Band 4 – Page 41

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].`

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.

Calculus, SPEC2 2023 VCAA 11 MC

The area of the curved surface generated by revolving part of the curve with equation $y=\cos ^{-1}(x)$ from $(0, \dfrac{\pi}{2})$ to $(1,0)$ about the $y$-axis can be found by evaluating

$2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}}\left(\cos ^{-1}(x) \sqrt{1+\dfrac{1}{x^2-1}}\right)} d x$
$2 \pi \displaystyle {\int_0^1\left(\cos ^{-1}(x) \sqrt{1+\dfrac{1}{x^2-1}}\right)} d x$
$2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}} \cos (y) \sqrt{1-\sin ^2(y)}} d y$
$2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}} \sqrt{1+u^2} \ d u}$, where $u=\sin (y)$
$2 \pi \displaystyle {\int_0^1 \sqrt{1+u^2} \ d u } $, where $u=\sin (y)$

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$E$

Show Worked Solution

$I = 2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}} \cos (y) \sqrt{1+\sin ^2(y)}}\ d y$

$\text{Let}\ \ u=\sin(y) $

$\dfrac{du}{dy}=\cos(y)\ \ \Rightarrow \ \ du=\cos(y)\ dy $

$\text{When}\ \ y=\dfrac{\pi}{2}\ \Rightarrow \ u=1,\ \ y=0\ \Rightarrow u=0 $

$\therefore I = 2 \pi \displaystyle {\int_0^{1} \sqrt{1+u^2}}\ d u$

$\Rightarrow E$

Calculus, SPEC2 2023 VCAA 10 MC

If $I_n=\displaystyle {\int_0^1\left((1-x)^n e^x\right) d x}$, where $n \in N$, then for $n \geq 1, I_n$ equals

$-1+n I_{n-1}$
$n I_{n-1}$
$-1-n I_{n-1}$
$-n I_{n-1}$
$(1-x)^n e^x+n I_{n-1}$

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$A$

Show Worked Solution

$\text{Using integration by parts:}$

\begin{aligned}
u &=(1-x)^n, & u^{′} &=n(1-x)^{n-1}\\
v^{′} & = e^x, & v & = e^x\ \\
\end{aligned}

\begin{aligned}
\int_0^1 ((1-x)^n e^x)\,dx\ &= uv-\int u^{′} v \ dx\\
& = \Bigr[(1-x)^n\,e^x\Bigr]_0 ^1- \int_0 ^1 n(1-x)^{n-1}\,e^x\ dx \\
& =-1-n\int_0 ^1 (1-x)^{n-1}\,e^x\ dx \\
& =-1-nI_{n-1} \\
\end{aligned}

$\Rightarrow A$

Calculus, SPEC2 2023 VCAA 8 MC

Initially a spa pool is filled with 8000 litres of water that contains a quantity of dissolved chemical. It is discovered that too much chemical is contained in the spa pool water. To correct this situation, 20 litres of well-mixed spa pool water is pumped out every minute while 15 litres of fresh water is pumped in each minute.

Let $Q$ be the number of kilograms of chemical that remains dissolved in the spa pool after $t$ minutes. The differential equation relating $Q$ to t is

$\dfrac{d Q}{d t}=\dfrac{4 Q}{t-1600}$
$\dfrac{d Q}{d t}=\dfrac{-Q}{400}$
$\dfrac{d Q}{d t}=\dfrac{3 Q}{t-1600}$
$\dfrac{d Q}{d t}=\dfrac{3 Q}{1600-t}$
$\dfrac{d Q}{d t}=\dfrac{4 Q}{1600-t}$

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$A$

Show Worked Solution

$\text{Volume}\ = 8000-5t $

$Q(t) = \dfrac{Q}{8000-5t} $

$\dfrac{dQ}{dt}=-\dfrac{20Q}{8000-5t} = \dfrac{4Q}{t-1600} $

$\Rightarrow A$

Calculus, SPEC2 2023 VCAA 7 MC

The direction field for a differential equation is shown above. On a certain solution curve of this differential equation, $y=2$ when $x=-1$.

The value of $y$ on the same solution curve when $x=1.5$ is closest to

$-0.5$
$0$
$0.5$
$1.0$
$1.5$

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$D$

Show Worked Solution

$\text{Trace a curve that does not cross slope lines from}\ (-1,2).$

$\text{Curve crosses}\ \ x=1\ \ \text{at}\ \ y\approx 1.$

$\Rightarrow D$

Calculus, SPEC2 2023 VCAA 6 MC

Consider the following pseudocode.

define $f(x, y)=e^{x y}$

$\begin{aligned} & x \leftarrow 0 \\ & y \leftarrow 0 \\ & h \leftarrow 0.5 \\ & n \leftarrow 0\end{aligned}$

while $n \geq 0$

$\begin{aligned} & y \leftarrow y+h \times f(x, y) \\ & x \leftarrow x+h \\ & n \leftarrow n+1\end{aligned}$

print $y$

end while

After how many iterations will the pseudocode print 2.709 ?

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$C$

Show Worked Solution

$\text{1st iteration:}\ 0+\dfrac{1}{2} \times e^0=\dfrac{1}{2} $

$\text{2nd iteration:}\ \dfrac{1}{2} +\dfrac{1}{2} \times e^{\frac{1}{4}}=1.142… $

$\text{3rd iteration:}\ 1.142 +\dfrac{1}{2} \times e^{1 \times 1.142}=2.7085… $

$\Rightarrow C$

Complex Numbers, SPEC2 2023 VCAA 5 MC

Let $z$ be a complex number where $\operatorname{Re}(z)>0$ and $\operatorname{Im}(z)>0$.

Given $|\bar{z}|=4$ and $\arg \left(z^3\right)=-\pi$, then $z^2$ is equivalent to

$ {4z} $
$ -2 \bar{z} $
$ 3z $
$\bar{z}^2$
$-4 \bar{z}$

Show Answers Only

$E$

Show Worked Solution

$\arg(z^3) = -\pi=\pi\ \ \Rightarrow \ \arg(z)=\dfrac{\pi}{3}$

$z$	$=4\text{cis}\Big{(}\dfrac{\pi}{3}\Big{)}\ \ (\abs{z}=\abs{\bar z})$
$z^2$	$=16\text{cis}\Big{(}\dfrac{2\pi}{3}\Big{)} $
	$=-4 \times 4\text{cis}\Big{(}-\dfrac{\pi}{3}\Big{)} $
	$=-4\,\bar z$

$\Rightarrow E$

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.

Calculus, MET1 2023 VCAA 5

Evaluate $\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx$. (1 mark)
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Hence, or otherwise, find all values of $k$ such that $\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx=\displaystyle \int_{0}^{\frac{\pi}{2}} \cos(x)\,dx$, where $-3\pi<k<2\pi$. (3 marks)
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a. $\dfrac{1}{2}$

b. $k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}$

Show Worked Solution

a.	$\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx$	$=\left[-\cos x\right]_0^\frac{\pi}{3}$
		$=-\cos\dfrac{\pi}{3}+\cos 0$
		$=-\dfrac{1}{2}+1$
		$=\dfrac{1}{2}$

b.	$\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx$	$=\left[\sin x\right]_k^\frac{\pi}{2}$
		$=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)$
		$=1-\sin (k)$

$\text{Using part (a):}$

$1-\sin (k)$	$=\dfrac{1}{2}$
$\sin (k)$	$=\dfrac{1}{2}$
$\therefore\ k$	$=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}$

Calculus, MET1 2023 VCAA 4

The graph of $y=x+\dfrac{1}{x}$ is shown over part of its domain.

Use two trapeziums of equal width to approximate the area between the curve, the $x$-axis and the lines $x=1$ and $x=3$. (2 marks)

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$5\dfrac{1}{6}$

Show Worked Solution

$\text{Trapezium rule approximation (see formula sheet):}$

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} x\rule[-1ex]{0pt}{0pt} & 1&2&3 \\
\hline
\rule{0pt}{2.5ex} f(x)\rule[-1ex]{0pt}{0pt} & 1+1=2 & 2+\dfrac{1}{2}=\dfrac{5}{2} & 3+ \dfrac{1}{3}=\dfrac{10}{3}\\
\hline
\end{array}

$\text{Area}$	$\approx \dfrac{3-1}{2\times 2}\Bigg[2+2\times\dfrac{5}{2}+\dfrac{10}{3}\Bigg]$
	$\approx\dfrac{1}{2}\Bigg[\dfrac{6}{3}+\dfrac{15}{3}+\dfrac{10}{3}\Bigg]$
	$\approx5\dfrac{1}{6}$

Calculus, MET1 2022 VCAA 1b

Find and simplify the rule of `f^{\prime}(x)`, where `f:R \rightarrow R, f(x)=\frac{\cos (x)}{e^x}`. (2 marks)

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`\frac{-(\sin x+\cos x)}{e^x}`

Show Worked Solution

Using the quotient rule

`f(x)`	`=\frac{\cos (x)}{e^x}`
`f^{\prime}(x)`	`=\frac{-e^x \sin x-e^x \cos x}{e^{2 x}}`
	`= \frac{-e^x(\sin x+\cos x)}{e^{2 x}}`
	`=\frac{-(\sin x+\cos x)}{e^x}`

Calculus, MET1 2022 VCAA 1a

Let `y=3xe^{2x}`.

Find `\frac{dy}{dx}`. (1 mark)

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`6xe^(2x) + 3e^(2x)`

Show Worked Solution

Using the product rule, given `y=3xe^{2x}`

`\frac{dy}{dx}`	`=3x xx 2e^(2x)+3 xx e^(2x)`
	`=6xe^(2x) + 3e^(2x)`

Algebra, MET1 2023 VCAA 2

Solve $e^{2x}-12=4e^{x}$ for $x\ \in\ R$. (3 marks)

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$x=\log_{e}6$

Show Worked Solution

$e^{2x}-12$	$=4e^{x}$
$e^{2x}-4e^{x}-12$	$=0$

$\text{Let}\ \ u=e^{x}:$

$u^2-4u-12$	$=0$
$(u-6)(u+2)$	$=0$

$\Rightarrow u=6\ \ \ \text{or}\ -2$

$\therefore e^{x}$	$=6\ \ \ \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ e^{x}=-2\ \text{(no solution)}$
$x$	$=\log_{e}6 $

Calculus, MET1 2023 VCAA 1a

Let $y=\dfrac{x^2-x}{e^x}$.

Find and simplify $\dfrac{dy}{dx}$. (2 marks)

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$-\Bigg(\dfrac{x^2-3x+1}{e^x}\Bigg) $

Show Worked Solution

$\text{Using the quotient rule:}$

$\dfrac{dy}{dx}$	$=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}$
	$=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}$
	$=\dfrac{-x^2+3x-1}{e^x}$

Trigonometry, SPEC2 2023 VCAA 3 MC

In the interval $-\pi \leq x \leq \pi$, the graph of $y=a+\sec (x)$, where $a \in R$, has two $x$-intercepts when

$0 \leq a \leq 1$
$-1<a<1$
$a \leq-1$ or $a>1$
$-1 \leq a<0$
$a<-1$ or $a \geq 1$

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$E$

Show Worked Solution

$\text{Sketch}\ \ y=\sec(x)\ \ \text{for}\ \ -\pi \leq x \leq \pi $

$\text{Translate vertically to test for values of}\ a\ \text{that satisfy}$

$\text{two intercepts on the}\ x\text{-axis.}$

$a \geq 1\ \ \text{or}\ \ a \lt -1$

$\Rightarrow E$

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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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}

Functions, SPEC2 2023 VCAA 2 MC

The graph of $y=\dfrac{x^3}{a x^2+b x+c}$ has asymptotes given by $y=2 x+1$ and $x=1$. The values of $a, b$ and $c$ are, respectively

$2,-4,2$
$ \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{1}{4} $
$ \dfrac{1}{2}, \dfrac{1}{4},-\dfrac{3}{4} $
$ \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{3}{4} $
$2,-4,-8$

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$B$

Show Worked Solution

$\text{Using partial fractions (by calc):}$

$\text{Remainder}\ = \dfrac{x}{a}-\dfrac{b}{a^2} = 2x+1$

$a=\dfrac{1}{2},\ \ b=-\dfrac{1}{4} $

$x=1\ \ \text{is a solution of}\ \ ax^2+bx+c=0$

$c=-\dfrac{1}{4} $

$\Rightarrow B$

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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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} $
		$=\sqrt{29}$

Functions, SPEC1 2023 VCAA 1

Consider the function $f$ with rule $f(x)=\dfrac{x^2+x-6}{x-1}$.

Show that the rule for the function $f$ can be written as $f(x)=x+2-\dfrac{4}{x-1}$. (1 mark)

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Sketch the graph of $f$ on the axes below, labelling any asymptotes with their equations. (3 marks)

--- 0 WORK AREA LINES (style=blank) ---

Show Answers Only

a.	$f(x)$	$=\dfrac{x^2+x-6}{x-1}$
		$=\dfrac{(x-1)(x+2)-4}{x-1}$
		$=x+2-\dfrac{4}{x-1} $

b. $\text{Asymptotes at:}\ x=1, \ y=x+2 $

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

Show Worked Solution

a.	$f(x)$	$=\dfrac{x^2+x-6}{x-1}$
		$=\dfrac{(x-1)(x+2)-4}{x-1}$
		$=x+2-\dfrac{4}{x-1} $

b. $\text{Asymptotes at:}\ x=1, \ y=x+2 $

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

Statistics, SPEC1 2023 VCAA 6

Josie travels from home to work in the city. She drives a car to a train station, waits, and then rides on a train to the city. The time, $X_c $ minutes, taken to drive to the station is normally distributed with a mean of 20 minutes $ (\mu_c=20) $ and standard deviation of 6 minutes $(\sigma_c=6) $. The waiting time, $ X_w $ minutes, for a train is normally distributed with a mean of 8 minutes $ (\mu_w=8) $ and standard deviation of $ \sqrt{3} $ minutes $ (\sigma_w=\sqrt{3}) $. The time, $ X_t $ minutes, taken to ride on a train to the city is also normally distributed with a mean of 12 minutes $ (\mu_t=12) $ and standard deviation of 5 minutes $ (\sigma_t=5) $. The three times are independent of each other.

Find the mean and standard deviation of the total time, in minutes, it takes for Josie to travel from home to the city. (2 marks)

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

Josie's waiting time for a train on each work day is independent of her waiting time for a train on any other work day. The probability that, for 12 randomly chosen work days, Josie's average waiting time is between 7 minutes 45 seconds and 8 minutes 30 seconds is equivalent to $ \text{Pr}(a<Z<b)$, where $Z \sim \text{N}(0,1)$ and $a$ and $b$ are real numbers.
Find the values of $a$ and $b$. (2 marks)

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

Show Answers Only

a. $E[X_c+X_w+X_t]=40$

$\sigma[X_c+X_w+X_t]=8$

b. $a=-0.5, \ b=1$

Show Worked Solution

a. $X_c ∼ N(20, 6^2), X_w ∼ N(8, (\sqrt{3})^2), X_t ∼ N(12, 5^2) $

$E[X_c+X_w+X_t]$	$=E[X_c]+E[X_w]+E[X_t]$
	$=20+8+12$
	$=40$

$\text{Var}[X_c+X_w+X_t]$	$=\text{Var}[X_c]+\text{Var}[X_w]+\text{Var}[X_t]$
	$=36+3+25$
	$=64$
$\sigma[X_c+X_w+X_t]$	$=8$

b. $E(\bar X)=\mu_w=8,\ \ \sigma(\bar X)=\dfrac{\sigma_w}{\sqrt{12}}=\dfrac{\sqrt3}{\sqrt{12}}=\dfrac{1}{2} $

$\text{Pr}(7.75<\bar X<8.5)$	$=\text{Pr} \Bigg{(} \dfrac{7.75-8}{\frac{1}{2}}<Z<\dfrac{8.5-8}{\frac{1}{2}}\Bigg{)} $
	$=\text{Pr} \Bigg{(} \dfrac{-0.25}{\frac{1}{2}}<Z<\dfrac{0.5}{\frac{1}{2}}\Bigg{)} $
	$=\text{Pr}(-0.5<Z<1)$

$\therefore a=-0.5, \ b=1$

Calculus, SPEC1 2023 VCAA 4

Consider the relation $x\, \arcsin \left(y^2\right)=\pi$.

Use implicit differentiation to find $\dfrac{d y}{d x}$ at the point $\left(6, \dfrac{1}{\sqrt{2}}\right)$.

Give your answer in the form $-\dfrac{\pi \sqrt{a}}{b}$, where $a, b \in Z^{+}$. (3 marks)

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

Show Answers Only

$\dfrac{dy}{dx}=-\dfrac{\pi \sqrt6}{144} $

Show Worked Solution

$\dfrac{d}{dx} (x\, \sin^{-1} y^2) $	$=0$
$\sin^{-1} y^2+ \dfrac{2xy}{\sqrt{1-y^4}} \cdot \dfrac{dy}{dx} $	$=0$

$\dfrac{dy}{dx} =-\dfrac{\sin^{-1}y^2 \times \sqrt{1-y^4}}{2xy} $

$\text{At}\ \ \Big{(} 6, \dfrac{1}{\sqrt{2}} \Big{)}, $

$\dfrac{dy}{dx}$	$= -\dfrac{\sin^{-1}(\frac{1}{2}) \times \sqrt{\frac{3}{4}}}{2 \times 6 \times \frac{1}{\sqrt 2}}$
	$=-\dfrac{\pi}{6} \times \dfrac{\sqrt3}{2} \times \dfrac{\sqrt2}{12} $
	$=-\dfrac{\pi \sqrt6}{144} $

Calculus, SPEC1 2023 VCAA 3

A particle moves along a straight line. When the particle is $x$ m from a fixed point $O$, its velocity, $ v$ m s$^{-1}$, is given by

$v=\dfrac{3 x+2}{2 x-1}$, where $x \geq 1$.

Find the acceleration of the particle, in m s$^{-2}$, when $x=2$. (2 marks)

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

Find the value that the velocity of the particle approaches as $x$ becomes very large. (1 mark)

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

Show Answers Only

a. $a= -\dfrac{56}{27} $

b. $\text{As}\ \ x \rightarrow \infty, \ v \rightarrow \dfrac{3}{2} $

Show Worked Solution

a.	$a$	$=v \cdot \dfrac{dv}{dx}$
		$=\dfrac{3x+2}{2x-1} \cdot \Bigg{(} \dfrac{3(2x-1)-2(3x+2)}{(2x-1)^2} \Bigg{)} $

$\text{When}\ \ x=2: $

$a=\dfrac{8}{3} \cdot \dfrac{(9-16)}{9}= -\dfrac{56}{27} $

b. $v=\dfrac{3x+2}{2x-1} = \dfrac{3+\frac{2}{x}}{2-\frac{1}{x}} $

$\text{As}\ \ x \rightarrow \infty, \ v \rightarrow \dfrac{3}{2} $

Complex Numbers, SPEC1 2023 VCAA 2

Consider the complex number $z=(b-i)^3$, where $b \in R^{+}$.

Find $b$ given that $\arg (z)=-\dfrac{\pi}{2}$. (3 marks)

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

Show Answers Only

$b=\sqrt{3} $

Show Worked Solution

$-\dfrac{\pi}{2} = \arg (z) = \arg[(b-i)^3]$

$b-i\ \Rightarrow \ \text{complex number that lives on}\ \ y=-i $

$3 \times \arg(b-i) = -\dfrac{\pi}{2} $

$\arg(b-i) = -\dfrac{\pi}{6} $

$\tan^{-1}\Big{(}\dfrac{-1}{b} \Big{)} $	$= -\dfrac{\pi}{6} $
$-\dfrac{1}{b}$	$=-\dfrac{1}{\sqrt{3}}$
$\therefore b$	$=\sqrt{3}$

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)

--- 1 WORK AREA LINES (style=lined) ---
What is the earliest start time, in days, for activity $J$ ? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
How many activities have a float time of zero? (1 mark)

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

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

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

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

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

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

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

\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)

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

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

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%.

Networks, GEN2 2023 VCAA 12

A country has five states, $A, B, C, D$ and $E$.

A graph can be drawn with vertices to represent each of the states.

Edges represent a border shared between two states.

What is the sum of the degrees of the vertices of the graph above? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Euler's formula, $v+f=e+2$, holds for this graph.
i. Complete the formula by writing the appropriate numbers in the boxes provided below. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
ii. Complete the sentence by writing the appropriate word in the space provided below. (1 mark)

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

Euler’s formula holds for this graph because the graph is connected and ______________.
The diagram below shows the position of state $A$ on a map of this country.
The four other states are indicated on the diagram as 1, 2, 3 and 4.

Use the information in the graph above to complete the table below. Match the state $(B, C, D$ and $E)$ with the corresponding state number $(1,2,3$ and 4$)$ given in the map above. (1 mark)

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

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

Show Answers Only

a. $\text{Sum of degrees}\ = 2+3+4+3+2 = 14$

b.i. $\text{Vertices = 5, Faces = 4, Edges = 7}$

$\Rightarrow 5 + 4 = 7 + 2$

b.ii. $\text{Planar}$

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 3\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 2 \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 1 \\
\hline
\end{array}

Show Worked Solution

a. $\text{Sum of degrees}\ = 2+3+4+3+2 = 14$

b.i. $\text{Vertices = 5, Faces = 4, Edges = 7}$

$\Rightarrow 5 + 4 = 7 + 2$

b.ii. $\text{Planar}$

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 3\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 2 \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 1 \\
\hline
\end{array}

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}

i. Complete matrix $H$ above by filling in the missing elements. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
ii. 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)

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

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

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)

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

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)

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

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%.

Financial Maths, 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)

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

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

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

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. $\text{Perpetuity}$

Financial Maths, 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)

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

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)

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

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

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)

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

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)

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

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

\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)

--- 2 WORK AREA LINES (style=lined) ---
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} $

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

Proof, EXT2 P2 2023 SPEC1 8

A function $f$ has the rule $f(x)=x\,e^{2x}$.

Use mathematical induction to prove that $f^{(n)}(x)=\big{(}2^{n}x+n\,2^{n-1}\big{)}e^{2x}$ for $n \in \mathbb{Z}^{+}$, where $f^{(n)}(x)$ represents the $n$th derivative of $f(x)$. That is, $f(x)$ has been differentiated $n$ times. (3 marks)

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

Show Answers Only

$\text{Proof (See Worked Solutions)}$

Show Worked Solution

$\text{If}\ \ n=1:$

$f(x)=x\,e^{2x}\ \ \Rightarrow \ \ f^{′}(x)=e^{2x} + 2x\,e^{2x} $

$f^{(1)}(x)=\big{(}2^{1}x + 1 \times 2^{1-1}\big{)}e^{2x} = e^{2x} + 2x\,e^{2x} $

$\therefore\ \text{True for}\ \ n=1.$

$\text{Assume true for}\ \ n=k $

$f^{(k)}(x)=\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}$

$\text{Prove true for}\ \ n=k+1 $

$\text{i.e.}\ f^{(k+1)}(x)=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}$

$\text{LHS}$	$= \dfrac{d}{dx} \big{(}f^{(k)}(x) \big{)} $
	$= \dfrac{d}{dx} \Big{[}\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\Big{]} $
	$= 2^{k}e^{2x}+(2^{k}x+k\,2^{k-1}) \times 2 e^{2x} $
	$=e^{2x} \big{(}2^{k}+2\cdot2^{k}x + 2k \cdot 2^{k-1}\big{)}$
	$=e^{2x}\big{(}2^{k+1}+2^{k}+k \cdot 2^{k} \big{)} $
	$=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}$
	$=\ \text{RHS}$

$\Rightarrow \text{True for}\ \ n=k+1$

$\therefore\ \text{Since true for}\ \ n=1\text{, by PMI, true for}\ n \in \mathbb{Z}^{+} $

Induction, SPEC1 2023 VCAA 8

A function $f$ has the rule $f(x)=x\,e^{2x}$.

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

Show Answers Only

$\text{See Worked Solutions}$

Show Worked Solution

$\text{If}\ \ n=1:$

$f(x)=x\,e^{2x}\ \ \Rightarrow \ \ f^{′}(x)=e^{2x} + 2x\,e^{2x} $

$f^{(1)}(x)=\big{(}2^{1}x + 1 \times 2^{1-1}\big{)}e^{2x} = e^{2x} + 2x\,e^{2x} $

$\therefore\ \text{True for}\ \ n=1.$

$\text{Assume true for}\ \ n=k $

$f^{(k)}(x)=\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}$

$\text{Prove true for}\ \ n=k+1 $

$\text{i.e.}\ f^{(k+1)}(x)=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}$

$\text{LHS}$	$= \dfrac{d}{dx} \big{(}f^{(k)}(x) \big{)} $
	$= \dfrac{d}{dx} \Big{[}\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\Big{]} $
	$= 2^{k}e^{2x}+(2^{k}x+k\,2^{k-1}) \times 2 e^{2x} $
	$=e^{2x} \big{(}2^{k}+2\cdot2^{k}x + 2k \cdot 2^{k-1}\big{)}$
	$=e^{2x}\big{(}2^{k+1}+2^{k}+k \cdot 2^{k} \big{)} $
	$=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}$
	$=\ \text{RHS}$

$\Rightarrow \text{True for}\ \ n=k+1$

$\therefore\ \text{Since true for n=1, by PMI, true for}\ n \in \mathbb{Z}^{+} $

Matrices, GEN1 2022 VCAA 5 MC

Matrix $E$ is a 2 × 2 matrix.

Matrix $F$ is a 2 × 3 matrix.

Matrix $G$ is a 3 × 2 matrix.

Matrix $H$ is a 3 × 3 matrix.

Which one of the following matrix products could have an inverse?

$EF$
$FH$
$GE$
$GF$
$HG$

Show Answers Only

$D$

Show Worked Solution

Matrix product requires: [m × n] × [n × m]

Matrix inverse requires a square matrix: [m × n] × [m × n] = [m × m]

Option A: $EF$ = [2 × 2] × [2 × 3] → [2 × 3] (not square, eliminate A)

Option B: $FH$ = [2 × 3] × [3 × 3] → [2 × 3] (not square, eliminate B)

Option C: $GE$ = [3 × 2] × [2 × 2] → [3 × 2] (not square, eliminate B)

Option D: $GF$ = [3 × 2] × [2 × 3] → [3 × 3] (correct)

Option E: $HG$ = [3 × 3] × [3 × 2] → [3 × 2] (not square, eliminate E)

$\Rightarrow D$

Matrices, GEN1 2022 VCAA 3 MC

Each day, members of a swim centre can choose to attend a morning session $(M)$, an afternoon session $(A)$ or no session $(N)$.

The transition diagram below shows the transition from day to day.

The transition diagram is incomplete.

Which one of the following transition matrices represents this transition diagram?

Show Answers Only

$C$

Show Worked Solution

$\text{By elimination}$

$\text{Option E: Eliminate as not all columns add to 1}$

$\text{Options A and D: Eliminate as N → M = 0.6, not 0.1 or 0.2}$

$\text{Option B: Eliminate as M → A = 0.2, not 0.5}$

$\text{Option C: Correct as N → M = 0.6, M → A = 0.2, A → N = 0.5 and leading diagonal 0.3, 0.4, 0.1 correct}$

$\Rightarrow C$

Matrices, GEN1 2022 VCAA 1-2 MC

A bike rental business rents road bikes $(R)$ and mountain bikes $(M)$ in three sizes: child $(C)$, junior $(J)$ and adult $(A)$.

Matrix $B$ shows the daily rental cost, in dollars, for each type of bike.

\begin{aligned} \\
& \quad R \ \quad \ \ M \\
B = & \begin{bmatrix}
80 & 95 \\
110 & 120 \\
120 & 125
\end{bmatrix}\begin{array}{l}
C\\
J\\
A
\end{array}
\end{aligned}

The element in row $i$ and column $j$ in matrix $B$ is $b_{ij}$.

Question 1

The daily cost of renting an adult mountain bike is shown in element

$b_{12}$
$b_{21}$
$b_{23}$
$b_{31}$
$b_{32}$

Question 2

On Sundays, the business increases the daily rental price for each type of bike by 10%.

To determine the rental cost for each type of bike on a Sunday, which one of the following matrix calculations needs to be completed?

$0.01B$
$0.1B$
$1.01B$
$1.1B$
$11B$

Show Answers Only

$\text{Question 1:}\ E$

$\text{Question 2:}\ D$

Show Worked Solution

$\text{Question 1}$

$\text{Adult Mountain Bike is in Row 3, Column 2}\ = b_{32}$

$\Rightarrow E$

$\text{Question 2}$

$\text{Increase of 10% to each type of bike rental charge.}$

$\text{Each element increases by 10%.}$

$\text{i.e. Multiply each element by a factor of 1.1.}$

$\Rightarrow D$

Recursion, GEN1 2022 VCAA 22 MC

Tim deposited $6000 into an investment account earning compound interest calculated monthly.

A rule for the balance, $T_n$, in dollars, after $n$ years is given by $T_n=6000 \times 1.003^{12n}$.

Let $R_n$ be a new recurrence relation that models the balance of Tim's account after $n$ months.

This recurrence relation is

$R_0=6000,\ \ \ R_{n+1}=R_n+18$
$R_0=6000,\ \ \ R_{n+1}=R_n+36$
$R_0=6000,\ \ \ R_{n+1}=1.003\,R_n$
$R_0=6000,\ \ \ R_{n+1}=1.0036\,R_n$
$R_0=6000,\ \ \ R_{n+1}=1.036\,R_n$

Show Answers Only

$C$

Show Worked Solution

$\text{Eliminate option A and B as they are linear.}$

$\text{Option D has a monthly interest rate = 0.36% (eliminate)}$

$\text{Option E has a monthly interest rate = 3.6% (eliminate)}$

$\text{Consider option C:}$

$T_1 = 6000 \times 1.003^{12 \times 1}\ \ \text{(after 1 year)}$

$R_1$	$= 6000 \times 1.003$	$= 1.003 \times R_0$
$R_2$	$= 6000 \times 1.003 \times 1.003$	$= 1.003 \times R_1$
$R_3$	$= 6000 \times 1.003 \times 1.003 \times 1.003$	$= 1.003 \times R_2$

$\therefore R_{n+1} = 1.003\,R_n\ \text{where } n\ \text{is in months.}$

$\Rightarrow C$

Recursion, GEN1 2022 VCAA 20 MC

Nidhi owns equipment that is used for 10 hours per day for all 365 days of the year.

The value of the equipment is depreciated by Nidhi using the unit cost method.

The value of the equipment, $E_n$, in dollars, after $n$ years can be modelled by the recurrence relation

$E_0=100\ 000,\ \ \ E_{n+1}=E_n-5475$

The value of the equipment is depreciated by

$1.50 per hour.
$10 per hour.
$15 per hour.
$1.50 per day.
$10 per day.

Show Answers Only

$A$

Show Worked Solution

$\text{1st year depreciation = \$5475}$

$\text{Hourly depreciation =\dfrac{5475}{365 \times 10}=$1.50$

$\Rightarrow A$

Data Analysis, GEN1 2022 VCAA 15 MC

The daily number of cups of coffee sold by a food truck over a three-week period is shown in the table below.

The six-mean smoothed number of cups of coffee, with centring, sold on Thursday in Week 2 is closest to

Show Answers Only

$B$

Show Worked Solution

$\text{Six-mean smoothed average (Thu, week 2)}$

$=[(98+104+145+163+134+128)\ ÷\ 6 + $

$(104+145+163+134+128+206)\ ÷\ 6]\ ÷\ 2 $

$=137.66…$

$\Rightarrow B$

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 4 MC

What is the IUPAC name of the molecule shown above?

3-hydroxy-3-ethyl-propan-1-amine
3-amino-1-methylpropan-1-ol
3-hydroxypentan-1-amine
1-aminopentan-3-ol

Show Answers Only

$D$

Show Worked Solution

Molecule has a 5-carbon chain with single bonds (“pentan”)
Alcohol functional group attached to 3rd carbon (“-3-ol”)

$\Rightarrow D$

CHEMISTRY M6 2016 VCE 21*

The ammonium ion $\ce{NH4+}$ acts as a weak acid according to the equation

$\ce{NH4+(aq) + H2O(l) \rightleftharpoons NH3(aq) + H3O+(aq)}$

Given the $K_a \ce{(NH4+) = 5.6 \times 10^{-10}}$, determine the [$\ce{H3O+}$] of a 0.200 M ammonium chloride solution. (2 marks)

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$1.06 \times 10^{-5}\ \text{mol L}^{-1}$

Show Worked Solution

$K_a = \dfrac{\ce{[NH3][H3O+]}}{\ce{[NH4+]}} $

$\text{Weak acid assumptions:}$

$\ce{[NH4+]_{eq} = 0.200\ M\ \ \text{and}\ \ [NH3] = [H3O+]} $

$5.6 \times 10^{-10}$	$= \dfrac{\ce{[H3O+]^2}}{0.200} $
$\ce{[H3O+]^2}$	$=0.200 \times 5.6 \times 10^{-10} $
$\ce{[H3O+]}$	$= \sqrt{1.12 \times 10^{-10}}$
	$=1.06 \times 10^{-5}\ \text{mol L}^{-1} $

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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b. 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%.

b. 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 2*

What is the correct systematic name for the compound shown above? (2 marks)

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3,4-dimethylheptane

Show Worked Solution

7 carbon string with single carbon and hydrogen bonds = heptane
Two methyl groups attached to carbon 3 and 4 (IUPAC convention)
Correct name is: 3,4-dimethylheptane

CHEMISTRY, M7 2015 VCE 5a

A reaction pathway is designed for the synthesis of the compound that has the structural formula shown below.

The table below gives a list of available organic reactants and reagents.

Complete the reaction pathway design flow chart below. Write the corresponding letter for the structural formula of all organic reactants in each of the boxes provided. The corresponding letter for the formula of other necessary reagents should be shown in the boxes next to the arrows. (5 marks)

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Show Worked Solution

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)

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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%.

PHYSICS, M1 EQ-Bank 1

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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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 2

A boy on a bike is travelling at 8 ms$^{-1}$ north when he sees a girl on a scooter travelling at 6 ms$^{-1}$ west, relative to the ground.

Draw a vector diagram to represent the velocity of the girl relative to the boy. (2 marks)

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Use the vector diagram to calculate the velocity of the girl relative to the boy. (2 marks)

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10 ms$^{-1}$, S36.9$^{\circ}$W

Show Worked Solution

a. $v_{\text{g}} = 6\ \text{ms}^{-1}\ \text{west} $

$-v_{\text{b}} = 8\ \text{ms}^{-1}\ \text{south} $

$v_{\text{g rel b}} = v_\text{g} + (-v_\text{b}) $

b. $v_{\text{g rel b}}=\sqrt{6^2+8^2} =10\ \text{ms}^{-1}$

	$\tan \theta$	$=\dfrac{8}{6}$
	$\theta$	$=\tan^{-1}\Big(\dfrac{8}{6}\Big)=53.1^{\circ}$

$\therefore$ Velocity of the girl relative to the boy is 10 ms$^{-1}$, S36.9°W.

PHYSICS, M1 EQ-Bank 11

Car A is travelling north at 40 ms$^{-1}$ and Car B is travelling at 30 ms$^{-1}$ east.

Determine the velocity of Car A relative to Car B. (3 mark)

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50 ms$^{-1}$, N36.9$^{\circ}$W

Show Worked Solution

$v_{\text{A rel B}}=v_A-v_B =v_A + (-v_B)$

$v_A = 40\ \text{ms}^{-1}\ \text{north}, \ -v_B = 30\ \text{ms}^{-1}\ \text{west} $

$v_{\text{A rel B}}=\sqrt{40^2+30^2}=50\ \text{ms}^{-1}$

$\tan \theta$	$=\dfrac{30}{40}$
$\theta$	$=\tan^{-1}\Big(\dfrac{30}{40}\Big)=36.9^{\circ}$

$\therefore$ Velocity of Car A relative to Car B is 50 ms$^{-1}$, N36.9°W.

PHYSICS, M1 EQ-Bank 10

Determine the magnitude and direction of the resultant velocity vector for an airplane that is simultaneously moving with a velocity of 350$^{-1}$ directly towards the west and 275 ms$^{-1}$ towards the northwest. (3 marks)

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578 ms$^{-1}$, N70.3°W

Show Worked Solution

Add vectors to show resultant vector:

Using the cosine rule to find the magnitude of $v$:

$v$	$=\sqrt{350^2+275^2-2 \times 350 \times 275 \times \cos 135°}$
	$=578.1376\dots\ \text{ms}^{-1}$

Using the sign rule to find $\theta$:

$\dfrac{\sin \theta}{275}$	$=\dfrac{\sin 135°}{578.1376}$
$\sin \theta$	$=\dfrac{\sin135° \times 275}{578.1376}$
$\theta$	$=\sin^{-1}\Big(\dfrac{\sin135° \times 275}{578.1376}\Big)$
	$=19.7^{\circ}$

$\therefore$ Resultant velocity of the aeroplane is 578 ms$^{-1}$, N70.3°W.

PHYSICS, M1 EQ-Bank 9

A car is travelling north and approaching an intersection at 50 kmh$^{-1}$.

While maintaining a constant speed, the car turns left and continues east at 50 kmh$^{-1}$.

Using a vector diagram, calculate the change in velocity of the car. (3 marks)

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$\text{70.7 kmh}^{-1}, \text{S45°E} $

Show Worked Solution

$v= 50\ \text{kmh}^{-1}\ \text{east},\ \ u= 50\ \text{kmh}^{-1}\ \text{north.}$

$\Delta v= v-u = v+(-u),\ \ \text{where}\ -u= 50\ \text{kmh}^{-1}\ \text{south}$

$\Delta v=\sqrt{50^2 + 50^2}=\sqrt{5000}=70.7\ \text{kmh}^{-1}$

$\tan \theta $	$=\dfrac{50}{50}$
$\theta$	$=\tan^{-1}\Big(\dfrac{50}{50}\Big)=45^{\circ}$

Change in velocity of the car = 70.7 $\text{kmh}^{-1}$, S45$^{\circ}$E.

Data Analysis, GEN1 2022 VCAA 9-11 MC

Table 1 summarises the results of a study that compared the effectiveness of individual and group instruction (instructional method) when training future basketball referees.

In this table, test grade is the response variable and instructional method is the explanatory variable.

Question 9

The variables test grade (A, B, C, D, E) and instructional method (individual, group) are

a numerical and a categorical variable respectively.
both nominal variables.
a nominal and an ordinal variable respectively.
both ordinal variables.
an ordinal and a nominal variable respectively.

Question 10

Of the students who received an A grade, the percentage who were instructed individually is closest to

Question 11

To become a qualified referee, a grade of A or B on the test is required. Those who receive a C, a D or an E will not qualify.

Using column percentages, a new two-way percentage frequency table is constructed from the data in Table 1.

In this new table, qualified to be a referee (yes, no) is the response variable and instructional method (individual, group) is the explanatory variable.

Which one of the following tables correctly displays the data from Table 1?

Show Answers Only

$\text{Question 9:} \ E$

$\text{Question 10:} \ C$

$\text{Question 11:} \ B$

Show Worked Solution

$\text{Question 9} $

$\text{Both variables are categorical.}$

$\text{Test grade (A, B, C, D, E) → ordinal, as they can be ranked.}$

$\text{Instructional method → nominal, has no ranking.}$

$\Rightarrow E$

$\text{Question 10} $

$\text{Students who received an A}\ =10+18 = 28 $

$\text{% instructed individually}\ = \dfrac{10}{28} \times 100 = 35.714…\% $

$\Rightarrow C$

Mean mark (Q10) 56%.

$\text{Question 11} $

$\text{yes (A or B):} $

$\text{Individual}\ =\dfrac{10+35}{115} \times 100 \approx 39\% $

$\text{Group}\ = \dfrac{18+30}{126} \times 100 \approx 38\% $

$\text{no (C, D or E):} $

$\text{Individual}\ = 100-39 \approx 61\% $

$\text{Group}\ = 100-38 \approx 62\% $

$\Rightarrow B$

Data Analysis, GEN1 2022 VCAA 7-8 MC

The association between the weight of a seal's spleen, spleen weight, in grams, and its age, in months, for a sample of seals is non-linear.

This association can be linearised by applying a $\log _{10}$ transformation to the variable spleen weight.

The equation of the least squares line for this scatterplot is

$\log _{10}$ (spleen weight) = 2.698 + 0.009434 × age

Question 7

The equation of the least squares line predicts that, on average, for each one-month increase in the age of the seals, the increase in the value of $\log _{10}$ (spleen weight) is

0.009434
0.01000
1.020
2.698
5.213

Question 8

Using the equation of the least squares line, the predicted spleen weight of a 30-month-old seal, in grams, is

3
511
772
957
1192

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$\text{Question 7:} \ A$

$\text{Question 8:} \ D$

Show Worked Solution

$\text{Question 7}$

$\text{Graph passes through (0, 2.7) and (130, 3.93)} $

$\text{Gradient}\ \approx \dfrac{3.93-2.7}{130} \approx 0.0946 $

$\Rightarrow A$

$\text{Question 8}$

$\text{Let the predicted spleen weight be}\ w:$

$\log_{10} w$	$=2.698 + 0.009434 \times 30$
$\log_{10} w$	$=2.98102$
$w$	$= 10^{2.98102}$
	$=957.2381527$

$\Rightarrow D$

Mean mark (Q8) 56%.

Data Analysis, GEN1 2022 VCAA 6 MC

The histogram below displays the distribution of spleen weight for a sample of 32 seals.

The histogram has a $\log _{10}$ scale.

The number of seals in this sample with a spleen weight of 1000 g or more is

Show Answers Only

$D$

Show Worked Solution

$\text{Given}\ \log_{10} 1000 = 3\ \ (10^3 = 1000) $

$\text{Number of seals} = 8 + 9 + 7 + 1 = 25$

$\Rightarrow D$

Data Analysis, GEN1 2022 VCAA 4 MC

The age, in years, of a sample of 14 possums is displayed in the dot plot below.

The mean and the standard deviation of age for this sample of possums are closest to

mean = 4.25 standard deviation = 2.6
mean = 4.8 standard deviation = 2.4
mean = 4.8 standard deviation = 2.5
mean = 4.9 standard deviation = 2.4
mean = 4.9 standard deviation = 2.5

Show Answers Only

$E$

Show Worked Solution

$\text{By calculator:}$

$\text{Sample standard deviation} = 2.525…$

$\text{Mean} = 4.928…$

$\Rightarrow E$

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

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$\text{Question 1:} \ C$

$\text{Question 2:} \ B$

$\text{Question 3:} \ D$

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.

$\Rightarrow D$

\(z\)	\(=4\text{cis}\Big{(}\dfrac{\pi}{3}\Big{)}\ \ (\abs{z}=\abs{\bar z})\)
\(z^2\)	\(=16\text{cis}\Big{(}\dfrac{2\pi}{3}\Big{)} \)
	\(=-4 \times 4\text{cis}\Big{(}-\dfrac{\pi}{3}\Big{)} \)
	\(=-4\,\bar z\)

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

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

a.	\(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\)	\(=\left[-\cos x\right]_0^\frac{\pi}{3}\)
		\(=-\cos\dfrac{\pi}{3}+\cos 0\)
		\(=-\dfrac{1}{2}+1\)
		\(=\dfrac{1}{2}\)

b.	\(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\)	\(=\left[\sin x\right]_k^\frac{\pi}{2}\)
		\(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\)
		\(=1-\sin (k)\)

\(1-\sin (k)\)	\(=\dfrac{1}{2}\)
\(\sin (k)\)	\(=\dfrac{1}{2}\)
\(\therefore\ k\)	\(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

\(\text{Area}\)	\(\approx \dfrac{3-1}{2\times 2}\Bigg[2+2\times\dfrac{5}{2}+\dfrac{10}{3}\Bigg]\)
	\(\approx\dfrac{1}{2}\Bigg[\dfrac{6}{3}+\dfrac{15}{3}+\dfrac{10}{3}\Bigg]\)
	\(\approx5\dfrac{1}{6}\)

\(e^{2x}-12\)	\(=4e^{x}\)
\(e^{2x}-4e^{x}-12\)	\(=0\)

\(u^2-4u-12\)	\(=0\)
\((u-6)(u+2)\)	\(=0\)