pure copper preferred over a copper alloy in telecommunications applications?
\( B \)
→ Pure copper offers lower stiffness than copper alloys (eliminate A), is work hardened (eliminate C), and generally has lower strength to weight ratios (eliminate D).
→ In telecommunications, where high conductivity is crucial for transmitting electrical signals, pure copper is the preferred choice to ensure efficient signal transmission.
\(\Rightarrow B \)
Roller coaster support structures can be made from either timber or steel. Compare the properties of the two materials in roller coaster support structures. (2 marks) --- 4 WORK AREA LINES (style=lined) --- → Unlike steel, timber has the ability to flex and bend, which can absorb some of the forces exerted by the roller coaster. → However, timber has less mechanical strength than steel and is more susceptible to rot and insect damage over time. Answers could include: → Steel frames are much more easily fabricated and assembled than timber frames, which can save time and costs in the construction process. → Steel is much more resistant to fire than timber, which makes it a safer material to use in roller coasters. → Steel has greater mechanical strength and is a more durable material than timber. It is able to withstand the higher stresses and forces exerted on a roller coaster. → Unlike steel, timber has the ability to flex and bend, which can absorb some of the forces exerted by the roller coaster. → However, timber has less mechanical strength than steel and is more susceptible to rot and insect damage over time. Answers could include: → Steel frames are much more easily fabricated and assembled than timber frames, which can save time and costs in the construction process. → Steel is much more resistant to fire than timber, which makes it a safer material to use in roller coasters. → Steel has greater mechanical strength and is a more durable material than timber. It is able to withstand the higher stresses and forces exerted on a roller coaster.
You are part of a team of engineers working collaboratively on the design of a new aircraft. Explain the benefits of collaboration when completing the engineering report. (3 marks) --- 6 WORK AREA LINES (style=lined) --- → A collaborative approach allows for the pooling of expert knowledge. This will produce a more comprehensive report with a lower risk of errors within the report. → Time efficiency – a division of tasks among team members that utilises their specific skill sets can expedite the writing of the report. → Validation of decisions across multiple team members who can independently verify and validate design calculations and recommendations is a more comprehensive approach. → Collaboration between engineers with differing years of experience is crucial for professional development and training the next generation of engineers. → A collaborative approach allows for the pooling of expert knowledge. This will produce a more comprehensive report with a lower risk of errors within the report. → Time efficiency – a division of tasks among team members that utilises their specific skill sets can expedite the writing of the report. → Validation of decisions across multiple team members who can independently verify and validate design calculations and recommendations is a more comprehensive approach. → Collaboration between engineers with differing years of experience is crucial for professional development and training the next generation of engineers.
How can computer graphics be utilised as a tool in aeronautical engineering? (2 marks) --- 4 WORK AREA LINES (style=lined) --- → Computer graphics can create a detailed and accurate visual models in three dimensions. → This technology can be extremely efficient in the design of aircraft components through simulation, allowing ideas to be tested and adjusted in short time frames. Answers could also include: → Computer graphics can create a detailed and accurate visual models in three dimensions. → This technology can be extremely efficient in the design of aircraft components through simulation, allowing ideas to be tested and adjusted in short time frames. Answers could also include:
A model car of mass 2.0 kg is propelled from rest by a rocket motor that applies a constant horizontal force of 4.0 N, as shown below. Assume that friction is negligible.
Which one of the following best gives the magnitude of the acceleration of the model car?
\(C\)
\(a\) | \(=\dfrac{F}{m}\) | |
\(=\dfrac{4.0}{2.0}\) | ||
\(=2\ \text{ms}^{-2}\) |
\(\Rightarrow C\)
The diagram below shows the electric field lines between four charged spheres: \(\text{P, Q, R}\) and \(\text{S}\). The magnitude of the charge on each sphere is the same.
Which of the following correctly identifies the type of charge (+ positive or – negative) that resides on each of the spheres \(\text{P, Q, R}\) and \(\text{S}\)?
\(\textbf{P}\) | \(\textbf{Q}\) | \(\textbf{R}\) | \(\textbf{S}\) | |
A. | \(\quad - \quad\) | \(\quad + \quad\) | \(\quad - \quad\) | \(\quad + \quad\) |
B. | \(\quad + \quad\) | \(\quad - \quad\) | \(\quad + \quad\) | \(\quad - \quad\) |
C. | \(\quad - \quad\) | \(\quad - \quad\) | \(\quad + \quad\) | \(\quad + \quad\) |
D. | \(\quad + \quad\) | \(\quad + \quad\) | \(\quad - \quad\) | \(\quad - \quad\) |
\(B\)
→ Electric field lines travel away from positive charges and towards negative charges.
→ Therefore, \(\text{P}\) and \(\text{R}\) must be positive charges.
\( \Rightarrow B\)
What effect does a catalyst have on a reaction?
\(A\)
→ A catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
\(\Rightarrow A\)
A catalyst
\(D\)
→ A catalyst increases the rate of a chemical reaction by reducing the activation energy of the reaction. Thus, by collision theory, particles require less energy than normal to react to form products.
\(\Rightarrow D\)
Standard solutions of sodium hydroxide, \(\ce{NaOH}\), must be kept in airtight containers. This is because \(\ce{NaOH}\) is a strong base and absorbs acidic oxides, such as carbon dioxide, \(\ce{CO2}\), from the air and reacts with them. As a result, the concentration of \(\ce{NaOH}\) is changed to an unknown extent.
\(\ce{CO2}\) in the air reacts with water to form carbonic acid, \(\ce{H2CO3}\). This can react with \(\ce{NaOH}\) to form sodium carbonate, \(\ce{Na2CO3}\).
--- 1 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
i. \(\ce{CO2(g) + H2O(l) \rightarrow H2CO3(aq)} \)
ii. \(\ce{2NaOH(aq) + H2CO3(aq) \rightarrow Na2CO3(aq) + 2H2O(l)}\)
i. \(\ce{CO2(g) + H2O(l) \rightarrow H2CO3(aq)} \)
ii. \(\ce{2NaOH(aq) + H2CO3(aq) \rightarrow Na2CO3(aq) + 2H2O(l)}\)
A pendulum is used to determine the value of acceleration due to gravity. The length of the pendulum is varied, and the time taken for the same number of oscillations is recorded.
Which of the following could increase the reliability of the results?
\(D\)
→ Reliability of data can be increased by repeating the same experiment numerous times and recording an average from the data as it eliminates random error.
\(\Rightarrow D\)
What is the safest method for disposing of a liquid hydrocarbon after an experiment?
\(D\)
→ Organic substances should be kept separate for safety (avoiding any possible other reactions).
\(\Rightarrow D\)
The gravitational field strength acting on a spacecraft decreases as its altitude increases.
This is due to a change in the
\(D\)
\(g=\dfrac{GM}{r^2}\ \ \Rightarrow\ \ g \propto \dfrac{1}{r^2}\)
\(\therefore\) As r increases, the gravitational field strength decreases
\( \Rightarrow D\)
Find the angle between the vectors
\(\underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k}\)
\(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\),
giving your answer to the nearest degree. (3 marks)
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\(87^{\circ} \)
\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right),\ \ \underset{\sim}{b}=\left(\begin{array}{c} -1 \\ 4 \\ 2 \end{array}\right) \]
\(\Big{|} \underset{\sim}{a} \Big{|} = \sqrt{1+4+9} = \sqrt{14} \)
\(\Big{|} \underset{\sim}{b} \Big{|} = \sqrt{1+16+4} = \sqrt{21} \)
\( \underset{\sim}{a} \cdot \underset{\sim}{b} = -1 + 8-6=1 \)
\(\cos\ \theta \) | \(=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\Big{|}\underset{\sim}{a}\Big{|} \cdot \Big{|}\underset{\sim}{b}\Big{|}} \) | |
\(=\dfrac{1}{\sqrt{294}} \) | ||
\( \theta\) | \(=\cos ^{-1} \Big{(}\dfrac{1}{\sqrt{294}}\Big{)} \) | |
\(=86.65…\) | ||
\(=87^{\circ} \) |
The temperature \(T(t)^{\circ} \text{C}\) of an object at time \(t\) seconds is modelled using Newton's Law of Cooling,
\(T(t)=15+4 e^{-3 t}\)
What is the initial temperature of the object?
\(D\)
\(\text{Initial temperature when}\ \ t=0:\)
\(T=15+4e^0=19\)
\(\Rightarrow D\)
Which of the following is equal to \((a+i b)^3\)?
\(C\)
\((a+i b)^3\) | \(=a^3+3a^2ib+3a(ib)^2+(ib)^3\) | |
\(=a^3+3a^2ib-3ab^2-ib^3\) | ||
\(=(a^3-3ab^2)+i(3a^2b-b^3) \) |
\(\Rightarrow C\)
The first three terms of an arithmetic sequence are 3, 7 and 11 .
Find the 15th term. (2 marks)
`59`
`a=T_1=3`
`d=T_2-T_1=7-3=4`
`T_15` | `=a+14xxd` | |
`=3+14xx4` | ||
`=59` |