Some isomers with the formula \( \ce{C4H8O} \) are shown.
Name ONE pair of functional group isomers and ONE pair of chain isomers from the structures above. (2 marks)
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Some isomers with the formula \( \ce{C4H8O} \) are shown.
Name ONE pair of functional group isomers and ONE pair of chain isomers from the structures above. (2 marks)
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Functional Group: Butan-2-one and butanal
Chain: Butanal and 2-methylpropanal
Functional Group: Butan-2-one and butanal
Chain: Butanal and 2-methylpropanal
The diagram shows apparatus that is used to investigate the interaction between the magnetic field produced by a coil and two copper rings \(X\) and \(Y\), when each is placed at position \(P\), as shown. Ring \(X\) is a complete circular ring, and a small gap has been cut in ring \(Y\). Each of the rings has a cross-sectional area of 4 × 10\(^{-4}\) m². The power supply connected to the coil produces an increasing current through the coil in the direction shown, when the switch is turned on. This produces a magnetic field at \(P\) that varies as shown in the graph.
--- 7 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. Between 0 – 0.03 seconds: → The magnetic field strength at point \(P\) is increasing at a constant rate. → Thus, ring \(X\) experiences a change in flux, which causes an EMF to be induced across the ring (Faraday’s Law) → This induced EMF causes a current to flow through ring \(X\) that will interact in the external magnetic field to reduce the change in flux that created it (Lenz’s Law). → Thus, a clockwise current will run through the ring, creating a north pole towards the solenoid, causing the ring to have a repulsive force away from the solenoid acting on it. Between 0.03 – 0.05 seconds: → There is no change in flux through the ring due to there being a constant magnetic field. → Therefore, there is no induced EMF or induced current. → Therefore, there is no force acting on the ring. b.i. Ring \(Y\) behaviour when placed at \(P\): → Ring \(Y\) will experience the same change in flux and hence the same induced EMF as ring \(X\) within 0 – 0.03 seconds. → However, due to the gap in ring \(Y\) (i.e. there is no closed circuit), no induced current will be able to flow through the ring. → Hence, there is no force exerted on ring \(Y\). ii. \(\varepsilon = 8 \times 10^{-5}\ \text{V}\) a. Between 0 – 0.03 seconds: → The magnetic field strength at point \(P\) is increasing at a constant rate. → Thus, ring \(X\) experiences a change in flux, which causes an EMF to be induced across the ring (Faraday’s Law) → This induced EMF causes a current to flow through ring \(X\) that will interact in the external magnetic field to reduce the change in flux that created it (Lenz’s Law). → Thus, a clockwise current will run through the ring, creating a north pole towards the solenoid, causing the ring to have a repulsive force away from the solenoid acting on it. Between 0.03 – 0.05 seconds: → There is no change in flux through the ring due to there being a constant magnetic field. → Therefore, there is no induced EMF or induced current. → Therefore, there is no force acting on the ring. b.i. Ring \(Y\) behaviour when placed at \(P\): → Ring \(Y\) will experience the same change in flux and hence the same induced EMF as ring \(X\) within 0 – 0.03 seconds. → However, due to the gap in ring \(Y\) (i.e. there is no closed circuit), no induced current will be able to flow through the ring. → Hence, there is no force exerted on ring \(Y\).
b.ii.
\(\varepsilon\)
\(=N\dfrac{\Delta \phi}{\Delta t}\)
\(=N \dfrac{A \Delta B}{\Delta t}\)
\(= 1 \times \dfrac{4 \times 10^{-4} \times (6 \times 10^{-3}-0)}{0.03-0}\)
\(=8 \times 10^{-5}\ \text{V}\)
Huntington's disease is caused by a misfolded protein 'Huntingtin'. It is caused by excess repeats of the DNA sequence CAG on the coding strand of DNA. The mRNA that is produced has the same sequence as the DNA. Use the codon chart, starting in the centre, to identify the amino acid that is repeated. (1 mark) → Glutamine → Glutamine
An ideal transformer is connected to a 240 V AC supply. It has 300 turns on the primary coil and 50 turns on the secondary coil. It is connected in the circuit with two identical light globes, \(X\) and \(Y\), as shown. --- 4 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) --- a. 40 V b. Closed switch \(\Rightarrow \) globe \(X\) and \(Y\) are connected in a parallel circuit. → Once the switch is closed, the total resistance through the circuit is less than when only light globe \(X\) is in the circuit. → Since \(V=IR\ \ \Rightarrow\ \ R \propto \dfrac{1}{I} \). Therefore, a decrease in resistance through the circuit leads to an increase in the current. → Transformer is ideal, so Power in = Power out ( \(V_{p}I_{p}=V_{s}I_{s}\) ). Hence, an increase in \(I_s\) corresponds to an increase in \(I_p\). → Therefore, the current in the primary coil is greater when the switch is closed. b. Closed switch \(\Rightarrow \) globe \(X\) and \(Y\) are connected in a parallel circuit. → Once the switch is closed, the total resistance through the circuit is less than when only light globe \(X\) is in the circuit. → Since \(V=IR\ \ \Rightarrow\ \ R \propto \dfrac{1}{I} \). Therefore, a decrease in resistance through the circuit leads to an increase in the current. → Transformer is ideal, so Power in = Power out ( \(V_{p}I_{p}=V_{s}I_{s}\) ). Hence, an increase in \(I_s\) corresponds to an increase in \(I_p\). → Therefore, the current in the primary coil is greater when the switch is closed.
a.
\( \dfrac{V_p}{V_s}\)
\(= \dfrac{N_p}{N_s}\)
\( V_s\)
\(=V_p \times \dfrac{N_s}{N_p}\)
\(=240 \times \dfrac{50}{300}\)
\(=40\ \text{V} \)
→ The voltage across light globe \(X\) is 40 V.
The James Webb Space Telescope observed an exoplanet emitting a peak wavelength of 1.14 \(\times\) 10\(^{-5}\) m. Calculate the temperature of the exoplanet. (2 marks) --- 4 WORK AREA LINES (style=lined) --- \(T= 254\) \( \text{K}\)
\(T\)
\(=\dfrac{b}{\lambda_{\text{max}}}\)
\(=\dfrac{2.898\times 10^{-3}}{1.14 \times 10^{-5}}\)
\(= 254\) \( \text{K}\)
The James Webb Space Telescope (JWST) has a mass of 6.1 × 10³ kg and orbits the Sun at a distance of approximately 1.52 × 10\(^{11}\) m. The Sun has a mass of 1.99 × 10\(^{30}\) kg. Calculate the magnitude of gravitational force the Sun exerts on the JWST. (2 marks) --- 4 WORK AREA LINES (style=lined) --- \(F= 35\) \( \text{N}\)
\(F\)
\(=\dfrac{GMm}{r^2}\)
\(=\dfrac{ 6.67 \times 10^{-11} \times 1.99 \times 10^{30} \times 6.1 \times 10^3}{(1.52 \times 10^{11} )^2}\)
\(=35\) \(\text{N}\)
A Hertzsprung–Russell diagram is shown. --- 4 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- a. The luminosity of a star can be determined by its size and temperature. Other variables could include: → Mass, colour and the power output of a star. b. Differences: → Star A is a main sequence star and is therefore fusing hydrogen to helium in its core via both the proton-proton chain and CNO cycle whereas Star B is a white dwarf star and therefore has no fusion taking place in its core. → Star A has a greater luminosity compared to Star B. → Star A is a younger star then Star B which is at the end of its lifecycle. Other differences could include: → Mass → Radius or size a. The luminosity of a star can be determined by its size and temperature. Other variables could include: → Mass, colour and the power output of a star. b. Differences between stars: → Star A is a main sequence star and is therefore fusing hydrogen to helium in its core via both the proton-proton chain and CNO cycle whereas Star B is a white dwarf star and therefore has no fusion taking place in its core. → Star A has a greater luminosity compared to Star B. → Star A is a younger star then Star B which is at the end of its lifecycle. Other differences could include: → Mass → Radius or size
Identify the components of a nucleotide. (1 mark) → A 5-carbon sugar, a phosphate group and a base pair. → A 5-carbon sugar, a phosphate group and a base pair.
An experiment was conducted to investigate the rate of binary fission in E.coli. The results of the experiment are shown.
Time (minutes) | Number of E.coli |
0 | 20 |
20 | 40 |
40 | 80 |
60 | 160 |
80 | 320 |
100 | 640 |
Which graph represents the data in the table?
\(B\)
By Elimination:
→ The independent variable (time) must be on the \(x\)-axis (Eliminate C and D).
→ The number of E. Coli starts at 20, not 0 (Eliminate A).
\(\Rightarrow B\)
A Punnett square is shown.
\(\text{B}\) | \(\text{b}\) | |
\(\text{B}\) | \(1\) | \(2\) |
\(\text{B}\) | \(3\) | \(4\) |
Which of the following options represents heterozygous offspring?
\(C\)
→ A heterozygous genotype is given by Bb, where-as a homozygous genotype is given by BB or bb.
→ Only options 2 and 4 show this.
\(\Rightarrow C\)
A cochlear implant is a device that is used to assist with hearing loss.
What does the cochlear implant electrode array stimulate?
\(D\)
→ The cochlear implant is able to bypass a dysfunctional cochlear and directly stimulate the auditory nerve.
\(\Rightarrow D\)
A projectile of mass \(M\) kg is launched vertically upwards from the origin with an initial speed \(v_0\) m s\(^{-1}\). The acceleration due to gravity is \( {g}\) ms\(^{-2}\).
The projectile experiences a resistive force of magnitude \(kMv^2\) newtons, where \(k\) is a positive constant and \(v\) is the speed of the projectile at time \(t\) seconds.
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i. \(\text{Taking up as positive:}\)
\(M\ddot x\) | \(=-Mg-kMv^2\) | |
\(\ddot x\) | \(=-g-kv^2\) | |
\(v \cdot \dfrac{dv}{dx}\) | \(=-(g+kv^2) \) | |
\(\dfrac{dv}{dx}\) | \(=-\dfrac{g+kv^2}{v} \) | |
\(\dfrac{dx}{dv}\) | \(=-\dfrac{v}{g+kv^2} \) | |
\(x\) | \(=-\dfrac{1}{2k} \displaystyle \int \dfrac{2kv}{g+kv^2}\, dv \) | |
\(=-\dfrac{1}{2k} \ln |g+kv^2|+c \) |
\(\text{When}\ \ x=o, \ v=v_0: \)
\(c=\dfrac{1}{2k} \ln |g+kv_0^2| \)
\(x\) | \(=\dfrac{1}{2k} \ln |g+kv_0^2|-\dfrac{1}{2k} \ln |g+kv^2| \) | |
\(=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g+kv_0^2}{g+kv^2} \Bigg{|} \) |
\(\text{When}\ \ v=0, x=H: \)
\(H=\dfrac{1}{2k} \ln \Bigg{(} \dfrac{g+kv_0^2}{g} \Bigg{)},\ \ \ \ (k>0) \)
ii. \(\text{When projectile travels downward:} \)
\(M \ddot x\) | \(=Mg-kMv^2\) | |
\(\ddot x\) | \(=g-kv^2\) | |
\(v \cdot \dfrac{dv}{dx}\) | \(=g-kv^2\) | |
\(\dfrac{dx}{dv}\) | \(=\dfrac{v}{g-kv^2}\) | |
\(x\) | \(=-\dfrac{1}{2k} \displaystyle \int \dfrac{-2kv}{g-kv^2}\,dv \) | |
\(=-\dfrac{1}{2k} \ln|g-kv^2|+c \) |
\(\text{When}\ \ x=0, \ v=0: \)
\(c=\dfrac{1}{2k} \ln g \)
\(x=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g}{g-kv^2} \Bigg{|} \)
\(\text{When}\ \ x=H, \ v=v_1: \)
\(\dfrac{1}{2k} \ln \Bigg{(} \dfrac{g+kv_0^2}{g} \Bigg{)}\) | \(=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g}{g-kv_1^2} \Bigg{|} \) | |
\(\dfrac{g+kv_0^2}{g} \) | \(=\dfrac{g}{g-kv_1^2} \) | |
\(g^2\) | \(=(g+kv_0^2)(g-kv_1^2) \) | |
\(g^2\) | \(=g^2-gkv_1^2+gkv_0^2-k^2v_0^2v_1^2 \) | |
\(gkv_0^2-gkv_1^2 \) | \(=k^2v_0^2v_1^2 \) | |
\(gk(v_0^2-v_1^2) \) | \(=k^2v_0^2 v_1^2 \) | |
\(g(v_0^2-v_1^2) \) | \(=kv_0^2v_1^2 \) |
Which of the following eukaryotic organelles is correctly matched with its function?
\(\textbf{Organelle}\) | \(\textbf{Function}\) | |
\(\text{A.}\) | Golgi body | mRNA production |
\(\text{B.}\) | Mitochondria | ATP synthesis |
\(\text{C.}\) | Ribosome | Waste storage |
\(\text{D.}\) | Lysosome | Movement |
\(B\)
By Elimination
→ Golgi bodies are involved in the modification, storage and distribution of the lipids, carbohydrates and proteins made in the E.R. (Eliminate A).
→ Ribosomes carry out protein synthesis (Eliminate C).
→ Lysosomes are involved in solid waste processing (Eliminate D).
\(\Rightarrow B\)
Let \(z\) be the complex number \(z=e^{\small{\dfrac{i \pi}{6}}} \) and \(w\) be the complex number \(w=e^{\small{\dfrac{3 i \pi}{4}}} \). --- 6 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- i. \(z=e^{\small\dfrac{i \pi}{6}} = \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i \) \(w=e^{\small\dfrac{3i \pi}{4}} = \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} \) \(\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} \) \( |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} \) \(\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC \) \(\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB \) \(\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} \) iii. \(\text{In}\ \triangle AOC: \) \( \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} \) \(\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. \) \(\text{Using the cos rule in}\ \triangle AOC: \)
\(|z+w|^2\)
\(=\Bigg{|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{|}\)
\(=\Bigg{|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{|}\)
\(=\Bigg{|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{|}\)
\(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)
\(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)
\(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)
\(=\dfrac{4-\sqrt6+\sqrt2}{2} \)
\(\cos\,\dfrac{7\pi}{24}\)
\(=\dfrac{|z|^2+|z+w|^2-|w|^2}{2|z||z+w|}\)
\(=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1 \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} \)
\(=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} \)
\(=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} \)
\(=\dfrac{8-2\sqrt6+2\sqrt2}{4} \)
A particle of mass 1 kg is projected from the origin with speed 40 m s\( ^{-1}\) at an angle 30° to the horizontal plane. --- 6 WORK AREA LINES (style=lined) --- The forces acting on the particle are gravity and air resistance. The air resistance is proportional to the velocity vector with a constant of proportionality 4 . Let the acceleration due to gravity be 10 m s \( ^{-2}\). The position vector of the particle, at time \(t\) seconds after the particle is projected, is \(\mathbf{r}(t)\) and the velocity vector is \(\mathbf{v}(t)\). --- 12 WORK AREA LINES (style=lined) --- --- 10 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) --- i. \(\underset{\sim}{v}(0)={\displaystyle\left(\begin{array}{cc} 40 \cos\ 30° \\ 40 \sin\ 30°\end{array}\right)} = {\displaystyle\left(\begin{array}{cc} 40 \times \frac{\sqrt3}{2} \\ 40 \times \frac{1}{2}\end{array}\right)} = {\displaystyle\left(\begin{array}{cc}20 \sqrt{3} \\ 20\end{array}\right)} \) ii. \(\text{Air resistance:} \) \(\underset{\sim}{F} = -4\underset{\sim}{v} = {\displaystyle\left(\begin{array}{cc} -4\dot{x} \\ -4\dot{y} \end{array}\right)} \) \(\text{Horizontally:}\) iv. \(\text{Range}\ \Rightarrow\ \text{Find}\ \ t\ \ \text{when}\ \ y=0: \) \(\Rightarrow \text{Solution when}\ \ t\approx 2.25\)
\(1 \times \ddot{x} \)
\(=-4 \dot{x} \)
\(\dfrac{d\dot{x}}{dt}\)
\(=-4\dot{x}\)
\(\dfrac{dt}{d\dot{x}}\)
\(= -\dfrac{1}{4\dot{x}} \)
\(t\)
\(=-\dfrac{1}{4} \displaystyle \int \dfrac{1}{\dot{x}} \ d\dot{x} \)
\(-4t\)
\(=\ln |\dot{x}|+c \)
\(\text{When}\ \ t=0, \ \dot{x}=20\sqrt3 \ \ \Rightarrow\ \ c=-\ln{20\sqrt3} \)
\(-4t\)
\(=\ln|\dot{x}|-\ln 20\sqrt3 \)
\(-4t\)
\(=\ln\Bigg{|}\dfrac{\dot{x}}{20\sqrt{3}} \Bigg{|} \)
\(\dfrac{\dot{x}}{20\sqrt{3}} \)
\(=e^{-4t} \)
\(\dot{x}\)
\(=20\sqrt{3}e^{-4t}\)
\(\text{Vertically:} \)
\(1 \times \ddot{y} \)
\(=-1 \times 10-4 \dot{y} \)
\(\dfrac{d\dot{y}}{dt}\)
\(=-(10+4\dot{y})\)
\(\dfrac{dt}{d\dot{y}}\)
\(= -\dfrac{1}{10+4\dot{y}} \)
\(t\)
\(=- \displaystyle \int \dfrac{1}{10+4\dot{y}} \ d\dot{y} \)
\(-4t\)
\(=- \displaystyle \int \dfrac{4}{10+4\dot{y}} \ d\dot{y} \)
\(-4t\)
\(=\ln |10+4\dot{y}|+c \)
\(\text{When}\ \ t=0, \ \dot{y}=20 \ \ \Rightarrow\ \ c=-\ln{90} \)
\(-4t\)
\(=\ln|10+4\dot{y}|-\ln 90 \)
\(-4t\)
\(=\ln\Bigg{|}\dfrac{10+4\dot{y}}{\ln{90}} \Bigg{|} \)
\(\dfrac{10+4\dot{y}}{90} \)
\(=e^{-4t} \)
\(4\dot{y}\)
\(=90e^{-4t}-10\)
\(\dot{y}\)
\(=\dfrac{45}{2} e^{-4t}-\dfrac{5}{2} \)
\(\therefore \underset{\sim}v={\displaystyle \left(\begin{array}{cc}20 \sqrt{3} e^{-4 t} \\ \dfrac{45}{2} e^{-4 t}-\dfrac{5}{2}\end{array}\right)}\)
iii. \(\text{Horizontally:}\)
\(x\)
\(= \displaystyle \int \dot{x}\ dx\)
\(= \displaystyle \int 20\sqrt3 e^{-4t}\ dt \)
\(=-5\sqrt3 e^{-4t}+c \)
\(\text{When}\ \ t=0, \ x=0\ \ \Rightarrow\ \ c=5\sqrt3 \)
\(x\)
\(=5\sqrt3-5\sqrt3 e^{-4t} \)
\(=5\sqrt3(1-e^{-4t}) \)
\(\text{Vertically:}\)
\(y\)
\(= \displaystyle \int \dot{y}\ dx\)
\(= \displaystyle \int \dfrac{45}{2} e^{-4t}-\dfrac{5}{2}\ dt \)
\(=-\dfrac{45}{8}e^{-4t}-\dfrac{5}{2}t+c \)
\(\text{When}\ \ t=0, \ y=0\ \ \Rightarrow\ \ c= \dfrac{45}{8} \)
\(y\)
\(=\dfrac{45}{8}-\dfrac{45}{8} e^{-4t}-\dfrac{5}{2}t \)
\(=\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2} \)
\(\therefore \underset{\sim}{r}=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \dfrac{45}{8}\left(1-e^{-4 t}\right)-\dfrac{5}{2} t\end{array}\right)\)
\(\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2}t \)
\(=0\)
\(\dfrac{45}{8}(1-e^{-4t}) \)
\(=\dfrac{5}{2}t \)
\(1-e^{-4t}\)
\(=\dfrac{4}{9}t \)
\(\text{Graph shows intersection of these two graphs.}\)
\(\therefore\ \text{Range}\)
\(=5\sqrt3(1-e^{(-4 \times 2.25)}) \)
\(=8.659…\)
\(=8.7\ \text{metres (to 1 d.p.)}\)
--- 3 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) --- i. \(\text{Proof (See Worked Solutions)} \) ii. \(\text{Proof (See Worked Solutions)} \) \(\text{Vertex (min) at}\ (1,-4) \) \(\text{Quadratic is monotonically increasing for}\ \ k \geq1 \) \(\text{At}\ \ k=3, \ k^2-2k-3=0 \) \( \therefore\ \ k^2-2 k-3 \geq 0\ \ \text{for}\ \ k \geq 3\) ii. \(\text{Prove}\ \ 2^n \geq n^2-2 \) \(\text{If}\ \ n=3: \) \(\text{LHS}\ = 2^3=8 \) \(\text{RHS}\ =3^3-2=7 \leq \text{LHS} \) \(\therefore \ \text{True for}\ \ n=3. \) \(\text{Assume true for}\ \ n=k: \) \(2^k \geq k^2-2 \ \ \ …\ (*) \) \(\text{Prove true for}\ \ n=k+1: \) \(\text{i.e.}\ \ 2^{k+1} \geq (k+1)^2-2 \) \(\therefore \text{Since true for}\ \ n=3,\ \text{by PMI, true for integers}\ \ n\geq3 \)
i.
\(k^2-2k-3\)
\(=0\)
\( (k-3)(k+1) \)
\(=0\)
\(\text{LHS}\)
\(=2^{k+1} \)
\(=2 \cdot 2^{k} \)
\( \geq 2(k^2-2) \ \ \ \text{(see (*) above)} \)
\( \geq 2k^2-4 \)
\( \geq k^2 + \underbrace{k^2-2k-3}_{\geq 0\ \ \text{(see part (i))}} +2k-1 \)
\(\geq k^2+2k-1 \)
\(\geq k^2+2k+1-2 \)
\(\geq (k+1)^2-2 \)
\(\Rightarrow \text{True for}\ \ n=k+1 \)
The complex number \(2+i\) is a zero of the polynomial
\(P(z)=z^4-3 z^3+c z^2+d z-30\)
where \(c\) and \(d\) are real numbers.
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i. \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)
\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)
ii. \(\text{Remaining zeros:}\ \ -3, 2 \)
i. \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)
\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)
ii. \(P(z)=z^4-3 z^3+c z^2+d z-30\)
\(\text{Let roots be:}\ \ 2+i, 2-i, \alpha, \beta \)
\( \sum\ \text{roots:}\)
\(2+i+2-i+\alpha + \beta\) | \(=-\dfrac{b}{a} \) | |
\(4+\alpha+\beta\) | \(=3\) | |
\(\alpha + \beta\) | \(=-1\ \ \ …\ (1) \) |
\(\text{Product of roots:} \)
\((2+i)(2-i)\alpha\beta \) | \(= \dfrac{e}{a} \) | |
\(5\alpha\beta\) | \(=-30\) | |
\(\alpha \beta \) | \(=-6\ \ \ …\ (2) \) |
\(\text{Substitute}\ \ \beta=-\alpha-1\ \ \text{into (2):} \)
\(\alpha(-\alpha-1) \) | \(=-6 \) | |
\(-\alpha^2-\alpha \) | \(=-6\) | |
\(\alpha^2+\alpha-6\) | \(=0\) | |
\( (\alpha+3)(\alpha-2) \) | \(=0\) |
\(\therefore\ \text{Remaining zeros:}\ \ -3, 2 \)
\(C\)
→ Energy from power stations passes through a step-up transformer to increase the voltage and decrease the current.
→ This energy then passes through a step-down transformer to be used in homes.
\(\Rightarrow C\)
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
\( \Rightarrow D\)
Caesium-137 has a half-life of 30 years.
What mass of caesium-137 will remain after 90 years, if the initial mass was 120 g?
\(B\)
\(t_\frac{1}{2} = \text{30 years,}\ \ \lambda =\dfrac{\ln2}{30}\)
\(N\) | \(=N_{0}e^{-\lambda t} \) | |
\(=120e^{-\dfrac{\ln2}{30} \times 90} \) | ||
\(=15\ \text{g}\) |
\(\Rightarrow B\)
Find \({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\). (4 marks)
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\(-\ln3 \)
\( \text{Let}\ \ \dfrac{5x-3}{(x+1)(x-3)} = \dfrac{A}{x+1}+\dfrac{B}{x-3}\ \ \ \text{for}\ \ A, B \in \mathbb{R} \)
\(\dfrac{5x-3}{(x+1)(x-3)}\) | \(=\dfrac{A}{x+1}+\dfrac{B}{x-3} \) | |
\(=\dfrac{A(x-3)+B(x+1)}{(x+1)(x-3)} \) |
\(\text{Equating numerators:}\)
\( A(x-3)+B(x+1)=5x-3 \)
\(\text{When}\ \ x=3, 4B=12\ \Rightarrow \ B=3 \)
\(\text{When}\ \ x=-1, -4A=-8\ \Rightarrow \ A=2 \)
\({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\) | \( =\displaystyle \int_0^2 \dfrac{2}{x+1}+\dfrac{3}{x-3}\ dx \) | |
\(= \Big{[} 2\ln|x+1|+3\ln|x-3| \Big{]}_0^2 \) | ||
\(= 2\ln3+3\ln1-2\ln1-3\ln3 \) | ||
\(=-\ln3 \) |
A particle moves in simple harmonic motion described by the equation
\( \ddot{x}=-9(x-4) . \)
Find the period and the central point of motion. (2 marks)
\(\text{Period}\ = \dfrac{2 \pi}{3} \)
\(\text{Centre of Motion:}\ x=4 \)
\( \ddot{x}=-9(x-4) \)
\( \Rightarrow\ n=3,\ \ c=4 \)
\(\text{Period}\ = \dfrac{2 \pi}{3} \)
\(\text{Centre of Motion:}\ x=4 \)
Find a vector equation of the line through the points \(A(-3,1,5)\) and \(B(0,2,3)\). (2 marks)
\[\underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]
\[\overrightarrow{AB}=\left(\begin{array}{c} 0-(-3) \\2-1 \\ 3-5 \end{array}\right) = \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right)\]
\[\therefore \underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]
Solve the quadratic equation
\(z^2-3 z+4=0\)
where \(z\) is a complex number. Give your answers in Cartesian form. (2 marks)
\(\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \)
\(z^2-3 z+4=0\)
\(z\) | \(=\dfrac{3 \pm \sqrt{(-3)^2-4 \times 1 \times 4}}{2} \) | |
\(=\dfrac{3 \pm \sqrt{-7}}{2} \) | ||
\(=\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \) |
A hemispherical water tank has radius \(R\) cm. The tank has a hole at the bottom which allows water to drain out. Initially the tank is empty. Water is poured into the tank at a constant rate of \(2 k R\) cm³ s\(^{-1}\), where \(k\) is a positive constant. After \(t\) seconds, the height of the water in the tank is \(h\) cm, as shown in the diagram, and the volume of water in the tank is \(V\) cm³. It is known that \(V= \pi \Big{(} R h^2-\dfrac{h^3}{3}\Big{)}. \) (Do NOT prove this.) While water flows into the tank and also drains out of the bottom, the rate of change of the volume of water in the tank is given by \(\dfrac{d V}{d t}=k(2 R-h)\). --- 5 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- --- 9 WORK AREA LINES (style=lined) --- i. \(V=\pi \Big{(}Rh^2-\dfrac{h^3}{3} \Big{)} \) \(\dfrac{dV}{dh} = \pi(2Rh-h^2) \) \(\dfrac{dV}{dt} = k(2R-h)\ \ \ \text{(given)} \) \(\Rightarrow c=0 \) \( t= \dfrac{\pi h^2}{2k} \) \( T= \dfrac{\pi R^2}{2k}\ \text{seconds} \) iii. \(\text{Net water flow}\ = k(2R-h)\ \ \text{(given)} \) \(\text{Flow in}\ =2kR\ \ \text{(given)} \) \(\text{Flow out}\ = k(2R-h)-2kR=-kh \) \( \dfrac{dh}{dt}= \dfrac{-kh}{\pi h(2R-h)} = \dfrac{-k}{\pi (2R-h)} \)
\(\dfrac{dh}{dt}\)
\(= \dfrac{dV}{dt} \cdot \dfrac{dh}{dV} \)
\(=k(2R-h) \cdot \dfrac{1}{\pi} \cdot \dfrac{1}{h(2R-h)} \)
\(= \dfrac{k}{\pi h} \)
ii. \(\dfrac{dt}{dh} = \dfrac{\pi h}{k} \)
\(t\)
\(= \displaystyle \int \dfrac{dt}{dh}\ dh \)
\(= \dfrac{\pi}{k} \displaystyle \int h\ dh \)
\(= \dfrac{\pi}{k} \Big{[} \dfrac{h^2}{2} \Big{]} +c \)
\(\text{When}\ \ t=0, h=0 \)
\(\text{Tank is full at time}\ T\ \text{when}\ \ h=R: \)
\(\dfrac{dt}{dh}\)
\(=\dfrac{- \pi (2R-h)}{k} \)
\( \displaystyle \int k\ dt\)
\(=- \pi \displaystyle \int (2R-h)\ dh \)
\(kt\)
\(=- \pi \Big{(} 2Rh-\dfrac{h^2}{2} \Big{)}+c \)
\(\text{When}\ \ t=0, \ h=R: \)
\(0\)
\(=- \pi \Big{(}2R^2-\dfrac{R^2}{2} \Big{)} + c\)
\(c\)
\(= \pi \Big{(} \dfrac{3R^2}{2} \Big{)} \)
\(\text{Find}\ t\ \text{when}\ h=0: \)
\(kt\)
\(=- \pi(0) + \pi \dfrac{3R^2}{2} \)
\(t\)
\(= \dfrac{3 \pi R^2}{2k} \)
\(= 3 \times \dfrac{\pi R^2}{2k} \)
\(\therefore\ \text{Tank takes 3 times longer to empty than fill.} \)
A network of running tracks connects the points \(A, B, C, D, E, F, G, H\), as shown. The number on each edge represents the time, in minutes, that a typical runner should take to run along each track. --- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) --- a. \(ABFGD\) b. \(\text{See worked solutions}\) a. \(\text{Using Djikstra’s Algorithm:}\) \(=3+11+1+2+4+5+5\) \(=31\) \(\text{Consider the MST below:}\) \(\text{Total time (MST)}= 3+1+2+4+5+5+9=29\) \(\therefore \text{ Given tree is NOT a MST.}\)
\(\text{Shortest route}\)
\(=ABFGD\)
\(=3+1+5+5\)
\(=14\)
b. \(\text{Total time of given spanning tree}\)
The table shows some of the flight distances (rounded to the nearest 10 km between various Australian cities.
--- 1 WORK AREA LINES (style=lined) ---
The distance-time graph shows the first two stages of a car journey from home to a holiday house.
--- 2 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
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A floor plan is shown. --- 4 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- a. \(5.2\times 5.94\) b. \(72\text{ tiles}\) c. \(8\text{ boxes}\) a. \(\text{Dimensions}=5.2\times (2.15+0.16+3.6)=5.2\times5.94\) b. \(\text{Method 1}\) c. \(\text{Boxes}=\dfrac{72}{10}=7.2\) \(\therefore\ \text{Boxes }=8\)
\(\text{Kitchen Area}\)
\(=3.2\times 3.6\)
\(=11.52\text{ m}^2\)
\(\text{Tile area}\)
\(=0.4\times 0.4\)
\(=0.16\text{ m}^2\)
\(\text{Tiles needed}\)
\(=\dfrac{11.52}{0.16}\)
\(=72\text{ tiles}\)
\(\text{Method 2}\)
\(\text{Tiles to fit width}\)
\(=\dfrac{3.6}{0.4}\)
\(=9\)
\(\text{Tiles to fit length}\)
\(=\dfrac{3.2}{0.4}\)
\(=8\)
\(\text{Tiles needed}\)
\(=9\times 8\)
\(=72\text{ tiles}\)
A company employs 50 people. The annual income of the employees is shown in the grouped frequency distribution table. \begin{array} {|c|c|c|c|} --- 4 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- a. \(A=$975\ 000\), \(B=$3\ 010\ 000\) b. \($60\ 200\) a. \(A=65\ 000\times 15 = $975\ 000\) \(B=540\ 000+715\ 000+975\ 000+525\ 000+255\ 000=$3\ 010\ 000\) b. \(\text{Mean}=\dfrac{\text{Total }fx}{\text{Total }f}=\dfrac{3\ 010\ 000}{50}=$60\ 200\)
\hline
\textit{Annual income} & \textit{Class centre} & \textit{Number of} & fx \\ \text{(\$)} & (x) & \textit{employees}\ (f) & \\
\hline
\rule{0pt}{2.5ex} \text{40 000 – 49 999} \rule[-1ex]{0pt}{0pt} & 45\ 000 & 12 & 540\ 000 \\
\hline
\rule{0pt}{2.5ex} \text{50 000 – 59 999} \rule[-1ex]{0pt}{0pt} & 55\ 000 & 13 & 715\ 000 \\
\hline\rule{0pt}{2.5ex} \text{60 000 – 69 999} \rule[-1ex]{0pt}{0pt} & 65\ 000 & 15 & A \\
\hline\rule{0pt}{2.5ex} \text{70 000 – 79 999} \rule[-1ex]{0pt}{0pt} & 75\ 000 & 7 & 525\ 000 \\
\hline\rule{0pt}{2.5ex} \text{80 000 – 89 999} \rule[-1ex]{0pt}{0pt} & 85\ 000 & 3 & 255\ 000 \\
\hline
\hline\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & & \textit{Total}\ = 50 & \textit{Total = B} \\
\hline
\end{array}
A gym has 9 pieces of equipment: 5 treadmills and 4 rowing machines. On average, each treadmill is used 65% of the time and each rowing machine is used 40% of the time. --- 2 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- i. \(P(T)=0.65, \ \ P(\overline{T})=1-0.65=0.35 \) \(P\text{(3 of 5 treadmills in use)}\ =\ ^5C_3 (0.65)^3(0.35)^2 \) ii. \(P(R)=0.4, \ \ P(\overline{R})=1-0.4=0.6 \) \(P\text{(no rowing machines in use)}\ =\ ^4C_0 (0.6)^4(0.4)^0=(0.6)^4 \) \(\text{Since 2 events are independent:} \) \(P\text{(3 treadmills and no rowing)}\ = \ ^5C_3 (0.65)^3(0.35)^2 \times (0.6)^4 \)
Use mathematical induction to prove that
\((1 \times 2)+\left(2 \times 2^2\right)+\left(3 \times 2^3\right)+\cdots+\left(n \times 2^n\right)=2+(n-1) 2^{n+1}\)
for all integers \(n \geq 1\). (3 marks)
--- 10 WORK AREA LINES (style=lined) ---
\(\text{Proof (See Worked Solutions)} \)
\(\text{Prove true for}\ \ n=1: \)
\(\text{LHS}\ = 1 \times 2=2 \)
\(\text{RHS}\ = 2+(1-1)2^2 = 2 = \text{LHS} \)
\(\text{Assume true for}\ \ n=k: \)
\((1 \times 2)+\left(2 \times 2^2\right)+\cdots+\left(k \times 2^k\right)=2+(k-1) 2^{k+1}\)
\(\text{Prove true for}\ \ n=k+1: \)
\((1 \times 2)+\left(2 \times 2^2\right)+\cdots+\left(k \times 2^k\right) + (k+1)2^{k+1}=2+k \times 2^{k+2}\)
\(\text{LHS}\) | \(=2+(k-1)2^{k+1} + (k+1) 2^{k+1} \) | |
\(=2+2^{k+1}(k-1+k+1) \) | ||
\(=2+2^{k+1}(2k) \) | ||
\(=2+k \times 2^{k+2} \) | ||
\(=\ \text{RHS} \) |
\(\Rightarrow\ \text{True for}\ \ n=k+1 \)
\(\therefore\ \text{Since true for}\ \ n=1, \text{by PMI, true for integers}\ \ n \geq 1. \)
Evaluate \(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\) using the substitution \(u=x-3\). (3 marks) --- 8 WORK AREA LINES (style=lined) --- \(\dfrac{56}{15}\) \(u=x-3\ \ \Rightarrow \ x=u+3 \) \(\dfrac{du}{dx}=1\ \ \Rightarrow \ du=dx \) \(\text{When}\ \ x=4, u=1 \) \(\text{When}\ \ x=3, u=0 \)
\(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\)
\(=\displaystyle \int_0^1(u+5) \sqrt{u}\ du\)
\(=\displaystyle \int_0^1 u^\frac{3}{2} +5u^\frac{1}{2}\ du\)
\(=\Big{[}\dfrac{2}{5} \times u^\frac{5}{2} + \dfrac{2}{3} \times 5u^\frac{3}{2}\Big{]}_0^1 \)
\(=\Big{[} \Big{(}\dfrac{2}{5} + \dfrac{10}{3}\Big{)}-0\Big{]}\)
\(=\dfrac{56}{15}\)
Find \( {\displaystyle \int} \dfrac{1}{\sqrt{4-9x^2}}\ dx\) (2 marks)
\(\dfrac{1}{3} \sin^{-1} \Big{(}\dfrac{3x}{2} \Big{)} +c \)
\({\displaystyle \int} \dfrac{1}{\sqrt{4-9x^2}}\ dx\) | \(=\dfrac{1}{3} {\displaystyle \int} \dfrac{3}{\sqrt{2^2-(3x)^2}}\ dx\) | |
\(=\dfrac{1}{3} \sin^{-1} \Big{(}\dfrac{3x}{2} \Big{)} +c \) |
Consider the polynomial
\(P(x)=x^3+a x^2+b x-12\),
where \(a\) and \(b\) are real numbers.
It is given that \(x+1\) is a factor of \(P(x)\) and that, when \(P(x)\) is divided by \(x-2\), the remainder is \(-18\) .
Find \(a\) and \(b\). (3 marks)
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\(a=2, \ b=-11\)
\((x+1)\ \ \text{is a factor of}\ P(x)\ \ \Rightarrow\ \ P(-1)=0 \)
\(0\) | \(=(-1)^3+a-b-12\) | |
\(a-b\) | \(=13\ \ …\ (1) \) |
\(P(x) ÷ (x-2) \ \text{has a remainder of}\ -18 \ \Rightarrow\ \ P(2)=-18 \)
\(-18\) | \(=8+4a+2b-12\) | |
\(4a+2b\) | \(=-14\) | |
\(2a+b\) | \(=-7\ \ …\ (2) \) |
\(\text{Add}\ \ (1) + (2): \)
\(3a=6\ \ \Rightarrow\ \ a=2\)
\(\text{Substitute}\ \ a=2\ \ \text{into (1):} \)
\(2-b=13\ \ \Rightarrow\ \ b=-11 \)
\(\therefore a=2, \ b=-11\)
In how many different ways can all the letters of the word CONDOBOLIN be arranged in a line? (2 marks)
\(302\ 400\)
\(\text{CONDOBOLIN → }\ 3 \times \text{O}, 2 \times \text{N}, 1 \times \text{C, D, B, L, I} \)
\(\text{Combinations}\) | \(=\dfrac{10!}{3! \times 2!} \) | |
\(=302\ 400\) |
The parametric equations of a line are given below. \begin{aligned} Find the Cartesian equation of this line in the form \(y=m x+c\). (2 marks) \(y=\dfrac{4}{3}x-\dfrac{4}{3} \) \(x=1+3t\ \ \Rightarrow \ \ t=\dfrac{x-1}{3} \)
& x=1+3 t \\
& y=4 t
\end{aligned}
\(y\)
\(=4t\)
\(y\)
\(=4\bigg{(}\dfrac{x-1}{3}\bigg{)} \)
\(y\)
\(=\dfrac{4}{3}x-\dfrac{4}{3} \)
The diagram shows the direction field of a differential equation. A particular solution to the differential equation passes through \((-2,1)\).
Where does the solution that passes through \((-2,1)\) cross the \(y\)-axis?
\(C\)
\(\text{Following gradients → cross y-axis slightly above 1.5}\)
\(\Rightarrow C\)
Consider the following statement.
'If an animal is a herbivore, then it does not eat meat.'
Which of the following is the converse of this statement?
\(D\)
\(\text{Statement:}\ \ P \Rightarrow Q \)
\(\text{Converse of statement:}\ \ Q \Rightarrow P \)
\(\Rightarrow D\)
--- 4 WORK AREA LINES (style=lined) ---
a. `g(x)\ text{cuts}\ xtext{-axis at 1 and}\ -3.`
`g(x)_max=g(-1)=4`
`text{Find intersection of graphs:}`
`(1-x)(3+x)` | `=x-1` | |
`3+x-3x-x^2` | `x-1` | |
`x^2+3x-4` | `=0` | |
`(x+4)(x-1)` | `=0` |
`text{Intersections at:}\ (1,0), (-4,-5)`
b. `text{From the graph:}`
`x-1<(1-x)(3+x)\ \ text{when}\ \ -4<x<1`
`text{Test}\ \ x=0:`
`0-1<(1-0)(3+0)\ \ =>\ \ -1<3\ \ text{(correct)}`
A university uses gas to heat its buildings. Over a period of 10 weekdays during winter, the gas used each day was measured in megawatts (MW) and the average outside temperature each day was recorded in degrees Celsius (°C). Using `x` as the average daily outside temperature and `y` as the total daily gas usage, the equation of the least-squares regression line was found. The equation of the regression line predicts that when the temperature is 0°C, the daily gas usage is 236 MW. The ten temperatures measured were: 0°, 0°, 0°, 2°, 5°, 7°, 8°, 9°, 9°, 10°, The total gas usage for the ten weekdays was 1840 MW. In any bivariate dataset, the least-squares regression line passes through the point `(bar x,bar y)`, where `bar x` is the sample mean of the `x`-values and `bary` is the sample mean of the `y`-values. --- 4 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- a. b. `y=-10.4x+236` c. `text{Answers could include one of the following:}` `text{→ 23°C is outside the range of the dataset and requires the trend}` `text{to be extrapolated.}` `text{→ At 23°C, the equation predicts negative daily gas usage.}` a. `barx=(0+0+0+2+5+7+8+9+9+10)/10=5^@text{C}` `bary=1840/10=184` `text{Regression line passes through:}\ (0,236) and (5,184)` b. `m=(y_2-y_1)/(x_2-x_1)=(184-236)/(5-0)=-10.4` `text{Equation of line}\ m=-10.4\ text{passing through}\ (0,236):` `text{→ 23°C is outside the range of the dataset and requires the trend}` `text{to be extrapolated.}` `text{→ At 23°C, the equation predicts negative daily gas usage.}`
`(y-y_1)`
`=m(x-x_1)`
`y-236`
`=-10.4(x-0)`
`y`
`=-10.4x+236`
c. `text{Answers could include one of the following:}`
A table of future value interest factors for an annuity of $1 is shown. --- 4 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- a. `text{Applicable interest rate}\ =6%` `text{Compounding periods}\ =10xx1=10` `=>\ text{Factor}\ = 13.181` `text{Compounding periods}\ =10xx4=40` `=>\ text{Factor}\ = 54.268`
`:.\ text{Contribution (annual)}`
`=(450\ 000)/13.181`
`=$34\ 140`
b. `text{Applicable interest rate}\ =(6%)/4=1.5%\ text{per quarter}`
`text{Total (after 10 years)}`
`=8535 xx 54.268`
`=$463\ 177.38`
Find the equation of the tangent to the curve `y=(2x+1)^3` at the point `(0,1)`. ( 3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`y=6x+1`
`y` | `=(2x+1)^3` | |
`dy/dx` | `=3xx2(2x+1)^2` | |
`=6(2x+1)^2` |
`text{At}\ x=0\ \ =>\ \ dy/dx=6xx1^2=6`
`text{Find equation of line}\ \ m=6,\ text{through}\ (0,1)`
`y-y_1` | `=m(x-x_1)` | |
`y-1` | `=6(x-0)` | |
`y` | `=6x+1` |
The table shows the probability distribution of a discrete random variable.
\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 1 & 2 & 3 & 4 \\
\hline
\rule{0pt}{2.5ex} P(X = x) \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ & \ \ 0.3\ \ & \ \ 0.5\ \ & \ \ 0.1\ \ & \ \ 0.1\ \ \\
\hline
\end{array}
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--- 4 WORK AREA LINES (style=lined) ---
a. | `E(X)` | `=0+1xx0.3+2xx0.5+3xx0.1+4xx0.1` |
`=0.3+1+0.3+0.4` | ||
`=2` |
b. | `text{Var}(X)` | `=E(X^2)-[E(X)]^2` |
`=(0+1^2xx0.3+2^2xx0.5+3^2xx0.1+4^2xx0.1)-2^2` | ||
`=(0.3+2+0.9+1.6)-4` | ||
`=0.8` |
`:. sigma` | `=sqrt(0.8)` | |
`=0.8944…` | ||
`=0.9\ \ text{(to 1 d.p.)}` |
A game involves throwing a die and spinning a spinner.
The sum of the two numbers obtained is the score.
The table of scores below is partially completed.
What is the probability of getting a score of 7 or more?
The number of bees leaving a hive was observed and recorded over 14 days at different times of the day.
Which Pearson's correlation coefficient best describes the observations?
`D`
`text{Correlation is positive and strong.}`
`text{Best option:}\ r=0.8`
`=>D`
Electricity provider \(A\) charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40. --- 1 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
a. c. \(A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}\) \(\therefore\ \text{Same charge at 400 kWh}\) d. \(\text{Cost at 800 kWh:}\) \(\text{Provider}\ A: \ 40 + 0.25 \times 800 = $240\) \(\text{Provider}\ B: \ 0.35 \times 800 = $280\) \(\therefore \text{Provider}\ A\ \text{is cheaper by \$40.}\)
\(\textit{Electricity used in a
month (kWh)}\)\(\ \ 0\ \ \)
\(400\)
\(1000\)
\(\textit{Monthly charge (\$)}\)
\(40\)
\(290\)
Provider \(B\) charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider \(B\)'s charges vary with the amount of electricity used in a month.
\(\textit{Electricity used in a
month (kWh)}\)\(\ \ 0\ \ \)
\(400\)
\(1000\)
\(\textit{Monthly charge (\$)}\)
\(40\)
\(140\)
\(290\)
A network of running tracks connects the points `A, B, C, D, E, F, G, H`, as shown. The number on each edge represents the time, in minutes, that a typical runner should take to run along each track. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. `ABFGD` b. `text{See worked solutions}`
On another planet, a ball is launched vertically into the air from the ground. The height above the ground, `h` metres, can be modelled using the function `h=-6 t^2+24t`, where `t` is measured in seconds. The graph of the function is shown. --- 1 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- a. `h_max = 24\ text{metres}` b. `t=1 and 3\ text{seconds}` a. `h_max = 24\ text{metres}` b. `3/4 xx h_max = 3/4xx24=18\ text{metres}` `text{From graph, ball is at at 18 metres when:}` `t=1 and 3\ text{seconds}`
The table shows some of the flight distances (rounded to the nearest 10 km) between various Australian cities.
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The graph shows Peta's heart rate, in beats per minute, during the first 60 minutes of a marathon.
--- 1 WORK AREA LINES (style=lined) ---
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a. `120\ text{beats/minute}`
b. `10:30\ text{am}`
a. `120\ text{beats/minute}`
b. `10:30\ text{am}`
A delivery truck was valued at $65 000 when new. The value of the truck depreciates at a rate of 22 cents per kilometre travelled.
What is the value of the truck after it has travelled a total distance of 132 600 km?
`A`
`text{Depreciation}` | `=0.22 xx 132\ 600` | |
`=$29\ 172` |
`text{Truck value}` | `=65\ 000-29\ 172` | |
`=$35\ 828` |
`=>A`
An amount of $2500 is invested at a simple interest rate of 3% per annum.
How much interest is earned in the first two years?
`B`
`I` | `=Prn` | |
`=2500 xx 3/100 xx 2` | ||
`=$150` |
`=>B`
The theory of evolution has been supported by studying the structures of vertebrate forelimbs from the fossil record.
This type of study is best described as
\(D\)
→ Palaeontology is simply, the study of fossils.
\(\Rightarrow D\)
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i. Scientists could include one of:
→ Van Helmont, Hales, Priestley, Ingen-Housz, Senebier, Saussure
ii. Priestly’s Experiment
→ In 1771, Joseph Priestly ignited a candle in a jar with some mint, in which he ignited the candle and then it went out. After a few trials, the candle wouldn’t ignite as all the oxygen in the air had been used (note that oxygen was not yet discovered). However after 27 days, by igniting the candle with a mirror and sunlight as to not open the jar, it reignited.
→ Similarly, by putting mice into a jar with and without a sprig of mint, the jar with the mint allowed the mouse to survive much longer.
→ These observations lead to questions about plant structure, and how they were somehow able to ‘restore air’ that is used by animals breathing and combustion reactions.
i. Scientists could include one of:
→ Van Helmont, Hales, Priestley, Ingen-Housz, Senebier, Saussure
ii. Priestly’s Experiment
→ In 1771, Joseph Priestly ignited a candle in a jar with some mint, in which he ignited the candle and then it went out. After a few trials, the candle wouldn’t ignite as all the oxygen in the air had been used (note that oxygen was not yet discovered). However after 27 days, by igniting the candle with a mirror and sunlight as to not open the jar, it reignited.
→ Similarly, by putting mice into a jar with and without a sprig of mint, the jar with the mint allowed the mouse to survive much longer.
→ These observations lead to questions about plant structure, and how they were somehow able to ‘restore air’ that is used by animals breathing and combustion reactions.
Refer to the diagram of plant tissue shown to answer Questions 8 and 9.
What is the width of Cell Y?
\(C\)
→ The scale interval = 0.1 mm = 100\( \mu \)m
→ The cell width ~ 1.3 scale intervals
\(\Rightarrow C\)
What is the name of the theory which describes evolution as patterns of rapid first appearances or extinctions followed by periods of little or no change?
\(D\)
→ Punctuated equilibrium describes when species adapt until they reach a stable or equilibrium state.
\(\Rightarrow D\)
Four vertebrate forelimbs are shown.
In which area of study do these forelimbs support the theory of evolution?
\(B\)
→ Comparative anatomy is the practise of comparing body structures of different species.
→ In this example we see how across vertebrate species there are similar forelimb structures, suggesting divergence from a common ancestor.
\(\Rightarrow B\)
Darwin and Wallace proposed the theory of evolution by natural selection.
Punctuated equilibrium differs from this theory in that punctuated equilibrium
\(A\)
→ Darwin and Wallace’s theory explains how species adapt overtime to their environment.
→ Punctuated equilibrium is a seperate theory which explains how species adapt until they reach a stable stage but this can be effected by rapid evolutionary change to their environment.
\(\Rightarrow A\)