Band 6 – Page 6

PHYSICS M8 2022 HSC 31

Following the Geiger-Marsden experiment, Rutherford proposed a model of the atom.

Bohr modified this model to explain the spectrum of hydrogen observed in experiments.

The Bohr-Rutherford model of the atom consists of electrons in energy levels around a positive nucleus.

How do features of this model account for all the experimental evidence above? Support your answer with a sample calculation and a diagram, and refer to energy, forces and photons. (9 marks)

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The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ Some alpha particles were deflected and some of these were deflected at very large angles. Rutherford concluded from this that the nucleus was dense and positively charged exerting a repulsive electromagnetic force on the fired alpha particles.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:

→ Bohr proposed that electrons orbited the atomic nucleus in quantised orbits at fixed energies. He proposed that electrons could move from a higher energy orbit (eg. n=1) to a lower energy orbit (n=3) by emitting a photon with energy `E=hf` equal to the energy difference between the two orbits.

→ Additionally, he proposed that electrons could move from a lower energy orbit to a higher energy orbit by absorbing a photon with energy `E=hf` equal to the energy difference between the two orbits.

→ This is able to account for the given emission spectra of hydrogen, where emission lines correspond to electron transitions from higher energy orbits to the second energy orbit which produce photons within the spectrum of visible light.

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:

`(1)/(lambda)`	`=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
	`=(1.097 xx10^7)((1)/(2^(2))-(1)/(6^(2)))`
	`=(2 xx1.097 xx10^7)/(9)`
	`=2.438 xx10^6`
`lambda`	`=410 text{nm}`

→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.

Show Worked Solution

The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:

`(1)/(lambda)`	`=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
	`=(1.097 xx10^7)((1)/(2^(2))-(1)/(6^(2)))`
	`=(2 xx1.097 xx10^7)/(9)`
	`=2.438 xx10^6`
`lambda`	`=410 text{nm}`

→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.

♦ Mean mark 51%.

PHYSICS, M7 2013 HSC 20 MC

The graph shows the maximum kinetic energy `E` with which photoelectrons are emitted as a function of frequency `f` for two different metals `X` and `Y`.

The metals are illuminated with light of wavelength 450 nm.

What would be the effect of doubling the intensity of this light without changing the wavelength?

For metal `X`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
For metal `X`, the number of photoelectrons emitted would increase but the maximum kinetic energy would remain unchanged.
For both metals `X` and `Y`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
For both metals `X` and `Y`, the number of photoelectrons emitted would increase but the maximum kinetic energy would remain unchanged.

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Show Worked Solution

PHYSICS M6 2022 HSC 19 MC

An AC generator is operated by turning a handle, which rotates a coil in a magnetic field.

The handle is turned at a constant speed and the AC voltage output of the generator causes a light globe connected to it to light up, as shown in Circuit 1.

A second identical light globe is then connected in series to the generator output, as shown in Circuit 2 . The handle is turned at the same constant speed.

Which statement describes and explains the effort required to turn the handle in Circuit 2, compared to Circuit 1 ?

The handle in Circuit 2 is easier to turn because the smaller current in Circuit 2 produces less opposing torque.
The handle in Circuit 2 is easier to turn because the voltage output is shared equally across the two identical light globes.
The handle in Circuit 2 is more difficult to turn because the larger current in Circuit 2 produces more opposing torque.
The handle in Circuit 2 is more difficult to turn because it takes more power to operate the two identical globes than it does to operate the single globe.

Show Answers Only

`A`

Show Worked Solution

→ Circuit 2 has a greater resistance due to the second light globe in series.

→ The current in circuit 2 is smaller.

→ As this current is the induced current which opposes the rotation of the handle, the handle in circuit 2 is easier to turn.

`=>A`

♦♦♦ Mean mark 17%.

PHYSICS M6 2022 HSC 18 MC

A charged oil droplet was observed between metal plates, as shown.

While the switch was open, the oil droplet moved downwards at a constant speed. After the switch was closed, the oil droplet moved upwards at the same constant speed.

Assume that the only three forces that may act on the oil droplet are the force of gravity, the force due to the electric field and the frictional force between the air and the oil droplet. The magnitudes of these forces are `F_G` (due to gravity), `F_E` (due to the electric field) and `F_F` (due to the frictional force).

Which row of the table shows all the forces affecting the motion of the oil droplet in the direction indicated, and the relationship between these 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}{|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Downwards motion}\rule[-1ex]{0pt}{0pt}& \text{Upwards motion} \\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}>F_{\text{F}}\rule[-1ex]{0pt}{0pt}&F_{\text{E}}>F_{\text{F}}\\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}>F_{\text{F}}\rule[-1ex]{0pt}{0pt}& F_{\text{E}}>F_{\text{G}}+F_{\text{F}}\\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}=F_{\text{F}}\rule[-1ex]{0pt}{0pt}& F_{\text{G}}=F_{\text{E}} \\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}=F_{\text{F}}\rule[-1ex]{0pt}{0pt}& F_{\text{E}}=F_{\text{G}}+F_{\text{F}} \\
\hline
\end{array}
\end{align*}

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$D$

Show Worked Solution

→ For the droplet to move at a constant speed, the net force acting on it must be zero.

→ For the downwards motion, this means the downwards gravitational force is equal in magnitude to the upwards frictional force (i.e. $F_{\text{G}}=F_{\text{F}}$).

→ For the upwards motion, this means the upwards electric force is equal in magnitude to the sum of the downwards frictional and gravitational forces (i.e. $F_{\text{E}}=F_{\text{G}}+F_{\text{F}}$).

$\Rightarrow D$

♦♦♦ Mean mark 26%.

PHYSICS, M8 2019 HSC 32

Describe how specific experiments have contributed to our understanding of the electron and ONE other fundamental particle. (5 marks)

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Millikan’s Oil Drop Experiment:

→ Millikan’s oil drop experiment involved first measuring the terminal velocity of charged oil droplets in a gravitational field and calculating their mass.

→ An electric field was applied to balance the gravitational field, allowing Millikan to find the electric force and hence, charge on an oil droplet.

→ This allowed him to find the charge on an electron as the smallest difference in charges between two oil drops.

Linear accelerator experiment discovering quarks:

→ An experiment involved using a linear accelerator to speed up and fire a beam of electrons at protons. The scattering pattern of the electrons was analysed and was consistent with protons having an internal structure with both positive and negative charges.

→ This contributed to our understanding of the existence of quarks.

Statistics, EXT1 S1 2022 HSC 14d

An airline company that has empty seats on a flight is not maximising its profit.

An airline company has found that there is a probability of 5% that a passenger books a flight but misses it. The management of the airline company decides to allow for overbooking, which means selling more tickets than the number of seats available on each flight.

To protect their reputation, management makes the decision that no more than 1% of their flights should have more passengers showing up for the flight than available seats.

Given management's decision and using the attached normal distribution probability table to find a suitable approximation, find the maximum number of tickets that can be sold for a flight which has 350 seats. (4 marks)

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`358`

Show Worked Solution

`text{Let}\ \ X=\ text{number of passengers taking a flight}`

`X ~ text{Bin}(n, 0.95)`

`E(X)=0.05,\ \ text{Var}(X)=n(0.95)(1-0.95)=0.0475n`

`X\ text{can be approximated by}\ \ Y ~ N(0.95n, 0.0475n)`

`text{Find}\ n\ text{such that}\ \ P(Y>350)=0.01:`

`text{Using the probability table}`

`=> ztext{-score of 2.33 corresponds to (closest) upper tail probability < 0.01}`

`2.33`	`=(350-0.95n)/sqrt(0.0475n)`
`2.33sqrt(0.0475)sqrtn`	`=350-0.95n`

`0.95n+2.33sqrt(0.0475)sqrtn-350=0`

`sqrtn`	`=(-2.33sqrt(0.0475)+-sqrt((2.33sqrt(0.0475))^2-4(0.95)(-350)))/(2(0.95))`
	`=18.9288…\ \ (n>0)`
	`=(18.9288…)^2`
	`~~358.30`

`:.\ text{Maximum tickets that can be sold = 358}`

♦♦♦ Mean mark 22%.

Vectors, EXT1 V1 2022 HSC 14b

The vectors `\vec{u}` and `\vec{v}` are not parallel. The vector `\vec{p}` is the projection of `\vec{u}` onto the vector `\vec{v}`.

The vector `\vec{p}` is parallel to `\vec{v}` so it can be written `\lambda_0 \vec{v}` for some real number `\lambda_0`. (Do NOT prove this.)

Prove that `|\vec{u}-\lambda \vec{v}|` is smallest when `\lambda=\lambda_0` by showing that, for all real numbers `\lambda,\|\vec{u}-\lambda_0 \vec{v}\| \leq|\vec{u}-\lambda \vec{v}|`. (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`overset(->)p`	`=text{proj}_(overset(->)v)overset(->)u`
`lambda_0 overset(->)v`	`=(overset(->)u*overset(->)v)/(\|overset(->)v\|^2) overset(->)v`
`lambda_0`	`=(overset(->)u*overset(->)v)/(\|overset(->)v\|^2 )\ \ \ …\ (1)`

`text{Show}\ \ |\vec{u}-\lambda_0 \vec{v}\| \leq|\vec{u}-\lambda \vec{v}| :`

`|\vec{u}-\lambda \vec{v}\|^2 -|\vec{u}-\lambda_0 \vec{v}|^2`

`=(vec{u}-\lambda \vec{v})*(vec{u}-\lambda \vec{v})-(vec{u}-\lambda_0 \vec{v})*(vec{u}-\lambda_0 \vec{v})`

`=vec{u}*vec{u}-2lambda vec{u}*vec{v}+lambda^2vec{v}*vec{v}-(vec{u}*vec{u}-2lambda_0vec{u}*vec{v}+lambda_0^2vec{v}*vec{v})`

`=-2lambdavec{u}*vec{v}+lambda^2|vec{v}|^2+2lambda_0vec{u}*vec{v}-lambda_0^2|vec{v}|^2`

`=|vec{v}|^2(lambda^2-lambda_0^2)-2vec{u}*vec{v}(lambda-lambda_0)`

`=|vec{v}|^2(lambda-lambda_0)[lambda+lambda_0-2(vec{u}*vec{v})/|vec{v}|^2]`

`=|vec{v}|^2(lambda-lambda_0)[lambda+lambda_0-2lambda_0]\ \ \ text{(see (1))}`

`=|vec{v}|^2(lambda-lambda_0)^2>=0`

`text{S}text{ince}\ \ |\vec{u}-\lambda \vec{v}\|^2 -|\vec{u}-\lambda_0 \vec{v}|^2>=0`

`=>\ |\vec{u}-\lambda_0 \vec{v}|^2<=|\vec{u}-\lambda \vec{v}\|^2 `

`=>\ |\vec{u}-\lambda_0 \vec{v}|<=|\vec{u}-\lambda \vec{v}\| \ \ text{… as required}`

`:. |\vec{u}-\lambda \vec{v}|\ \ text{is smallest when}\ \ lambda=\lambda_0`

♦♦♦ Mean mark 22%.

Calculus, EXT1 C3 2022 HSC 10 MC

Which of the following could be the graph of a solution to the differential equation

`(dy)/(dx)=sin y+1?`

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`B`

Show Worked Solution

`text{One Strategy}`

`text{When}\ \ (dy)/(dx)=0:`

`siny=-1\ \ =>\ \ y=(3pi)/2 + 2kpi\ \ (kinZZ)`

`text{Graphically,}\ \ y=(3pi)/2 + 2kpi\ \ text{are horizontal asymptotes.}`

`=>B`

♦♦♦ Mean mark 27%.

Calculus, EXT1 C2 2022 HSC 9 MC

A given function `f(x)` has an inverse `f^{-1}(x)`.

The derivatives of `f(x)` and `f^{-1}(x)` exist for all real numbers `x`.

The graphs `y=f(x)` and `y=f^{-1}(x)` have at least one point of intersection.

Which statement is true for all points of intersection of these graphs?

All points of intersection lie on the line `y=x`.
None of the points of intersection lie on the line `y=x`.
At no point of intersection are the tangents to the graphs parallel.
At no point of intersection are the tangents to the graphs perpendicular.

Show Answers Only

`D`

Show Worked Solution

`text{By Elimination:}`

`text{Consider}\ \ f(x)=x\ \ =>\ \ f^(-1)(x)=x:`

`text{All POI lie on}\ \ y=x\ \ text{and all tangents are parallel}`

`text{→ Eliminate B and C}`

`text{Consider}\ \ f(x)=-x\ \ =>\ \ f^(-1)(x)=-x:`

`text{All POI lie on}\ \ y=-x`

`text{→ Eliminate A}`

`=>D`

♦♦♦ Mean mark 13%.

Financial Maths, 2ADV M1 2022 HSC 32

In a reducing-balance loan, an amount `$P` is borrowed for a period of `n` months at an interest rate of 0.25% per month, compounded monthly. At the end of each month, a repayment of `$M` is made. After the `n`th repayment has been made, the amount owing, `$A_n`, is given by

`A_(n)=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`

(Do NOT prove this.)

Jane borrows $200 000 in a reducing-balance loan as described.
The loan is to be repaid in 180 monthly repayments.
Show that `M` = 1381.16, when rounded to the nearest cent. (2 marks)

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After 100 repayments of $1381.16 have been made, the interest rate changes to 0.35% per month.
At this stage, the amount owing to the nearest dollar is $100 032. (Do NOT prove this.)
Jane continues to make the same monthly repayments.
For how many more months will Jane need to make full monthly payments of $1381.16? (3 marks)

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The final payment will be less than $1381.16.
How much will Jane need to pay in the final payment in order to pay off the loan? (2 marks)

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`text{Proof (See Worked Solutions)}`
`83`
`$931.54`

Show Worked Solution

a. `text{Show}\ \ M=$1381.16`

`A_(n)`	`=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`
`0`	`=200\ 000(1.0025)^180-M underbrace((1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^179))_(text(GP where)\ a = 1,\ r = 1.0025,\ n = 180)`
`0`	`=200\ 000(1.0025)^180-M((1(1.0025^180- 1))/(1.0025-1))`

`M((1.0025^180-1)/(1.0025-1))=200\ 000(1.0025)^180`

`:.M`	`=200\ 000(1.0025)^180 -: (1.0025^180-1 )/(1.0025-1)`
	`=1381.163…`
	`=$1381.16\ \ text{… as required}`

b. `P=$100\ 032,\ \ r=1.0035 and M=$1381.16`

`text{Find}\ \ n\ \ text{when}\ \ A_n=0:`

`A_(n)`	`=P(1.0035)^(n)-1381.16(1+(1.0035)^(1)+(1.0035)^(2)+cdots+(1.0035)^(n-1))`
`0`	`=100\ 032(1.0035)^n-1381.16 ((1.0035^n- 1)/(1.0035-1))`
`0`	`=100\ 032(1.0035)^n-1381.16/0.0035 (1.0035^n- 1)`
	`=100\ 032(1.0035)^n-394\ 617(1.0035)^n- 394\ 617`

`294\ 585(1.0035)^n`	`=394\ 617`
`1.0035^n`	`=(394\ 617)/(294\ 585)`
`n`	`=ln((394\ 617)/(294\ 585))/ln1.0035`
	`=83.674…`
	`=83\ text{more months with full payment}`

♦♦ Mean mark part (b) 38%.

c. `text{Find}\ \ A_83:`

`A_83`	`=100\ 032(1.0035)^83-1381.16 ((1.0035^83- 1)/(1.0035-1))`
	`=928.291…`

`text{Interest will be added for the last month:}`

`:.\ text{Final payment}`	`=928.291… xx 1.0035`
	`=$931.54`

♦♦♦ Mean mark part (c) 14%.

Calculus, 2ADV C3 2022 HSC 31

A line passes through the point `P(1,2)` and meets the axes at `X(x, 0)` and `Y(0, y)`, where `x>1`.

Show that `y=(2x)/(x-1)`. (2 marks)

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Find the minimum value of the area of triangle `XOY`. (4 marks)

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`text{Proof (See Worked Solutions)}`
`4\ text{u}^2`

Show Worked Solution

a. `text{Show}\ \ y=(2x)/(x-1)`

`text{S}text{ince}\ \ m_(YP)=m_(PX):`

`(y-2)/(0-1)`	`=(2-0)/(1-x)`
`y-2`	`=(-2)/(1-x)`
`y`	`=2-2/(1-x)`
	`=(2(1-x)-2)/(1-x)`
	`=(-2x)/(1-x)`
	`=(2x)/(x-1)\ \ text{… as required}`

♦♦♦ Mean mark part (a) 17%.
COMMENT: `y=(2x)/(x-1)` is the expression of a relationship between the intercepts and not the equation of the line.

b.	`A`	`=1/2 xx b xxh`
		`=1/2x((2x)/(x-1))`
		`=(x^2)/(x-1)`

`(dA)/dx`	`=((x-1)*2x-x^2(1))/((x-1)^2)`
	`=(2x^2-2x-x^2)/((x-1)^2)`
	`=(x(x-2))/((x-1)^2)`

`text{SP’s occur when}\ \ (dA)/dx=0:`

`x=0\ \ text{or}\ \ 2`

`text{Use 1st derivative test to find max/min:}`

`=>\ text{MIN at}\ \ x=2`

`:.A_min`	`=1/2 xx 2 xx (2xx2)/(2-1)`
	`=4\ text{u}^2`

♦♦ Mean mark part (b) 29%.

Statistics, 2ADV S3 2022 HSC 30

A continuous random variable $X$ has cumulative distribution function given by

$F(x)= \begin{cases}
1 & x>e^3 \\
\ \\
\dfrac{1}{k}\, \ln x & 1 \leq x \leq e^3 . \\
\ \\
0 & x<1\end{cases}$

Show that $k = 3$. (1 mark)

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Given that $P(X < c)=2P(X > c)$, find the exact value of $c$. (2 marks)

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$\text{Proof (See Worked Solutions)}$
$e^2$

Show Worked Solution

a. $\text{Show} \ \ k=3$

\begin{aligned}
{\left[\dfrac{1}{k} \, \ln x\right]_1^{e^3} } & =1 \\
\dfrac{1}{k}\, \ln \left(e^3\right)-\dfrac{1}{k}\, \ln 1 & =1 \\
\dfrac{1}{k}(3)-\frac{1}{k}(0) & =1 \\
k & =3 \ldots \text{as required}
\end{aligned}

♦♦ Mean mark (a) 38%.

b.	$P(X<c)$	$=\left[\dfrac{1}{3}\, \ln x\right]_0^c$
		$=\dfrac{1}{3}\, \ln c$

\begin{aligned}
2 P(X>c) & =2 P(1-P(X<c)) \\
& =2\left(1-\frac{1}{3} \ln c\right)
\end{aligned}

$\text { Given } P(X<c)=2 P(X>c)$

\begin{aligned}
\dfrac{1}{3}\, \ln c & =2-\dfrac{2}{3}\, \ln c \\
\ln c & =2 \\
\therefore c & =e^2
\end{aligned}

♦♦♦ Mean mark (b) 19%.

Calculus, 2ADV C4 2022 HSC 29

The diagram shows the graph of `y=2^{-x}`. Also shown on the diagram are the first 5 of an infinite number of rectangular strips of width 1 unit and height `y=2^{-x}` for non-negative integer values of `x`. For example, the second rectangle shown has width 1 and height `(1)/(2)`.

The sum of the areas of the rectangles forms a geometric series.
Show that the limiting sum of this series is 2. (1 mark)

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Show that `int_(0)^(4)2^(-x)\ dx=(15)/(16 ln 2)`. (2 marks)

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Use parts (a) and (b) to show that `e^(15) < 2^(32)`. (2 marks)

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`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

a. `text{Consider the rectangle heights:}`

`2^0=1, \ 2^(-1)=1/2, \ 2^(-2)= 1/4, \ 2^(-3)= 1/8, …`

`=>\ text{Rectangle Areas}\ = 1, \ 1/2, \ 1/4, \ 1/8, …`

`a=1,\ \ r=1/2`

`S_oo=a/(1-r)=1/(1-1/2)=2\ \ text{… as required}`

b. `text{Show}\ \ int_0^4 2^(-x)\ dx = 15/(16ln2)`

`int_0^4 2^(-x)\ dx `	`=(-1)/ln2[2^(-x)]_0^4`
	`=(-1)/ln2(1/16-1)`
	`=1/ln2-1/(16ln2)`
	`=(16-1)/(16ln2)`
	`=15/(16ln2)\ \ text{… as required}`

Mean mark (b) 56%.

c. `text{Show}\ \ e^15<2^32`

`text{Area under curve < Sum of rectangle areas}`

`15/(16ln2)`	`<2`
`15`	`<32ln2`
`15/32`	`<ln2`
`e^(15/32)`	`<e^(ln2)`
`root(32)(e^15)`	`<2`
`e^15`	`<2^32\ \ text{… as required}`

♦♦♦ Mean mark (c) 9%.

Probability, 2ADV S1 2022 HSC 9 MC

Liam is playing two games. He is equally likely to win each game. The probability that Liam will win at least one of the games is 80%.

Which of the following is closest to the probability that Liam will win both games?

Show Answers Only

`A`

Show Worked Solution

`Ptext{(at least 1 W)}\ = 1-Ptext{(LL)}\ =0.8`

`Ptext{(LL)}`	`=0.2`
`Ptext{(L)}`	`=sqrt0.2`
	`=0.447`

`Ptext{(W)}`	`=1-0.447=0.553`
`Ptext{(WW)}`	`=(0.553)^2`
	`=0.31`

`=>A`

… Mean marks/comments here

♦♦♦ Mean mark 27%.

Financial Maths, STD2 F4 2022 HSC 36

Frankie borrows $200 000 from a bank. The loan is to be repaid over 23 years at a rate of 7.2% per annum, compounded monthly. The repayments have been set at $1485 per month.

The interest charged and the balance owing for the first three months of the loan are shown in the spreadsheet below.

What are the values of `A` and `B`? (2 marks)
After 50 months of repaying the loan, Frankie decides to make a lump sum payment of $ 40 000 and to continue making the monthly repayments of $1485. The loan will then be fully repaid after a further 146 monthly repayments.
How much less will Frankie pay overall by making the lump sum payment? (3 marks)

Show Answers Only

`A=$1198.29,\ \ B=$199\ 139.86`
`$78\ 800`

Show Worked Solution

a. `text{Monthly interest rate}\ =7.2/12=0.6text{%}`

`A`	`=199\ 715 xx 0.6/100`
	`=$1198.29`

`B`	`=P+I-R`
	`=199\ 428.29 + 1196.57-1485`
	`=$199\ 139.86`

b. `text{Total payments if lump sum not paid}`

`= (23xx12) xx 1485`

`=$409\ 860`

`text{Total payments if lump sum paid}`

`=40\ 000 + (50 + 146) xx 1485`

`=$331\ 060`

`text{Savings by paying the lump sum}`

`=409\ 860-331\ 060`

`=$78\ 800`

♦♦♦ Mean mark (b) 17%.

Statistics, STD2 S1 2022 HSC 15 MC

The cumulative frequency graph shows the distribution of the number of movie downloads made by 100 people in one month.

Which box-plot best represents the same data as displayed in the cumulative frequency graph?

Show Answers Only

`C`

Show Worked Solution

`text{1st quartile}\ ~~ 3`

`text{Median}\ ~~ 6`

`text{3rd quartile}\ ~~ 7`

`=>C`

♦♦♦ Mean mark 30%.

Statistics, STD2 S5 2022 HSC 13 MC

A random variable is normally distributed with mean 0 and standard deviation 1 . The table gives the probability that this random variable lies below `z` for some positive values of `z`.

The probability values given in the table are represented by the shaded area in the following diagram.

What is the probability that a normally distributed random variable with mean 0 and standard deviation 1 lies between 0 and 1.94 ?

0.0262
0.4738
0.5262
0.9738

Show Answers Only

`B`

Show Worked Solution

`P(z<1.94) = 0.9738`

`P(z<0) = 0.5`

`:. P(0.5<z<1.94) = 0.9738-0.5 = 0.4738`

`=> B`

♦♦♦ Mean mark 22%.

PHYSICS, M6 2019 HSC 18 MC

A circular loop of wire is connected to a battery and a lamp. The apparatus is moved from `P` to `Q` along the path shown at a constant velocity through a region containing a uniform magnetic field.

Which graph shows the brightness of the lamp as the apparatus moves between `P` and `Q` ?

Show Answers Only

`B`

Show Worked Solution

→ Initially, the current travels clockwise through the loop of wire. As it enters the magnetic field, an anticlockwise current is induced in the loop in order to induce a magnetic field out of page, opposing the external magnetic field (Lenz’s Law).

→ This decreases the net current through the loop, causing a decrease in brightness.

→ As the loop exits the magnetic field, a clockwise current is induced to create a magnetic field into the page, opposing the decrease in magnetic flux passing through it (Lenz’s Law).

→ This increases the net current in the loop, causing an increase in brightness.

`=>B`

♦♦♦ Mean mark 25%.

PHYSICS, M8 2019 HSC 8 MC

A typical galaxy has a diameter of 100 000 light years (∼30 000 pc).

Which graph is consistent with Hubble's measurements of the recessional velocity of galaxies?

Show Answers Only

`A`

Show Worked Solution

By elimination:

Hubble found that the recessional velocity of a galaxy was directly proportional to its distance (linear graph).

→ Eliminate C and D

Hubble measured distance using parsecs.

→ Eliminate B

`=>A`

♦ Mean mark 25%.

CHEMISTRY, M6 2021 HSC 20 MC

The trimethylammonium ion, $\ce{[({CH_3)_3NH}]^+}$, is a weak acid. The acid dissociation equation is shown.

$\ce{[(CH3)3NH]+($aq$)+H2O($l$)\rightleftharpoons H3O+($aq$)+(CH3)3N($aq$)} \quad K_a = 1.55 \times 10^{-10}$

At 20°C, a saturated solution of trimethylammonium chloride, $\ce{[(CH_3)_3NH]Cl}$, has a pH of 4.46.

What is the $K_{sp}$ of trimethylammonium chloride?

$1.26 \times 10^{-9}$
$7.76$
$60.2$
$5.01 \times 10^{10}$

Show Answers Only

$C$

Show Worked Solution

$\ce{\left[\left(CH3\right)_3 NH \right]^{+}(aq)+ H2O(l) \leftrightharpoons H3O ^{+}(aq)+\left(CH3\right)_3 N(aq)}$

$K_a=\dfrac{\left[\left(\text{CH}_3\right)_3 \text{N}\right]\left[ \text{H}_3 \text{O} ^{+}\right]}{\left[\left( \text{CH} _3\right)_3 \text{NH} \right]^{+}}$

$\text{To calculate \(K_{sp}$ equation:}\)

$\ce{\left[\left(CH _3\right)_3 NH \right] Cl (s) \leftrightharpoons\left[\left( CH _3\right)_3 NH \right]^{+}(aq)+ Cl ^{-}(aq)}$

$K_{sp}=\ce{[(CH3)_3NH)^+] [Cl^-]}$

$\text{pH} = \ce{4.46 \rightarrow \left[H3O^+\right] = 10^{-4.46}}$

$\text{Using stoichiometry;}$

$\ce{[(CH3)_3N)^+]=[H3O^+] = 10^{-4.46}}$

$\text{Using}\ K_{a}:$

$1.55 \times 10^{-10}=\dfrac{\left(10^{-4.46} \times 10^{-4.46}\right)}{\ce{\left(CH3\right)3NH^{+}}}$

$\ce{\left[\left(\left(CH3\right)_3NH \right)^{+}\right]}=\dfrac{\left(10^{-4.46} \times 10^{-4.46}\right)}{1.55 \times 10^{-10}}=7.7565 \ldots \text{mol L}^{-1}$

$\ce{\left[Cl^{-}\right]=\left[\left(\left( CH3\right)_3NH\right)^{+}\right]}=7.7565 \ldots \text{mol L}^{-1}$

$\therefore K_{sp}=\ce{\left[\left(\left(CH3\right)_3NH\right)^{+}\right]\times\left[Cl^{-}\right]=7.7565 \ldots \times 7.7565 \ldots=60.2}$

$\Rightarrow C$

♦♦♦ Mean mark 19%.

CHEMISTRY, M8 2021 HSC 9 MC

The amount of paracetamol in a sample needs to be determined.

The UV absorption spectrum of paracetamol is shown.

Based on the absorption spectra provided, which solvent should be used to determine the amount of paracetamol?

Show Answers Only

`D`

Show Worked Solution

→ The solvent used shouldn’t have an absorption spectrum with a maximum that corresponds to that of the paracetamol.

→ This is to ensure that the absorption of solvent will have little to no impact on the measured absorption of the paracetamol sample.

`=> D`

♦♦♦ Mean mark 24%.

PHYSICS, M6 2020 HSC 19 MC

A conductor `P Q` is in a uniform magnetic field. The conductor rotates around the end `P` at a constant angular velocity.

Which graph shows the induced emf between `P` and `Q` as the conductor completes one revolution from the position shown?

Show Answers Only

`C`

Show Worked Solution

At the starting position shown, electrons in the rod are moving to the right, parallel to the magnetic field lines. So, there is no force acting on the.

→ EMF of zero.

After a quarter of a rotation, electrons in the rod are moving up the page. Using the right hand palm rule, they experience a force out of the page. This will not induce an EMF between `P` and `Q.`

→ the graph will show an EMF of zero at both `t=0` and after a quarter of a rotation.

`=>C`

♦♦ Mean mark 10%.

PHYSICS, M7 2020 HSC 18 MC

An observer sees Io complete one orbit of Jupiter as Earth moves from `P_1` to `P_2`, and records the observed orbital period as `t_p`. Similarly, the time for one orbit of Io around Jupiter was measured as Earth moved between the pairs of points at `Q`, `R` and `S`, with the corresponding measured periods of Io being `t_Q`, `t_R` and `t_S`.

Which measurement of the orbital period would be the longest?

`t_P`
`t_Q`
`t_R`
`t_S`

Show Answers Only

`B`

Show Worked Solution

When the Earth is travelling between the pairs of points at `Q `, it is moving away from Jupiter:-

→ light must travel further to signal the end of an orbit than it does to signal the start of an orbit.

→ `t_(Q)` would be the longest measured orbital period.

`=>B`

♦ Mean mark 21%.

PHYSICS, M8 2020 HSC 16 MC

A model of the core of a nuclear fission reactor is shown.

When the reactor is operating normally, the moderator, control rods and coolant work in combination to maintain a controlled nuclear reaction in the fuel rods.

The moderator is a liquid which slows down neutrons to increase the rate of fission. The control rods absorb free neutrons. The coolant reduces the core temperature.

A fault causes some of the moderator to leak out of the core.

Which action would compensate for the effect of the loss of moderator?

Withdraw the control rods from the core.
Lower the control rods further into the core.
Pump the coolant through the core at a faster rate.
Reduce the temperature of the coolant before pumping it into the core.

Show Answers Only

`A`

Show Worked Solution

Loss of moderator leads to a reduction in the rate of fission. Withdrawing the control rods decreases the amount of neutrons absorbed, increasing the rate of fission.

`=>A`

♦ Mean mark 25%.

PHYSICS, M7 2021 HSC 33

Two experiments are performed with identical light sources having a wavelength of 400 nm.

In experiment `A`, the light is incident on a pair of narrow slits `5.0 xx 10^(-5)` m apart, producing a pattern on a screen located 3.0 m behind the slits.

In experiment `B`, the light is incident on different metal samples inside an evacuated tube as shown. The kinetic energy of any emitted photoelectrons can be measured.

Some results from experiment `B` are shown.

How do the results from Experiment `A` and Experiment `B` support TWO different models of light? In your answer, include a quantitative analysis of each experiment. (9 marks)

--- 20 WORK AREA LINES (style=lined) ---

Show Answers Only

→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

`d sin theta`	`=m lambda`
`5xx10^(-5)sin theta`	`=1xx400 xx10^(-9)`
`theta`	`=0.46^(@)`
`s`	`=3 xx tan(0.46) = 0.024` m

→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`

Show Worked Solution

→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

`d sin theta`	`=m lambda`
`5xx10^(-5)sin theta`	`=1xx400 xx10^(-9)`
`theta`	`=0.46^(@)`
`s`	`=3 xx tan(0.46) = 0.024` m

`f=c/lambda =(3.00 xx10^(8))/(400 xx10^(-9))=7.50 xx10^(14)` Hz

`E=hf=6.626 xx10^(-34)xx7.50 xx10^(14)=4.97 xx10^(-19)` J

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

`K_(max)=hf-phi=4.97 xx10^(-19)-4.60 xx10^(-19)=3.70 xx10^(-20)\ \text{J}`

♦ Mean mark 52%.

PHYSICS, M7 2021 HSC 20 MC

A metal cylinder is located in a uniform magnetic field. The work function of the metal is `phi`.

Photons having an energy of 2`phi` strike the side of the cylinder, liberating photoelectrons which travel perpendicular to the magnetic field in a circular path. The maximum radius of the path is `r`.

If the photon energy is doubled, what will the maximum radius of the path become?

`2r`
`3r`
`sqrt2r`
`sqrt3r`

Show Answers Only

`D`

Show Worked Solution

`K_max`	`=(1)/(2)mv_(max)^2`
`v_(max)`	`=sqrt((2K_(max))/(m))\ \ `… (1)

`r=(mv)/(qB)`

Substitute into (1):

`r=(m)/(qB)sqrt((2K_(max))/(m))`

`r prop sqrt(K_(max))`

When the photon energy is `2phi, K_max=phi.`

When the photon energy is doubled to `4phi, K_max=3phi.`

∴ As `r prop sqrt(K_(max)) ` the radius increases by a factor of `sqrt(3).`

`=>D`

♦ Mean mark 24%.

PHYSICS, M5 2021 HSC 14 MC

Which of the following statements correctly describes the gravitational interaction between the Earth and the Moon?

The Earth accelerates towards the Moon.
The net force acting on the Earth is zero.
The Moon and Earth experience equal and opposite accelerations.
The force acting on the Moon is smaller than the force acting on the Earth.

Show Answers Only

`A`

Show Worked Solution

→ According to Newton’s Third Law, the earth and moon experience equal forces in opposite directions. This causes them to accelerate towards each other.

`=>A`

♦ Mean mark 19%.

PHYSICS, M6 2021 HSC 10 MC

A strong magnet is moved past a copper block at a constant speed as shown.

What is the direction of the force acting on the copper block?

To the left
To the right
Into the page
Out of the page

Show Answers Only

`B`

Show Worked Solution

→ Eddy currents will be induced in the copper block. According to Lenz’s Law, this will produce a force that opposes the motion of the magnet.

→ This is done by minimising the relative motion between the block and the magnet, producing a force on the copper block to the right.

`=>B`

♦♦♦ Mean mark 18%.

CHEMISTRY, M5 2021 HSC 11 MC

Consider this system in a fixed volume at constant temperature.

$\ce{PCl5}(s)\rightleftharpoons \ce{PCl3}(l) + \ce{Cl2}(g)$

This system is initially at equilibrium. A small amount of solid $\ce{PCl5}$ is added.

Which statement is correct?

The amount of $\ce{Cl2}$ will increase.
The amount of $\ce{PCl3}$ will decrease.
The amount of $\ce{Cl2}$ will not change.
The amount of $\ce{PCl5}$ will increase then decrease.

Show Answers Only

`C`

Show Worked Solution

The equilibrium constant expression is: `text{K}_(eq) = [text{Cl}_(2)(g)]`

→ As we can see from the equilibrium constant expression, the value of `text{K}_(eq)` is only dependent on the concentration of `text{Cl}_(2)`.

→ As a result, the addition of solid `text{PCl}_5\` will have no affect on the value of `text{K}_(eq)` and thus has no impact on the equilibrium position.

→ Thus, the amount of `text{Cl}_(2)` will not change.

`=> C`

♦♦♦ Mean mark 19%.

CHEMISTRY, M6 2021 HSC 6 MC

Which row of the table describes what happens when a solution of a weak acid is diluted? (Assume constant temperature.)

Show Answers Only

`C`

Show Worked Solution

A weak acid has the following equilibrium:

`text{HA} (aq) + text{H}_2 text{O} (l) ⇋ text{A}^(-) (aq) + text{H}_3 text{O}^(+) (aq)`

`text{K}_a = [[text{A}^(-)][text{H}_3text{O}^(+)] ]/[[text{HA}]]`

→ The value of `text{K}_a` is only affected by temperature, and thus the value of `text{K}_a` will remain the same.

→ When the solution is diluted, water is added. According to Le Chatelier’s Principle, the equilibrium will shift to the right to counteract the change.

→ Thus, the equilibrium will shift to the right and increase the extent of ionisation.

`=>C`

♦♦♦ Mean mark 25%.

BIOLOGY, M5 2021 HSC 10 MC

Cystic fibrosis is an autosomal recessive disorder caused by mutations in the CFTR gene. Many different recessive alleles cause cystic fibrosis.

The four most common alleles of the CFTR gene and their frequencies in the Australian population are shown in the table.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \quad \textit{Allele} \quad \quad& \ \ \textit{Frequency of allele (%)}\ \ \\
\hline \rule{0pt}{2.5ex}\text{A} \rule[-1ex]{0pt}{0pt}& 98.33 \\
\hline \rule{0pt}{2.5ex}\text{al} \rule[-1ex]{0pt}{0pt}& 1.13 \\
\hline \rule{0pt}{2.5ex}\text{a2}\rule[-1ex]{0pt}{0pt}& 0.08 \\
\hline \rule{0pt}{2.5ex}\text{a3} \rule[-1ex]{0pt}{0pt}& 0.07 \\
\hline
\end{array}

What will be the most common genotype of cystic fibrosis patients in Australia?

a1/a1
a1/a2
A/a1
A/A

Show Answers Only

$A$

Show Worked Solution

→ A (dominant) where a single copy produces a normal phenotype.

→ a1 and a2 produce cystic fibrosis, with a1 having the highest frequency.

→ Most likely genotype of cystic fibrosis patient is a1/a1.

$\Rightarrow A$

♦♦♦ Mean mark 22%.

Calculus, MET2 2020 VCAA 5

Let `f: R to R, \ f(x)=x^{3}-x`.

Let `g_{a}: R to R` be the function representing the tangent to the graph of `f` at `x=a`, where `a in R`.

Let `(b, 0)` be the `x`-intercept of the graph of `g_{a}`.

Show that `b= {2a^{3}}/{3 a^{2}-1}`. (3 marks)
State the values of `a` for which `b` does not exist. (1 mark)
State the nature of the graph of `g_a` when `b` does not exist. (1 mark)
i. State all values of `a` for which `b=1.1`. Give your answer correct to four decimal places. (1 mark)
ii. The graph of `f` has an `x`-intercept at (1, 0).
State the values of `a` for which `1 <= b <= 1.1`.
Give your answers correct to three decimal places. (1 mark)

The coordinate `(b, 0)` is the horizontal axis intercept of `g_a`.

Let `g_b` be the function representing the tangent to the graph of `f` at `x=b`, as shown in the graph below.

Find the values of `a` for which the graphs of `g_a` and `g_b`, where `b` exists, are parallel and where `b!=a`. (3 marks)

Let `p:R rarr R, \ p(x)=x^(3)+wx`, where `w in R`.

Show that `p(-x)=-p(x)` for all `w in R`. (1 mark)

A property of the graphs of `p` is that two distinct parallel tangents will always occur at `(t, p(t))` and `(-t,p(-t))` for all `t!=0`.

Find all values of `w` such that a tangent to the graph of `p` at `(t, p(t))`, for some `t > 0`, will have an `x`-intercept at `(-t, 0)`. (1 mark)
Let `T:R^(2)rarrR^(2),T([[x],[y]])=[[m,0],[0,n]][[x],[y]]+[[h],[k]]`, where `m,n in R text(\{0})` and `h,k in R`.

State any restrictions on the values of `m`, `n`, `h`, and `k`, given that the image of `p` under the transformation `T` always has the property that parallel tangents occur at `x = -t` and `x = t` for all `t!=0`. (1 mark)

Show Answers Only

`text(See Worked Solutions.)`
`a=+-sqrt3/3`
`text(Horizontal line.)`
i. `a=-0.5052, 0.8084, 1.3468`
ii. `a in (-0.505,-0.500]uu(0.808,1.347)`
`a=+- sqrt5/5`
`text(See Worked Solutions.)`
`w=-5t^2`
`h=0`

Show Worked Solution

a. `f^(‘)(a) = 3a^2-1`

`g_a(x)\ \ text(has gradient)\ \ 3a^2-1\ \ text(and passes through)\ \ (a, a^3-a)`

`g_a(x)-(a^3-a)`	`=(3a^2- 1)(x-a)`
`g_a(x)`	`=(3a^2-1)(x-a)+a^3-a`

`x”-intercept occurs at”\ (b,0):`

`0=(3a^2-1)(b-a) + a^3-a`

`(3a^2-1)(b-a)`	`=a-a^3`
`3a^2b-3a^3-b+a`	`=a-a^3`
`b(3a^2-1)`	`=a-a^3+3a^3-a`
`:.b`	`=(2a^3)/(3a^2-1)`

b. `b\ text{does not exist when:}`

♦ Mean mark part (b) 46%.

`(3a^2-1)=0`

`a=+-sqrt3/3`

♦♦ Mean mark part (c) 23%.

c. `text{If}\ \ a=+-sqrt3/3,\ \ g_a^(‘)(x) = 0`

`=>\ text{the graph is a horizontal line (does not cross the}\ xtext{-axis).}`

d.i. `”Solve “{2a^{3}}/{3 a^{2}-1}=1.1” for “a:`

`a=-0.5052″ or “a=0.8084” or “a=1.3468\ \ text{(to 4 d.p.)}`

d.ii. `” Solve “1 <= (2a^(3))/(3a^(2)-1) < 1.1” for “a:`

♦♦♦ Mean mark part (d)(ii) 13%.

`a in (-0.505,-0.500]uu(0.808,1.347)\ \ text{(to 3 d.p.)}`

e. `f^(‘)(b) = 3b^2-1`

`g_b(x)\ \ text(has gradient)\ \ 3b^2-1\ \ text(and passes through)\ \ (b, b^3-b)`

`g_b(x)-(b^3-b)`	`=(3b^2- 1)(x-b)`
`g_b(x)`	`=(3b^2-1)(x-b)+b^3-b`

`g_a(x)\ text{||}\ g_b(x)\ \ text{when}`

♦♦♦ Mean mark part (e) 13%.

`3a^2-1`	`=3b^2-1`
	`=3(2a^3)/(3a^2-1)-1`

`=> a=+-1, +- sqrt5/5, 0`

`text(Test each solution so that)\ \ b!=a :`

`text(When)\ \ a=+-1, 0 \ => \ b=a`

`:. a=+- sqrt5/5`

f.	`p(-x)`	`=(-x)^3-wx`
		`=-x^3-wx`
		`=-(x^3+wx)`
		`=-p(x)`

g. `p^(‘)(t) = 3t^2+w`

♦♦♦ Mean mark part (g) 3%.

`p(t)\ \ text(has gradient)\ \ 3t^2+w\ \ text(and passes through)\ \ (t, t^3+wt)`

`p(t)-(t^3+wt)`	`=(3t^2+w)(x-t)`
`p(t)`	`=(3t^2+w)(x-t) + t^3+wt`

`text{If}\ p(t)\ text{passes through}\ \ (-t, 0):`

`0=(3t^2+w)(-2t) + t^3+wt`

`=>w=-5t^2\ \ (t>0)`

h. `text{Property of parallel tangents is retained under transformation}`

♦♦♦ Mean mark part (h) 2%.

`text{if rotational symmetry remains (odd function).}`

`=>h=0`

`text(No further restrictions apply to)\ m, n\ \ text{or}\ \ k.`

NETWORKS, FUR2 2020 VCAA 5

The Sunny Coast cricket clubroom is undergoing a major works project.

This project involves nine activities: `A` to `I`.

The table below shows the earliest start time (EST) and duration, in months, for each activity.

The immediate predecessor(s) is also shown.

The duration for activity `C` is missing.

The information in the table above can be used to complete a directed network.

This network will require a dummy activity.

Complete the following sentence by filling in the boxes provided. (1 mark)

This dummy activity could be drawn as a directed edge from the end of activity

the start of activity

What is the duration, in months, of activity `C`? (1 mark)
Name the four activities that have a float time. (1 mark)
The project is to be crashed by reducing the completion time of one activity only.

What is the minimum time, in months, that the project can be completed in? (1 mark)

Show Answers Only

`B\ text{to the start of activity}\ C.`
`text{2 months}`
`A, E, F, H`
`text{17 months}`

Show Worked Solution

a. `B\ text{to the start of activity}\ C.`

♦♦ Mean mark part (a) 26%.

b. `text{Sketch network diagram.}`

♦ Mean mark part (b) 49%.

`text{Duration of Activity C = 2 months}`

c. `text{Critical path:}\ BCDGI`

♦♦ Mean mark part (c) 23%.

`text{Activities with a float time are activities}`

`text{not on critical path.}`

`:. \ text{Four activities are:}\ \ A, E, F, H`

d. `text{Completion time of}\ BCDGI = 20\ text{months}`

♦♦♦ Mean mark part (d) 12%.

`text{Reduce the completion of}\ B\ text{by 3 months to create}`

`text{a new minimum completion time of 17 months.}`

Statistics, MET2 2020 VCAA 3

A transport company has detailed records of all its deliveries. The number of minutes a delivery is made before or after its schedule delivery time can be modelled as a normally distributed random variable, `T`, with a mean of zero and a standard deviation of four minutes. A graph of the probability distribution of `T` is shown below.

If `"Pr"(T <= a)=0.6`, find `a` to the nearest minute. (1 mark)
Find the probability, correct to three decimal places, of a delivery being no later than three minutes after its scheduled delivery time, given that it arrives after its scheduled delivery time. (2 marks)
Using the model described above, the transport company can make 46.48% of its deliveries over the interval `-3 <= t <= 2`.
It has an improved delivery model with a mean of `k` and a standard deviation of four minutes.
Find the values of `k`, correct to one decimal place, so that 46.48% of the transport company's deliveries can be made over the interval `-4.5 <= t <= 0.5` (3 marks)

A rival transport company claims that there is a 0.85 probability that each delivery it makes will arrive on time or earlier.

Assume that whether each delivery is on time or earlier is independent of other deliveries.

Assuming that the rival company's claim is true, find the probability that on a day in which the rival company makes eight deliveries, fewer than half of them arrive on time or earlier. Give your answer correct to three decimal places. (2 marks)
Assuming that the rival company's claim is true, consider a day in which it makes `n` deliveries.
i. Express, in terms of `n`, the probability that one or more deliveries will not arrive on time or earlier. (1 mark)
ii. Hence, or otherwise, find the minimum value of `n` such that there is at least a 0.95 probability that one or more deliveries will not arrive on time or earlier. (1 mark)
An analyst from a government department believes the rival transport company's claim is only true for deliveries made before 4 pm. For deliveries made after 4 pm, the analyst believes the probability of a delivery arriving on time or earlier is `x`, where `0.3 <=x <= 0.7`
After observing a large number of the rival transport company's deliveries, the analyst believes that the overall probability that a delivery arrives on time or earlier is actually 0.75
Let the probability that a delivery is made after 4 pm be `y`.
Assuming that the analyst's belief are true, find the minimum and maximum values of `y`. (2 marks)

Show Answers Only

`a= 1\ text(minute)`
`0.547`
`k=-1.5, -2.5`
`0.003`
i. `1-0.85^n`
ii. `19`
`2/3`

Show Worked Solution

a. `T\ ~\ N(0, 4^2)`

`text(Solve (by CAS): Pr)(T<=a) = 0.6`

`:. a= 1\ text(minute)`

b.	`text{Pr}(T <= 3∣T > 0)`	`=(text{Pr}(0 < T <= 3))/(text{Pr}(T > 0))`
		`=(0.27337 dots)/(0.5)`
		`=0.547\ \ text{(to 3 d.p.)}`

c. `text(Given)\ \ text{Pr}(-3 <= T <= 2) = 0.4648`

`sigma = 4 text{minutes}`

`=> \ text{Pr}(-4.5 <= T – 1.5 <= 0.5) = 0.4648`

`=> k=-1.5`

`text(By symmetry of the normal distribution)`

`text{Pr}(-2 <= T <= 3) = text{Pr}(-3 <= T <= 2) = 0.4648`

`=> \ text{Pr}(-4.5 <= T – 2.5 <= 0.5) = 0.4648`

`=> k=-2.5`

`:. k=-1.5, -2.5`

d. `text{Let}\ \ X\ ~\ text{Bi}(8, 0.85)`

`text(Solve (by CAS):)`

`text{Pr}(X<=3) = 0.003\ \ text{(to 3 d.p.)}`

e.i. `text{Pr(at least 1 delivery is late)}`

`= 1-\ text{Pr(all deliveries are on time)}`

`=1-0.85^n`

e.ii. `text{Solve for}\ n:`

`1-0.85^n`	`<0.95`
`n`	`>18.43…`

`:.n_min=19`

f. `text{Pr(delivery made after 4pm)} = y`

`=>\ text{Pr(delivery made before 4pm)} = 1-y`

`0.85(1-y)+xy`	`=0.75`
`y`	`=-(0.1)/(x-0.85)`
	`=(2)/(17-20 x)`

`text(Given ) 0.3<=x<=0.7:`

`y_min = (2)/(17-20 xx 0.3) = 2/11`

`y_max = (2)/(17-20 xx 0.7) = 2/3`

GEOMETRY, FUR2 2020 VCAA 3

Khaleda manufacturers the face cream in Dhaka, Bangladesh.

Dhaka is located at latitude 24° N and longitude 90° E.

Assume that the radius of Earth is 6400 km.

Write a calculation that shows that the radius of the small circle of Earth at latitude 24° N is 5847 km, rounded to the nearest kilometre. (1 mark)

Khaleda receives an order from Abu Dhabi, United Arab Emirates (24° N, 54° E).

Find the shortest small circle distance between Dhaka and Abu Dhabi.
Round your answer to the nearest kilometre. (1 mark)

Khaleda sends the order by plane from Dhaka (24° N, 90° E) to Abu Dhabi (24° N, 54° E).

The flight departs Dhaka at 1.00 pm and arrives in Abu Dhabi 11 hours later.

The time difference between Dhaka and Abu Dhabi is two hours.

What time did the flight arrive in Abu Dhabi? (1 mark)

A helicopter takes the order from the airport to the customer's hotel.

The hotel is 27 km south and 109 km east of the airport.

Show that the bearing of the hotel from the airport is 104°, correct to the nearest degree. (1 mark)
After the delivery to the hotel, the helicopter returns to its hangar.
The hangar is located due south of the airport.
The helicopter flies directly from the hotel to the hangar on a bearing of 282° .
How far south of the airport is the hangar?
Round your answer to the nearest kilometre. (1 mark)

Show Answers Only

`text{See Worked Solution}`
`3674 \ text{km}`
`10:00 \ text{pm}`
`text{See Worked Solution}`
`4 \ text{km}`

Show Worked Solution

a. `text(Let)\ \ r= \ text(radius of small circle)`

`cos \ 24^@`	`= r/6400`
`r`	`=6400 xx cos 24^@`
	`= 5846.69 …`
	`=5847 \ text{(nearest km)}`

b. `text{Small circle through Dhaka and Abu Dhabi has radius of 5847 km.}`

`text{Longitudinal difference}`	`= 90 – 54`
	`= 36^@`

`:. \ text{Shortest distance}`	`= 36/360 xx 2 xx pi xx 5847`
	`= 3673.77 …`
	`= 3674 \ text{km (nearest km)}`

c. `text{Dhaka} \ (24^@ text{N}, 90^@ \ text{E}) \ text{is further east than Abu Dhabi} \ (24^@ text{N}, 54^@ \ text{E})`

`=> \ text{Dhaka is 2 hours ahead.}`

`:. \ text{Flight arrival time (Abu Dhabi time)}`

`= 1:00 \ text{pm} + 11 \ text{hours} – 2 \ text{hours}`

`= 10:00 \ text{pm}`

`tan theta`	`= 109/27`
`theta`	`=tan^(-1) (109/27) = 76.1^@`

`:.\ text{Bearing of hotel from airport}`

`=180 – 76.1`

`=104^@\ \ text{(nearest degree)}`

`text{Let X = position of hangar}`

`text{Find OX:}`

`tan 12^@`	`= text{OX}/109`
`text{OX}`	`= 109 xx tan 12^@`
	`= 23.17 \ text{km}`

`:. \ text{AX (distance hangar is south of airport)}`

`= 27 – 23.17`

`= 4 \ text{km (nearest km)}`

GEOMETRY, FUR2 2020 VCAA 2

Khaleda has designed a logo for her business.

The logo contains two identical equilateral triangles,

The side length of each triangle is 4.8 cm, shown in the diagram below.

Write a calculation to show that the area of one of the triangles, rounded to the nearest centimetre, is 10 cm². (1 mark)

In the logo, the two triangles overlap, as shown below. Part of the logo is shaded and part of the logo is not shaded.

What is the area of the entire logo?
Round your answer to the nearest square centimetre. (1 mark)
What is the ratio of the area of the shaded region to the area of the non-shaded region of the logo? (1 mark)
The logo is enlarged and printed on the boxes for shipping.
The enlarged logo and the original logo are similar shape.
The area of the enlarged logo is four times the area of the original logo.
What is the height, in centimetres, of the enlarged logo? (1 mark)

Show Answers Only

`10 \ text{cm}^2 \ (text{nearest cm}^2)`
`15 \ text{cm}^2`
`1:2`
`9.6 \ text{cm}`

Show Worked Solution

a. `text{Triangle is equilateral (all angles = 60}^@)`

`text{Using the sine rule:}`

`A`	`= 1/2 a b sin c`
	`= 1/2 xx 4.8 xx 4.8 xx sin 60^@`
	`= 9.976 …`
	`= 10 text{cm}^2 (text{nearest cm}^2)`

♦♦ Mean mark part (b) 32%.

b. `text{L} text{ogo is made up of 2 identical triangles.}`

`text{Divide each triangle into 4 equal smaller triangles.}`

`text{Total shading = 2 small triangles}\ = 1/2 xx \ text{area of 1 triangle}`

`:. \ text{Area of logo}`

`= 2 xx 10 – 1/2 xx 10`

`= 15 \ text{cm}^2`

♦ Mean mark part (c) 48%.

c.	`text{Shaded region}`	`: \ text{non-shaded}`
	`text{2 triangles}`	`: 4 \ text{triangles}`
	`1`	`: 2`

d. `text{Area scale factor = 4 (given)}`

♦♦♦ Mean mark part (d) 17%.

`text{Length scale factor} = sqrt4 = 2`

`:. \ text{Height of enlarged logo}`

`= 2 xx \ text{height of original logo}`

`= 2 xx 4.8`

`= 9.6 \ text{cm}`

Calculus, MET2 2020 VCAA 2

An area of parkland has a river running through it, as shown below. The river is shown shaded.

The north bank of the river is modelled by the function `f_(1):[0,200]rarr R, \ f_(1)(x)=20 cos((pi x)/(100))+40`.

The south bank of the river is modelled by the function `f_(2):[0,200]rarr R, \ f_(2)(x)=20 cos((pi x)/(100))+30`.

The horizontal axis points east and the vertical axis points north.

All distances are measured in metres.

A swimmer always starts at point `P`, which has coordinates (50, 30).

Assume that no movement of water in the river affects the motion or path of the swimmer, which is always a straight line.

The swimmer swims north from point `P`.
Find the distance, in metres, that the swimmer needs to swim to get to get to the north bank of the river. (1 mark)
The swimmer swims east from point `P`.
Find the distance, in metres, that the swimmer needs to swim to get to the north bank of the river. (2 marks)
On another occasion, the swimmer swims the minimum distance from point `P` to the north bank of the river.
Find this minimum distance. Give your answer in metres, correct to one decimal place. (2 marks)
Calculate the surface area of the section of the river shown on the graph in square metres. (1 mark)
A horizontal line is drawn through point `P`. The section of the river that is south of the line is declared a no "no swimming" zone.
Find the area of the "no swimming" zone, correct to the nearest square metre. (3 marks)
Scientists observe that the north bank of the river is changing over time. It is moving further north from its current position. They model its predicted new location using the function with rule `y=kf_(1)(x)`, where `k >= 1`.
Find the values of `k` for which the distance north across the river, for all parts of the river, is strictly less than 20 m. (2 marks)

Show Answers Only

`10\ text{m}`
`16 2/3\ text{m}`
`8.5\ text{m}`
`2000\ text{m}^2`
`837\ text{m²}`
`k in [1, 7/6)`

Show Worked Solution

a. `text{Since swimmer swims due north,}`

`text{Distance}\ = 40-30=10\ text{m}`

b. `text{Solve} \ f_(1)(x)=30 \ text{or} \ x in[50,100]`

`=> x=200/3`

`:.\ text{Distance to swim (east) to reach north bank}`

`=200/3-50`

`=16 2/3\ text{m}`

c. `text{Let swimmer arrive at north bank at the point}\ \ (x,f_(1)(x))`

`text{By Pythagoras,}`

♦ Mean mark part (c) 39%.

`d(x)=sqrt((x-50)^(2)+(f_(1)(x)-30)^(2))`

`text{Solve} \ d/dx(d(x))=0 \ text{for} \ x:`

`x=54.47…`

`:. d_min=8.5\ text{m (to 1 d.p.)}`

d. `text{Shaded Area}`

`=int_(0)^(200)(f_(1)(x)-f_(2)(x))\ dx`

`=2000\ text{m}^2`

e. `text{Find}\ \ f_(1)(x) = 30 \ text{for} x in [50,150]:`

♦♦ Mean mark part (e) 35%.

`=>x=200/3, 400/3`

`text{Find}\ \ f_(2)(x) = 30 \ text{for} \ x in [50,150]:`

`=>x=50, 150`

`:.\ text{Area}`	`=int_(50)^(150)(30-f_(2)(x))\ dx-int_((200)/(3))^((400)/(3))(30-f_(1)(x))\ dx`
	`=837\ text{m² (to nearest m²)}`

♦♦♦Mean mark part (f) 15%.

f. `text{Let}\ \ D(x)=\ text{vertical distance between banks}`

`D(x)`	`=kf_(1)(x)-f_(2)(x)`
	`=20k cos((pi x)/(100))+40k-(20 cos((pi x)/(100))+30)`
	`=(20k-20)cos((pi x)/(100)) +40k-30`

`text{Given}\ \ D(x)<20 \ text{for}\ x in[0,200]`

`text{Maximum} \ cos((pi x)/(100)) = 1\ text{when}\ \ x=0, 200`

`text{Solve} \ 20k-20+40k-30<20\ \ text{for}\ k:`

`=> k<7/6`

`:. k in [1,7/6)`

GRAPHS, FUR2 2020 VCAA 4

Another section of Kyla's business services and details cars and trucks.

Every vehicle is both serviced and detailed.

Each car takes two hours to service and one hour to detail.

Each truck takes three hours to service and three hours to detail.

Let `x` represent the number of cars that are serviced and detailed each day.

Let `y` represent the number of trucks that are serviced and detailed each day.

Past records suggest there are constraints on the servicing and detailing of vehicles each day.

These constraints are represented by Inequalities 1 to 4 below.

`text{Inequality 1}`	`x >= 16`
`text{Inequality 2}`	`y >= 10`
`text{Inequality 3}`	`2x + 3y <= 96`	`text{(servicing department)}`
`text{Inequality 4}`	`x + 3y <= 72`	`text{(detailing department)}`

Explain the meaning of Inequality 1 in the context of this problem. (1 mark)
Each employee at the business works eight hours per day.
What is the maximum number of employees who can work in the servicing department each day? (1 mark)

The graph below shows the feasible region (shaded) that satisfies Inequalities 1 to 4 .

On a day when 20 cars are serviced and detailed, what is the maximum number of trucks that can be serviced and detailed? (1 mark)
When servicing and detailing, the business makes a profit of $150 per car and $225 per truck.
List the points within the feasible region that will result in a maximum profit for the day. (2 marks)

Show Answers Only

`text{At least 16 cars are serviced and detailed each day.}`
`12`
`17 \ text{trucks}`
`(24, 16),(27, 14),(30, 12), \ text{and} (33, 10)`

Show Worked Solution

a. `text{At least 16 cars are serviced and detailed each day.}`

♦♦ Mean mark part (b) 34%.

b. `text{Consider Inequality 3:} \ 2x + 3y ≤ 96`

`text{Total time} \ ≤ 96 \ text{hours and each employee works 8 hours.}`

`:. \ text{Maximum employees in servicing}`

`=96/8`

`=12`

c. `text{Find} \ y_text{max} \ text{when} \ x= 20`

♦ Mean mark part (c) 42%.

`text{With reference to the feasible region,}`

`y_text{max} \ text{lies on the line of Inequality 4}`

`x + 3y`	`= 72`
`20 + 3y`	`= 72`
`3y`	`= 52`
`y`	`= 17.33`

`:. y_text{max} = 17 \ text{trucks (highest integer within the feasible region)}`

d. `text{Profits: $150 per car, $225 per truck}`

♦♦♦ Mean mark part (d) 19%.

`P = 150x + 225y`

`=> \ m_P = – 150/225 = – 2/3`

`text{The objective function} \ (P) \ text{is parallel to}\ \ 2x+3y=96\ \ text{(Inequality 3)}`

`=> text{Maximum profit occurs at integer co-ordinates on line} \ AB.`

`A\ text(occurs at intersection of:)`

`2x+3y=96 and x+3y=72\ \ =>\ \ x=24`

`B\ text(occurs at intersection of:)`

`2x+3y=96 and y=10\ \ =>\ \ x=33`

`text(Max profit requires)\ \ x in [24,33]`

`text{Test each integer}\ xtext{-value for an integer}\ ytext{-value:}`

`:.\ text{Maximum profit occurs at}`

`text{(24, 16), (27, 14), (30, 12) and (33, 10)}`

GRAPHS, FUR1 2020 VCAA 10 MC

The feasible region for a linear programming problem is shaded in the diagram below.

The line through points `A` and `B` is horizontal and the line through points `C` and `D` is vertical.

The equation for the objective function for this problem is of the form

`Z = bx + 4.5y` where `b > 0`

The value(s) of `b` such that the objective function is maximised only at point `C` is

`0 < b < 4.5`
`b > 4.5`
`b = 9`
`0 < b < 9`
`b > 9`

Show Answers Only

`E`

Show Worked Solution

`m_(BC) = (14 – 0)/(0 – 7) = -2`

♦♦ Mean mark 31%.

`text(Objective function maximized at)\ C\ text(only)`

`=>\ text(gradient is)\ < -2`

`Z = bx + 4.5y`

`y = -b/4.5x + Z/4.5`

`text{The maximum will occur at}\ C\ text{only if the objective function}`

`text{has the steeper of the two gradients.}`

`-b/4.5`	`< -2`
`-b`	`< -9`
`b`	`> 9`

`=> E`

Graphs, MET2 2020 VCAA 20 MC

Let `f:R→R, \ f(x)=cos(ax)`, where `a in R text(\{0})`, be a function with the property

`f(x)=f(x+h),` for all `h in Z`

Let `g:D rarr R, \ g(x)=log_(2)(f(x))` be a function where the range of `g` is `[-1,0]`.

A possible interval for `D` is

`[(1)/(4),(5)/(12)]`
`[1,(7)/(6)]`
`[(5)/(3),2]`
`[-(1)/(3),0]`
`[-(1)/(12),(1)/(4)]`

Show Answers Only

`B`

Show Worked Solution

`f(x)=cos(alpha x)=f(x+h)=cos(a(x+h))`

♦♦♦ Mean mark 18%.

`=>a=2pi`

`g(x)=log_(2)(f(x))=log_(2)(f(x+h))=log_(2)(cos(a(x+h)))`

`-1leqlog_(2)(cos(2pi x))leq0`

`(1)/(2)leq cos(2pi x)leq1`

`text(Sketch) \ y=cos(2pi x)\ \ text(by CAS.)`

`text(By inspection of graph,)\ \ (1)/(2)leq cos(2pi x)leq1\ \text{for}\ x in [1,(7)/(6)]`

`=>B`

Probability, MET2 2020 VCAA 19 MC

Shown below is the graph of `p`, which is the probability function for the number of times, `x`, that a ' 6 ' is rolled on a fair six-sided die in 20 trials.

Let `q` be the probability function for the number of times, `w`, that a ' 6 ' is not rolled on a fair six-sided die in 20 trials. `q(w)` is given by

`p(20-w)`
`p(1-(w)/( 20))`
`p((w)/( 20))`
`p(w-20)`
`1-p(w)`

Show Answers Only

`A`

Show Worked Solution

`q∼text(Bi)(20,(5)/(6)),quad p∼text(Bi)(20,(1)/(6))`

♦ Mean mark 45%.

`q(19)=([20],[19])((5)/(6))^(19)((1)/(6))=p(1)=([20],[1])((1)/(6))((5)/(6))^(19)`

`q(18)=([20],[18])((5)/(6))^(18)((1)/(6))^(2)=p(2)=([20],[2])((1)/(6))^(2)((5)/(6))^(18)`

`text(Generally,)`

`g(w)=([20],[w])((5)/(6))^(w)((1)/(6))^(20-w)=p(20-w)=([20],[20-w])((1)/(6))^(20-w)((5)/(6))^(w)`

`q(w)=p(20-w)`

`=>A`

Graphs, MET2 2020 VCAA 13 MC

The transformation `T:R^(2)rarrR^(2)` that maps the graph of `y=cos(x)` onto the graph of `y=cos(2x+4)` is

`T([[x],[y]])=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-4],[0]])`
`T([[x],[y]])=[[(1)/(2),0],[0,1]][[x],[y]]+[[-4],[0]]`
`T([[x],[y]])=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-2],[0]])`
`T([[x],[y]])=[[2,0],[0,1]]([[x],[y]]+[[2],[0]])`
`T([[x],[y]])=[[2,0],[0,1]][[x],[y]]+[[2],[0]]`

Show Answers Only

`A`

Show Worked Solution

`y=cos(2x+4)=cos(2(x+2))`

♦♦♦ Mean mark 26%.

`text{Dilation of factor} \ 1/2 \ text{from} \ y text{-axis}.`

`text{Translation of 2 units to the left.}`

`T([[x],[y]])=[[(1)/(2),0],[0,1]][[x],[y]]+[[-2],[0]]=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-4],[0]])`

`=> A`

NETWORKS, FUR1 2020 VCAA 7 MC

Four friends go to an ice-cream shop.

Akiro chooses chocolate and strawberry ice cream.

Doris chooses chocolate and vanilla ice cream.

Gohar chooses vanilla ice cream.

Imani chooses vanilla and lemon ice cream.

This information could be presented as a graph.

Consider the following four statements:

The graph would be connected.
The graph would be bipartite.
The graph would be planar.
The graph would be a tree.

How many of these four statements are true?

Show Answers Only

`E`

Show Worked Solution

`text(True statements :)`

♦♦♦ Mean mark 11%.

`text{The graph is connected (a path exists between all vertices).}`

`text{The graph is bipartite (vertices can be divided into two groups).}`

`text{The graph is planar (no edges cross if strawberry is moved to the top).}`

`text{The graph forms a tree (no cycles).}`

`=> E`

Networks, STD2 N3 SM-Bank 50

Roadworks planned by the local council require 13 activities to be completed.

The network below shows these 13 activities and their completion times in weeks.

What is the earliest start time, in weeks, of activity `K`? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
How many of these activities have zero float time? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
It is possible to reduce the completion time for activities `A, E, F, L` and `K`.
The reduction in completion time for each of these five activities will incur an additional cost.
The table below shows the five activities that can have their completion time reduced and the associated weekly cost, in dollars.
The completion time for each these five activities can be reduced by a maximum of two weeks.
The overall completion time for the roadworks can be reduced to 16 weeks.
What is the minimum cost, in dollars, of this change in completion time? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`14 \ text{weeks}`
`7`
`$ 380 \ 000`

Show Worked Solution

a. `text{Scan forwards:}`

`EST \ (text{activity} \ K)`

`= A \ E \ J`

`= 6 + 5 + 3`

`= 14 \ text{weeks}`

b. `text{Scan backwards:}`

`text{Critical paths:}\ ADGLM\ text(and)\ AEHLM`

`:. \ text{7 activities have no float time:} \ ADEGHLM`

c. `text{There are 9 possible paths}`

`ADGLM\ (19), AEHLM\ (19) , AEIM\ (14)`

`AEJK\ (17), BCEIM \ (13), BCEJK\ (16)`

`BCEHLM\ (18), BCDGLM \ (18), BFK (11)`

`text{Consider the 5 paths with completions over 16 weeks}`

` to \ text{all contain}\ A\ text{or}\ L \ text{or both.}`

`text{Consider} \ ADGLM, AEHLM \ (text{contains both} \ A\ text{and} \ L )`

`to \ text{reduce} \ L xx 2 \ , A xx 1 \ text{to reach 16 weeks}`

`to \ text{cheaper than} \ L xx 1 \ , \ A xx 2`

`text{Consider} \ BCEHLM , BCDGLM\ (text{both contain} \ L \ text{only})`

`to \ text{reduce} \ L xx 2 \ text{to reach 16 weeks}`

`text{Consider} \ AEJK \ ( text{contains} \ A \ text{only} )`

`to \ text{reduce} \ A xx 1 \ text{reach 16 weeks}.`

`:. \ text{Minimum cost to reduce time to 16 weeks}`

`= 2 xx 120\ 000 + 1 xx 140\ 000`

`= $ 380\ 000`

GRAPHS, FUR2 2021 VCAA 4

Health and training sessions are held each day at the new community centre.

Let `x` be the number of sessions for children each day.
Let `y` be the number of sessions for adults each day.
The total number of all sessions each day must be at least 10.
The number of sessions for adults must not be less than the number of sessions for children.
It takes 30 minutes for each session for children and 40 minutes for each session for adults.

The constraints on the health and training sessions each day can be represented by the following five inequalities.

Inequality 1 `x >= 0`

Inequality 2 `y >= 0`

Inequality 3 `x + y >= 10`

Inequality 4 `y >= x`

Inequality 5 `30x + 40y <= 600`

Explain the meaning of Inequality 5 in the context of this situation. (1 mark)

The graph below shows the feasible region (shaded) that satisfies Inequality 1 to 5.

What is the maximum number of sessions for children each day? (1 mark)
The community profits from these sessions.
Each session for children makes a profit of $45 and each session for adults makes a profit of $60.
i. Determine the maximum profit per day from all health and training sessions. (1 mark)
ii. List all the points within the feasible region that result in this maximum profit. (1 mark)

Show Answers Only

`text{The time of all adult and children sessions each day, in total,}`
`text{must be less than 600 minutes (10 hours).}`
`8`
i. `$900`
ii. `(0, 15), (4, 12) and (8, 9)`

Show Worked Solution

a. `text{The time of all adult and children sessions each day, in total,}`

`text{must be less than 600 minutes (10 hours).}`

b. `text{Maximum children sessions = 8}`

`text{(largest integer value of} \ x \ text{in the feasible region)}`

c.i. `P = 45 x + 60 y`

`y = – 3/4 x + P/60`

`text{Objective function’s gradient is the same as} \ 30x + 40y = 600`

`=> \ text{maximum profit occurs at integer points on graph of}`

`30x+40y=600\ \ text{in feasible region.}`

`text{Using (0, 15):}`

`P_text{max}`	`= 45 xx 0 + 60 xx 15`
	`= $900`

c.ii.

`P_text{max} \ text{occurs at (0, 15), (4, 12) and (8, 9)}`

Statistics, SPEC2 2021 VCAA 6

The maximum load of a lift in a chocolate company's office building is 1000 kg. The masses of the employees who use the lift are normally distributed with a mean of 75 kg and a standard deviation of 8 kg. On a particular morning there are `n` employees about to use the lift.

What is the maximum possible value of `n` for there to be less than a 1% chance of the lift exceeding the maximum load? (2 marks)

Clare, who is one of the employees, likes to have a hot drink after she exits the lift. The time taken for the drink machine to dispense a hot drink is normally distributed with a mean of 2 minutes and a standard deviation of 0.5 minutes. Times taken to dispense successive hot drinks are independent.

Clare has a meeting at 9.00 am and at 8.52 am she is fourth in the queue for a hot drink. Assume that the waiting time between hot drinks dispensed is negligible and that it takes Clare 0.5 minutes to get from the drink machine to the meeting room.
What is the probability, correct to four decimal places, that Clare will get to her meeting on time? (2 marks)

Clare is a statistician for the chocolate company. The number of chocolate bars sold daily is normally distributed with a mean of 60 000 and a standard deviation of 5000. To increase sales, the company decides to run an advertising campaign. After the campaign, the mean daily sales from 14 randomly selected days was found to be 63 500.

Clare has been asked to investigate whether the advertising campaign was effective, so she decides to perform a one-sided statistical test at the 1% level of significance.

i. Write down suitable null and alternative hypotheses for this test. (1 mark)
ii. Determine the `p` value, correct for decimal places, for this test. (1 mark)
iii. Giving a reason, state whether there is any evidence for the success of the advertising campaign. (1 mark)
Find the range of values for the mean daily sales of another 14 randomly selected days that would lead to the null hypothesis being rejected when tested at the 1% level of significance. Give your answer correct to the nearest integer. (1 mark)
The advertising campaign has been successful to the extent that the mean daily sales is now 63 000.
A statistical test is applied at the 5% level of significance.
Find the probability that the null hypothesis would be incorrectly accepted, based on the sales of another 14 randomly selected days and assuming a standard deviation of 5000. Give your answer correct to three decimal places. (2 marks)

Show Answers Only

`12`
`0.3085`
i. `H_0: \ mu = 60 \ 000, \ H_1: \ mu > 60 \ 000`
ii. `0.0044`
iii. `text{Successful as} \ p text{-value is below 0.01.}`
`n ≥ 63\ 109`
`0.274`

Show Worked Solution

a. `text{Method 1}`

♦♦ Mean mark 22%.

`text{Let} \ \ M_i\ ~\ N (75, 8^2) \ \text{for} \ \ i = 1, 2, 3, … , n`

`W_n = W_1 + W_2 + … + W_n`

`E (W_n) = E(M_1 + … + M_n) = 75n`

`text(s.d.) (W_n) = text{s.d.} (M_1 + … M_n) = 8 sqrtn`

`W_n\ ~\ N (75n, 8^2 n)`

`text{Using} \ Z\ ~\ N (0, 1)`

`text(Pr) (W_n > 1000) = 0.01`

`text(Pr) (Z > {1000 – 75 n}/{8 sqrtn}) = 0.01`

`n = 12.5`

`:. \ text{Largest} \ n = 12`

`text{Method 2}`

`text{By trial and error}`

`text(Pr) (M_1 + … + M_11 > 1000) ≈ 0`

`text(Pr) (M_1 + … + M_12 > 1000) ≈ 0.0002`

`text(Pr) (M_1 + … + M_13 > 1000) ≈ 0.193`

`:. \ text{Largest} \ n = 12`

b. `T_i\ ~\ N (2, 0.5^2) \ \ text{for} \ \ i = 1, 2, …`

Mean mark part (b) 51%.

`text{Wait time} \ (T) = T_1 + T_2 + T_3 + T_4`

`E(T) = 4 xx 2 = 8`

`text{s.d.}(T) = text{s.d.}(T_1 + T_2 + T_3 + T_4) = sqrt4 xx 0.5 = 1`

`T\ ~\ N (8, 1)`

`text(Pr) (T < 7.5) = 0.3085`

`text{By CAS: normCdf} (0, 7.5, 8, 1)`

c.i. `H_0: \ mu = 60 \ 000`

`H_1: \ mu > 60 \ 000`

c.ii. `text(Pr) (barX > 63\ 500 | mu = 60 \ 000) = 0.004407`

`:. \ p \ text{value} = 0.0044`

`text{By CAS: norm Cdf} (63\ 500, oo, 60\ 000, 5000/sqrt14)`

c.iii. `text{S} text{ince the} \ p text{-value is below 0.01, there is strong evidence}`

`text{the advertising was effective (against the null hypothesis).}`

d. `text(Pr) (barX > n | mu = 60 \ 000) < 0.01`

♦♦♦ Mean mark part (d) 16%.

`n ≥ 63\ 109`

`text{By CAS: inv Norm} (0.99, 60 \ 000, 5000/sqrt14)`

e. `text{Similar to part (d):}`

♦♦♦ Mean mark part (e) 10%.

`text(Pr) (barX > n | mu = 60 \ 000) < 0.05`

`n ≤ 62198`

`text{By CAS: invNorm} \ (0.95, 60 \ 000, 5000/sqrt14)`

`text{If} \ \ mu = 63\ 000, text{find probability null hypothesis incorrectly accepted:}`

`text(Pr) (barX < 62\ 198 | mu = 63\ 000) = 0.274`

`text{By CAS: normCdf} (0, 62\ 198, 63\ 000, 5000/sqrt14)`

NETWORKS, FUR2 2021 VCAA 4

Roadworks planned by the local council require 13 activities to be completed.

The network below shows these 13 activities and their completion times in weeks.

What is the earliest start time, in weeks, of activity `K`? (1 mark)
How many of these activities have zero float time? (1 mark)
It is possible to reduce the completion time for activities `A, E, F, L` and `K`.
The reduction in completion time for each of these five activities will incur an additional cost.
The table below shows the five activities that can have their completion time reduced and the associated weekly cost, in dollars.
The completion time for each these five activities can be reduced by a maximum of two weeks.
The overall completion time for the roadworks can be reduced to 16 weeks.
What is the minimum cost, in dollars, of this change in completion time? (1 mark)

Show Answers Only

`14 \ text{weeks}`
`7`
`$ 380 \ 000`

Show Worked Solution

a. `text{Scan forwards:}`

`EST \ (text{activity} \ K)`

`= A \ E \ J`

`= 6 + 5 + 3`

`= 14 \ text{weeks}`

b. `text{Scan backwards:}`

`text{Critical paths:}\ ADGLM\ text(and)\ AEHLM`

`:. \ text{7 activities have no float time:} \ ADEGHLM`

c. `text{There are 9 possible paths}`

`ADGLM\ (19), AEHLM\ (19) , AEIM\ (14)`

`AEJK\ (17), BCEIM \ (13), BCEJK\ (16)`

`BCEHLM\ (18), BCDGLM \ (18), BFK (11)`

`text{Consider the 5 paths with completions over 16 weeks}`

` to \ text{all contain}\ A\ text{or}\ L \ text{or both.}`

`text{Consider} \ ADGLM, AEHLM \ (text{contains both} \ A\ text{and} \ L )`

`to \ text{reduce} \ L xx 2 \ , A xx 1 \ text{to reach 16 weeks}`

`to \ text{cheaper than} \ L xx 1 \ , \ A xx 2`

`text{Consider} \ BCEHLM , BCDGLM\ (text{both contain} \ L \ text{only})`

`to \ text{reduce} \ L xx 2 \ text{to reach 16 weeks}`

`text{Consider} \ AEJK \ ( text{contains} \ A \ text{only} )`

`to \ text{reduce} \ A xx 1 \ text{reach 16 weeks}.`

`:. \ text{Minimum cost to reduce time to 16 weeks}`

`= 2 xx 120\ 000 + 1 xx 140\ 000`

`= $ 380\ 000`

NETWORKS, FUR2 2021 VCAA 3

The network diagram below shows the local road network of Town `M`.

The number on the edges indicate the maximum number of vehicles per hour that can travel along each road in this network.

The arrows represent the permitted direction of travel.

The vertices `A, B, C, D, E` and `F` represent the intersections of the roads.

Determine the maximum number of vehicles that can travel from the entrance to the exit per hour. (1 mark)
The local council plans to increase the number of vehicles per hour that can travel from the entrance to the exit by increasing the capacity of only one road.
i. Complete the following sentence by filling in the boxes provided.

The road that should have its capacity increased is the road from vertex

to
ii. What should be the minimum capacity of this road to maximise the flow of vehicles from the entrance to the exit? (1 mark)

Show Answers Only

`1330`
i. `A\ text(to)\ D`
ii. `780`

Show Worked Solution

`text{Minimum cut}`	`= 680 + 650`
	`= 1330`

`:. \ text{Maximum number of vehicles} = 1330`

b.i. `text{Vehicles from} \ A \ text{to} \ B \ text{restricted to 700 per hour.}`

`:. \ text{Road to increase capacity is from} \ A \ text{to} \ D.`

b.ii. `text{Minimum cut} = 1330 \ text{(partial)}`

`text{Next lowest cut} = 620 + 840 = 1460`

`text{Extra capacity} = 1460 – 1330 = 130`

`:. \ text{Minimum capacity of road} \ A \ text{to}\ D \ text{should be}`

`= 650 + 130`

`= 780`

MATRICES, FUR2 2021 VCAA 4

Five staff members in Elena's office played a round-robin video game tournament, where each employee played each of the other employees once. In each game there was a winner and a loser.

A table of their one-step and two-step dominances was prepared to summarise the results.

Consider the results matrix shown below.

A '1' in this matrix shows that the player named in that row defeated the player named in that column.

A '0' in this matrix shows that the player named in that row lost to the player named in that column.

Use all of the information provided to complete the results matrix. (2 marks)

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad loser),(quadqquadqquadqquadqquadqquad \ \ I qquad\ J qquad \ K qquad\ L qquad M),(wi\n\n\er qquad{:(I),(J),(K),(L),(M):}[(0,…,…,…,…),(…,0,…,…,…),(0,0,0,1,0),(…,…,…,0,…),(…,…,…,…,0)]):}`

Show Answers Only

`{:(qquadqquadqquadqquadqquadqquadqquadqquadqquad ),(quadqquadqquad \ \ \ I \ \ \ J \ \ K \ \ L \ \ \ M),(qquad{:(I),(J),(K),(L),(M):}[(0,1,1,1,0),(0,0,1,1,1),(0,0,0,1,0),(0,0,0,0,1),(1,0,1,0,0)]):}`

Show Worked Solution

`text{Katie only beat Leslie} \ => \ text{everyone else beat Katie}`

`{:(qquadqquadqquadqquadqquad),(quadqquadqquadqquad I qquad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,…,…,…),(…,0,…,…,…),(0,0,0,1,0),(…,…,…,0,…),(…,…,…,…,0)]):} \ to \ {:(qquadqquadqquadqquadqquadqquadqquadqquadqquad),(quadqquadqquadqquad I qquad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,1,…,…),(…,0,1,…,…),(0,0,0,1,0),(…,…,0,0,…),(…,…,1,…,0)]):}`

`text{Katie only has one 2-step dominances. She only beat Leslie.}`

`=>\ text{Leslie only won 1 game (note she has two 2-step dominances)}`

`=>\ text{Leslie must have beaten Mikki (she is the only person with two 1-step dominances)}`

`to \ {:(qquadqquadqquadqquadqquad),(quadqquadqquadqquad I quad\ J qquad \ K qquad\ L qquad M),(qquad{:(I),(J),(K),(L),(M):}[(0,…,1,1,…),(…,0,1,1,…),(0,0,0,1,0),(0,0,0,0,1),(…,…,1,1,0)]):} `

`text{Ike beat either Jolene or Mikki}`

`=> \ text{Jolene has more wins than Mikki and since}`

` qquadqquad text{Ike has the most two-step dominances}`

`=> \ text{Ike beat Jolene and Jolene beat Mikki}`

`to \ {:(qquadqquadqquadqquadqquadqquadqquadqquadqquad),(quadqquadqquad \ \ \ I \ \ J \ \ K \ \ L \ \ M),(qquad{:(I),(J),(K),(L),(M):}[(0,1,1,1,0),(0,0,1,1,1),(0,0,0,1,0),(0,0,0,0,1),(1,0,1,0,0)]):}`

MATRICES, FUR2 2021 VCAA 3

A market research study of shoppers showed that the buying preferences for the three olive oils, Carmani (`C`), Linelli (`L`) and Ohana (`O`), change from month to month according to the transition matrix `T` below.

`qquadqquadqquadqquad \ text(this month)`

`T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} qquad text(next month)):}`

The initial state matrix `S_0` below shows the number of shoppers who bought each brand of olive oil in July 2021.

`S_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`

Let `S_n` represent the state matrix describing the number of shoppers buying each brand `n` months after July 2021.

How many of these 8000 shoppers bought a different brand of olive oil in August 2021 from the brand bought in July 2021? (1 mark)
Using the rule `S_(n+1) = T xx S_(n)`, complete the matrix `S_1` below. (1 mark)

`S_1 = {:[(3060),(text{_____}),(text{_____})]{:(C),(L),(O):} :}`
Consider the shoppers who were expected to buy Carmani olive oil in August 2021.
What percentage of these shoppers also bought Carmani olive oil in July 2021?
Round your answer to the nearest percentage. (1 mark)
Write a calculation that shows Ohana olive oil is the brand bought by 50% of these shoppers in the long run. (1 mark)
Further research suggests more shoppers will buy olive oil in the coming months.
A rule to model this situation is `R_(n+1) = T xx R_n + B`, where `R_n` represents the state matrix describing the number of shoppers `n` months after July 2021.

`qquadqquadqquadqquad \ text(this month)`
`T = {:(qquad\ C quadquadqquad \ L quadquad \ O ),([(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:(C),(L),(O):} ):} qquad text(next month) \ , \ B = {:[(200),(100),(k)]{:(C),(L),(O):} :}, \ R_0 = {:[(3200),(2000),(2800)]{:(C),(L),(O):} :}`

`k` represents the extra number of shoppers expected to buy Ohana olive oil each month.

If `R_2 = {:[(3333),(2025),(3642)]:}`, what is the value of `k`? (1 mark)

Show Answers Only

`1160`
`L = 1900 \ , \ O = 3040`
`89text(%)`
`text(See Worked Solutions)`
`200`

Show Worked Solution

a. `text{Consider matrix} \ T`

`text{Carmani – 15% bought new brand}`

`text{Linelli – 20%, Ohana – 10%}`

`:. text{Shoppers}`	`= 0.15 xx 3200 xx + 0.2 xx 2000 + 0.1 xx 2800`
	`= 1160`

b. `S_1 = [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)]{:[(3200),(2000),(2800)]:} = {:[(3060),(1900),(3040)]:}`

`:. \ L = 1900 \ , \ O = 3040`

c. `text{Carmani purchasers in August)} \ = 3060 \ text{(see part b)}`

`text{Carmani purchasers in July} = 3200`

`text{Carmani purchasers in both July and August}`

`= 0.85 xx 3200`

`= 2720`

`:.\ text{% of August purchasers who bought in July}`

`= 2720/3060 xx 100`

`= 88.88 …`

`=89text(%)`

d. `S = T^50 xx S_0 = {:[(2400),(1600),(4000)]:}`

`text{Total shoppers} = 8000`

`:.\ text{Ohana purchasers (in long run)}`

`= 4000/8000 xx 100`

`= 50text(%)`

e.	`R_1`	`= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3200),(2000),(2800)] + [(200),(180),(k)]`
	`R_2`	`= [(0.85,0.10, 0.05),(0.05,0.80,0.05),(0.10,0.10,0.90)] [(3260),(2000),(k+3040)] + [(200),(100),(k)]`
		`= [(3171 + 0.05 (k + 3040)),(text{not required}),(text{not required})]`

`text{Equating matrices, solve for} \ k:`

`3171 + 0.05 (k + 3040) = 3333`

`:. k = 200`

CORE, FUR2 2021 VCAA 9

Sienna invests $152 431 into an annuity from which she will receive a regular monthly payment of $900 for 25 years. The interest rate for this annuity is 5.1% per annum, compounding monthly.

Let `V` be the balance of the annuity after `n` monthly payments. A recurrence relation written in terms of `V_0 , V_{n + 1}` and `V_n` can model the value of this annuity from month to month.
Showing recursive calculations, determine the value of the annuity after two months.
Round your answer to the nearest cent. (2 marks)
After two years, the interest rate for this annuity will fail to 4.6%.
To ensure that she will still receive the same number of $900 monthly payments, Sienna will add an extra one-pff amount into the annuity at this time.
Determine the value of this extra amount that Sienna will add.
Round your answer to the nearest cent. (1 mark)

Show Answers Only

`$151 \ 925.59`
`$7039.20`

Show Worked Solution

a. `text{Payments are monthly} \ => \ r = 5.1/12 = 0.425text(%) \ text{per month}`

`R`	`= 1 + r/100 = 1.00425`
`V_1`	`= RV_0 – text{payment}`
	`= 1.00425 xx 152 \ 431 – 900`
	`= $152\ 178.83`
`:.V_2`	`= 1.00425 xx 152\ 178.83 – 900`
	`= $ 151 \ 925.59`

b. `text{Find} \ V_24 \ text{(annuity value after 2 years) by TVM Solver:}`

`N`	`= 24`
`I text{(%)}`	`= 5.1`
`PV`	`= -152 \ 431`
`PMT`	`= 900`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = 146 \ 073.7405`

`text{Find}\ PV\ text{of annuity needed (by TVM solver):}`

`N`	`= 243 xx 12 = 276`
`Itext{(%)}`	`= 4.6`
`PV`	`= ?`
`PMT`	`= 900`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PV = -153\ 112.9399`

`:. \ text{Amount to add}`	`= 153 \ 112.94 – 146 \ 073.74`
	`= $ 7039.20`

CORE, FUR2 2021 VCAA 8

For renovations to the coffee shop, Sienna took out a reducing balance loan of $570 000 with interest calculated fortnightly.

The balance of the loan, in dollars, after `n` fortnights, `S_n` can be modelled by the recurrence relation

`S_0 = 570 \ 000,` `S_{n+1} = 1.001 S_n - 1193`

Calculate the balance of this loan after the first fortnightly repayment is made. (1 mark)
Show that the compound interest rate for this loan is 2.6% per annum. (1 mark)
For the loan to be fully repaid, to the nearest cent, Sienna's final repayment will be a larger amount.
Determine this final repayment amount.
Round your answer to the nearest cent. (1 mark)

Show Answers Only

`$ 569 \ 377`
`2.6text(%)`
`$ 1198.56`

Show Worked Solution

a.	`S_1`	`= 1.001 xx 570 \ 000 – 1193`
		`= $ 569 \ 377`

b. `text{Fortnights in 1 year} = 26`

`text{Rate per fortnight} = (1.001 – 1) xx 100text(%) = 0.1text(%)`

`:.\ text(Annual compound rate)`	`= 26 xx 0.1text(%)`
	`= 2.6text(%)`

c. `text{Find}\ N \ text{by TVM Solver:}`

`N`	`= ?`
`I text{(%)}`	`= 2.6`
`PV`	`= 570 \ 000`
`PMT`	`= -1193`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 26`

`=> N = 650.0046 …`

`text{Find} \ FV \ text{after exactly 650 payments:}`

`N`	`= 650`
`Itext{(%)}`	`= 2.6`
`PV`	`= 570 \ 000`
`PMT`	`= -1193`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 26`

`=> FV = -5.59`

`:. \ text{Final repayment}`	`= 1193 + 5.59`
	`= $ 1198.59`

Measurement, STD1 M3 2021 HSC 10 MC

The compass bearing of `B` from `A` is N38°E.

What is the true bearing of A from B?

128°
218°
232°
322°

Show Answers Only

`B`

Show Worked Solution

♦♦♦ Mean mark 20%.

`text(Bearing)\ (A\ text(from)\ B)`	`= 180 + 38`
	`= 218^@`

`=> B`

Mechanics, SPEC2 2021 VCAA 5

A mass of `m_1` kilograms is placed on a plane inclined at 30° to the horizontal. It is connected by a light inextensible string to a second mass of `m_2` kilograms that hangs below a frictionless pulley situated at the top end of the incline, over which the string passes.

Given that the inclined plane is smooth, find the relationship between `m_1` and `m_2` if the mass `m_1` moves down the plane at constant speed. (2 marks)

The masses are now placed on a rough plane inclined at 30°, with the light inextensible string passing over a frictionless pulley in the same way, as shown in the diagram above. Let `N` be the magnitude of the normal force exerted on the mass `m_1` by the plane. A resistance force of magnitude `lambdaN` acts on and opposes the motion of the mass `m_1`.

The mass `m_1` moves up the plane.
i. Mark and label all forces acting on this mass on the diagram above. (1 mark)
ii. Taking the direction up the plane as positive, find the acceleration of the mass `m_1` in terms of `m_1`, `m_2` and `lambda`. (2 marks)

Some time after the masses have begun to move, the mass `m_2` hits the ground at 4.5 ms`\ ^(-1)` and the string becomes slack. At this instant, the mass `m_1` is at the point `P` on the plane, which is 2 m from the pulley. Take the value of `lambda` to be 0.1

How far from point `P` does the mass `m_1` travel before it starts to slide back down the plane?
Give your answer in metres, correct to two decimal places. (2 marks)
Find the time taken, from when the string becomes slack, for the mass `m_1` to return to point `P`.
Give your answer correct to the nearest tenth of a second. (3 marks)

Show Answers Only

`2m_2`
i.
ii. `a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(m_1 + m_2)`
`s = 1.76\ text{m}`
`1.7\ text{seconds}`

Show Worked Solution

a. `m_1g sin30 – m_2g = (m_1 + m_2)a`

`text(S)text(ince)\ m_1\ text(moves at constant speed,)\ a = 0`

`m_1 g · 1/2 – m_2 g`	`= 0`
`m_1`	`= 2m_2`

b.i.

b.ii. `text(S)text(ince)\ m_1\ text(is moving up the slope)`

♦ Mean mark part (b)(ii) 43%.

`m_2g – m_1 g · 1/2 – lambdam_1 g · cos 30`	`= (m_1 + m_2)a`
`m_2g – m_1 g · 1/2 – lambdam_1 g · cos sqrt3/2`	`= (m_1 + m_2)a`

`:. a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(2(m_1 + m_2))`

c. `text(After)\ m_2\ text(hits the ground)`

♦♦♦ Mean mark part (c) 17%.

`m_1a`	`=-m_1g*1/2 – lambda m_1 g sqrt3/2`
`a`	`= -g/2(1 + lambda sqrt3)`
	`= -g/2(1 + 0.1 xx sqrt3)`

`text(By CAS, solve)\ \ v^2 = u^2 + 2as,\ text(for)\ \ s:`

`0 = 4.5^2 – g(1 + 0.1 xx sqrt3)s`

`s = 1.76\ text{m (to 2 d.p.)}`

d. `u = 4.5\ \ text(ms)^(-1), s = 1.76\ text(m),\ a = -g/2(1 + 0.1sqrt3)`

♦♦♦ Mean mark part (d) 7%.

`text(Let)\ \ t_1 = text(Time travelling up slope until stopping)`

`s = ut_1 + 1/2at^2`

`1.76 = 4.5t_1 – 1/2 · g/2(1 + 0.1sqrt3)t_1^2`

`t_1 = 0.78\ text(seconds)`

`text(Let)\ t_2 =\ text(time travelling down the slope)`

`=>\ text(friction is reversed)`

`a`	`= g/2(1 – 0.1sqrt3)`
`1.76`	`= 0 xx t_2 + 1/2 · g/2(1 – 0.1sqrt3) t_2^2`
`t_2`	`= 0.93\ text(seconds)`

`:.\ text(Total time)`	`= t_1 + t_2`
	`= 0.78 + 0.93`
	`= 1.7\ text{seconds (to 1 d.p.)}`

Vectors, SPEC2 2021 VCAA 4

A car that performs stunts moves along a track, as shown in the diagram below. The car accelerates from rest at point `A`, is launched into the air by the ramp `BO` and lands on a second section of track at or beyond point `C`. This second section of track is inclined at `10^@` to the horizontal.

Due to tailwind, the effect of air resistance is negligible. Point `O` is taken as the origin of a cartesian coordinate system and all displacements are measured in metres. Point `C` has the coordinates `(16, 4)`.

At point `O`, the speed of the car is `u` ms`\ ^(-1)` and it takes off at an angle of `theta` to the horizontal direction. After the car passes point `O`, it follows a trajectory where the position of the car's rear wheels relative to point `O`, is given by

`underset~r(t) = ut cos(theta) underset~i + (ut sin(theta) - 1/2 g t^2)underset~j` until the car lands on the second section of track that starts at point `C`.

Show that the path of the rear wheels of the car, while in the air, is given in cartesian form by
`y = x tan(theta) - (4.9x^2)/(u^2cos^2(theta))`. (1 mark)
If `theta = 30^@`, find the minimum speed, in ms`\ ^(-1)`, that the car must reach at point `O` for the rear wheels to land on the second section of track at or beyond point `C`. Give your answer correct to two decimal places. (2 marks)
The ramp `BO` is constructed so that the angle `theta` can be varied.
For what values of `theta` and `u` will the path of the rear wheels of the car join up smoothly with the beginning of the second section of track at point `C`? Give your answer for `theta` in degrees, correct to the nearest degree, and give your answer for `u` in ms`\ ^(-1)`, correct to one decimal place. (3 marks)

The car accelerates from rest along the horizontal section of track `AB`, where its acceleration, `a` ms`\ ^(-2)`, after it has travelled `s` metres from point `A`, is given by `a = 60/v`, where `v` is its speed at `s` metres.

Show that `v` in terms of `s` given by `v = (180x)^(1/3)`. (2 marks)
After the car leaves point `A`, it accelerates to reach a speed of 20 ms`\ ^(-1)` at point `B`. However, if the stunt is called off, the car immediately brakes and reduces its speed at a rate of 9 ms`\ ^(-2)`. It is only safe to call off the stunt if the car can come to rest at or before point `B`. Point `W` is the furthest point along the section `AB` at which the stunt can be called off.
How far is point `W` from point `B`? Give your answer in metres, correct to one decimal place. (3 marks)

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`text(Show Worked Solutions)`
`17.87\ text(ms)^-1`
`u = 16.4\ text(ms)^-1, \ theta = 34.1^@`
`text(Show Worked Solutions)`
`16.4\ text{m}`

Show Worked Solution

a. `x = ut costheta\ …\ (1)`

`y=ut sin theta – 1/2 g t^2\ …\ (2)`

`text{Using (1):}`

`t = x/(u costheta)\ …\ (3)`

`text{Substitute (3) into (2):}`

`y`	`= u · x/(u costheta) sintheta – 4.9 · (x^2)/(u^2cos^2theta)`
	`= x tan theta – (4.9 x^2)/(u^2 cos^2 theta)`

b. `text(When)\ \ theta = 30^@`

♦ Mean mark part (b) 45%.

`y = x\ tan 30^@ – (4.9 x^2)/(u^2 cos^2 30^@)`

`y = x/sqrt3 – (4.9x^2)/(u^2) · 4/3`

`text{Substitute (16, 4) into equation and solve for}\ u:`

`4 = 16/sqrt3 – (4.9 xx 16^2)/(u^2) · 4/3`

`u = 17.87\ text(ms)^-1\ \ text{(to 2 d.p.)}`

c. `text(Smooth join will occur when)`

♦♦♦ Mean mark part (c) 12%.

`y(16) = 4\ …\ (1), \ \ text{and}`

`text(Gradient of motion equals the gradient of landing ramp)`

`(dy)/(dx)|_(x = 16) = -tan10^@\ …\ (2)`

`text{Solve (1) and (2) by CAS for}\ \ u, theta:`

`u = 16.4\ text(ms)^-1`

`theta = 34.1^@`

d. `a = 60/v`

♦ Mean mark part (d) 40%.

`v · (dv)/(ds) = 60/v`

`int v^2 dv = int 60\ ds`

`1/3 v^3 = 60s + c`

`text(When)\ \ s = 0, v = 0 \ => \ c = 0`

`1/3 v^3`	`= 60s`
`v^3`	`= 180s`
`v`	`= (180s)^(1/3)`

e. `text(Find)\ s\ text{at point}\ B\ text{(by CAS):}`

♦♦♦ Mean mark part (e) 8%.

`(180s)^(1/3)`	`= 20`
`s`	`= (400)/9\ text(m)`

`text(Car decelerates at 9 ms)^(-1)`

`d/(ds)(1/2v^2)`	`= -9`
`1/2 v^2`	`= -9s + c`

`text(When)\ \ s = 400/9, v = 0 \ => \ c = 400`

`1/2 v^2`	`= 400 – 9s`
`v`	`= sqrt(800 – 18s)`

`text{Solve for}\ s\ text{(by CAS):}`

`(180s)^(1/3)`	`= sqrt(800 – 18s)`
`s`	`= 28.08`

`text(Distance of)\ W\ text(from)\ B`

`= 400/9 – 28.08`

`= 16.4\ text{m (to 1 d.p.)}`

Calculus, SPEC2 2021 VCAA 3

A thin-walled vessel is produced by rotating the graph of `y = x^3 - 8` about the `y`-axis for `0 <= y <= H`.

All lengths are measured in centimetres.

1. Write down a definite integral in terms of `y` and `H` for the volume of the vessel in cubic centimetres. (1 mark)
2. Hence, find an expression for the volume of the vessel in terms of `H`. (1 mark)

Water is poured into the vessel. However, due to a crack in the base, water leaks out at a rate proportional to the square root of the depth `h` of water in the vessel, that is `(dV)/(dt) = -4sqrth`, where `V` is the volume of water remaining in the vessel, in cubic centimetres, after `t` minutes.

1. Show that `(dh)/(dt) = (-4sqrth)/(pi(h + 8)^(2/3))`. (2 marks)
2. Find the maximum rate, in centimetres per minute, at which the depth of water in the vessel decreases, correct to two decimal places, and find the corresponding depth in centimetres. (2 marks)
3. Let `H = 50` for a particular vessel. The vessel is initially full and water continues to leak out at a rate of `4 sqrth` cm³ min`\ ^(-1)`.
4. Find the maximum rate at which water can be added, in cubic centimetres per minute, without the vessel overflowing. (1 mark)
The vessel is initially full where `H = 50` and water leaks out at a rate of `4sqrth` cm³ min`\ ^(-1)`. When the depth of the water drops to 25 cm, extra water is poured in at a rate of `40sqrt2` cm³ min`\ ^(-1)`.
Find how long it takes for the vessel to refill completely from a depht of 25 cm. Give your answer in minutes, correct to one decimal place. (3 marks)

Show Answers Only

1. `V = pi int_0^H (y + 8)^(2/3)\ dy`
2. `V = (3pi)/5 [(H + 8)^(5/3) – 32]`
1. `text(See Worked Solutions)`
2. `0.62\ text(cm/min when)\ \ h = 24.`
3. `4sqrt50\ \ text(cm³/min)`
`31.4\ text(mins)`

Show Worked Solution

a.i. `y = x^3 – 8 \ => \ x = root3(y + 8) \ => \ x^2 = (y + 8)^(2/3)`

`:. V = pi int_0^H (y + 8)^(2/3)\ dy`

a.ii. `V = (3pi)/5 [(H + 8)^(5/3) – 32]`

b.i. `(dV)/(dt) = -4sqrth`

`(dV)/(dh) = pi(h + 8)^(2/3) \ => \ (dh)/(dV) =1/(pi(h + 8)^(2/3))`

`(dh)/(dt)`	`= (dV)/(dt) * (dh)/(dV)`
	`= (-4sqrth)/(pi(h + 8)^(2/3))`

b.ii. `text(Solve)\ (d^2h)/(dt^2) = 0\ \ text(for)\ \ h\ \ text{(by CAS):}`

♦ Mean mark part (b)(ii) 42%.

`(d^2h)/(dt^2) = (2(h – 24))/(3sqrth pi (h + 8)^(5/3))=0`

`=>h = 24`

`text(At)\ \ h=24, \ (dh)/(dt) = -0.62\ text(cm/min)`

`:.\ text(Max rate at which depth decreases is)`

`0.62\ text(cm/min when)\ \ h = 24.`

♦♦ Mean mark part (b)(iii) 24%.

b.iii. `text(At)\ \ H = 50, text(vessel is full and losing water at)\ \ 4sqrt50\ \ text(cm³/min)`

`:. text(Water can be added at a max-rate of)\ \ 4sqrt50\ \ text(cm³/min and)`

`text(vessel will not overflow.)`

c. `(dV)/(dt) = 40sqrt2 – 4sqrth`

♦♦♦ Mean mark part (c) 16%.

`(dV)/(dh) · (dh)/(dt) = 40sqrt2 – 4sqrth`

`pi(h + 8)^(2/3) · (dh)/(dt)`	`= 40sqrt2 – 4sqrth`
`(dh)/(dt)`	`= (40sqrt2 – 4sqrth)/(pi(h + 8)^(2/3)`
`(dt)/(dh)`	`= (pi(h + 8)^(2/3))/(40sqrt2 – 4sqrth)`
`t`	`= int (pi(h + 8)^(2/3))/(40sqrt2 – 4sqrth)\ dh`

`text(Time of vessel to refill from)\ \ h = 25\ \ text(to)\ \ h = 50:`

`t`	`= int_25^50 (pi(h + 8)^(2/3))/(40sqrt2 – 4sqrth)\ dh`
	`~~ 31.4\ text(mins)`

Calculus, SPEC2 2021 VCAA 1

Let `f(x) = ((2x - 3)(x + 5))/((x - 1)(x + 2))`.

Express `f(x)` in the form `A + (Bx + C)/((x - 1)(x + 2))`, where `A`, `B` an `C` are real constants. (1 mark)
State the equation of the asymptotes of the graph of `f`. (2 marks)
Sketch the graph of `f` on the set of axes below. Label the asymptotes with their equations, and label the maximum turning point and the point of inflection with their coordinates, correct to two decimal places. Label the intercepts with the coordinate axes. (3 marks)
Let `g_k(x) = ((2x - 3)(x + 5))/((x - k)(x + 2))`, where `k` is a real constant.
i. For what values of `k` will the graph of `g_k`, have two asymptotes? (2 marks)
ii. Given that the graph of `g_k` has more than two asymptotes, for what values of `k` will the graph of `g_k` have no stationary points? (2 marks)

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`f(x) = 2 + (5x – 11)/((x – 1)(x + 2))`
`text(Horizontal asymptote:)\ y = 2`
i. `k = 3/2 => g_k(x) = 2 + 6/(x + 2)`
ii. `k , -5\ \ text(or)\ \ k > 3/2`

Show Worked Solution

a. `text{By CAS (prop Frac}\ f(x)):`

`f(x) = 2 + (5x – 11)/((x – 1)(x + 2))`

b. `text(Vertical asymptotes:)\ x = 1, x = –2`

`text(As)\ \ x -> ∞, y -> 2`

`text(Horizontal asymptote:)\ y = 2`

♦ Mean mark part (c) 48%.

d.i. `text(Two asymptotes only when:)`

♦♦ Mean mark part (d)(i) 24%.

`k = -2 \ => \ g_k(x) = 2 – (23 + x)/((x + 2)^2)`

`k = -5 \ => \ g_k(x) = 2 – 7/(x + 2)`

`k = 3/2 \ => \ g_k(x) = 2 + 6/(x + 2)`

d.ii. `text(By CAS, solve)\ \ d/(dx)(g_k(x)) = 0\ \ text(for)\ \ x:`

♦♦♦ Mean mark part (d)(ii) 16%.

`x = (-4k + 15 ± sqrt(-21(2k^2 + 7k – 15)))/(2k + 3)`

`text(No solutions occur when:)`

`k = -3/2\ \ text(or)`

`2k^2 + 7k – 15 < 0`

`=> k < -5\ \ text(or)\ \ k > 3/2`

Calculus, MET2 2021 VCAA 5

Part of the graph of `f: R to R , \ f(x) = sin (x/2) + cos(2x)` is shown below.

State the period of `f`. (1 mark)
State the minimum value of `f`, correct to three decimal places. (1 mark)
Find the smallest positive value of `h` for which `f(h - x) = f(x)`. (1 mark)

Consider the set of functions of the form `g_a : R to R, \ g_a (x) = sin(x/a) + cos(ax)`, where `a` is a positive integer.

State the value of `a` such that `g_a (x) = f(x)` for all `x`. (1 mark)
i. Find an antiderivative of `g_a` in terms of `a`. (1 mark)
ii. Use a definite integral to show that the area bounded by `g_a` and the `x`-axis over the interval `[0, 2a pi]` is equal above and below the `x`-axis for all values of `a`. (3 marks)
Explain why the maximum value of `g_a` cannot be greater than 2 for all values of `a` and why the minimum value of `g_a` cannot be less than –2 for all values of `a`. (1 mark)
Find the greatest possible minimum value of `g_a`. (1 mark)

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`4 pi`
`-1.722`
`2 pi`
`text{See Worked Solutions}`
i. `text{See Worked Solutions}`
ii. `text{See Worked Solutions}`
`text{See Worked Solutions}`
`- sqrt2`

Show Worked Solution

a. `text{By inspection, graph begins to repeat after 4pi.}`

`text{Period}\ = 4 pi`

b. `text{By CAS: Sketch}\ \ f(x) = sin (x/2) + cos(2x)`

`f_min = -1.722`

♦ Mean mark part (c) 21%.

c. `text{If} \ \ f(x)\ \ text{is reflected in the} \ y text{-axis and translated} \ 2 pi \ text{to the right} => text{same graph}`

`f(x) = f(-x + h) = f(2 pi – x)`

`:. h = 2 pi`

d. `f(x) = sin(x/2) + cos(2x)`

`g_a(x) = sin(x/a) + cos(2a)`

`g_a(x) = f(x) \ \ text{when} \ \ a = 2`

e.i. `int g_a (x)\ dx`

♦ Mean mark part (e)(i) 50%.

`= -a cos (x/a) + {sin (ax)}/{a} , \ (c = 0)`

e.ii. `int_0^{2a pi} g_a(x)\ dx`

♦ Mean mark part (e)(ii) 29%.

`= {sin (2a^2 pi)}/{a}`

`= 0 \ \ (a ∈ ZZ^+)`

`text{When integral = 0, areas above and below the} \ x text{-axis are equal.}`

f. `g_a (x) = sin(x/a) + cos (ax)`

♦ Mean mark part (f) 13%.

`-1 <= sin(x/a) <= 1 \ \ text{and}\ \ -1 <= cos(ax) <= 1`

`:. -2 <= g_a (x) <= 2`

g. `text{Sketch}\ \ g_a (x) \ \ text{by CAS}`

♦ Mean mark part (g) 2%.

`text{Minimum access at}\ \ a = 1`

`g_a(x)_min = – sqrt2`

Probability, MET2 2021 VCAA 4

A teacher coaches their school's table tennis team.

The teacher has an adjustable ball machine that they use to help the players practise.

The speed, measured in metres per second, of the balls shot by the ball machine is a normally distributed random variable `W`.

The teacher sets the ball machine with a mean speed of 10 metres per second and standard deviation of 0.8 metres per second.

Determine `text(Pr) (W ≥11)`, correct to three decimal places. (1 mark)
Find the value of `k`, in metres per second, which 80% of ball speeds are below. Give your answer in metres per second, correct to one decimal place. (1 mark)

The teacher adjusts the height setting for the ball machine. The machine now shoots balls high above the table tennis table.

Unfortunately, with the new height setting, 8% of balls do not land on the table.

Let `overset^P` be the random variable representing the sample proportion of the balls that do not land on the table in random samples of 25 balls.

Find the mean and the standard deviation of `overset^P`. (2 marks)
Use the binomial distribution to find `text(Pr) (overset^P > 0.1)`, correct to three decimal places. (2 marks)

The teacher can also adjust the spin setting on the ball machine.

The spin, measured in revolutions per second, is a continuous random variable `X` with the probability density function

`f(x) = {(x/500, 0 <= x < 20), ({50-x}/{750}, 20 <= x <= 50), (\ 0, text(elsewhere)):}`

Find the maximum possible spin applied by the ball machine, in revolutions per second. (1 mark)
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`0.106`
`0.8`
`sqrt46/125`
`0.323`
`50 \ text{revolutions per second}`
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Show Worked Solution

a. `W\ ~\ N (10,0.8^2)`

`text{By CAS: norm Cdf} \ (11, ∞, 10, 0.8)`

`text(Pr) (W >= 11) = 0.106`

b. `text(Pr) (W < k) = 0.8 \ \ \ text{By CAS: inv Norm} \ (0.8, 10, 0.8)`

c. `E(overset^P) = 0.08 = 2/25`

♦ Mean mark part (c) 45%.

`text(s.d.) (overset^P) = sqrt{{0.08 xx 0.92}/{25}} = sqrt46/125`

d. `X\ ~\ text(Bi) (25, 0.08)`

♦ Mean mark part (d) 49%.

`text{By CAS: binomCdf} \ (25, 0.08, 3, 25)`

`text(Pr) (overset^P > 0.1)`	`= text(Pr) (X > 0.1 xx 25)`
	`= text(Pr) (X > 2.5)`
	`= text(Pr) (X >= 3)`
	`= 0.323`

e. `text{Maximum spin = 50 revolutions per second}`

♦♦ Mean mark part (e) 21%.

f. This content is no longer in the Study Design.

g. This content is no longer in the Study Design.

h. This content is no longer in the Study Design.

Calculus, MET2 2021 VCAA 3

Let `q(x) = log_e (x^2 - 1) - log_e (1 - x)`.

State the maximal domain and the range of `q`. (2 marks)
i. Find the equation of the tangent to the graph of `q` when `x = –2`. (1 mark)
ii. Find the equation of the line that is perpendicular to the graph of `q` when `x = –2` and passes through the point (–2, 0). (1 mark)

Let `p(x) = e^{-2x} - 2e^-x + 1.`

Explain why `p` is not a one-to one function. (1 mark)
Find the gradient of the tangent to the graph of `p` at `x = a`. (1 mark)

The diagram below shows parts of the graph of `p` and the line `y = x + 2`.

The line `y = x + 2` and the tangent to the graph of `p` at `x = a` intersect with an acute angle of `theta` between them.

Find the value(s) of `a` for which `theta = 60^@`. Give your answer(s) correct to two decimal places. (3 marks)
Find the `x`-coordinate of the point of intersection between the line `y = x + 2` and the graph of `p`, and hence find the area bounded by `y = x + 2`, the graph of `p` and the `x`-axis, both correct to three decimal places. (3 marks)

Show Answers Only

`text{Domain:} \ x ∈ (-∞, -1)`
`text{Range:} \ y ∈ R`
i. `-x – 2`
ii. `x + 2`
`text{Not a one-to-one function as it fails the horizontal line test.}`
`-2e^{-2a} + 2e^{-a}`
`-0.67`
`1.038\ text(u)^2`

Show Worked Solution

a. `text(Method 1)`

`x^2 – 1 > 0 \ \ => \ \ x > 1 \ \ ∪ \ \ x < -1`

`1 – x > 0 \ \ => \ \ x < 1`

`:. \ x ∈ (-∞, -1)`

`text(Method 2)`

`text{Sketch graph by CAS}`

`text{Asymptote at} \ x = -1`

`text{Domain:} \ x ∈ (-∞, -1)`

`text{Range:} \ y ∈ R`

b. i. `text{By CAS (tanLine} \ (q (x), x, –2)):`

`y = -x – 2`

ii. `text{By CAS (normal} \ (q (x), x, –2)):`

`y = x + 2`

c. `text{Sketch graph by CAS.}`

`p(x) \ text{is not a one-to-one function as it fails the horizontal line test}`

`text{(i.e. it is a many-to-one function)}`

d. `p′(x) = -2e^{-2x} + 2e^-x`

`p′(a) = -2e^{-2a} + 2e^{-a}`

`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a} -2e^{-2a} = – tan (15^@) \ \ text{for}\ a:`

`a = -0.11`

`text{Case 2}`

`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a} – 2e^{-2a} = -tan 75^@\ \ text{for}\ a:`

`a = -0.67`

f. `text{At intersection,}`

`x + 2 = e^{-2x} -2e^{-x} + 1`

`x = -0.750`

`text{Area}`	`= int_{-2}^{-0.750} x + 2\ dx + int_{-0.750}^0 e^{-2x} – 2e^{-x} + 1\ dx`
	`= 1.038 \ text(u)^2`

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b.	\(P(X<c)\)	\(=\left[\dfrac{1}{3}\, \ln x\right]_0^c\)
		\(=\dfrac{1}{3}\, \ln c\)