Band 5 – Page 7

Functions, MET2 2023 VCAA 9 MC

The function $f$ is given by

$f(x) = \begin {cases}
\tan\Bigg(\dfrac{x}{2}\Bigg) &\ \ 4 \leq x \leq 2\pi \\
\sin(ax) &\ \ \ 2\pi\leq x\leq 8
\end{cases}$

The value of $a$ for which $f$ is continuous and smooth at $ x$ = $2\pi$ is

$-2$
$-\dfrac{\pi}{2}$
$-\dfrac{1}{2}$
$\dfrac{1}{2}$
$2$

Show Answers Only

$C$

Show Worked Solution

$\text{Solve for}\ a\ \text{given}\ \ x=2\pi:$

$\tan\Bigg(\dfrac{2\pi}{2}\Bigg)=\sin(a2\pi)=0$

$a=\pm \dfrac{1}{2}$

$\text{For smoothness, solve for}\ a\ \text{given}\ \ x=2\pi:$

$\dfrac{d}{dx}\tan\Bigg(\dfrac{x}{2}\Bigg)=\dfrac{d}{dx}\sin(ax)$

$\therefore a=-\dfrac{1}{2}$

$\Rightarrow C$

♦ Mean mark 46%.

Probability, MET2 2023 VCAA 8 MC

A box contains $n$ green balls and $m$ red balls. A ball is selected at random, and its colour is noted. The ball is then replaced in the box.

In 8 such selections, where $n\neq m$, what is the probability that a green ball is selected at least once?

$8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7$
$1-\Bigg(\dfrac{n}{n+m}\Bigg)^8$
$1-\Bigg(\dfrac{m}{n+m}\Bigg)^8$
$1-\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7$
$1-8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7$

Show Answers Only

$C$

Show Worked Solution

$\text{Let}\ \ X=\ \text{choosing a green ball}$

$\text{Pr}(X\geq 1)$	$=1-\text{Pr}(X=0)$
	$=1-\Bigg(\dfrac{m}{n+m}\Bigg)^8$

$\Rightarrow C$

♦ Mean mark 49%.

Calculus, MET2 2023 VCAA 6 MC

Suppose that $\displaystyle \int_{3}^{10} f(x)\,dx=C$ and $\displaystyle \int_{7}^{10} f(x)\,dx=D$. The value of $\displaystyle \int_{7}^{3} f(x)\,dx$ is

$C+D$
$C+D-3$
$C-D$
$D-C$
$CD-3$

Show Answers Only

$D$

Show Worked Solution

$\text{Given }\displaystyle \int_{3}^{10} f(x)\,dx=C\ \ \text{and}\ \displaystyle \int_{7}^{10} f(x)\,dx=D$

$\text{We can deduce:}$

$\displaystyle \int_{3}^{10} f(x)\,dx$	$=\displaystyle \int_{3}^{7} f(x)\,dx+\displaystyle \int_{7}^{10} f(x)\,dx$
$C$	$=\displaystyle \int_{3}^{7} f(x)\,dx+D$
$C-D$	$=\displaystyle \int_{3}^{7} f(x)\,dx$
$\therefore\ \displaystyle \int_{7}^{3} f(x)\,dx$	$=D-C$

$\Rightarrow D$

♦ Mean mark 49%.
MARKER’S COMMENT: 41% of students chose option C incorrectly assuming $\int_{7}^3 f(x)\,dx=\int_{3}^7 f(x)\,dx$.

Graphs, MET2 2023 VCAA 3 MC

Two function, $p$ and $q$, are continuous over their domains, which are $[-2, 3)$ and $(-1, 5]$, respectively.

The domain of the sum function $p+q$ is

$[-2, 5]$
$[-2, -1)\cup (3, 5]$
$[-2, -1)\cup (-1, 3)\cup(3, 5]$
$[-1, 3]$
$(-1, 3)$

Show Answers Only

$E$

Show Worked Solution

$\text{Domain of sum function = intersection of two domains.}$

$[-2, 3)\cap(-1, 5]=(-1, 3)$

$\Rightarrow E$

♦ Mean mark 47%.

Graphs, MET1 2022 VCAA 6

The graph of `y=f(x)`, where `f:[0,2 \pi] \rightarrow R, f(x)=2 \sin(2x)-1`, is shown below.

On the axes above, draw the graph of `y=g(x)`, where `g(x)` is the reflection of `f(x)` in the horizontal axis. (2 marks)
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Find all values of `k` such that `f(k)=0` and `k \in[0,2 \pi]`. (3 marks)
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Let `h: D \rightarrow R, h(x)=2 \sin(2x)-1`, where `h(x)` has the same rule as `f(x)` with a different domain.
The graph of `y=h(x)` is translated `a` units in the positive horizontal direction and `b` units in the positive vertical direction so that it is mapped onto the graph of `y=g(x)`, where `a, b \in(0, \infty)`.

1. Find the value for `b`. (1 mark)
  
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2. Find the smallest positive value for `a`. (1 mark)
  
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3. Hence, or otherwise, state the domain, `D`, of `h(x)`. (1 mark)
  
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a. Graph `y=g(x)`

b. `\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`

c.i. `b=2`

cii. `=\frac{\pi}{2}`

ciii. `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

Show Worked Solution

b. `2 \sin (2 k)-1`	`=0` `0<=k<=2\pi`
`sin (2 k)`	`=1/2`
`2k`	`=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}`
`k`	`=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`

c.i ` 2 \sin 2(x-a)-1+b`	`=-(2 \sin 2 x-1)`
`:.\ -1+b`	`=1`
`b`	`=2`

c.ii `2 \sin 2(x-a)`	`=-(2 \sin 2 x-1)`
`\sin (2 x-2 a)`	`=-\sin 2 x`
`\therefore 2 a`	`=\pi`
`a`	`=\frac{\pi}{2}`

♦ Mean mark (c.ii) 50%.
MARKER’S COMMENT: Students confused vertical and horizontal translations. Common error `a=\frac{\pi}{4}.`

c.iii The domain for `f(x)` is `[0,2pi]`

`:. \ D` is `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

♦♦♦ Mean mark (c.iii) 10%.
MARKER’S COMMENT: Common error was translating in the wrong direction. A common incorrect answer was `\left[-\frac{\pi}{2}, \frac{5 \pi}{2}\right].`

Functions, MET1 2022 VCAA 5b

Find the maximal domain of `f`, where `f(x)=\log _e\left(x^2-2 x-3\right)`. (3 marks)

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`x \in(-\infty,-1) \` ∪ `(3, \infty)`

Show Worked Solution

`f(x)` exists where `x^2-2 x-3>0`

`:.\ (x – 3)(x + 1) > 0`

From the graph `x < -1` and `x > 3`

`x \in(-\infty,-1) \` ∪ `(3, \infty)`

♦ Mean mark (b) 53%
MARKER’S COMMENT: Students often used incorrect notation. Drawing the graph of the parabola assisted with recognition of correct domain.

Probability, MET1 2022 VCAA 4

A card is drawn from a deck of red and blue cards. After verifying the colour, the card is replaced in the deck. This is performed four times.

Each card has a probability of `\frac{1}{2}` of being red and a probability of `\frac{1}{2}` of being blue.

The colour of any drawn card is independent of the colour of any other drawn card.

Let `X` be a random variable describing the number of blue cards drawn from the deck, in any order.

Complete the table below by giving the probability of each outcome. (2 marks)

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Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red. (1 mark)

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The deck is changed so that the probability of a card being red is `\frac{2}{3}` and the probability of a card being blue is `\frac{1}{3}`.
Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red. (2 marks)

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a. Complete table

\begin{array} {|c|c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline {Pr}(X=x) & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 \\
\hline \end{array}

b. `3/8`

c. `4/9`

Show Worked Solution

a. Using binomial distribution where `n=4` and `p= 1/2`

`text{Pr}(X=1)`	`=\ ^4 C_1(1/2)^1(1/2)^3 = 41/21/8`	`=4/16`
`text{Pr}(X=3)`	`=\ ^4 C_3(1/2)^3(1/2)^1= 41/81/2`	`=4/16`
`text{Pr}(X=4)`	`=\ ^4 C_4(1/2)^4(1/2)^0= 11/161`	`=1/16`

\begin{array} {|c|c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline {Pr}(X=x) & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 \\
\hline \end{array}

b. Trials are independent of first trial.

Binomial where `n=3, p=1/2`

`text{Pr}(X=2) = \ ^3 C_2*(1/2)^2*(1/2)^1 = 3*1/8 = 3/8`

♦♦ Mean mark (b) 40%.

c. Trials are independent.

Binomial where `n=3, p=2/3`

`text{Pr}(X=2) = \ ^3 C_2*(2/3)^2*(1/3)^1 = 3*4/9*1/3 = 4/9`

♦ Mean mark 55%.
MARKER’S COMMENT: Many students wrongly treated this question as conditional probability.

Algebra, MET1 2022 VCAA 3

Consider the system of equations

`kx-5y=4+k`

`3x+(k+8) y=-1`

Determine the value of `k` for which the system of equations above has an infinite number of solutions. (3 marks)

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`k=-3` for infinite solutions

Show Worked Solution

Making `y` the subject of each equation:

`k x-5 y`	`=4+k `
`y`	`=\frac{k}{5} x+\frac{(-4-k)}{5}`	(1)

`3x+(k+8) y`	`=-1`
`y`	`=\frac{-3}{k+8} \times-\frac{1}{k+8}`	(2)

From equations (1) and (2)

`m_1=\frac{k}{5}` and `m_2=-\frac{3}{k+8}`

An infinite number of solutions occur when `m_1=m_2`

`\frac{k}{5}`	`=-\frac{3}{k+8}`
`k(k+8)`	`=-15`
`k^2+8 k+15`	`=0`
`(k+3)(k+5)`	`=0`

`\therefore k=-5,-3`

Let `c_1` and `c_2` be the `y`-intercepts of equations (1) and (2)

then `c_1 =\frac{-k-4}{5}` and `c_2 =-\frac{1}{k+8}`

`\therefore \frac{-k-4}{5}=-\frac{1}{k+8}`

`(k+4)(k+8)`	`=5`
`k^2+12 k+27`	`=0`
`(k+3)(k+9)`	`=0`
`k=-3 \text {, }`	`k=-9`

`\therefore k=-3` for infinite solutions and only value to satisfy both equations.

♦ Mean mark 50%.
MARKER’S COMMENT: Students are reminded to always use the correct variables.

Calculus, MET1 2023 VCAA 9

The shapes of two walking tracks are shown below.

Track 1 is described by the function $f(x)=a-x(x-2)^2$.

Track 2 is defined by the function $g(x)=12x-bx^2$.

The unit of length is kilometres.

Given that $f(0)=12$ and $g(1)=9$, verify that $a=12$ and $b=-3$. (1 mark)
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Verify that $f(x)$ and $g(x)$ both have a turning point at $P$.
Give the co-ordinates of $P$. (2 marks)
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A theme park is planned whose boundaries will form the triangle $\Delta OAB$ where $O$ is the origin, $A$ is at $(k, 0)$ and $B$ is at $(k, g(k))$, as shown below, where $k \in (0, 4)$.
Find the maximum possible area of the theme park, in km². (3 marks)

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a. $\text{See worked solution}$

b. $P(2, 12)\ \text{Also see worked solution}$

c. $\dfrac{128}{9}$

Show Worked Solution

a.	$f(0)$	$=a-0(0-2)^2=12$
	$\therefore\ a$	$=12$

$g(1)$	$=12\times 1+b\times 1^2$
$12+b$	$=9$
$\therefore b$	$=-3$

b.	$f(x)$	$=12-x(x-2)^2$
		$=-x^3+4x^2-4x+12$
	$f^{′}(x)$	$=-(3x^2-8x+4)$

$\text{For turning point }f^{′}(x)=0:$

$ -(3x^2-8x+4)$	$=0$
$-(3x-2)(x-2)$	$=0$

$\therefore\ x=\dfrac{2}{3}\ \text{or}\ 2$

$f(2)$	$=12-2(2-2)^2=12$
$f\Big{(}\dfrac{2}{3}\Big{)}$	$=12-\dfrac{2}{3}\Big(\dfrac{2}{3}-2\Big)^2<12$

$\therefore\ \text{Given graph of }f(x)\ \text{relevant turning point is }(2, 12).$

$g(x)=12x-3x^2\ \Rightarrow\ g^{′}(x)=12-6x$

$\text{For turning point}\ \ g^{′}(x)=0:$

$12-6x=0\ \ \Rightarrow \ x=2$

$g(x)$	$=12x-3x^2$
$g(2)$	$=12\times 2-3\times 2^2=12$

$\therefore\ g(x)\ \text{also has a turning point at }(2, 12).$

♦ Mean mark (b) 48%.

c. $\text{Area of triangle }\Delta OAB$

$A(k)$	$=\dfrac{1}{2}\times k\times g(k)$
	$=\dfrac{k}{2}\times (12k-3k^2)$
	$=6k^2-\dfrac{3k^3}{2}$

$\text{Max area when }A^{′}(k)=0:$

$A^{′}(k)=12k-\dfrac{9k^2}{2}=\dfrac{1}{2}(24k-9k^2)$

$\dfrac{1}{2}k(24-9k)=0$

$\therefore\ k=\dfrac{24}{9}=\dfrac{8}{3}$

$\text{Maximum area occurs when}\ k=\dfrac{8}{3}:$

$A(k)$	$=\dfrac{1}{2}k\times (12k-3k^2)$
$A\bigg(\dfrac{8}{3}\bigg)$	$=\dfrac{1}{2}\times\dfrac{8}{3}\times\bigg (12\bigg(\dfrac{8}{3}\bigg)-3\bigg(\dfrac{8}{3}\bigg)^2\bigg)$
	$=\dfrac{4}{3}\bigg(32-\dfrac{64}{3}\bigg)$
	$=\dfrac{4}{3}\bigg(\dfrac{96}{3}-\dfrac{64}{3}\bigg)$
	$=\dfrac{4}{3}\times\dfrac{32}{3}$
	$=\dfrac{128}{9}$

♦♦♦ Mean mark (c) 27%.
MARKER’S COMMENT: Many students made arithmetic errors substituting the fractional values of $k$ into $A(k)$ to find max area.

Calculus, SPEC2 2023 VCAA 9 MC

The position of a particle moving in the Cartesian plane, at time $t$, is given by the parametric equations

$x(t)=\dfrac{6 t}{t+1}$ and $y(t)=\dfrac{-8}{t^2+4}$, where $t \geq 0$.

What is the slope of the tangent to the path of the particle when $t=2$ ?

$-\dfrac{1}{3}$
$-\dfrac{1}{4}$
$\dfrac{1}{3}$
$\dfrac{3}{4}$
$\dfrac{4}{3}$

Show Answers Only

$D$

Show Worked Solution

$\text{At}\ \ t=2\ \text{(by calc):}$

$\dfrac{dx}{dt}=\dfrac{2}{3}, \ \dfrac{dy}{dt}=\dfrac{1}{2} $

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4} $

$\Rightarrow D$

Calculus, MET1 2023 VCAA 7

Consider $f:(-\infty, 1]\rightarrow R, f(x)=x^2-2x$. Part of the graph of $y=f(x)$ is shown below.

State the range of $f$. (1 mark)
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Sketch the graph of the inverse function $y=f^{-1}(x)$ on the axes above. Label any endpoints and axial intercepts with their coordinates. (2 marks)
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Determine the equation of the domain for the inverse function $f^{-1}$. (2 marks)
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Calculate the area of the regions enclosed by the curves of $f,\ f^{-1}$ and $y=-x$. (2 marks)
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a. $[-1, \infty)$

c. $f^{-1}(x)=1-\sqrt{x+1}$

$\text{Domain}\ [-1, \infty)$

d. $A=\dfrac{1}{3}$

Show Worked Solution

a. $[-1, \infty)$

c. $\text{When }f(x)\ \text{is written in turning point form}$

$y=(x-1)^2-1$

$\text{Inverse function: swap}\ x \leftrightarrow y$

$x$	$=(y-1)^2-1$
$x+1$	$=(y-1)^2$
$-\sqrt{x+1}$	$=y-1$
$f^{-1}(x)$	$=1-\sqrt{x+1}$

$\text{Domain}\ [-1, \infty)$

♦ Mean mark (c) 48%.
MARKER’S COMMENT: Common error → writing the function as $f^{-1}(x)=1+\sqrt{x+1}$.

d. $\text{One strategy of many possibilities:}$

	$A$	$=2\displaystyle \int_{0}^{1} \big(-x-(x^2-2x)\big)\,dx$
		$=2\displaystyle \int_{0}^{1} \big(x-x^2\big)\,dx$
		$=2\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1$
		$=2\bigg(\dfrac{1}{2}-\dfrac{1}{3}-(0)\bigg)$
		$=\dfrac{1}{3}$

♦♦♦ Mean mark (d) 24%.
MARKER’S COMMENT: Using the symmetry properties of the graph and its inverse helped answer this question efficiently.

Statistics, MET1 2023 VCAA 6

Let $\hat{P}$ be the random variable that represents the sample proportion of households in a given suburb that have solar panels installed.

From a sample of randomly selected households in a given suburb, an approximate 95% confidence interval for the proportion $p$ of households having solar panels installed was determined to be (0.04, 0.16).

Find the value of $\hat{p}$ that was used to obtain this approximate 95% confidence interval. (1 mark)

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Use $z=2$ to approximate the 95% confidence interval.

Find the size of the sample from which this 95% confidence interval was obtained. (2 marks)
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A larger sample of households is selected, with a sample size four times the original sample.
The sample proportion of households having solar panels installed is found to be the same.
By what factor will the increased sample size affect the width of the confidence interval? (1 mark)
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a. $\hat{p}=0.1$

b. $n=100$

c. $\text{The confidence interval width is halved.}$

Show Worked Solution

a. $\text{The span of a confidence interval is symmetrical about}\ \hat{p}.$

$\hat{p}=\dfrac{0.04+0.16}{2}=0.1$

♦ Mean mark (a) 52%.

b.	$0.06$	$=2\times\sqrt{\dfrac{0.1\times0.9}{n}}$
	$0.03$	$=\sqrt{\dfrac{0.1\times0.9}{n}}$
	$(0.03)^2$	$=\dfrac{0.1\times0.9}{n}$
	$0.0009$	$=\dfrac{9}{100n}$
	$n$	$=\dfrac{9}{100 \times 0.0009}$
	$n$	$=100$

♦♦ Mean mark (b) 38%.

c. $\text{C.I. width} \propto \dfrac{1}{\sqrt{n}}$

→ $\text{If}\ \ n\ \Rightarrow \ 4n,\ \ \text{C.I. width}\ \propto \dfrac{1}{\sqrt{4n}} = \dfrac{1}{2\sqrt{n}}$

→ $\text{The confidence interval is halved.}$

♦♦♦ Mean mark (c) 23%.

Probability, MET1 2023 VCAA 8

Suppose that the queuing time, $T$ (in minutes), at a customer service desk has a probability density function given by

$f(t) = \begin {cases}
kt(16-t^2) &\ \ 0 \leq t \leq 4 \\
\\
0 &\ \ \text{elsewhere}
\end{cases}$

for some $K \in R$.

Show that $k=\dfrac{1}{64}$. (1 mark)

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Find $\text{E}(T)$. (2 marks)

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What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute? (3 marks)
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a.	$\displaystyle \int_{0}^{4} (16-t^2)\,dt$	$=1$
	$k\left[8t^2-\dfrac{t^4}{4}\right]_0^4$	$=1$
	$k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)$	$=1$
	$64k$	$=1$
	$\therefore\ k$	$=\dfrac{1}{64}$

b. $E(T)=\dfrac{32}{15}$

c. $\dfrac{16}{25}=0.64$

Show Worked Solution

a.	$\displaystyle \int_{0}^{4} kt(16-t^2)\,dt$	$=1$
	$k\left[8t^2-\dfrac{t^4}{4}\right]_0^4$	$=1$
	$k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)$	$=1$
	$64k$	$=1$
	$\therefore\ k$	$=\dfrac{1}{64}$

♦ Mean mark (a) 43%.

b.	$E(T)$	$=\dfrac{1}{64}\displaystyle \int_{0}^{4} (16t^2-t^4)\,dt$
		$=\dfrac{1}{64}\left[\dfrac{16t^3}{3}-\dfrac{t^5}{5}\right]_0^4$
		$=\dfrac{1}{64}\Bigg(\dfrac{1024}{3}-\dfrac{1024}{5}-0\Bigg)$
		$=\dfrac{1}{64}\times\dfrac{2048}{15}$
		$=\dfrac{32}{15}$

♦♦ Mean mark (b) 38%.

c.	$\text{Pr}(2<T<4\|T>1)$	$=\dfrac{\text{Pr}(2<T<4)}{\text{Pr}(T>1)}$
		$=\dfrac{\dfrac{1}{64}\displaystyle \int_{2}^{4} (16t-t^3)\,dt}{\dfrac{1}{64}\displaystyle \int_{1}^{4} (16t-t^3)\,dt}$
		$=\dfrac{\left[8t^2-\dfrac{t^4}{4}\right]_2^4}{\left[8t^2-\dfrac{t^4}{4}\right]_1^4}$
		$=\dfrac{(64-(32-4))}{\bigg(64-\bigg(8-\dfrac{1}{4}\bigg)\bigg)}$
		$=\dfrac{36}{\bigg(\dfrac{225}{4}\bigg)}=\dfrac{144}{225}=\dfrac{16}{25}=0.64$

♦♦♦ Mean mark (c) 24%.
MARKER’S COMMENT: Simplifying fractions caused problems. Cancelling factors will assist with calculations.

Calculus, MET1 2022 VCAA 2b

Evaluate `\int_0^1(f(x)(2 f(x)-3))dx`, where `\int_0^1[f(x)]^2 dx=\frac{1}{5}` and `\int_0^1 f(x) dx=\frac{1}{3}`. (3 marks)

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`-\frac{3}{5}`

Show Worked Solution

`\int_0^1(f(x)(2 f(x)-3)) d x`	`=\int_0^1\left(2[f(x)]^2-3 f(x)\right) d x`
	`=2 \int_0^1[f(x)]^2 d x-3 \int_0^1 f(x) d x`
	`=2 \cdot \frac{1}{5}-3 \cdot \frac{1}{3}`
	`=-\frac{3}{5}`

♦ Mean mark 50%.

Calculus, MET1 2022 VCAA 2a

Let `g:\left(\frac{3}{2}, \infty\right) \rightarrow R, g(x)=\frac{3}{2 x-3}`

Find the rule for an antiderivative of `g(x)`. (1 mark)

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`\frac{3}{2} \log _e(2 x-3)`

Show Worked Solution

`\int \frac{3}{2 x-3} d x`	`=frac{3}{2}\int \frac{2}{2 x-3} d x`
	`=\frac{3}{2} \log _e(2 x-3)`

Notes:

→ As the domain is `\left(\frac{3}{2}, \infty\right)` the natural log to the base `e` is used

→ A constant is not required in the solution as the question only asks for AN antiderivative not a family of solutions.

♦ Mean mark 50%.

Graphs, MET1 2023 VCAA 3

Sketch the graph of $f(x)=2-\dfrac{3}{x-1}$ on the axes below, labelling all asymptotes with their equation and axial intercepts with their coordinates. (3 marks)

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Find the values of $x$ for which $f(x)\leq1$. (1 mark)
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b. $1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]$

Show Worked Solution

a. $\text{Vertical asymptote when}\ \ x=1$

$y\text{-int:}\ y=2-(-3)\ \ \Rightarrow\ \ y=5$

$x\text{-int:}\ 2-\dfrac{3}{x-1}=0\ \ \Rightarrow\ \ x=\dfrac{5}{2} $

$\text{As}\ \ x \rightarrow \infty, \ \ y \rightarrow 2^{-}; \ \ x \rightarrow -\infty, \ \ y \rightarrow 2^{+} $

b. $\text{From the graph:}$

$f(x)=1\ \text{when }x=4$

$x>1\ \text{to the right of the vertical asymptote}$

$\therefore\ f(x)\leq1\ \text{when}\ \ 1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]$

♦♦ Mean mark (b) 38%.
MARKER’S COMMENT: Many students did not use their graph from part (a).

Vectors, SPEC1 2023 VCAA 10

The position vector of a particle at time $t$ seconds is given by

$\underset{\sim}{\text{r}}(t)=\big{(}5-6 \ \sin ^2(t) \big{)} \underset{\sim}{\text{i}}+(1+6 \ \sin (t) \cos (t)) \underset{\sim}{\text{j}}$, where $t \geq 0$.

Write $5-6\, \sin ^2(t)$ in the form $\alpha+\beta\, \cos (2 t)$, where $\alpha, \beta \in Z^{+}$. (1 mark)

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Show that the Cartesian equation of the path of the particle is $(x-2)^2+(y-1)^2=9.$ (2 marks)

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The particle is at point $A$ when $t=0$ and at point $B$ when $t=a$, where $a$ is a positive real constant.
If the distance travelled along the curve from $A$ to $B$ is $\dfrac{3 \pi}{4}$, find $a$. (1 mark)

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Find all values of $t$ for which the position vector of the particle, $\underset{\sim}{\text{r}}(t)$, is perpendicular to its velocity vector, $\underset{\sim}{\dot{\text{r}}}(t)$. (2 marks)

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Show Answers Only

a. $2+3\, \cos (2 t) $

b. $ x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)$

$y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)$

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

c. $a=\dfrac{\pi}{8}$

d. $t =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…) $

Show Worked Solution

a. $5-6\, \sin ^2(t)=5-6 \times \dfrac{1}{2}(1-\cos (2 t))$

$\ \ \ \quad \quad \quad \quad \quad \quad \begin{aligned}
& =5-3+3\, \cos (2 t) \\
& =2+3\, \cos (2 t)
\end{aligned}$

b. $ x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)$

$y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)$

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

c. $\text { Motion is circular, centre }(2,1) \text {, radius }=3$

\begin{aligned}
\text { Arc length } & = r \theta \\
3 \theta & =\dfrac{3 \pi}{4} \\
\theta & =\dfrac{\pi}{4}
\end{aligned}

$\therefore a=\dfrac{\pi}{8}$

d. $\underset{\sim}{r}=(2+3\, \cos (2 t)) \underset{\sim}{i}+(1+3\, \sin (2 t)) \underset{\sim}{j}$

$\underset{\sim}{\dot{r}}=-6\, \sin (2 t) \underset{\sim}{i}+6\, \cos (2 t) \underset{\sim}{j}$

$\text { Find } t \text { when } \underset{\sim}{r} \cdot \underset{\sim}{\dot{r}}=0 \text { : }$

\begin{aligned}
\underset{\sim}{r} \cdot \underset{\sim}{\dot{r}} & =6\left(\begin{array}{l}
2+3\, \cos (2 t) \\
1+3\, \sin (2 t)
\end{array}\right)\left(\begin{array}{l}
-\sin (2 t) \\
\cos (2 t)
\end{array}\right) \\
0 &=-2\,\sin (2 t)-3\, \cos (2 t) \sin (2 t)+\cos (2 t)+3\, \cos (2 t) \sin (2 t)\\
0 &=-2\, \sin (2 t)+\cos (2 t)\\
2\, \sin (2 t) &=\cos (2 t)\\
\tan (2 t) & =\dfrac{1}{2} \\
2 t & =\tan ^{-1}\left(\dfrac{1}{2}\right)+k \pi \\
t & =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…)
\end{aligned}

Vectors, SPEC1 2023 VCAA 9

A plane contains the points $ A(1,3,-2), B(-1,-2,4)$ and $ C(a,-1,5)$, where $a$ is a real constant. The plane has a $y$-axis intercept of 2 at the point $D$.

Write down the coordinates of point $D$. (1 mark)

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Show that $\overrightarrow{A B}$ and $\overrightarrow{A D}$ are $-2 \underset{\sim}{\text{i}}-5 \underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}$ and $-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2 \underset{\sim}{\text{k}}$, respectively. (1 mark)

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Hence find the equation of the plane in Cartesian form. (2 marks)

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Find $a$. (1 mark)

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$\overline{A B}$ and $\overline{A D}$ are adjacent sides of a parallelogram. Find the area of this parallelogram. (1 mark)

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Show Answers Only

a. $D(0,2,0)$

b. $\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}$

$\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}$

c. $4 x+2 y+3 z =4$

d. $a=-\dfrac{9}{4}$

e. $A=\sqrt{29}$

Show Worked Solution

a. $D(0,2,0)$

c. $\overrightarrow{A B} \times \overrightarrow{A D}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ -2 & -5 & 6 \\ -1 & -1 & 2\end{array}\right|$

$\ \quad \quad \quad \quad \ \begin{aligned} & =(-10+6)\underset{\sim}{\text{i}}-(-4+6)\underset{\sim}{\text{j}}+(2-5) \underset{\sim}{\text{k}} \\ & =-4 \underset{\sim}{i}-2 \underset{\sim}{j}-3 \underset{\sim}{k}\end{aligned}$

$\text{Plane:}\ \ -4 x-2 y-3 z=k$

$\text{Substitute}\ D(0,2,0)\ \text{into equation}\ \ \Rightarrow \ \text{k}=-4$

\begin{aligned}
\therefore-4 x-2 y-3 z & =-4 \\
4 x+2 y+3 z & =4
\end{aligned}

d. $\text{Find}\ a\ \Rightarrow \ \text{substitute}\ (a, 1,-5)\ \text{into plane equation:}$

\begin{aligned}
-4 & =-4a+2-15 \\
4 a & =-9 \\
a & =-\dfrac{9}{4}
\end{aligned}

e.	$\text{Area}$	$=\|\overrightarrow{A B} \times \overrightarrow{A D}\|$
		$=\|-4\underset{\sim}{i}-2\underset{\sim}{j}-3\underset{\sim}{k}\|$
		$=\sqrt{16+4+9} $
		$=\sqrt{29}$

Calculus, SPEC1 2023 VCAA 7

The curve defined by the parametric equations

$x=\dfrac{t^2}{4}-1, \ y=\sqrt{3} t$, where $0 \leq t \leq 2 \text {, }$

is rotated about the $x$-axis to form an open hollow surface of revolution.

Find the surface area of the surface of revolution.

Give your answer in the form $\pi\left(\dfrac{a \sqrt{b}}{c}-d\right)$, where $a, b, c$ and $d \in Z^{+}$.

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Show Answers Only

$\pi\Bigg{(} \dfrac{64\sqrt3}{3}-24\Bigg{)} $

Show Worked Solution

$x=\dfrac{t^2}{4}-1 \ \Rightarrow \dfrac{dx}{dt}=\dfrac{t}{2} $

$y=\sqrt{3} t \ \Rightarrow \dfrac{dy}{dt} = \sqrt3$

$\text{S.A.}$	\[=2\pi \int_0^2 \sqrt3 t \times \sqrt{\Big{(}\dfrac{dx}{dt}\Big{)}^2+\Big{(}\frac{dy}{dt}\Big{)}^2}\ dt\]
	\[=2\sqrt3 \pi \int_0^2 t\sqrt{\dfrac{t^2}{4}+3}\ dt\]

$\text{Let}\ \ u=\dfrac{t^2}{4}\ \ \Rightarrow \ \dfrac{du}{dt}=\dfrac{t}{2}\ \ \Rightarrow \ 2\,du=t\,dt$

$\text{When}\ \ t=2, u=4; \ t=0, u=3$

$\text{S.A.}$	\[=4\sqrt3\pi \times \dfrac{2}{3}\Big{[}u^{\frac{3}{2}}\Big{]}_3^4 \]
	$=\dfrac{8\sqrt3 \pi}{3}\big{(}4^{\frac{3}{2}}-3^{\frac{3}{2}}\big{)} $
	$=\dfrac{8\sqrt3 \pi}{3}(8-3\sqrt3) $
	$=\pi\Bigg{(} \dfrac{64\sqrt3}{3}-24\Bigg{)} $

Networks, GEN2 2023 VCAA 14

One of the landmarks in state $A$ requires a renovation project.

This project involves 12 activities, $A$ to $L$. The directed network below shows these activities and their completion times, in days.

The table below shows the 12 activities that need to be completed for the renovation project.

It also shows the earliest start time (EST), the duration, and the immediate predecessors for the activities.

The immediate predecessor(s) for activity $I$ and the EST for activity $J$ are missing.

\begin{array} {|c|c|c|}
\hline
\quad \textbf{Activity} \quad & \quad\quad\textbf{EST} \quad\quad& \quad\textbf{Duration}\quad & \textbf{Immediate} \\
& & & \textbf{predecessor(s)} \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 0 & 6 & - \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 0 & 4 & - \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 6 & 7 & A \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 & 5 & B \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 4 & 10 & B \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 13 & 4 & C \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 9 & 3 & D \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 9 & 7 & D \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 13 & 6 & - \\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & - & 6 & E, H \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 19 & 4 & F, I \\
\hline
\rule{0pt}{2.5ex} L \rule[-1ex]{0pt}{0pt} & 23 & 1 & J, K \\
\hline
\end{array}

Write down the immediate predecessor(s) for activity $I$. (1 mark)

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What is the earliest start time, in days, for activity $J$ ? (1 mark)

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How many activities have a float time of zero? (1 mark)

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The managers of the project are able to reduce the time, in days, of six activities.

These reductions will result in an increase in the cost of completing the activity.

The maximum decrease in time of any activity is two days.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \quad \quad A \quad \quad & \quad \quad B \quad \quad& \quad \quad F \quad \quad & \quad \quad H \quad \quad & \quad \quad I \quad \quad & \quad \quad K \quad \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{Daily cost (\$)} \rule[-1ex]{0pt}{0pt} & 1500 & 2000 & 2500 & 1000 & 1500 & 3000 \\
\hline
\end{array}

If activities $A$ and $B$ have their completion time reduced by two days each, the overall completion time of the project will be reduced.
What will be the maximum reduction time, in days? (1 mark)
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The managers of the project have a maximum budget of $15 000 to reduce the time for several activities to produce the maximum reduction in the project's overall completion time.
Complete the table below, showing the reductions in individual activity completion times that would achieve the earliest completion time within the $ 15 000 budget. (1 mark)

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\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

Show Answers Only

a. $\text{Immediate predecessors of}\ I:\ C, G$

b. $\text{EST}(J) = 4+5+7 = 16\ \text{days}$

c. $\text{5 activities have a float time of zero.}$

d. $\text{Maximum reduction time = 2 days}$

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 1\\
\hline
\end{array}

Show Worked Solution

a. $\text{Immediate predecessors of}\ I:\ C, G$

$\text{Dummy activity before activity}\ C\ \text{does not effect this.}$

b. $\text{Scan network:}$

$\text{EST}(J) = 4+5+7 = 16\ \text{days}$

c. $\text{Critical Path:}\ A\ C\ I\ K\ L$

$\text{Activities on the critical path have a float time of zero.}$

$\Rightarrow \ \text{5 activities have a float time of zero.}$

d. $\text{If activities}\ A\ \text{and}\ B\ \text{are reduced by 2 days,}$

$\text{the critical path remains:}\ A\ C\ I\ K\ L\ \text{(22 days)}$

$\text{Maximum reduction time = 2 days}$

♦♦ Mean mark (c) 35%.
♦♦ Mean mark (d) 33%.

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 1\\
\hline
\end{array}

♦♦♦ Mean mark (e) 9%.

Networks, GEN2 2023 VCAA 13

The state $A$ has nine landmarks, $G, H, I, J, K, L, M, N$ and $O$.

The edges on the graph represent the roads between the landmarks.

The numbers on each edge represent the length, in kilometres, along each road.

Three friends, Eden, Reynold and Shyla, meet at landmark $G$.

Eden would like to visit landmark $M$.
What is the minimum distance Eden could travel from $G$ to $M$ ? (1 mark)

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Reynold would like to visit all the landmarks and return to $G$.
Write down a route that Reynold could follow to minimise the total distance travelled. (1 mark)

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Shyla would like to travel along all the roads.
To complete this journey in the minimum distance, she will travel along two roads twice.
Shyla will leave from landmark $G$ but end at a different landmark.
Complete the following by filling in the boxes provided.
The two roads that will be travelled along twice are the roads between: (1 mark)
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Show Answers Only

a. $\text{Shortest path:}\ G\ K\ I\ M$

$\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}$

b. $\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:$

$G\ H\ K\ I\ J\ M\ O\ L\ N\ G$

$G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}$

c. $\text{vertex}\ L\ \text{and vertex}\ N$

$\text{vertex}\ J\ \text{and vertex}\ M$

Show Worked Solution

a. $\text{Shortest path:}\ G\ K\ I\ M$

$\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}$

b. $\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:$

$G\ H\ K\ I\ J\ M\ O\ L\ N\ G$

$G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}$

♦ Mean mark (b) 44%.

c. $\text{Using an educated guess and check methodology:}$

$\text{vertex}\ L\ \text{and vertex}\ N$

$\text{vertex}\ J\ \text{and vertex}\ M$

♦♦♦ Mean mark (c) 12%.

Matrices, GEN2 2023 VCAA 11

The circus requires 180 workers to put on each show.

From one show to the next, workers can either continue working $(W)$ or they can leave the circus $(L)$.

Once workers leave the circus, they do not return.

It is known that 95% of the workers continue working at the circus.

This situation can be modelled by the matrix recurrence relation

$S_0=\begin{bmatrix}180\\ 0\end{bmatrix}, \quad \quad S_{n+1}=T S_n+B$

Write down matrix $T$, the transition matrix, for this recurrence relation. (1 mark)

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${\displaystyle}
\begin{aligned}
& \quad \quad\quad \ \ \ \textit{this show}\\
& \quad \quad \quad \ \ \ W \quad \quad L \\
& T=\begin{bmatrix}
\ \rule[-3ex]{1cm}{0.15mm} & \ \rule[-3ex]{1cm}{0.15mm} \ \\
\ \rule[-3ex]{1cm}{0.15mm} & \ \rule[-3ex]{1cm}{0.15mm} \ \\
\rule[1ex]{0pt}{0pt}
\end{bmatrix} \begin{array}{ll}
&\rule[0ex]{0pt}{0pt}\\
\rule[-3ex]{0pt}{0pt}W\\
\rule[-3ex]{0pt}{0pt}L
\end{array} \ \textit{ next show}
&
\end{aligned}$

Write down matrix $B$ for this recurrence relation to ensure that the circus always has 180 workers. (1 mark)
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${\displaystyle}
\begin{aligned}
& B=\begin{bmatrix}
\ \rule[-3ex]{1cm}{0.15mm}\ \\
\ \rule[-3ex]{1cm}{0.15mm} \ \\
\rule[1ex]{0pt}{0pt}
\end{bmatrix}
&
\end{aligned}$

Show Answers Only

a. $ T = \begin{bmatrix} 0.95 & 0 \\ 0.05 & 1 \end{bmatrix} $

b. $ B = \begin{bmatrix} 0 \\ 9 \end{bmatrix} $

Show Worked Solution

a. $ T = \begin{bmatrix} 0.95 & 0 \\ 0.05 & 1 \end{bmatrix} $

$\text{Note that}\ \ t_{22}=1\ \ \text{so all employees who have left are counted.}$

b. $\text{First transition:}$

$S_1= T \times S_0+B = \begin{bmatrix} 0.95 & 0 \\ 0.05 & 1 \end{bmatrix} \times \begin{bmatrix}180\\ 0\end{bmatrix} +B = \begin{bmatrix} 171\\ 9\end{bmatrix} + B$

$\Rightarrow\ \text{9 extra workers are needed each month to ensure a total of 180 workers.}$

$ \therefore B = \begin{bmatrix} 9 \\ 0 \end{bmatrix} $

♦♦♦ Mean mark (a) 25%.
♦♦ Mean mark (b) 39%.

Matrices, GEN2 2023 VCAA 10

Within the circus, there are different types of employees: directors $(D)$, managers $(M)$, performers $(P)$ and sales staff $(S).$ Customers $(C)$ attend the circus.

Communication between the five groups depends on whether they are customers or employees, and on what type of employee they are.

Matrix $G$ below shows the communication links between the five groups.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \ \ M \ \ P \ \ \ S \ \ \ C \\
& G=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0
\end{bmatrix}\\
&
\end{aligned}

In this matrix:

- The ' 1 ' in row $D$, column $M$ indicates that the directors can communicate directly with the managers.
- The ' 0 ' in row $P$, column $D$ indicates that the performers cannot communicate directly with the directors.

A customer wants to make a complaint to a director.
What is the shortest communication sequence that will successfully get this complaint to a director? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Matrix $H$ below shows the number of two-step communication links between each group. Sixteen elements in this matrix are missing.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \quad M \quad P \quad \ S \quad \ C \\
& H=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix} {\displaystyle}
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\
&
\end{aligned}

1. Complete matrix $H$ above by filling in the missing elements. (1 mark)
  --- 0 WORK AREA LINES (style=lined) ---
2. What information do elements $g_{21}$ and $h_{21}$ provide about the communication between the circus employees? (1 mark)
  --- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $C\ S\ M\ D$

b.i. $ \quad $ $\begin{aligned} & \begin{array}{cccccc}\quad \ \ \ D \ \ M \ \ \ P \ \ \ S \ \ \ C\end{array} \\ &
\begin{array}{c}D \\ M \\P \\ S \\ C\end{array}
\begin{bmatrix}
1 & 2 & 1 & 2 & 2 \\
0 & 3 & 1 & 2 & 2 \\
1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\ & \end{aligned}$

b.ii. $g_{21}\ \text{indicates managers can communicate directly with directors.}$

$h_{21}\ \text{indicates managers, however, cannot communicate with}$

$\text{directors through another person who speaks directly to a director.}$

Show Worked Solution

a. $C\ S\ M\ D$

b.ii. $g_{21}\ \text{indicates managers can communicate directly with directors.}$

$h_{21}\ \text{indicates managers, however, cannot communicate with}$

$\text{directors through another person who speaks directly to a director.}$

♦♦♦ Mean mark (b)(ii) 25%.

Matrices, GEN2 2023 VCAA 9

The circus is held at five different locations, $E, F, G, H$ and $I$.

The table below shows the total revenue for the ticket sales, rounded to the nearest hundred dollars, for the last 20 performances held at each of the five locations.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Location} \rule[-1ex]{0pt}{0pt} & E & F & G & H & I \\
\hline
\rule{0pt}{2.5ex} \textbf{Ticket Sales} \rule[-1ex]{0pt}{0pt} & \$960\ 000 & \$990\ 500 & \$940\ 100 & \$920\ 800 & \$901\ 300 \\
\hline
\end{array}

The ticket sales information is presented in matrix $R$ below.

$R=\begin{bmatrix}
960\ 000 & 990\ 500 & 940\ 100 & 920\ 800 & 901\ 300
\end{bmatrix}$

Complete the matrix equation below that calculates the average ticket sales per performance at each of the five locations. (1 mark)

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$\begin {bmatrix}\rule{2cm}{0.25mm} \end {bmatrix}\times R = \begin {bmatrix}\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} \end {bmatrix}$

The circus would like to increase its total revenue from the ticket sales from all five locations.

The circus will use the following matrix calculation to target the next 20 performances.

$ [t] \times R \times \begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{bmatrix}$

Determine the value of $t$ if the circus would like to increase its revenue from ticket sales by 25%. (1 mark)

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The circus moves from one location to the next each month. It rotates through each of the five locations, before starting the cycle again.

The following matrix displays the movement between the five locations.

\begin{aligned}
& \quad \ \ \ this \ month\\
& \ \ \ E \ \ \ F \ \ \ G \ \ \ H \ \ \ I \\
& \begin{bmatrix}
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0
\end{bmatrix} \begin{array}{ll}
E & \\
F\\
G & \ \ next \ month \\
H & \\
I
\end{array}\\
&
\end{aligned}

The circus started in town $I$.
What is the order in which the circus will visit the five towns? (1 mark)

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Show Answers Only

a. $\Big{[} \dfrac{1}{20} \Big{]} \times R = [48\ 000\ \ \ 49\ 525\ \ \ 47\ 005\ \ \ 46\ 040\ \ \ 45\ 065] $

b. $t=1.25$

c. $I\ H\ E\ G\ F$

Show Worked Solution

a. $\Big{[} \dfrac{1}{20} \Big{]} \times R = [48\ 000\ \ \ 49\ 525\ \ \ 47\ 005\ \ \ 46\ 040\ \ \ 45\ 065] $

b. $t=1.25$

c. $I\ H\ E\ G\ F$

$\text{Process: look for a 1 in “this month” column}\ I $

$\Rightarrow\ \text{corresponds to “next month” letter}\ H$

$\text{Repeat by then looking for a 1 in “this month” column}\ H $

$\Rightarrow\ \text{corresponds to “next month” letter}\ E\ \text{etc…}$

♦♦♦ Mean mark (a) 25%
♦ Mean mark (b) 47%.

Matrices, GEN2 2023 VCAA 8

A circus sells three different types of tickets: family $(F)$, adult $(A)$ and child $(C)$.

The cost of admission, in dollars, for each ticket type is presented in matrix $N$ below.

$N=\begin{bmatrix}
36 \\
15 \\
8
\end{bmatrix}\begin{aligned}
F \\
A \\
C
\end{aligned}$

The element in row $i$ and column $j$ of matrix $N$ is $n_{i j}$.

Which element shows the cost for one child ticket? (1 mark)

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A family ticket will allow admission for two adults and two children.
Complete the matrix equation below to show that purchasing a family ticket could give families a saving of $10. (1 mark)

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$\displaystyle{\begin {bmatrix} 0 &2&2 \end {bmatrix} \times N - \begin{bmatrix} \rule{1cm}{0.25mm} & \rule{1cm}{0.25mm} & \rule{1cm}{0.25mm} \end {bmatrix} \times N = \left[ 10\right]}$

On the opening night, the circus sold 204 family tickets, 162 adult tickets and 176 child tickets.
The owners of the circus want a 3 × 1 product matrix that displays the revenue for each ticket type: family, adult and child.
This product matrix can be achieved by completing the following matrix multiplication.

$K \times N=\begin{bmatrix}
7344 \\
2430 \\
1408
\end{bmatrix}$

Write down matrix $K$. (1 mark)

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

Show Answers Only

a. $n_{31}$

b. $ \begin{bmatrix} 0 & 2 & 2 \end{bmatrix} \times \begin{bmatrix} 36 \\ 15 \\ 8 \end{bmatrix} – \ \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 36 \\ 15 \\ 8 \end{bmatrix} $

$ = [0 \times 36 + 2 \times 15 + 2 \times 8]-[1 \times 36 + 0 \times 15 + 0 \times 8] $

\[ = \left[\begin{array}{c} 46 \end{array}\right]-\left[\begin{array}{c} 36 \end{array}\right] \]

\[ = \left[\begin{array}{c} 10 \end{array}\right] \]

c. $ \begin{bmatrix}204 & 0 & 0 \\ 0 & 162 & 0 \\ 0 & 0 & 176\end{bmatrix} \begin{bmatrix}36 \\ 15 \\ 8\end{bmatrix} = \begin{bmatrix}204 \times 36 \\ 162 \times 15 \\ 176 \times 8\end{bmatrix} = \begin{bmatrix}7344 \\ 2430 \\ 1408\end{bmatrix}$

Show Worked Solution

a. $\text{Cost of one child ticket is in row 3, column 1}$

$\Rightarrow n_{31}$

♦ Mean mark (a) 48%.

b. $ \begin{bmatrix} 0 & 2 & 2 \end{bmatrix} \times \begin{bmatrix} 36 \\ 15 \\ 8 \end{bmatrix} – \ \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 36 \\ 15 \\ 8 \end{bmatrix} $

$ = [0 \times 36 + 2 \times 15 + 2 \times 8]-[1 \times 36 + 0 \times 15 + 0 \times 8] $

\[ = \left[\begin{array}{c} 46 \end{array}\right]-\left[\begin{array}{c} 36 \end{array}\right] \]

\[ = \left[\begin{array}{c} 10 \end{array}\right] \]

♦ Mean mark (b) 44%.

♦♦ Mean mark 34%.

Recursion and Finance, GEN2 2023 VCAA 7

Arthur takes out a new loan of $60 000 to pay for an overseas holiday.

Interest on this loan compounds weekly.

The balance of the loan, in dollars, after $n$ weeks, $V_n$, can be determined using a recurrence relation of the form

$V_0=60\ 000, \quad V_{n+1}=1.0015\,V_n-d$

Show that the interest rate for this loan is 7.8% per annum. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Determine the value of $d$ in the recurrence relation if
i. Arthur makes interest-only repayments (1 mark)

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ii. Arthur fully repays the loan in five years. Round your answer to the nearest cent. (1 mark)

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Arthur decides that the value of $d$ will be 300 for the first year of repayments.
If Arthur fully repays the loan with exactly three more years of repayments, what new value of $d$ will apply for these three years? Round your answer to the nearest cent. (1 mark)

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For what value of $d$ does the recurrence relation generate a geometric sequence? (1 mark)

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Show Answers Only

Show Worked Solution

a. $\text{Weekly interest rate factor}\ = 1.0015-1 = 0.0015 = 0.15\% $

$I\%(\text{annual}) = 52 \times 0.15 = 7.8\%$

♦♦ Mean mark (a) 32%.

b.i. $d= 0.15\% \times 60\ 000 = \dfrac{0.15}{100} \times 60\ 000 = 90$

b.ii. $\text{By TVM solver:} $

$N$	$=5 \times 52 = 260$
$I\%$	$=7.8$
$PV$	$= -60\ 000$
$PMT$	$= ?$
$FV$	$=0$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow PMT = d= $278.86$

♦ Mean mark (b)(i) 47%.
♦ Mean mark (b)(ii) 40%.

c. $d=300\ \text{for the 1st 52 weeks.}$

$\text{Find}\ FV\ \text{after 52 weeks:}$

$N$	$=52$
$I\%$	$=7.8$
$PV$	$= -60\ 000$
$PMT$	$= 300$
$FV$	$=?$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow FV = $48\ 651.67$

$\text{Find}\ PMT\ \text{given}\ FV=0\ \text{after 156 more weeks:}$

$N$	$=156$
$I\%$	$=7.8$
$PV$	$= -48\ 651.67$
$PMT$	$= ?$
$FV$	$=0$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow PMT = d= $350.01$

♦♦ Mean mark (c) 32%.

d. $\text{Geometric sequence when}\ \ d=0$

$V_0=60\ 000, V_1=60\ 000(1.0015), V_2=60\ 000(1.0015)^2, …$

♦♦♦ Mean mark (d) 11%.

Recursion and Finance, GEN2 2023 VCAA 6

Arthur invests $600 000 in an annuity that provides him with a monthly payment of $3973.00.

Interest is calculated monthly.

Three lines of the amortisation table for this annuity are shown below.

\begin{array} {|c|c|}
\hline
\textbf{Payment} & \textbf{Payment} & \textbf{Interest} & \textbf{Principal reduction} & \textbf{Balance} \\
\textbf{number} & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & 0.00 & 0.00 & 0.00 & 600\ 000.00 \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2520.00 & 1453.00& 598\ 547.00\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2513.90 & 1459.10 & 597\ 087.90 \\
\hline
\end{array}

The interest rate for the annuity is 0.42% per month.
Determine the interest rate per annum. (1 mark)

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Using the values in the table, complete the next line of the amortisation table.
Write your answers in the spaces provided in the table below.
Round all values to the nearest cent. (1 mark)

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Let $V_n$ be the balance of Arthur's annuity, in dollars, after $n$ months.
Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$ that can model the value of the annuity from month to month. (1 mark)

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The amortisation tables above show that the balance of the annuity reduces each month.
If the balance of an annuity remained constant from month to month, what name would be given to this type of annuity? (1 mark)

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Show Answers Only

a. $I\% (\text{annual}) = 12 \times 0.42 = 5.04\% $

b. $\text{Row 3 calculations are as follows:}$

$\text{Payment}\ = \$3973.00\ \text{(remains constant)}$

$\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 $

$\text{Principal reduction}\ = 3973.00-2507.77 = \$1465.23 $

$\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67$

c. $V_0 = 600\ 000$

$V_{n+1} = 1.0042 \times V_n-3973$

d. $\text{Perpetuity}$

Show Worked Solution

a. $I\% (\text{annual}) = 12 \times 0.42 = 5.04\% $

b. $\text{Row 3 calculations are as follows:}$

$\text{Payment}\ = \$3973.00\ \text{(remains constant)}$

$\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 $

$\text{Principal reduction}\ = 3973.00+2507.77 = \$1465.23 $

$\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67$

♦♦ Mean mark (b) 40%.

c. $V_0 = 600\ 000$

$V_{n+1} = 1.0042 \times V_n-3973$

♦ Mean mark (c) 41%.

d. $\text{Perpetuity}$

Recursion and Finance, GEN2 2023 VCAA 5

Arthur borrowed $30 000 to buy a new motorcycle.

Interest on this loan is charged at the rate of 6.4% per annum, compounding quarterly.

Arthur will repay the loan in full with quarterly repayments over six years.

How many repayments, in total, will Arthur make? (1 mark)

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The balance of the loan, in dollars, after $n$ quarters, $A_n$, can be modelled by the recurrence relation

$A_0=30\ 000, \quad A_{n+1}=1.016 A_n-1515.18$

Showing recursive calculations, determine the balance of the loan after two quarters.
Round your answer to the nearest cent. (1 mark)

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The final repayment required will differ slightly from all the earlier repayments of $1515.18
Determine the value of the final repayment.
Round your answer to the nearest cent. (1 mark)

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Show Answers Only

a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

c. $ \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Show Worked Solution

a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

Mean mark (b) 51%.
♦♦ Mean mark (c) 25%

c. $\text{Solve for}\ N\ \text{using TMV calculator:}$

$N$	$=24$
$I\%$	$=6.4$
$PV$	$= -30\ 000$
$PMT$	$= 1515.18$
$FV$	$=?$
$\text{P/Y}$	$=\ \text{C/Y}\ = 4$

$FV = -0.14$

$\therefore \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Data Analysis, GEN2 2023 VCAA 4

The time series plot below shows the average monthly ice cream consumption recorded over three years, from January 2010 to December 2012.

The data for the graph was recorded in the Northern Hemisphere.

In this graph, month number 1 is January 2010, month number 2 is February 2010 and so on.

Identify a feature of this plot that is consistent with this time series having a seasonal component. (1 mark)

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The long-term seasonal index for April is 1.05
Determine the deseasonalised value for average monthly ice cream consumption in April 2010 (month 4).
Round your answer to two decimal places. (1 mark)

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The table below shows the average monthly ice cream consumption for 2011 .

Consumption (litres/person)
Year	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sept	Oct	Nov	Dec
2011	0.156	0.150	0.158	0.180	0.200	0.210	0.183	0.172	0.162	0.145	0.134	0.154

Show that, when rounded to two decimal places, the seasonal index for July 2011 estimated from this data is 1.10. (2 marks)

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Show Answers Only

a. $\text{Maximum values are 12 months apart.}$

b. $\text{Deseasonalised (Apr)}\ = \dfrac{0.180}{1.05} = 0.1714… = 0.17\ \text{(2 d.p.)}$

c. $\text{2011 monthly mean}\ =\dfrac{2.004}{12}=0.167$

$\text{Seasonal index (July)}$	$=\dfrac{0.183}{0.167}$
	$=1.095…$
	$=1.10\ \text{(2 d.p.)}$

Show Worked Solution

a. $\text{Maximum values are 12 months apart.}$

♦♦♦ Mean mark (a) 26%.

b. $\text{Deseasonalised (Apr)}\ = \dfrac{0.180}{1.05} = 0.1714… = 0.17\ \text{(2 d.p.)}$

c. $\text{2011 monthly mean}$

$=(0.156+0.15+0.158+0.18+0.2+0.21+0.183+0.172+$

$0.162+0.145+0.134+0.154)\ ÷\ 12 $

$=\dfrac{2.004}{12}$

$=0.167$

$\text{Seasonal index (July)}$	$=\dfrac{0.183}{0.167}$
	$=1.095…$
	$=1.10\ \text{(2 d.p.)}$

♦♦ Mean mark (c) 36%.

Data Analysis, GEN2 2023 VCAA 3

The scatterplot below plots the average monthly ice cream consumption, in litres/person, against average monthly temperature, in °C. The data for the graph was recorded in the Northern Hemisphere.

When a least squares line is fitted to the scatterplot, the equation is found to be:

consumption = 0.1404 + 0.0024 × temperature

The coefficient of determination is 0.7212

Draw the least squares line on the scatterplot graph above. (1 mark)

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Determine the value of the correlation coefficient $r$.
Round your answer to three decimal places. (1 mark)

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Describe the association between average monthly ice cream consumption and average monthly temperature in terms of strength, direction and form. (1 mark)

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\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \quad \quad \quad \quad \quad \quad \quad \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}
Referring to the equation of the least squares line, interpret the value of the intercept in terms of the variables consumption and temperature. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Use the equation of the least squares line to predict the average monthly ice cream consumption, in litres per person, when the monthly average temperature is –6°C. (1 mark)

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Write down whether this prediction is an interpolation or an extrapolation. (1 mark)

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

Show Answers Only

b. $r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \text{strong} \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \text{linear} \\
\hline
\end{array}

d. $\text{At 0°C, the predicted average consumption is:}$

$\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}$

e. $\text{Find consumption} (c)\ \text{when temperature} (t) = -6:$

$c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} $

f. $\text{Extrapolation (even though the axes extend to –6°C, the data set} $

$\text{range finishes with a lower limit around –4.5°C.)}$

Show Worked Solution

b. $r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}$

d. $\text{At 0°C, the predicted average consumption is:}$

$\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}$

e. $\text{Find consumption} (c)\ \text{when temperature} (t) = -6:$

$c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} $

f. $\text{Extrapolation (even though the axes extend to –6°C, the data set} $

$\text{range finishes with a lower limit around –4.5°C.)}$

Networks, GEN1 2022 VCAA 7-8 MC

A project involves 11 activities, $A$ to $K$.

The table below shows the earliest start time and duration, in days, for each activity.

The immediate predecessor(s) of each activity is also shown.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \textbf{Activity}\ \ & \textbf{Earliest} & \ \ \textbf{Duration}\ \ & \textbf{Immediate}\\
& \textbf{start time} \rule[-1ex]{0pt}{0pt} & &\textbf{predecessor}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \text{0} & \text{6} & \text{-}\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \text{0} & \text{7} & \text{-}\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & \text{6} & \text{10} & A\\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & \text{6} & \text{7} & A\\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & \text{7} & \text{8} & B\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & \text{15} & \text{2} & D,\ E\\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & \text{15} & \text{2} & E\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & \text{17} & \text{3} & G\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & \text{20} & \text{6} & C,\ F,\ H\\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & \text{17} & \text{5} & G\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & \text{26} & \text{2} & I,\ J\\
\hline
\end{array}

Question 7

A directed network for this project will require a dummy activity.

The dummy activity will be drawn from the end of

activity $A$ to the start of activity $D$.
activity $E$ to the start of activity $F$.
activity $F$ to the start of activity $I$.
activity $G$ to the start of activity $H$.
activity $I$ to the start of activity $J$.

Question 8

When this project is completed in the minimum time, the sum of all the float times, in days, will be

Show Answers Only

$\text{Question 7:}\ B$

$\text{Question 8:}\ D$

Show Worked Solution

$\text{Question 7}$

Draw network from table:

→ $F$ starts after completion of both $D$ and $E$.

→ $G$ starts after activity $E$ only.

$\Rightarrow B$

♦ Mean mark (Q7) 49%.

$\text{Question 8}$

Scanning forwards and backwards on network diagram:

→ The critical path is $B E G H I K$

→ Float times occur at all points not on the critical path.

$\text{Total Float Times}$	$= A + C + D + F + J$
	$= 4 + 4 + 5 + 3 + 4$
	$= 20$

$\Rightarrow D$

♦♦ Mean mark (Q8) 34%.

Networks, GEN1 2022 VCAA 6 MC

A landscaping project has 12 activities. The network below gives the time, in hours, that it takes to complete each activity.

The earliest start time, in hours, for activity $G$ is

Show Answers Only

$C$

Show Worked Solution

$\text{Critical Path}$	$= CEGIL\ \ \text{(see diagram above)}$
$\text{EST for}\ G$	$= 5 + 7 = 12\ \text{hours}$

$\Rightarrow C$

♦ Mean mark 48%.

Networks, GEN1 2022 VCAA 5 MC

A connected graph consists of five vertices and four edges.

Which one of the following statements is not true?

The graph could be a tree.
The graph could be planar.
The graph could be bipartite.
The graph could contain a path.
The graph could contain a cycle.

Show Answers Only

$E$

Show Worked Solution

By elimination:

By definition the graph could be a tree or a path (eliminate A and D).

Consider B: Planar graphs satisfy $f + v = e + 2.$ In the connected graph described, $f = 1,\ v = 5,\ e = 4$. Satisfies equation (eliminate B).

Consider C: The graph could be bipartite – see below (eliminate C).

Consider E: The graph could not contain a cycle and remain connected.

$\Rightarrow E$

♦ Mean mark 54%.

Networks, GEN1 2022 VCAA 3 MC

An athletics club needs to select one team of four athletes.

The team is required to have one long jump, one high jump, one shot put and one javelin competitor.

The following table shows the best distances, in metres, for each athlete for each event.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \textbf{Athlete}\ \ & \textbf{Long jump} & \textbf{High jump} & \ \ \textbf{Shot put} \ \ & \quad \textbf{Javelin} \quad \\
\rule[-1ex]{0pt}{0pt}& \textbf{(m)}& \textbf{(m)}& \textbf{(m)}& \textbf{(m)}\\
\hline
\rule{0pt}{2.5ex} \text{Eve} \rule[-1ex]{0pt}{0pt} & \text{4.8} & \text{1.7} & \text{13.1} & \text{40.9} \\
\hline
\rule{0pt}{2.5ex} \text{Harsha} \rule[-1ex]{0pt}{0pt} & \text{4.8} & \text{1.6} & \text{13.9} & \text{39.5} \\
\hline
\rule{0pt}{2.5ex} \text{Shona} \rule[-1ex]{0pt}{0pt} & \text{5.1} & \text{1.8} & \text{14.4} & \text{41.2} \\
\hline
\rule{0pt}{2.5ex} \text{Taylor} \rule[-1ex]{0pt}{0pt} & \text{4.8} & \text{1.7} & \text{12.8} & \text{39.8} \\
\hline
\end{array}

The athletics club will allocate each athlete to one event in order to maximise the total distance that the team jumps and throws.

Which allocation of athlete to event must occur in order to maximise the total distance?

Show Answers Only

$D$

Show Worked Solution

$\text{Consider each option:}$

$\text{Option A: Total distance = 59.6 m} $

$\text{Option B: Total distance = 61.6 m} $

$\text{Option C: Total distance = 60.4 m} $

$\text{Option D: Total distance = 61.8 m} $

$\text{Option E: Total distance = 60.8 m} $

$\Rightarrow D$

♦♦ Mean mark 39%.
COMMENT: The Hungarian Algorithm is likely to be a less efficient strategy for many here.

Networks, STD2 N2 SM-Bank 40

The map below shows seven countries within Central America.

Draw a network diagram of the map where seven vertices represent each of the countries on the map and edges represent a border shared between two countries. (2 marks)

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Show Worked Solution

Networks, GEN1 2022 VCAA 2 MC

The map below shows seven countries within Central America.

A network diagram was drawn with seven vertices to represent each of the countries on the map of Central America. Edges were drawn to represent a border shared between two countries.

The number of edges that this network has is

Show Answers Only

$C$

Show Worked Solution

There are 7 edges.

$\Rightarrow C$

♦ Mean mark 49%.

Matrices, GEN1 2022 VCAA 6 MC

Consider the following system of simultaneous linear equations.

$y+z=4$

$x-y+z=1$

$-x+y=2$

The solution to these simultaneous equations can be found by calculating

Show Answers Only

$E$

Show Worked Solution

$
\begin{bmatrix}
0 & 1 & 1 \\ 1 & -1 & 1 \\
-1 & 1 & 0\end{bmatrix}
\times \begin{bmatrix}
x \\ y \\ z\end{bmatrix}
= \begin{bmatrix}
4 \\ 1 \\ 2\end{bmatrix}
$

$
\begin{bmatrix}
x \\ y \\ z\end{bmatrix}
= \begin{bmatrix}
0 & 1 & 1 \\
1 & -1 & 1 \\
-1 & 1 & 0
\end{bmatrix}^{-1}
\times \begin{bmatrix}
4 \\ 1 \\ 2
\end{bmatrix}
$

$
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
= \begin{bmatrix}
1 & -1 & -2 \\
1 & -1 & -1 \\
0 & 1 & 1
\end{bmatrix}
\times\begin{bmatrix}
4 \\ 1 \\ 2\end{bmatrix}
$

$\Rightarrow E$

♦♦ Mean mark 31%.

MATRICES, FUR1 2022 VCAA 7 MC

Matrix `K` is a permutation matrix.

`K = [(0,0,1,0,0),(0,1,0,0,0),(0,0,0,1,0),(0,0,0,0,1),(1,0,0,0,0)]`

Matrix `M` is a column matrix that is multiplied once by matrix `K` to obtain matrix `P`.

When matrix `M` is multiplied by matrix `K`, the element `m_31` moves to element

`p_11`
`p_21`
`p_31`
`p_41`
`p_51`

Show Answers Only

`A`

Show Worked Solution

`text{The matrix product}\ KM\ text{is column matrix}\ P.`

`text{Row 1 of matrix}\ K\ text{moves the 3rd row of matrix}\ M\ text{to the}`

`text{first row of matrix}\ P.`

`=>A`

♦ Mean mark 42%.

Recursion and Finance, GEN1 2022 VCAA 24 MC

On 1 January 2020, Dion invested $10 500 into an investment account paying compound interest of 0.52% quarterly.

At the end of each quarter, after the interest was credited, Dion added an additional amount of money.

Let $D_n$ represent the additional amount, in dollars, added at the end of quarter $n$.

This additional amount per quarter is modelled by the recurrence relation

$D_1=C,\ \ \ D_{n+1}=D_n$

The balance of Dion's investment account on 1 January 2022 was $12 700.95

The value of $C$ is

$71.69
$215.55
$260.22
$270.15
$275.12

Show Answers Only

$B$

Show Worked Solution

$\text{Annual interest rate}\ = 0.52 \times 4 = 2.08\%$

$D_n = D_{n+1}\ \text{indicates the additional payment is constant}$

$\text{By TVM Solver:}$

$N$	$=4 \times 2 = 8$
$I\%$	$=2.08$
$PV$	$=-10\ 500$
$PMT$	$=?$
$FV$	$=12\ 700.95$
$\text{P/Y}$	$=4$
$\text{C/Y}$	$=4$

$PMT = -215.55$

$\Rightarrow B$

♦♦ Mean mark 32%.

Recursion and Finance, GEN1 2022 VCAA 23 MC

Li invests $4000 for five years at 3.88% per annum, compounding annually.

Joseph invests a sum of money for five years, which earns simple interest paid annually.

Let $J_n$ be the value, in dollars, of Joseph's investment after $n$ years.

The two investments will finish at the same value, rounded to the nearest cent, if Joseph's investment is modelled by which one of the following recurrence relations?

$J_0=2000,\ \ \ J_{n+1}=J_n+467.72$
$J_0=2500,\ \ \ J_{n+1}=J_n+367.72$
$J_0=3000,\ \ \ J_{n+1}=J_n+317.72$
$J_0=3500,\ \ \ J_{n+1}=J_n+267.72$
$J_0=4000,\ \ \ J_{n+1}=J_n+67.72$

Show Answers Only

$D$

Show Worked Solution

$L_n$	$ = L_0 \times 1.0388^n$
$L_5$	$ = 4000 \times 1.0388^5 \approx 4838.60\ \ \text{(2 d.p.)} $

$\text{Check Li’s total versus each option:}$

$J_{n+1}$	$= J_n + 267.72$
$J_5$	$= 3500 + 267.72 \times 5$
	$= 4838.60$

$\Rightarrow D$

♦♦ Mean mark 49%.

Recursion and Finance, GEN1 2022 VCAA 21 MC

Consider the following four statements regarding nominal and effective interest rates as they apply to compound interest investments and loans:

An effective interest rate is the same as a nominal interest rate if interest compounds annually.
Effective interest rates increase as the number of compounding periods per year increases.
A nominal rate of 12% per annum is equivalent to a nominal rate of 1% per month.
An effective interest rate can be lower than a nominal interest rate.

How many of these four statements are true?

Show Answers Only

$D$

Show Worked Solution

$\text{Statement 1}$

$\text{Let}\ \ n = 1\ \ \text{and}\ \ r = 12\% :$

$R_{eff}=\Big{(}1 + \dfrac{12}{100 \times 1}\Big{)}^1-1=12\%\ \ \checkmark$

$\text{Statement 2}$

$\text{Let}\ \ n = 2\ \ \text{and}\ \ r = 12\% :$

$R_{eff}=\Big{(}1 + \dfrac{12}{100 \times 2}\Big{)}^2-1=12.36\%\ \ \checkmark $

$\text{Statement 3}$

$\dfrac{12\text{%}}{12\ \text{months}} = 1\text{%} \ \text{per month}\ \ \checkmark$

$\text{Statement 4}$

$\text{Statements 1 and 2 make the 4th statement incorrect.}$

$\Rightarrow D$

♦♦ Mean mark 33%.

Recursion and Finance, GEN1 2022 VCAA 18-19 MC

The balance of a loan, $V_n$, in dollars, after $n$ months is modelled by the recurrence relation

$V_0=400\ 000,\ \ \ V_{n+1}=1.003\,V_n-2024$

Question 18

The balance of the loan first falls below $398 000 after how many months?

Question 19

With a small change to the final payment, the loan is expected to be repaid in full in

25 years.
26 years.
28 years.
29 years.
30 years.

Show Answers Only

$\text{Question 18: C}$

$\text{Question 19: A}$

Show Worked Solution

$\text{Question 18}$

$V_1$	$=1.003 \times 400\ 000-2024 = $399 176$
$V_2$	$=1.003 \times 399\ 176-2024= $398 349.528$
$V_3$	$=1.003 \times 398\ 349.528-2024= $397 520.58$

$\Rightarrow C$

$\text{Question 19}$

$\text{Monthly interest rate = 0.3%}$

$\text{Annual interest rate}\ (r) = 12 \times 0.3 = 3.6\%$

$\text{Solve for}\ N\ \text{using TMV calculator:}$

$I\%$	$=3.6$
$PV$	$= 400\ 000$
$PMT$	$= -2024$
$FV$	$=0$
$\text{P/Y}$	$=12$
$\text{C/Y}$	$=12$

$N = 300.002\ \text{months}\ \approx \dfrac{300}{12} \approx 25\ \text{years}$

$\Rightarrow A$

♦ Mean mark (Q19) 47%.

Recursion and Finance, GEN1 2022 VCAA 17 MC

A sequence of numbers is generated by the recurrence relation shown below.

$R_0 = 2,\ \ \ R_{n+1} = 2-R_n$

The value of $R_2$ is

$-4$
$-2$
$0$
$2$
$4$

Show Answers Only

$D$

Show Worked Solution

$R_0$	$=2\ \ \ \ \ R_{n+1} = 2-R_n$
$R_1$	$= 2-2 = 0$
$R_2$	$= 2-0 = 2$

$\Rightarrow D$

♦ Mean mark 44%.

Data Analysis, GEN1 2022 VCAA 16 MC

The seasonal index for sales of sunscreen in summer is 1.25

To correct for seasonality, the actual sunscreen sales for summer should be

reduced by 20%
reduced by 25%
reduced by 80%
increased by 20%
increased by 25%

Show Answers Only

$A$

Show Worked Solution

$\text{Deseasonalised Sales}$

	$= \dfrac{\text{Actual Sales}}{\text{Seasonal Index}}$
	$=\dfrac{\text{Actual Sales}}{1.25}$
	$=0.8 \times \text{Actual Sales}$

$\therefore \ \text{Summer sales should be reduced by 20%.}$

$\Rightarrow A$

♦ Mean mark 40%.

Data Analysis, GEN1 2022 VCAA 12-14 MC

The scatterplot below displays the body length, in centimetres, of 17 crocodiles, plotted against their head length, in centimetres. A least squares line has been fitted to the scatterplot. The explanatory variable is head length.

Question 12

The equation of the least squares line is closest to

head length = –40 + 7 × body length
body length = –40 + 7 × head length
head length = 168 + 7 × body length
body length = 168 – 40 × head length
body length = 7 + 168 × head length

Question 13

The median head length of the 17 crocodiles, in centimetres, is closest to

Question 14

The correlation coefficient $r$ is equal to 0.963

The percentage of variation in body length that is not explained by the variation in head length is closest to

0.9%
3.7%
7.3%
92.7%
96.3%

Show Answers Only

$\text{Question 12}:\ B$

$\text{Question 13}:\ B$

$\text{Question 12}:\ C$

Show Worked Solution

$\text{Question 12}$

$\text{Gradient}\ = \dfrac{550-170}{85-30}=6.9\ \ \text{(eliminate D and E)}$

$\text{Head length is the explanatory variable (eliminate A and C)}$

$\Rightarrow B$

$\text{Question 13}$

$\text{Median score}\ =\dfrac{17+1}{2} = 9\text{th score}$

$\text{Median head length = 51 cm}$

$\Rightarrow B$

♦ Mean mark (Q13) 51%.

$\text{Question 14}$

$\text{Percentage explained by variation in head length}$

$r^2 = 0.963^2 = 0.9273 \approx 92.7\% $

$\text{Percentage not explained by variation in head length}$

$100-92.7 \approx 7.3\%$

$\Rightarrow C$

♦ Mean mark (Q14) 48%.

CHEMISTRY, M7 2020 VCE 3

Below is a reaction pathway beginning with hex-3-ene.

Write the IUPAC name of Compound J in the box provided. (1 mark)

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State the reagent(s) required to convert hex-3-ene to hexan-3-ol in the box provided. (1 mark)

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Draw the structural formula for a tertiary alcohol that is an isomer of hexan-3-ol. (1 mark)

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Hexan-3-ol is reacted with Compound M under acidic conditions to produce Compound L.
Draw the semi-structural formula for Compound M in the box provided on the image above. (1 mark)

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i. Draw the semi-structural formula for Compound K in the box provided on the image above. (1 mark)

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ii. Name the class of organic compound (homologous series) to which Compound K belongs. (1 mark)

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What type of reaction produces Compound K from hexan-3-ol? (1 mark)

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a. 3-bromohexane

b. Steam and any specific inorganic strong acid (although not $\ce{HCl}$) is correct.

eg. $\ce{H2O, H+}$

d. Correct answers included one of:

$\ce{CH3COOH}$ or $\ce{HOOCCH3}$

e.i. Correct answers included one of the following:

$\ce{CH3CH2COCH2CH2CH3}$

$\ce{CH3CH2CH2COCH2CH3}$

$\ce{CH3CH2CO(CH2)2CH3}$

e.ii. Ketone

f. Oxidation

Show Worked Solution

a. 3-bromohexane

b. Steam and any specific inorganic strong acid (although not $\ce{HCl}$) is correct.

eg. $\ce{H2O, H+}$

♦ Mean mark (b) 38%.

♦ Mean mark (c) 49%.

d. Correct answers included one of:

$\ce{CH3COOH}$ or $\ce{HOOCCH3}$

e.i. Correct answers included one of the following:

$\ce{CH3CH2COCH2CH2CH3}$

$\ce{CH3CH2CH2COCH2CH3}$

$\ce{CH3CH2CO(CH2)2CH3}$

e.ii. Ketone

f. Oxidation

♦ Mean mark e(i) 39%.

CHEMISTRY, M7 2020 VCE 7 MC

How many structural isomers have the molecular formula $\ce{C3H6BrCl}$?

Show Answers Only

$B$

Show Worked Solution

Isomers are:

$\ce{CH3CH2CHBrCl}$: 1-bromo-1-chloropropane

$\ce{CH3CHClCH2Br}$: 1-bromo-2-chloropropane

$\ce{CH2ClCH2CH2Br}$: 1-bromo-3-chloroopropane

$\ce{CH3CHBrCH2Cl}$: 2-bromo-1-chloropropane

$\ce{CH3CBrClCH3}$: 2-bromo-2-chloropropane

$\Rightarrow B$

♦ Mean mark 46%.

CHEMISTRY, M6 2016 VCE 20 MC

How does diluting a 0.1 M solution of lactic acid, $\ce{HC3H5O3}$, change its pH and percentage ionisation?

	pH	Percentage ionisation
A.	increase	decrease
B.	increase	increase
C.	decrease	increase
D.	decrease	decrease

Show Answers Only

$B$

Show Worked Solution

→ Ionisation of lactic acid: $\ce{HC3H5O3(aq) + H2O(l) \rightleftharpoons C3H5O3-(aq) + H3O+(aq)} $

→ Adding water decreases $\ce{[H3O+]}$. The equilibrium then shifts to partially compensate for this change, favouring the forward reaction.

→ $\ce{[H3O+]}$ increases in this process although the new equilibrium $\ce{[H3O+]}$ is lower (pH higher) than the original system.

→ Overall, the pH increases and the percentage ionisation increases.

$\Rightarrow B$

♦♦ Mean mark 34%.

CHEMISTRY, M7 2016 VCE 7a

Butanoic acid is the simplest carboxylic acid that is also classified as a fatty acid. Butanoic acid may be synthesised as outlined in the following reaction flow chart.

Draw the structural formula of but-1-ene in the box provided. (1 mark)

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State the reagent(s) needed to convert but-1-ene to Compound Y in the box provided. (1 mark)

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Write the systematic name of Compound Y in the box provided. (1 mark)

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Write the semi-structural formula of butanoic acid in the box provided. (1 mark)

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Write a balanced half-equation for the conversion of $\ce{Cr2O7^2– to Cr3+}$. (2 marks)

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ii. $\ce{H2O}$ and $\ce{H3PO4}$ (catalyst)

iii. butan-1-ol or 1-butanol

iv. $\ce{CH3CH2CH2COOH}$

v. $\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- \rightarrow 2Cr^{3+}(aq) + 7H2O(l)} $

Show Worked Solution

ii. $\ce{H2O}$ and $\ce{H3PO4}$ (catalyst)

iii. butan-1-ol or 1-butanol

iv. $\ce{CH3CH2CH2COOH}$

v. $\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- \rightarrow 2Cr^{3+}(aq) + 7H2O(l)} $

♦♦ Mean mark (ii) 29%.

CHEMISTRY, M7 2015 VCE 5c

A student mixed salicylic acid with ethanoic anhydride (acetic anhydride) in the presence of concentrated sulfuric acid. The products of this reaction were the painkilling drug aspirin (acetyl salicylic acid) and ethanoic acid.

An incomplete structure of the aspirin molecule is shown above.
Complete the structure by filling in the two boxes provided in the diagram. (2 marks)

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Sulfuric acid is used as a catalyst in this reaction.
Explain how a catalyst increases the rate of this reaction. (2 marks)

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ii. Sulphuric acid increases the rate of reaction by:

→ providing an alternative reaction pathway that involves a lower activation energy for the reagents.

→ this increases the likelihood of successful collisions.

Show Worked Solution

♦ Mean mark (i) 48%.

ii. Sulphuric acid increases the rate of reaction by:

→ providing an alternative reaction pathway that involves a lower activation energy for the reagents.

→ this increases the likelihood of successful collisions.

CHEMISTRY, M7 2016 VCE 24 MC

Methanol is a liquid fuel that is often used in racing cars. The thermochemical equation for its complete combustion is

$\ce{2CH3OH(l) + 3O2(g)\rightarrow 2CO2(g) + 4H2O(l) \quad \quad \ \ \ \Delta H = –1450 kJ mol^{–1}}$

Octane is a principal constituent of petrol, which is used in many motor vehicles. The thermochemical equation for
the complete combustion of octane is

$\ce{2C8H18(l) + 25O2(g)\rightarrow 16CO2(g) + 18H2O(l) \quad \quad \Delta H = –10\ 900 kJ mol^{-1}}$

The molar mass of methanol is 32 g mol$^{-1}$ and the molar mass of octane is 114 g mol$^{–1}$. Which one of the following statements is the most correct?

Burning just 1.0 g of octane releases almost 96 kJ of heat energy.
Burning just 1.0 g of methanol releases almost 23 kJ of heat energy.
Octane releases almost eight times more energy per kilogram than methanol.
The heat energy released by methanol will not be affected if the oxygen supply is limited.

Show Answers Only

$B$

Show Worked Solution

Consider option B:

→ 1 mole of $\ce{CH3OH}$ produces 725 kJ of heat energy

→ $\ce{MM (CH3OH)} = 32.0\ \text{grams}$

→ Heat energy of 1 gram $\ce{CH3OH} = \dfrac{725}{32.0} = 22.7$ kJ

$\Rightarrow B$

♦ Mean mark 40%.

CHEMISTRY, M7 2016 VCE 22 MC

When ethene is mixed with chlorine in the presence of UV light, the following reaction takes place.

$\ce{CH2CH2(g) + Cl2(g) \xrightarrow{\text{UV light}} CH2ClCH2Cl(l)}$

Reactions of organic compounds can be classified in a number of ways. The following list shows four possible classifications:

1. addition
2. substitution
3. redox
4. condensation

Which classification(s) applies to the reaction between ethene and chlorine?

1
1 and 2
1 and 3
4

Show Answers Only

$C$

Show Worked Solution

Addition reaction:

→ $\ce{Cl2}$ has added across the $\ce{C=C}$ double bond (eliminate D).

Redox reaction:

→ The oxidation state of $\ce{Cl}$ decreases from zero in $\ce{Cl2(g)}$ to –1 $\ce{CH2ClCH2Cl(l).}$

→ In $\ce{CH2=CH2}$, the oxidation state of each $\ce{H}$ is +1, which means that the oxidation state of each $\ce{C}$ is –2 .

→ In $\ce{CH2ClCH2Cl}$, the oxidation states are $\ce{H}:$ +1, $\ce{Cl}:$ –1, $\ce{C}:$ –1

→ The oxidation state of $\ce{C}$ has increased (from –2 to –1 ) and the oxidation state of $\ce{Cl}$ has decreased from 0 to –1 , confirming it is a redox reaction.

$\Rightarrow C$

♦♦ Mean mark 30%.

CHEMISTRY, M6 2015 VCE 22 MC

What is the pH of a 0.0500 M solution of barium hydroxide, $\ce{Ba(OH)2}$?

1.00
1.30
12.7
13.0

Show Answers Only

$D$

Show Worked Solution

$\ce{Ba(OH)2(aq) \rightarrow Ba^{2+}(aq) + 2OH-(aq)} $

$\ce{[OH-] = 2 \times [Ba(OH)2] = 2 \times 0.0500 = 0.100 M} $

$\ce{[H3O+]} = \dfrac{10^{-14}}{\ce{[OH-]}} = \dfrac{10^{-14}}{0.100} = 10^{-13}\ \text{M} $

$\text{pH}\ = -\log_{10} 10^{-13} = 13$

$\Rightarrow D$

♦ Mean mark 49%.

CHEMISTRY, M6 2015 VCE 8

Hydrogen sulfide, in solution, is a diprotic acid and ionises in two stages.

$\ce{H2S(aq) + H2O(l)\rightleftharpoons HS-(aq) + H3O+(aq)}$ $\quad K_{a1} = 9.6 × 10^{–8} \text{ M}$

$\ce{HS–(aq) + H2O(l)\rightleftharpoons S^{2-}(aq) + H3O+(aq)}$ $\quad K_{a2} = 1.3 × 10^{–14} \text{ M}$

A student made two assumptions when estimating the pH of a $0.01 \text{ M}$ solution of $\ce{H2S}$:

Assumption 1: The pH can be estimated by considering only the first ionisation reaction.

Assumption 2: The concentration of $\ce{H2S}$ at equilibrium is approximately equal to $0.01 \text{ M}$.

Explain why these two assumptions are justified. (2 marks)

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Use the two assumptions given above to calculate the pH of a 0.01 M solution of $\ce{H2S}$. (3 marks)

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Some solid sodium hydrogen sulfide, $\ce{NaHS}$, is added to a 0.01 M solution of $\ce{H2S}$.
Predict the effect of this addition on the pH of the hydrogen sulfide solution. Justify your prediction. (2 marks)

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

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a. 1st assumption:

→ $K_{a2}$ is significantly smaller than the first ionisation $\ce{(K_{a1})}$, making its impact on the $\ce{[H3O+]}$ / pH level negligible.

2nd assumption:

→ $K_{a1}$ is very small, making the extent of the ionisation of $\ce{H2S}$ very small and hence a minimal change in $\ce{[H2S]}$ results.

b. $\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 $

c. Adding $\ce{NaHS}$:

→ Increases the $\ce{[HS-]}$.

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the $\ce{[H3O+]}$ and the pH will therefore increase.

Show Worked Solution

a. 1st assumption:

→ $K_{a2}$ is significantly smaller than the first ionisation $\ce{(K_{a1})}$, making its impact on the $\ce{[H3O+]}$ / pH level negligible.

2nd assumption:

→ $K_{a1}$ is very small, making the extent of the ionisation of $\ce{H2S}$ very small and hence a minimal change in $\ce{[H2S]}$ results.

♦♦ Mean mark (a) 27%.

b.	$K_{a1}$	$=\dfrac{\ce{[HS-][H3O+]}}{\ce{[H2S]}} $
	$9.6 \times 10^{-8}$	$=\dfrac{\ce{[H3O+]^2}}{0.01}$
	$\ce{[H3O+]^2}$	$=0.01 \times 9.6 \times 10^{-8} $
	$\ce{[H3O+]}$	$=\sqrt{9.6 \times 10^{-10}}$
		$=3.1 \times 10^{-5}\ \text{M} $

$\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 $

c. Adding $\ce{NaHS}$:

→ Increases the $\ce{[HS-]}$.

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the $\ce{[H3O+]}$ and the pH will therefore increase.

♦♦ Mean mark (c) 35%.

CHEMISTRY, M6 2015 VCE 23 MC

The following table shows the value of the ionisation constant of pure water at various temperatures and at a constant pressure.

$\text{Temperature (°C)}$	$0$	$25$	$50$	$75$	$100$
$K_W$	$ 1.1 \times 10^{-15}$	$ 1.0 \times 10^{-14}$	$ 5.5 \times 10^{-14}$	$ 2.0 \times 10^{-13}$	$ 5.6 \times 10^{-13}$

Given this data, which one of the following statements about pure water is correct?

The $\ce{[OH–]}$ will decrease with increasing temperature.
The $\ce{[H3O+]}$ will increase with increasing temperature.
Its pH will increase with increasing temperature.
Its pH will always be exactly 7 at any temperature.

Show Answers Only

$B$

Show Worked Solution

$\ce{2H2O(l) \rightleftharpoons H3O+(aq) + OH-(aq)} $

→ Table shows that $K_W$ increases as temperature increases.

→ Ionisation equation shifts right as temperature increases, causing an increase in $\ce{[H3O+]}$ and $\ce{[OH-]}$ (eliminate A).

→ pH decreases as $\ce{[H3O+]}$ and temperature increase (eliminate C).

→ pH of water = 7 at 25°C only (eliminate D).

$\Rightarrow B$

♦ Mean mark 55%.

PHYSICS, M1 EQ-Bank 18

A physics student comes across a river which runs north to south and has a current of 3 ms$^{-1}$ running south.

The student starts on the west side of the river at point A and paddles a kayak at 5 ms$^{-1}$ directly across the river to finish at point B.

Calculate the angle which he must position the boat to travel in a straight line across the river. (2 mark)

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If the river is 100 metres wide, determine the time it takes for the student to cross the river. (2 mark)

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Show Answers Only

a. $36.9^{\circ}$

b. $\text{25 seconds}$

Show Worked Solution

$\sin \theta$	$=\dfrac{3}{5}$
$\theta$	$=\sin^{-1}\Big(\dfrac{3}{5}\Big)=36.9^{\circ}$

The student must turn 36.9$^{\circ}$ into the current as shown on the diagram.

b. Using Pythagoras:

$v=\sqrt{5^2-3^2}=4\ \text{ms}^{-1}$

$t=\dfrac{d}{s}=\dfrac{100}{4}=25\ \text{s}$

$\therefore$ It will take the student 25 seconds to travel from A to B.

PHYSICS, M1 EQ-Bank 12

Below is a description of the motion of a runner. The motion can be divided into three stages.

Stage 1: Runner travels 120 metres south taking 20 seconds.

Stage 2: Runner turns west and travels at 5 ms$^{-1}$ for half a minute.

Stage 3: Runner travels directly back to their starting position.

Determine the distance that the runner ran during Stage 2 of their journey. (1 mark)

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Find the displacement of the start point from the runner Stage 2 is completed. (3 marks)

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a. $\text{150 m}$

b. $\text{192.1 m, N51.3°E.}$

Show Worked Solution

a.	$d$	$=v \times t$
		$=5 \times 30$
		$=150\ \text{m}$

b. Stage 1 and Stage 2 displacement diagram:

Using Pythagoras:

$d$	$=\sqrt{150^2 + 120^2} $
	$=\sqrt{36900}$
	$=192.1\ \text{m}$

$\tan\,\theta$	$=\dfrac{120}{150} $
$\theta$	$=\tan^{-1}\Big(\dfrac{120}{150}\Big)$
	$=38.7^{\circ}$

$\therefore$ Displacement of the start point from the runner is 192.1 m, N51.3°E.

Data Analysis, GEN1 2022 VCAA 1-3 MC

The histogram below displays the distribution of skull width, in millimetres, for 46 female possums.

Question 1

The shape of the distribution is best described as

negatively skewed.
approximately symmetric.
negatively skewed with a possible outlier.
positively skewed with a possible outlier.
approximately symmetric with a possible outlier.

Question 2

The percentage of the 46 possums with a skull width of less than 55 mm is closest to

Question 3

The third quartile $(Q_3)$ for this distribution, in millimetres, could be

55.8
56.2
56.9
57.7
58.3

Show Answers Only

$\text{Question 1:} \ C$

$\text{Question 2:} \ B$

$\text{Question 3:} \ D$

Show Worked Solution

$\text{Question 1}$

The distribution’s centre is in the 56–57 group and if the possible outlier is disregarded, the tail of the distribution is spread more to the left → i.e. negatively skewed with possible outlier.

$\Rightarrow C$

♦♦ Mean mark (Q1) 36%.
COMMENT: 54% of students chose $E$. The graph distribution however, shows that more than 50% of the data is to the left of the median.

$\text{Question 2}$

$\text{Percentage}$	$= \dfrac{1+1+2+1+2+5}{46} \times 100$
	$=\dfrac{12}{46} \times 100$
	$=26.086…$%

$\Rightarrow B$

$\text{Question 3}$

$ Q_3\ =0.75 \times 46 = 34.5 $

The 34.5th score lies between 57 and 58, therefore, 57.7.

$\Rightarrow D$

PHYSICS, M1 EQ-Bank 11

A plane is travelling at 315 ms$^{-1}$ north when it passes through a dense cloud and slows down to a velocity of 265 ms$^{-1}$ for safety precautions.

The plane did not change direction and travelled 2.5 km while it was slowing down.

Using north as the positive direction for all calculations, determine:

the change in velocity of the plane. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
the plane's acceleration. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
the time over which the plane slowed down. (2 marks)

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

Show Answers Only

a. $\text{50 ms}^{-1}\ \text{south}$

b. $\text{5.8 ms}^{-2}\ \text{south}$

c. $\text{8.62 s}$

Show Worked Solution

a.	$\Delta v$	$=v-u$
		$=265-315$
		$=-50\ \text{ms}^{-1}$
		$=50\ \text{ms}^{-1}\ \text{south}$

b. Using $v^2=u^2 +2as$ (time is not given):

$a$	$=\dfrac{v^2-u^2}{2s}$
	$=\dfrac{(265)^2-(315)^2}{2 \times 2500}$
	$=-5.8\ \text{ms}^{-2}$
	$=5.8\ \text{ms}^{-2}$ to the south.

c. Using $v=u+at$:

$t$	$=\dfrac{v-u}{a}$
	$=\dfrac{265-315}{-5.8}$
	$=8.62\ \text{s}$

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\(\displaystyle \int_{3}^{10} f(x)\,dx\)	\(=\displaystyle \int_{3}^{7} f(x)\,dx+\displaystyle \int_{7}^{10} f(x)\,dx\)
\(C\)	\(=\displaystyle \int_{3}^{7} f(x)\,dx+D\)
\(C-D\)	\(=\displaystyle \int_{3}^{7} f(x)\,dx\)
\(\therefore\ \displaystyle \int_{7}^{3} f(x)\,dx\)	\(=D-C\)

\(\text{Axis of symmetry}\)	\(=-\dfrac{b}{2a}\)	\((b = 2b)\)
	\(=-\dfrac{2b}{2a}\)
	\(=-\dfrac{b}{a}\)

a.	\(f(0)\)	\(=a-0(0-2)^2=12\)
	\(\therefore\ a\)	\(=12\)

\(g(1)\)	\(=12\times 1+b\times 1^2\)
\(12+b\)	\(=9\)
\(\therefore b\)	\(=-3\)

b.	\(f(x)\)	\(=12-x(x-2)^2\)
		\(=-x^3+4x^2-4x+12\)
	\(f^{′}(x)\)	\(=-(3x^2-8x+4)\)

\(\text{Pr}(X\geq 1)\)	\(=1-\text{Pr}(X=0)\)
	\(=1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)

\( -(3x^2-8x+4)\)	\(=0\)
\(-(3x-2)(x-2)\)	\(=0\)

\(f(2)\)	\(=12-2(2-2)^2=12\)
\(f\Big{(}\dfrac{2}{3}\Big{)}\)	\(=12-\dfrac{2}{3}\Big(\dfrac{2}{3}-2\Big)^2<12\)

\(g(x)\)	\(=12x-3x^2\)
\(g(2)\)	\(=12\times 2-3\times 2^2=12\)