Band 3 – Page 3

PHYSICS, M7 2025 HSC 1 MC

Which of the following did Maxwell contribute to the understanding of the nature of light?

Explanation of atomic emission spectra
Prediction of the speed of electromagnetic waves
Experimental support for the particle model of light
Experimental confirmation of light beyond the visible spectrum

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$B$

Show Worked Solution

Maxwell’s contribution was the prediction of the speed of electromagnetic waves, showing that light is an electromagnetic wave.

$\Rightarrow B$

PHYSICS, M5 2025 HSC 29

A mass moves around a vertical circular path of radius $r$, in Earth's gravitational field, without loss of mechanical energy. A string of length $r$ maintains the circular motion of the mass.

When the mass is at its highest point $B$, the tension in the string is zero.

Show that the speed of the mass at the highest point, $B$, is given by $v=\sqrt{r g}$. (2 marks)

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Compare the speed of the mass at point $A$ to that at point $B$. Support your answer using appropriate mathematical relationships. (3 marks)

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a. $\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}$

$\dfrac{mv_B^2}{r}$	$=T+mg$
$v_B^2$	$=rg \ \ (T=0)$
$v_B$	$=\sqrt{r g}$

b. $\text {Total} \ ME=E_k+GPE$

$\text{At point} \ B :$

$\text {Total} \ ME$	$=\dfrac{1}{2} m v_B^2+mg(2r)$
	$=\dfrac{1}{2} m \times r g+2 mrg$
	$=\dfrac{5}{2} m r g$

$\text{At point} \ A :$

$\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg$

$\text{Since \(ME$ is conserved:}\)

$\dfrac{5}{2} mrg$	$=\dfrac{1}{2} m v_A^2+m r g$
$\dfrac{1}{2} mv_A^2$	$=\dfrac{3}{2} m r g$
$v_A^2$	$=3 rg$
$v_A$	$=\sqrt{3rg}=\sqrt{3} \times v_B$

Show Worked Solution

a. $\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}$

$\dfrac{mv_B^2}{r}$	$=T+mg$
$v_B^2$	$=rg \ \ (T=0)$
$v_B$	$=\sqrt{r g}$

b. $\text {Total} \ ME=E_k+GPE$

$\text{At point} \ B :$

$\text {Total} \ ME$	$=\dfrac{1}{2} m v_B^2+mg(2r)$
	$=\dfrac{1}{2} m \times r g+2 mrg$
	$=\dfrac{5}{2} m r g$

$\text{At point} \ A :$

$\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg$

♦♦ Mean mark (b) 34%.

$\text{Since \(ME$ is conserved:}\)

$\dfrac{5}{2} mrg$	$=\dfrac{1}{2} m v_A^2+m r g$
$\dfrac{1}{2} mv_A^2$	$=\dfrac{3}{2} m r g$
$v_A^2$	$=3 rg$
$v_A$	$=\sqrt{3rg}=\sqrt{3} \times v_B$

The starting position of a simple AC generator is shown. It consists of a single rectangular loop of wire in a uniform magnetic field of 0.5 T. This loop is connected to two slip rings and the slip rings are connected via brushes to a voltmeter.

The loop is rotated at a constant rate through an angle of 90 degrees from the starting position in the direction indicated, in 0.1 seconds.
Calculate the magnitude of the average emf generated during this rotation. (2 marks)

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The same coil was then rotated at 10 revolutions per second from the starting position. The voltage varies with time, as shown in the graph.

On the same axes, sketch a graph that shows the variation of voltage with time if the rotational speed is 20 revolutions per second in the opposite direction, beginning at the original starting position. (3 marks)

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a. $\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}$

$\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}$

Show Worked Solution

a. $\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}$

$\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}$

BIOLOGY, M7 2025 HSC 21

The diagram shows components of the innate immune system in humans.

State the role of TWO components that protect against infection. (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \quad \quad \textit{How it protects against infection}\quad \quad \rule[-1ex]{0pt}{0pt}\\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\end{array}

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Any TWO of the following components

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \textit{How it protects against infection}\quad\rule[-1ex]{0pt}{0pt}\\
\hline
\text{Skin} & \text{Acts as a physical barrier}\\
\ & \text{preventing pathogen entry into} \\
\ & \text{tissue}\\
\hline
\text{Stomach} &\text{Destroys ingested pathogens} \\
\text{acid} & \text{through low pH chemical}\\
\ & \text{environment.}\\
\hline
\text{Mucus} &\text{Traps pathogens and prevents} \\
\text{lining} & \text{their entry into underlying}\\
\ & \text{tissues.}\\
\hline
\text{Nasal} &\text{Filters and traps airborne} \\
\text{hair} & \text{pathogens, preventing}\\
\ & \text{respiratory entry.}\\
\hline
\text{Tear glands} &\text{Produce lysozyme enzyme that} \\
\ & \text{destroys bacterial cell walls.}\\
\hline
\text{Urinary} &\text{Flushes pathogens from the} \\
\text{tract} & \text{urethra preventing infection.}\\
\hline
\end{array}

Show Worked Solution

Any TWO of the following components

BIOLOGY, M8 2025 HSC 24

The following flow chart represents the control of body temperature in humans.

Complete the flow chart to give an example of mechanism A and an example of mechanism B. (2 marks)
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Outline how mechanism B maintains homeostasis. (2 marks)
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a. Mechanism A (Decreases temperature):

Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

Shivering → Vasoconstriction

b. Maintaining homeostasis

When body temperature drops below normal range, thermoreceptors detect the change.
The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
Body temperature increases back to normal range, restoring homeostasis.

Show Worked Solution

a. Mechanism A (Decreases temperature):

Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

Shivering → Vasoconstriction

b. Maintaining homeostasis

When body temperature drops below normal range, thermoreceptors detect the change.
The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
Body temperature increases back to normal range, restoring homeostasis.

Functions, 2ADV EQ-Bank 8

Solve for $x$, giving your answers in the simplest form $a+b\sqrt{c}$ where $a, b$ and $c$ are real:

$5 x^2-20 x+4=0$ (2 marks)

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$x=2 \pm \dfrac{4}{5} \sqrt{5}$

Show Worked Solution

$5 x^2-20 x+4=0$

$x$	$=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
	$=\dfrac{20 \pm \sqrt{20^2-4 \times 5 \times 4}}{2 \times 5}$
	$=\dfrac{20 \pm \sqrt{320}}{10}$
	$=2 \pm \dfrac{8 \sqrt{5}}{10}$
	$=2 \pm \dfrac{4}{5} \sqrt{5}$

Functions, 2ADV EQ-Bank 7 MC

What are the solutions to $3x^2+2x-4=0$?

$x=\dfrac{-1 \pm \sqrt{13}}{3}$
$x=\dfrac{1 \pm \sqrt{13}}{3}$
$x=\dfrac{-1 \pm \sqrt{5}}{2}$
$x=\dfrac{1 \pm \sqrt{5}}{2}$

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$A$

Show Worked Solution

$3 x^2+2 x-4=0$

$x$	$=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2a}$
	$=\dfrac{-2 \pm \sqrt{2^2-4 \times 3 \times-4}}{2 \times 3}$
	$=\dfrac{-2 \pm \sqrt{52}}{6}$
	$=\dfrac{-1 \pm \sqrt{13}}{3}$

$\Rightarrow A$

Functions, 2ADV EQ-Bank 10

Express $y=x^2-4 x+6$ in the form $y=(x-a)^2+c$. (1 mark)

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Graph the parabola, labelling its vertex and $y$-intercept. (2 marks)

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a. $y=(x-2)^2+2$

b. $\text {Vertex}=(2,2)$

$y \text{-intercept}=(0,6)$

Show Worked Solution

a.	$y$	$=x^2-4 x+6$
		$=x^2-4 x+4+2$
		$=(x-2)^2+2$

b. $\text {Vertex}=(2,2)$

$y \text{-intercept}=(0,6)$

Functions, 2ADV EQ-Bank 2

Find the equation of the line that passes through $(2,1)$ and $(-3,4)$. (2 marks)

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Determine whether $(7,-2)$ lies on the line. (1 mark)

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a. $y=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

b. $\text {Substitute}\ (7,-2) \ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Show Worked Solution

a. $(2,1),(-3,4)$

$\text{Gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

$\text{Find equation with} \ \ m=-\dfrac{3}{5} \ \ \text{through}\ \ (2,1):$

$y-1$	$=-\dfrac{3}{5}(x-2)$
$y$	$=-\dfrac{3}{5} x+\dfrac{11}{5}$

b. $\text {Substitute}\ (7,-2)\ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Functions, 2ADV EQ-Bank 1

A set of ordered pairs $(x, y)$ on the coordinate plane are represented by set $A$ below:

$A=\{(1,3),(4,6),(5,6),(0,1),(1,7)\}$

Explain if Set $A$ a function or a relation? (2 marks)

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$\text {If set \(A$ is a function, there can only be one $y$-value}\)

$\text{for any given } x \text{-value.}$

$\text{This is not the case} \ \Rightarrow\ (1,3),(1,7)$

$\therefore\ \text{Set} \ A \ \text {is a relation.}$

Show Worked Solution

$\text {If set \(A$ is a function, there can only be one $y$-value}\)

$\text{for any given } x \text{-value.}$

$\text{This is not the case} \ \Rightarrow\ (1,3),(1,7)$

$\therefore\ \text{Set} \ A \ \text {is a relation.}$

Functions, 2ADV EQ-Bank 6

Simplify $\dfrac{x^2}{x^2-2 x-15}-\dfrac{x}{x+3}$. (2 marks)

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$\dfrac{5 x}{(x+3)(x-5)}$

Show Worked Solution

$\dfrac{x^2}{x^2-2 x-15}-\dfrac{x}{x+3}$	$=\dfrac{x^2}{(x+3)(x-5)}-\dfrac{x}{x+3}$
	$=\dfrac{x^2-x(x-5)}{(x+3)(x-5)}$
	$=\dfrac{x^2-x^2+5 x}{(x+3)(x-5)}$
	$=\dfrac{5 x}{(x+3)(x-5)}$

Functions, 2ADV EQ-Bank 4

Simplify $\dfrac{a^3 b-a b^3}{a^2+2 a b+b^2}$. (2 marks)

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$\dfrac{a b(a-b)}{a+b}$

Show Worked Solution

$\dfrac{a^3 b-a b^3}{a^2+2 a b+b^2}$	$=\dfrac{a b\left(a^2-b^2\right)}{(a+b)^2}$
	$=\dfrac{a b(a+b)(a-b)}{(a+b)^2}$
	$=\dfrac{a b(a-b)}{a+b}$

L&E, 2ADV EQ-Bank 5

The mass `M` kg of a baby pig at age `x` days is given by `M = A(1.1)^x` where `A` is a constant. The graph of this equation is shown.

What is the value of `A`? (1 mark)

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What is the daily growth rate of the pig’s mass? Write your answer as a percentage. (1 mark)

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i. `1.5\ text(kg)`

ii. `10text(%)`

Show Worked Solution

i. `text(When)\ x = 0:`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!

`1.5`	`= A(1.1)^0`
`:. A`	`= 1.5\ text(kg)`

ii. `text(Daily growth rate)\ = 0.1 = 10text(%)`

L&E, 2ADV EQ-Bank 4

The number of bacteria in a culture grows from 100 to 114 in one hour.

What is the percentage increase in the number of bacteria? (1 mark)

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The bacteria continue to grow according to the formula `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

What is the number of bacteria after 15 hours? (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\; & \;\; 5 \;\; & \;\; 10 \;\; & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\; & \;\; 193 \;\; & \;\; 371 \;\; & \;\; ? \;\; \\
\hline
\end{array}

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Use the values of `n` from `t = 0` to `t = 15` to draw a graph of `n = 100(1.14)^t`.

Use about half a page for your graph and mark a scale on each axis. (2 marks)

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Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300. (1 mark)

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i. `text(14%)`

ii. `714`

iii. `text(Proof)\ \ text{(See Worked Solutions)}`

iv. `text(8.4 hours)`

Show Worked Solution

i. `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

ii. `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n= 100(1.14)^15= 713.793\ …\ = 714\ \ \ text{(nearest whole)}`

iii.

iv. `text(Using the graph)`

`text(The number of bacteria reaches 300 after ~ 8.4 hours.)`

Functions, 2ADV EQ-Bank 6

Identify where the graph $f(x)=\dfrac{\abs{x}}{x}$ is not continuous. (1 mark)

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Sketch the graph of $f(x)$. (2 marks)

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a. $\text {Denominator} \neq 0$

$f(x)\ \text{is not continuous when} \ \ x=0.$

b. $\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1$

$\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1$

Show Worked Solution

a. $\text {Denominator} \neq 0$

$f(x)\ \text{is not continuous when} \ \ x=0$

b. $\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1$

$\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1$

Functions, 2ADV EQ-Bank 5

Identify where the graph $f(x)=\dfrac{x^2-1}{x-1}$ is not continuous. (1 mark)

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Sketch the graph of $f(x)$. (2 marks)

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

$\text{Since denominator} \neq 0$

$f(x) \ \ \text{is not continuous when} \ \ x=1.$

Show Worked Solution

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

$\text{Since denominator} \neq 0$

$f(x) \ \ \text{is not continuous when} \ \ x=1.$

Functions, 2ADV EQ-Bank 4

Consider the function $y=f(x)$ where

$f(x)= \begin{cases}x^2+6, & \text { for } x \leqslant 0 \\ 6, & \text { for } 0<x \leqslant 3 \\ 2^x, & \text { for } x>3\end{cases}$

Sketch $y=f(x)$ (3 marks)

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For what value of $x$ is $y=f(x)$ NOT continuous? (1 mark)

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b. $f(x)\ \ \text {is NOT continuous at}\ \ x=3.$

Show Worked Solution

b. $f(x)\ \ \text {is NOT continuous at}\ \ x=3.$

Functions, 2ADV EQ-Bank 3

Graph the function $y=f(x)$ where:

$f(x)= \begin{cases}x^2, & \text { for } x \leq-1 \\ x-1, & \text { for }-1<x \leq 1 \\ -x^3, & \text { for } x>1 \end{cases}$. (3 marks)

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Show Worked Solution

Measurement, STD2 EQ-Bank 5 MC

The distance from the earth to the sun is approximately 150 million kilometres.

This distance expressed in scientific notation is:

$1.5\times 10^{9}$
$1.5\times 10^{8}$
$1.5\times 10^{7}$
$1.5\times 10^{6}$

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$B$

Show Worked Solution

$150\ \text{million}\ =\ 150\,000\,000=1.5\times 10^8$

$\Rightarrow B$

Statistics, STD2 EQ-Bank 17

A Physics class of 12 students is going on a 4 day excursion by bus.

The students are asked to each pack one bag for the trip. The bags are weighed, and the weights (in kg) are listed in order as follows:

$8,\ \ 9, \ \ 10,\ \ 10, \ \ 15, \ \ 18, \ \ 22, \ \ 25, \ \ 29, \ \ 35, \ \ 38, \ \ 41 $

Use the above data to produce a five number summary for the weights of the bags. (2 marks)
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Using your five number summary from part (a), calculate the interquartile range of the weights. (2 marks)
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a. $\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 8\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 20\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 32 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 41\\
\hline
\end{array}

b. $22$

Show Worked Solution

a. $\text{Five number Summary}$

b.	$\text{IQR}$	$=Q_3-Q_1$
		$=32-10=22$

Algebra, STD2 EQ-Bank 3 MC

If $d=\sqrt{\dfrac{h}{5}}$, what is the value of $d$, correct to one decimal place, when $h=28$?

Show Answers Only

$B$

Show Worked Solution

$\text{When}\ h=28:$

$d$	$=\sqrt{\dfrac{h}{5}}$
$d$	$=\sqrt{\dfrac{28}{5}}$
	$=2.366…$
	$=2.4\ (\text{1 deccimal place})$

$\Rightarrow B$

Measurement, STD2 EQ-Bank 2 MC

Arrange the numbers $5.6\times 10^{-2}$, $4.8\times 10^{-1}$, $7.2\times 10^{-2}$ from smallest to largest.

$5.6\times 10^{-2}$, $7.2\times 10^{-2}$, $4.8\times 10^{-1}$
$4.8\times 10^{-1}$, $5.6\times 10^{-2}$, $7.2\times 10^{-2}$
$7.2\times 10^{-2}$, $5.6\times 10^{-2}$, $4.8\times 10^{-1}$
$4.8\times 10^{-1}$, $7.2\times 10^{-2}$, $5.6\times 10^{-2}$

Show Answers Only

$A$

Show Worked Solution

$5.6\times 10^{-2}=0.052$

$4.8\times 10^{-1}=0.48$

$7.2\times 10^{-2}=0.072$

$\therefore\ \text{Correct order is: }\ 5.6\times 10^{-2}$, $7.2\times 10^{-2}$, $4.8\times 10^{-1}$

$\Rightarrow A$

Algebra, STD2 EQ-Bank 16

Jerico is the manager of a weekend market in which there are 220 stalls for rent. From past experience, Jerico knows that if he charges $d$ dollars to rent a stall. then the number of stalls, $s$, that will be rented is given by:

$s=220-4d$

How many stalls will be rented if Jerico charges $7.50 per stall . (1 mark)

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Complete the following table for the function $s=220-4d$. (1 mark)

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\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad d\quad \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 10\quad\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 30\quad & \rule{0pt}{2.5ex} \quad 50\quad \\
\hline
\rule{0pt}{2.5ex} \quad s\quad \rule[-1ex]{0pt}{0pt} & \ & \ & \\
\hline
\end{array}

Using an appropriate vertical scale and labelled axes, graph the function $s=220-4d$ on the grid below. (2 marks)

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Does it make sense to use the formula $s=220-4d$ to calculate the number of stalls rented if Jerico charges $60 per stall? Explain your answer. (2 marks)

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a. $190\ \text{stalls will be rented}$

d. $\text{When}\ d=60, s=220-4\times 60=-20$

$\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}$

$\text{as you cannot have a negative number of stalls.}$

Show Worked Solution

a. $s=220-4d=220-4\times 7.50=190$

$\therefore 190\ \text{stalls will be rented}$

d. $\text{When}\ d=60, s=220-4\times 60=-20$

$\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}$

$\text{as you cannot have a negative number of stalls.}$

Functions, 2ADV EQ-Bank 2

Rationalise the denominator in $\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$, and express in the simplest form. (2 marks)

Show Answers Only

$\dfrac{\sqrt{15}-\sqrt{6}-\sqrt{10}+2}{3} $

Show Worked Solution

$\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\dfrac{(\sqrt{3}-\sqrt{2})(\sqrt{5}-\sqrt{2})}{5-2}$

$=\dfrac{\sqrt{15}-\sqrt{6}-\sqrt{10}+2}{3}$

CHEMISTRY, M8 2025 HSC 30

Phosgene is used in industry as a starting material to synthesise useful polymers. Phosgene $\ce{(Cl2CO)}$ is a gas at room temperature and is highly toxic.

Justify a suitable precaution when using phosgene. (2 marks)

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Phosgene is synthesised by the reaction of carbon monoxide $\ce{(CO)}$ and chlorine $\ce{(Cl2)}$ in the gas phase.
1. 1. $\ce{Cl2(g) + CO(g) \rightleftharpoons Cl2CO(g)}$
Explain why an excess of carbon monoxide and a catalyst are used in the industrial synthesis of phosgene. (3 marks)

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a. Precaution when using phosgene inside a fume hood:

As phosgene is a highly toxic gas at room temperature. working in a fume hood prevents inhalation by safely removing the gas from the breathing zone and venting it outside the laboratory.
This measure significantly reduces the risk of poisoning and exposure.

b. Role of excess carbon monoxide:

An excess of carbon monoxide is used to increase the rate of production of phosgene.
According to Le Chatelier’s principle, increasing the concentration of $\ce{CO}$ shifts the equilibrium to the right, favouring the formation of $\ce{Cl2CO}$ and increasing the overall yield of phosgene.

Role of catalyst:

A catalyst is used to increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
This allows phosgene to be produced more rapidly and efficiently without affecting the equilibrium position.
In industrial settings, this increases production speed and reduces energy costs.

Show Worked Solution

a. Precaution when using phosgene inside a fume hood:

As phosgene is a highly toxic gas at room temperature. working in a fume hood prevents inhalation by safely removing the gas from the breathing zone and venting it outside the laboratory.
This measure significantly reduces the risk of poisoning and exposure.

b. Role of excess carbon monoxide:

An excess of carbon monoxide is used to increase the rate of production of phosgene.
According to Le Chatelier’s principle, increasing the concentration of $\ce{CO}$ shifts the equilibrium to the right, favouring the formation of $\ce{Cl2CO}$ and increasing the overall yield of phosgene.

Role of catalyst:

A catalyst is used to increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
This allows phosgene to be produced more rapidly and efficiently without affecting the equilibrium position.
In industrial settings, this increases production speed and reduces energy costs.

CHEMISTRY, M7 2025 HSC 27

Mixtures of hydrocarbons can be obtained from crude oil by the process of fractional distillation. Examples include petrol, diesel and natural gas.

Outline an environmental implication for a use of a named hydrocarbon mixture that is obtained from crude oil. (2 marks)

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Ethene is a simple hydrocarbon obtained from crude oil.
Using structural formulae, write the chemical equation for the conversion of ethene to ethanol, including any other necessary reagents. (3 marks)

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When ethanol is reacted with ethanoic acid, ethyl ethanoate is formed, as shown by the equation.
1. 1. $\text{ethanol} \ + \ \text{ethanoic acid } \ \rightleftharpoons \ \text{ethyl ethanoate} \ +\ \text{water}$
The graph below shows the concentration of ethanol from the start of the reaction, $t_0$, up to a time $t_1$.
At time $t_1$, an additional amount of ethanol is added to the system.
Sketch on the graph the changes that occur in the concentration of ethanol between time $t_1$, and when the system reaches a new equilibrium before time $t_2$. (3 marks)
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a. Environmental implication:

The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
Increased $\ce{CO2}$ levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
A typical component of petrol, octane, burns according to:
$\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}$

Show Worked Solution

a. Environmental implication:

The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
Increased $\ce{CO2}$ levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
A typical component of petrol, octane, burns according to:
$\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}$

There will be a sudden increase at $t_1$ and then the concentration of ethanol will decrease smoothly until a new equlibrium concentration (greater than the original equilibrium concentration) is reached.

CHEMISTRY, M7 2025 HSC 24

65.0 g of ethyne gas reacts with an excess of gaseous hydrogen chloride to produce chloroethene.

Draw the full structural formula of ethyne and identify the shape of the molecule. (2 marks)

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \text{Structural formula } \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \quad\text{ Shape of molecule } \quad \quad\\
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

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The molar masses of the compounds in the reaction are provided.

\begin{array}{|l|c|}
\hline \rule{0pt}{2.5ex}\text{Compound} \rule[-1ex]{0pt}{0pt}& \text{ Molar mass } \\
\hline \rule{0pt}{2.5ex}\text{Ethyne} \rule[-1ex]{0pt}{0pt}& 26.04 \\
\hline \rule{0pt}{2.5ex}\text{Hydrogen chloride} \rule[-1ex]{0pt}{0pt}& 36.46 \\
\hline \rule{0pt}{2.5ex}\text{Chloroethene} \rule[-1ex]{0pt}{0pt}& 62.50 \\
\hline
\end{array}

Calculate the mass of chloroethene produced, using the molar masses provided. Include a relevant chemical equation in your answer. (3 marks)

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a. Structural formula:

The shape of ethyne is linear.

b. 156 grams

Show Worked Solution

a. Structural formula:

The shape of ethyne is linear.

b. The chemical equation for the reacton occuring is shown below:

$\ce{C2H2(g) + HCl(g) -> C2H3Cl(g)}$

$n\ce{(C2H2)} = \dfrac{m}{MM} = \dfrac{65.0}{26.04} = 2.496\ \text{mol}$
As the reactants and products are in a $1:1:1$ ratio, $n\ce{(C2H2)}_{\text{reacted}} = n\ce{(C2H3Cl)}_{\text{produced}}$
$m\ce{(C2H3Cl)} = n \times MM = 2.496 \times 62.50 = 156\ \text{grams (3 sig.fig.)}$

CHEMISTRY, M7 2025 HSC 15 MC

Consider the following sequence of reactions.

Prop-2-en-1-ol was reacted with hydrogen gas to form liquid $X$.
$X$ was oxidised, producing liquid $Y$ that formed bubbles of a gas when reacted with aqueous sodium carbonate.
$Y$ was heated under reflux with methanol and a drop of concentrated sulfuric acid, producing an organic liquid, $Z$.

This process has been presented in the flow chart below.

Which option correctly identifies the structures for $X$, $Y$ and $Z$?

Show Answers Only

$D$

Show Worked Solution

The first reaction that occurs is a hydrogenation addition reaction in which prop-2-en-1-ol reacts under a Pd catalyst to produce propan-1-ol.
The primary alcohol propan-1-ol then undergoes oxidation to produce the carboxylic acid propanoic acid. This is confirmed as when it is reacted with sodium carbonate, it undergoes an acid-carbonate reaction to produce carbon dixoide which is observed through the bubbles.
The third reaction is an estification reaction in which propanic acid reacts with methanol under relfex to produce methyl-propanoate.
All three of the correctly drawn compounds can be observed in $D$

$\Rightarrow D$

CHEMISTRY, M5 2025 HSC 12 MC

Consider the following reaction.

$\ce{3AgNO3(aq) + FeCl3(aq) \rightleftharpoons 3AgCl(s) + Fe(NO3)3(aq)}$

What is the correct equilibrium expression for this reaction?

$\dfrac{\left[\ce{Fe(NO3)3}\right]}{\left[\ce{AgNO3}\right]\left[\ce{FeCl3}\right]}$
$\dfrac{\left[\ce{Fe(NO3)3}\right]}{\left[\ce{AgNO3}\right]^3\left[\ce{FeCl3}\right]}$
$\dfrac{\left[\ce{AgNO3}\right]^3 \left[\ce{FeCl3}\right]}{\ce{[AgCl]^3}\left[ \ce{Fe}\ce{(NO3)_3}\right]}$
$\dfrac{\ce{[AgCl]^3}\left[ \ce{Fe}\ce{(NO3)_3}\right]}{\left[\ce{AgNO3}\right]^3\left[\ce{FeCl3}\right]}$

Show Answers Only

$B$

Show Worked Solution

The equilibrium expression does not include reactants or products that are in the solid or liquid state, hence $\ce{3AgCl(s)}$ is not included in the equilibrium expression.
$\therefore K_{eq} = \dfrac{\ce{[Fe(NO3)3]}}{\ce{[AgNO3]^3[FeCl3]}}$

$\Rightarrow B$

CHEMISTRY, M7 2025 HSC 10 MC

Which of the following options lists ALL the forces that are present between molecules of butanoic acid?

Covalent bonding
Dispersion and dipole-dipole
Covalent bonding and hydrogen bonding
Dispersion, dipole-dipole and hydrogen bonding

Show Answers Only

$D$

Show Worked Solution

Hydrogen bonding occurs due to the $\ce{OH}$ group in the carboxylic acid, which can form strong hydrogen bonds between molecules.
Dipole-dipole forces arise from the polar $\ce{C=O}$ bond in the carboxyl group (carbonyl oxygen is highly electronegative).
Dispersion forces are always present between molecules due to temporary dipoles, especially from the nonpolar hydrocarbon chain.

$\Rightarrow D$

CHEMISTRY, M7 2025 HSC 9 MC

Some Torres Strait Islander Peoples pound the leaves of the vine Derris uliginosa to extract a chemical called saponin. Saponin is a relatively large molecule that contains both a water-soluble carbohydrate chain and a fat-soluble side chain.

Which is the most likely use of saponin?

As a cleaning agent
As a neutraliser of insect bites
As a dye for pigmentation and painting
As an additive to food for flavouring and tenderising

Show Answers Only

$A$

Show Worked Solution

Saponin has both water-soluble and fat-soluble parts, allowing it to dissolve fats in water and act like a soap or detergent.
The fat-soluble side chain bonds with dirts and fats while the water soluble component bonds with the water molecules to form micelle.
The micelles can then be washed away leaving surfaces clean.

$\Rightarrow A$

CHEMISTRY, M5 2025 HSC 5 MC

$\ce{PCl3}$ and $\ce{Cl2}$ were introduced to an empty sealed vessel.

$\ce{PCl3}$ reacted with $\ce{Cl2}$ to produce $\ce{PCl5}$.

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

Which graph best represents the changing concentration of $\ce{Cl2}$ as the system approached the equilibrium point?

Show Answers Only

$C$

Show Worked Solution

The initial concentration of $\ce{Cl2}$ is postive it is introduced into the system.
The concentration of $\ce{Cl2}$ will decrease with the rate of decrease slowing down over time as the system reaches a dynamic equilibrium.
Hence the concentration of $\ce{Cl2}$ will be non-zero when the system reaches dynamic equilibrium.

$\Rightarrow C$

CHEMISTRY, M7 2025 HSC 2 MC

Consider this reaction.

Which reaction type is shown?

Addition
Oxidation
Reduction
Substitution

Show Answers Only

$B$

Show Worked Solution

The reaction shows butan-2-ol being converted to butanone.
This occurs via oxidation where a secondary alcohol is converted to a ketone.

$\Rightarrow B$

CHEMISTRY, M6 2025 HSC 1 MC

An aqueous solution of an unknown acid $\ce{(HA)}$ is represented below.

Which row of the table best describes this solution?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \textit{Strong}\quad \quad \rule[-1ex]{0pt}{0pt}& \ \ \textit{Concentrated} \ \ \\
\hline
\rule{0pt}{2.5ex}\checkmark\rule[-1ex]{0pt}{0pt}&\checkmark\\
\hline
\rule{0pt}{2.5ex}\checkmark\rule[-1ex]{0pt}{0pt}& \large{\times}\\
\hline
\rule{0pt}{2.5ex}\large{\times}\rule[-1ex]{0pt}{0pt}& \checkmark \\
\hline
\rule{0pt}{2.5ex}\large{\times}\rule[-1ex]{0pt}{0pt}& \large{\times} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

A strong acid is one that fully ionises when placed in solution:
$\ce{HA(aq) -> H^+(aq) + A^-(aq)}$
As there are only a small number of molecules in the solution the acid concentration is dilute.

$\Rightarrow B$

Mechanics, EXT2 M1 2025 HSC 14b

The acceleration of a particle is given by $\ddot{x}=32 x\left(x^2+3\right)$, where $x$ is the displacement of the particle from a fixed-point $O$ after $t$ seconds, in metres. Initially the particle is at $O$ and has a velocity of 12 m s$^{-1}$ in the negative direction.

Show that the velocity of the particle is given by $v=-4\left(x^2+3\right)$. (2 marks)

--- 10 WORK AREA LINES (style=lined) ---
Find the time taken for the particle to travel 3 metres from the origin. (2 marks)

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

Show Answers Only

i. $\text{See Worked Solutions}$

ii. $t=\dfrac{\pi}{12 \sqrt{3}} \ \text{sec}$

Show Worked Solution

i. $\ddot{x}=32 x\left(x^2+3\right)$

$\text{Show} \ \ v=-4\left(x^2+3\right)$

$\text{Using} \ \ \ddot{x}=v \cdot \dfrac{dv}{dx}:$

$v \cdot \dfrac{dv}{dx}$	$=32 x\left(x^2+3\right)$
$\displaystyle \int v \, dv$	$=\displaystyle \int 32 x^3+96 x\, dx$
$\dfrac{v^2}{2}$	$=8 x^4+48 x^2+c$

$\text{When} \ \ x=0, v=-12 \ \Rightarrow \ c=72$

$\dfrac{v^2}{2}=8 x^4+48 x^2+72$

$v^2=16\left(x^4+6 x^2+9\right)$

$v=-4\left(x^2+3\right) \quad (V=-12 \ \ \text {when} \ \ x=0)$

ii. $\dfrac{dx}{dt}=-4\left(x^2+3\right)$

$\dfrac{dt}{dx}=-\dfrac{1}{4\left(x^2+3\right)}$

$t=-\dfrac{1}{4} \displaystyle \int \dfrac{1}{3+x^2} d x=-\frac{1}{4} \times \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$

$\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=0$

$\text{Since particle is moving left at} \ \ t=0,$

$\text{Find \(t$ when} \ \ x=-3:\)

$t$	$=-\dfrac{1}{4 \sqrt{3}} \times \tan ^{-1}\left(-\dfrac{3}{\sqrt{3}}\right)$
	$=-\dfrac{1}{4 \sqrt{3}} \times \tan ^{-1}(-\sqrt{3})$
	$=-\dfrac{1}{4 \sqrt{3}} \times-\dfrac{\pi}{3}$
	$=\dfrac{\pi}{12 \sqrt{3}} \ \text{sec}$

Proof, EXT2 P1 2025 HSC 13c

For positive real numbers $a$ and $b$, prove that $\dfrac{a+b}{2} \geq \sqrt{a b}$. (2 marks)

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Hence, or otherwise, show that $\dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}$ for any integer $n \geq 0$. (2 marks)

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

Show Answers Only

i. $\text {Using}\ \ (\sqrt{a}-\sqrt{b})^2 \geqslant 0:$

$a-2 \sqrt{a b}+b$	$\geqslant 0$
$a+b$	$\geqslant 2 \sqrt{a b}$
$\dfrac{a+b}{2}$	$\geqslant \sqrt{a b}$

ii. $\text{Show}\ \ \dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}$

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

$\text{Let} \ \ a=2 n+1, b=2 n+3$

$\dfrac{2 n+1+2 n+3}{2}$	$\geqslant \sqrt{2 n+1} \sqrt{2 n+3}$
$2 n+2$	$\geqslant \sqrt{2 n+1} \sqrt{2 n+3}$
$\dfrac{1}{2 n+2}$	$\leqslant \dfrac{1}{\sqrt{2 n+1} \sqrt{2 n+3}}$
$\dfrac{2 n+1}{2 n+2}$	$\leqslant \dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}$

$\text{Consider part (i):}\ (\sqrt{a}-\sqrt{b})^2 \geqslant 0$

$\Rightarrow \ \text{Equality only holds when} \ \ a=b$

$\text{Since}\ \ a=2 n+1 \neq b=2 n+3$

$\dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}$

Show Worked Solution

i. $\text {Using}\ \ (\sqrt{a}-\sqrt{b})^2 \geqslant 0:$

$a-2 \sqrt{a b}+b$	$\geqslant 0$
$a+b$	$\geqslant 2 \sqrt{a b}$
$\dfrac{a+b}{2}$	$\geqslant \sqrt{a b}$

ii. $\text{Show}\ \ \dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}$

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

$\text{Let} \ \ a=2 n+1, b=2 n+3$

$\dfrac{2 n+1+2 n+3}{2}$	$\geqslant \sqrt{2 n+1} \sqrt{2 n+3}$
$2 n+2$	$\geqslant \sqrt{2 n+1} \sqrt{2 n+3}$
$\dfrac{1}{2 n+2}$	$\leqslant \dfrac{1}{\sqrt{2 n+1} \sqrt{2 n+3}}$
$\dfrac{2 n+1}{2 n+2}$	$\leqslant \dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}$

♦ Mean mark (ii) 43%.

$\text{Consider part (i):}\ (\sqrt{a}-\sqrt{b})^2 \geqslant 0$

$\Rightarrow \ \text{Equality only holds when} \ \ a=b$

$\text{Since}\ \ a=2 n+1 \neq b=2 n+3$

$\dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}$

BIOLOGY, M8 2025 HSC 25

The graph shows the changes in UV level in a single day.

The Cancer Council suggests that sun protection is needed whenever the UV level is 3 or above. The information provided on a sunscreen product suggests that sunscreen should be used between 10 am and 4 pm.

Using the graph, evaluate the information provided on the sunscreen product with regard to the Cancer Council suggestion. (3 marks)

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

Show Answers Only

Evaluation Judgment:

The sunscreen product recommendation is partially effective but requires improvement for optimal sun protection.

Supporting Evidence:

The graph shows UV levels exceed 3 from approximately 8 am to 6 pm.
The sunscreen recommendation (10 am to 4 pm) misses two critical hours of UV exposure.
Between 8-10 am and 4-6 pm, UV levels remain above 3, requiring protection.

Conclusion:

The product advice inadequately protects users during morning and late afternoon exposure periods when UV damage still occurs.

Show Worked Solution

Evaluation Judgment:

The sunscreen product recommendation is partially effective but requires improvement for optimal sun protection.

Supporting Evidence:

The graph shows UV levels exceed 3 from approximately 8 am to 6 pm.
The sunscreen recommendation (10 am to 4 pm) misses two critical hours of UV exposure.
Between 8-10 am and 4-6 pm, UV levels remain above 3, requiring protection.

Conclusion:

The product advice inadequately protects users during morning and late afternoon exposure periods when UV damage still occurs.

Mechanics, EXT2 M1 2025 HSC 12e

A particle of mass $m$ kg moves along a horizontal line with an initial velocity of $V_0 \ \text{ms}^{-1}$.

The motion of the particle is resisted by a constant force of $m k$ newtons and a variable force of $m v^2$ newtons, where $k$ is a positive constant and $v \ \text{ms}^{-1}$ is the velocity of the particle at $t$ seconds.

Show that the distance travelled when the particle is brought to rest is $\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)$ metres. (3 marks)

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$ma$	$-m k-m v^2$
$a$	$=-k-v^2$
$v \cdot \dfrac{d v}{d x}$	$=-\left(k+\nu^2\right)$
$\dfrac{d x}{d v}$	$=-\dfrac{v}{k+v^2}$
$x$	$=-\displaystyle\frac{1}{2} \int \frac{2 v}{k+v^2}\, d v=-\frac{1}{2} \ln \left(k+v^2\right)+c$

$\text{At} \ \ x=0, v=v_0 \ \ \Rightarrow\ \ c=\dfrac{1}{2} \ln \left(k+V_0^2\right)$

$x=\dfrac{1}{2} \ln \left(k+V_0^2\right)-\dfrac{1}{2} \ln \left(k+v^2\right)=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k+v^2}\right)$

$\text{Find \(x$ when $\ v=0\ $ (distance travelled):}\)

$x=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\ \text{metres}$

Show Worked Solution

$ma$	$-m k-m v^2$
$a$	$=-k-v^2$
$v \cdot \dfrac{d v}{d x}$	$=-\left(k+\nu^2\right)$
$\dfrac{d x}{d v}$	$=-\dfrac{v}{k+v^2}$
$x$	$=-\displaystyle\frac{1}{2} \int \frac{2 v}{k+v^2}\, d v=-\frac{1}{2} \ln \left(k+v^2\right)+c$

$\text{At} \ \ x=0, v=v_0 \ \ \Rightarrow\ \ c=\dfrac{1}{2} \ln \left(k+V_0^2\right)$

$x=\dfrac{1}{2} \ln \left(k+V_0^2\right)-\dfrac{1}{2} \ln \left(k+v^2\right)=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k+v^2}\right)$

$\text{Find \(x$ when $\ v=0\ $ (distance travelled):}\)

$x=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\ \text{metres}$

Calculus, EXT2 C1 2025 HSC 12d

Find $\displaystyle \int \frac{x^2-2 x+9}{(4-x)\left(x^2+1\right)} \, dx$. (4 marks)

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

Show Answers Only

$2\, \tan ^{-1} x-\ln \abs{4-x}+C$

Show Worked Solution

$\dfrac{x^2-2 x+9}{(4-x)\left(x^2+1\right)}=\dfrac{A}{(4-x)}+\dfrac{B x+C}{x^2+1}$

$A\left(x^2+1\right)+(B x+C)(4-x)=x^2-2 x+9$

$\text{If}\ \ x=4:$

$17 A=16-8+9=17 \ \ \Rightarrow\ \ A=1$

$\text{If}\ \ x=0:$

$1+c \times 4=9 \ \ \Rightarrow\ \ C=2$

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

$2+3 B+6=8 \ \ \Rightarrow\ \ B=0$

$\displaystyle\int \frac{x^2-2 x+9}{(4-x)\left(x^2+1\right)}\, d x$	$=\displaystyle\int \frac{1}{4-x}\, d x+\int \frac{2}{x^2+1}\, d x$
	$=2\, \tan ^{-1} x-\ln \abs{4-x}+C$

Vectors, EXT2 V1 2025 HSC 11d

Force ${\underset{\sim}{F}}_1$ has magnitude 12 newtons in the direction of vector $2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}$.
Show that ${\underset{\sim}{F}}_1=8 \underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Force ${\underset{\sim}{F}}_1$ from part (i) and a second force, ${\underset{\sim}{F}}_2=-6 \underset{\sim}{i}+12 \underset{\sim}{j}+4 \underset{\sim}{k}$, both act upon a particle.
Show that the resultant force acting on the particle is given by:
${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}.$ (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Calculate ${\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}$, where ${\underset{\sim}{F}}_3$ is the resultant force from part (ii) and $\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}$. (1 mark)

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

Show Answers Only

i. $\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}=\sqrt{2^2+(-2)^2+1^2}=3$

${\underset{\sim}{F}}_1=\dfrac{12}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)$

${\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$

ii. ${\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)$

${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}$

iii. ${\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22$

Show Worked Solution

i. $\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}=\sqrt{2^2+(-2)^2+1^2}=3$

${\underset{\sim}{F}}_1=\dfrac{12}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)$

${\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$

${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}$

iii. ${\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22$

Proof, EXT2 P2 2025 HSC 12b

Given the function $y=x e^{2 x}$, use mathematical induction to prove that $\dfrac{d^n y}{d x^n}=\left(2^n x+n 2^{n-1}\right) e^{2 x}$ for all positive integers $n$, where $\dfrac{d^n y}{d x^n}$ is the
$n$th derivative of $y$ and $\dfrac{d}{d x}\left(\dfrac{d^n y}{d x^n}\right)=\dfrac{d^{n+1} y}{d x^{n+1}}$. (3 marks)

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

Show Answers Only

$\text{Proof (See worked solutions)}$

Show Worked Solution

$\text{Prove} \ \ \dfrac{d^n y}{d x^n}=\left(2^n x+n 2^{n-1}\right) e^{2 x}$

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

$\operatorname{LHS}=\dfrac{d}{d x}\left(x e^{2 x}\right)=x \cdot 2 e^{2 x}+1 \cdot e^{2 x}=(2 x+1) e^{2 x}$

$\operatorname{RHS}=\left(2^{1} \cdot x+1 \cdot 2^{0}\right) e^{2 x}=(2 x+1) e^{2 x}=\operatorname{RHS}$

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

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

$\dfrac{d^k y}{d x^k}=\left(2^k x+k\, 2^{k-1}\right) e^{2 x}$

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

$\text{i.e.}\ \dfrac{d^{k+1} y}{d x^{k+1}}=\left(2^{k+1} x+(k+1) 2^k\right) e^{2 x}$

$\dfrac{d^{k+1} y}{d x^{k+1}}$	$=\dfrac{d}{d x}\left(2^k x+k\cdot 2^{k-1}\right) e^{2 x}$
	$=\dfrac{d}{d x}\left(2^k x e^{2 x}+k\cdot 2^{k-1} e^{2 x}\right)$
	$=2^k x \cdot 2 e^{2 x}+2^k \cdot e^{2 x}+k \cdot 2^{k-1} \cdot 2 e^{2 x}$
	$=2^{k+1} x e^{2 x}+2^k \cdot e^{2 x}+k \cdot 2^k \cdot e^{2 x}$
	$=2^{k+1} x e^{2 x}+(k+1) 2^k \cdot e^{2 x}$
	$=\left(2^{k+1} x+(k+1) 2^k\right) e^{2 x}$

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

$\therefore \ \text{Since true for \(\ \ n=1$, by PMI, true for integers} \ \ n \geqslant 1.\)

Complex Numbers, EXT2 N2 2025 HSC 11a

The location of the complex number $z$ is shown on the diagram below.

On the diagram, indicate the locations of $\bar{z}$ and $i \bar{z}$. (2 marks)

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Show Worked Solution

Mechanics, EXT2 M1 2025 HSC 4 MC

A particle in simple harmonic motion has speed $v \ \text{ms}^{-1}$, given by $v^2=-x^2+2 x+8$ where $x$ is the displacement from the origin in metres.

What is the amplitude of the motion?

Show Answers Only

$B$

Show Worked Solution

$\text{Using} \ \ v^2=n^2\left(a^2-(x-c)^2\right):$

$v^2$	$=-x^2+2 x+8$
	$=9-\left(x^2-2 x+1\right)$
	$=9-(x-1)^2$

$\therefore a^2 = 9\ \ \Rightarrow\ \ a=3$

$\Rightarrow B$

BIOLOGY, M6 2025 HSC 20 MC

Researchers studying T-cell acute lymphoblastic leukaemia (T-ALL) examined a section of DNA in individuals A and B.

In individual B, they found proteins that can regulate the expression of oncogene LMO 2. The following diagram represents the sections of DNA from the two individuals.

Based on this information, which of the following statements is the best explanation of the cause of T-ALL?

T-ALL is caused by an inactive Oncogene LMO 2.
The regulating proteins are the direct cause of T-ALL.
The regulating proteins have no effect as they are made on non-coding DNA.
The regulating proteins can activate the Oncogene LMO 2, causing the condition.

Show Answers Only

$D$

Show Worked Solution

D is correct: Regulating proteins activate oncogene LMO 2, leading to T-ALL development.

Other Options:

A is incorrect: T-ALL is caused by an active oncogene, not an inactive one.
B is incorrect: Proteins themselves aren’t the cause; they activate the oncogene.
C is incorrect: Regulating proteins can affect gene expression regardless of DNA location.

Complex Numbers, EXT2 N1 2025 HSC 3 MC

What are the square roots of $3-4 i$ ?

$1-2 i$ and $-1+2 i$
$1+2 i$ and $-1-2 i$
$2-i$ and $-2+i$
$-2-i$ and $2+i$

Show Answers Only

$C$

Show Worked Solution

$\text{Let} \ \ z=\sqrt{3-4 i}$ :

$z^2=3-4 i$

$z^2=a^2-b^2+2 a b\,i$

$\text{Equate real/imaginary parts:}$

$a^2-b^2=3\ \ldots\ (1)$

$2 a b=-4 \ \ \Rightarrow \ \ a b=-2\ \ldots\ (2)$

$\text{By inspection:}$

$a=2, b=-1 \ \Rightarrow \ z_1=2-i$

$a=-2, b=1 \ \Rightarrow \ z_2=-2+i$

$\Rightarrow C$

BIOLOGY, M8 2025 HSC 15 MC

Cochlear implants assist with hearing. The following are five steps involved in the process.

The sound signal is turned into electrical impulses.
The implant sends electrical impulses to the electrodes in the cochlea.
Sounds are picked up by the microphone.
The auditory nerve picks up and sends electrical impulses to the brain.
The electrical impulses are transmitted across the skin to the implant.

Which is the correct order for this process?

1, 4, 3, 2, 5
1, 3, 2, 4, 5
3, 1, 5, 2, 4
3, 1, 2, 5, 4

Show Answers Only

$C$

Show Worked Solution

C is correct: Sound → converted to electrical → transmitted to implant → sent to cochlea → auditory nerve.

Other Options:

A is incorrect: Sound must be picked up first, not converted first.
B is incorrect: Signal must be transmitted across skin before reaching cochlea.
D is incorrect: Impulses transmitted to implant before being sent to electrodes.

BIOLOGY, M7 2025 HSC 11 MC

In 2018, the Victorian government reported the target of vaccinating more than 95% of children below the age of five had been achieved.

What is the benefit of achieving a 95% vaccination rate for an infectious disease?

Only 5% of individuals will catch the disease.
Only 5% of individuals will be protected by the vaccine.
This will protect the remaining 5% who are not vaccinated.
The vaccinated individuals can transfer immunity to the remaining 5%.

Show Answers Only

$C$

Show Worked Solution

C is correct: High vaccination rate provides herd immunity protecting unvaccinated individuals.

Other Options:

A is incorrect: Herd immunity means fewer than 5% will catch disease.
B is incorrect: 95% are protected by vaccine, not 5%.
D is incorrect: Vaccinated individuals cannot transfer immunity to others through vaccination.

Calculus, EXT2 C1 2025 HSC 12a

Using integration by parts, evaluate $\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx$. (3 marks)

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$\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx = 1$

Show Worked Solution

$u$	$=x$	$u^{\prime}$	$=1$
$v^{\prime}$	$=\sin\,x$	$v$	$=-\cos\,x$

$\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx$	$=\Big[-x\,\cos\,x\Big]_0^{\small{\dfrac{\pi}{2}}} + \displaystyle \int_0^{\small{\dfrac{\pi}{2}}} \cos x \, dx$
	$=(0-0)+\Big[\sin\,x\Big]_0^{\small{\dfrac{\pi}{2}}} $
	$=\sin\,\dfrac{\pi}{2}-\sin\,0$
	$=1$

Calculus, EXT2 C1 2025 HSC 11f

Find $\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx$. (2 marks)

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$5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c$

Show Worked Solution

$\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx$	$=\displaystyle \int \dfrac{5}{\sqrt{16-(x^2+6x+9)}} \, dx$
	$=\displaystyle \int \dfrac{5}{\sqrt{16-(x+3)^2}} \, dx$
	$=5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c$

BIOLOGY, M7 2025 HSC 10 MC

The graph shows the number of recorded deaths, due to measles, before and after the measles vaccine was included in the National Immunisation Program (NIP).

Which of the following is a trend shown in the graph?

Most people have been immunised against measles since 1975.
Over the last 100 years the number of deaths has consistently fallen.
The number of measles-induced deaths has fallen to near zero since measles vaccine was added to the NIP.
The National Immunisation Program was the primary reason for the great reduction in measle-induced deaths.

Show Answers Only

$C$

Show Worked Solution

C is correct: Graph clearly shows deaths fell to near zero after 1975.

Other Options:

A is incorrect: Graph shows deaths, not immunisation rates; cannot infer this.
B is incorrect: Deaths fluctuated considerably before 1975, not consistent decline.
D is incorrect: Deaths were already declining before NIP; cannot claim it’s primary reason.

BIOLOGY, M5 2025 HSC 8 MC

When a red camellia flower is crossed with a white camellia flower, all the offspring are covered in both red and white petals.

What is the reason for this occurrence?

One gene is controlling multiple characteristics.
Environmental factors affect the phenotype of camellia flowers.
Alleles for both red and white colour in camellia flowers are recessive.
Petal colour in camellia flowers is controlled by a co-dominance pattern of inheritance.

Show Answers Only

$D$

Show Worked Solution

D is correct: Both alleles are expressed equally producing red and white petals.

Other Options:

A is incorrect: One gene controls one characteristic; this is pleiotropy not co-dominance.
B is incorrect: Environmental factors don’t explain both parental colours appearing in offspring.
C is incorrect: Both alleles are expressed, indicating co-dominance not recessive inheritance.

BIOLOGY, M8 2025 HSC 7 MC

An animal's body temperature and the air temperature of the animal's environment were measured every 4 hours, and the following data were recorded.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ \ \ \ \ \ \ {Time} \ \ \ \ \ \ \ \ \ \rule[-1ex]{0pt}{0pt} & {Body \ temperature} \rule[-1ex]{0pt}{0pt} & {Air \ temperature} \\
{} & \text{(°C)} & \text{(°C)} \\
\hline
\rule{0pt}{2.5ex} \text{4 am} \rule[-1ex]{0pt}{0pt} & \text{41.3} \rule[-1ex]{0pt}{0pt} & \text{19.2}\\
\hline
\rule{0pt}{2.5ex} \text{8 am} \rule[-1ex]{0pt}{0pt} & \text{41.1} \rule[-1ex]{0pt}{0pt} & \text{18.8}\\
\hline
\rule{0pt}{2.5ex} \text{12 pm} \rule[-1ex]{0pt}{0pt} & \text{40.8} \rule[-1ex]{0pt}{0pt} & \text{21.5}\\
\hline
\rule{0pt}{2.5ex} \text{4 pm} \rule[-1ex]{0pt}{0pt} & \text{41.4} \rule[-1ex]{0pt}{0pt} & \text{26.4}\\
\hline
\rule{0pt}{2.5ex} \text{8 pm} \rule[-1ex]{0pt}{0pt} & \text{41.2} \rule[-1ex]{0pt}{0pt} & \text{27.5}\\
\hline
\rule{0pt}{2.5ex} \text{12 am} \rule[-1ex]{0pt}{0pt} & \text{41.5} \rule[-1ex]{0pt}{0pt} & \text{23.0}\\
\hline
\end{array}

Based on this data, which row of the table indicates what type of animal it is and why?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\ & \\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\ & \\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\ & \\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\ & \\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex} {Type \ of \ animal}\rule[-1ex]{0pt}{0pt}& {Reason } \\
\hline
\rule{0pt}{2.5ex}\text{Ectotherm}\rule[-1ex]{0pt}{0pt}&\text{Body temperature is around 41°C and varies} \\ & \text{with the air temperature. }\\
\hline
\rule{0pt}{2.5ex}\text{Ectotherm}\rule[-1ex]{0pt}{0pt}& \text{Body temperature is around 41°C and is} \\ & \text{always above the air temperature. }\\
\hline
\rule{0pt}{2.5ex}\text{Endotherm }\rule[-1ex]{0pt}{0pt}& \text{Body temperature is relatively constant} \\ & \text{despite changes in air temperature. } \\
\hline
\rule{0pt}{2.5ex}\text{Endotherm}\rule[-1ex]{0pt}{0pt}& \text{Body temperature is relatively constant, and} \\ & \text{air temperature is relatively constant. } \\
\hline
\end{array}
\end{align*}

Show Answers Only

$C$

Show Worked Solution

C is correct: Body temperature remains constant around 41°C despite air temperature fluctuations.

Other Options:

A is incorrect: Body temperature doesn’t vary with air temperature; remains constant.
B is incorrect: Ectotherms cannot maintain constant body temperature above air temperature.
D is incorrect: Air temperature varies significantly from 18.8°C to 27.5°C.

BIOLOGY, M7 2025 HSC 5 MC

An infectious disease is spread through direct contact between hosts.

Under which conditions will this disease spread most rapidly?

Low population density with infected individuals rarely dying from the disease
High population density with infected individuals rarely dying from the disease
Low population density with infected individuals quickly dying from the disease
High population density with infected individuals quickly dying from the disease

Show Answers Only

$B$

Show Worked Solution

B is correct: High density increases contact rate and slow death maintains infectious individuals.

Other Options:

A is incorrect: Low population density reduces contact frequency between hosts.
C is incorrect: Low density and quick death both limit disease transmission.
D is incorrect: Quick death reduces time for infected individuals to spread disease.

Complex Numbers, EXT2 N1 2025 HSC 11c

The complex number $z$ is given by $x+i y$.

Find, in Cartesian form:

$z^2$ (1 mark)

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$\dfrac{1}{z}$. (2 marks)

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i. $z^2=x^2-y^2+2xy\,i$

ii. $\dfrac{1}{z}=\dfrac{x}{x^2-y^2}-\dfrac{y}{x^2-y^2}\,i$

Show Worked Solution

i.	$z^{2}$	$=(x+iy)^2$
		$=x^2-y^2+2xy\,i$

ii.	$\dfrac{1}{z}$	$=\dfrac{1}{x+iy}$
		$=\dfrac{x-iy}{(x+iy)(x-iy)}$
		$=\dfrac{x-iy}{x^2+y^2}$
		$=\dfrac{x}{x^2+y^2}-\dfrac{y}{x^2+y^2}\,i$

Complex Numbers, EXT2 N1 2025 HSC 11b

The complex numbers $w$ and $z$ are given by $w=2 e^{\small{\dfrac{i \pi}{6}}}$ and $z=3 e^{\small{\dfrac{i \pi}{6}}}$. Find the modulus and argument of $w z$. (2 marks)

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$\text{Modulus = 6, Argument}\ = \dfrac{\pi}{3} $

Show Worked Solution

$w=2 e^{\small{\dfrac{i \pi}{6}}}, \ \ z=3 e^{\small{\dfrac{i \pi}{6}}}$

$wz=2 e^{\small{\dfrac{i \pi}{6}}} \times 3 e^{\small{\dfrac{i \pi}{6}}} = 6 e^{\small{\dfrac{i \pi}{3}}}$

$\text{Modulus = 6, Argument}\ = \dfrac{\pi}{3} $

Proof, EXT2 P1 2025 HSC 2 MC

Consider the statement:

$\exists\, x \in Z$, such that $x^2$ is odd.

Which of the following is the negation of the statement?

$\forall\, x \in Z , x^2$ is odd
$\forall\, x \in Z , x^2$ is even
$x^2$ is even $\Rightarrow x \in Z$
$\exists\, x \in Z$, such that $x^2$ is even

Show Answers Only

$B$

Show Worked Solution

$x^2\ \text{is odd is negated by}\ x^2\ \text{is even.}$

$\therefore\ \text{Negation of statement:}\ \forall\, x \in Z , x^2\ \text{is even}$

$\Rightarrow B$

Measurement, STD1 M4 2025 HSC 17

Paint is sold in two sizes at a local shop.

Four-litre cans at $90 per can
Ten-litre cans at $205 per can

Mina needs to purchase 80 litres of paint.

Calculate the amount of money Mina will save by purchasing only ten-litre cans of paint rather than only four-litre cans of paint. (2 marks)

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$\text{Saving}=$160$

Show Worked Solution

$\text{4 litre cans needed}\ =\dfrac{80}{4}=20$

$\text{Cost using 4 litre cans}\ =20\times $90=$1800$

$\text{10 litre cans needed}\ =\dfrac{80}{10}=8$

$\text{Cost using 10 litre cans}\ =8\times $205=$1640$

$\therefore\ \text{Saving}\ =$1800-$1640=$160$

Networks, STD1 N1 2025 HSC 14

The time, in minutes, it takes to travel by road between six towns is recorded and shown in the network diagram below.

In this network the shortest path corresponds to the minimum travel time.
What is the minimum travel time between towns $A$ and $F$, and what is the corresponding path? (2 marks)

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New roads are built to connect a town $G$ to towns $A$ and $D$. The table gives the times it takes to travel by the new roads.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Town} \rule[-1ex]{0pt}{0pt} & \textit{Time} \text{(minutes)} \rule[-1ex]{0pt}{0pt} & \textit{Town} \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 8 \rule[-1ex]{0pt}{0pt} & G \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 22 \rule[-1ex]{0pt}{0pt} & D \\
\hline
\end{array}

Add the new roads and times to the network diagram below. (2 marks)

--- 0 WORK AREA LINES (style=lined) ---
Explain whether the path in part (a) is still the shortest path from $A$ to $F$ after the new roads are added. (1 mark)

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a. $\text{Minimum travel time}\ = 15+20+10+5+8=58\ \text{minutes}$

$\text{Path: }A → B → C → D → E → F$

b. $\text{New roads}\ A → G\ \text{and }G → D $

c. $\text{Using the new roads }A → G\ \text{and }G → D:$

$\text{Minimum travel time}\ =8+22+5+8=43\ \text{minutes.}$

$\text{Therefore the original path is no longer the shortest path.}$

Show Worked Solution

a. $\text{Minimum travel time}\ = 15+20+10+5+8=58\ \text{minutes}$

$\text{Path: }A → B → C → D → E → F$

b. $\text{New roads}\ A → G\ \text{and }G → D $

c. $\text{Using the new roads }A → G\ \text{and }G → D:$

$\text{Minimum travel time}\ =8+22+5+8=43\ \text{minutes.}$

$\text{Therefore the original path is no longer the shortest path.}$

Measurement, STD1 M4 2025 HSC 2 MC

Mario drives from his home to his friend’s house. He watches a movie at his friend’s house and then drives home.

Which distance−time graph best represents Mario’s complete journey?

Show Answers Only

$B$

Show Worked Solution

$\text{Mario’s trip starts at home (zero) and ends at home (zero).}$

$\Rightarrow B$

Trigonometry, EXT1 T3 2025 HSC 12e

Express $\sqrt{3} \, \sin x-\cos x$ in the form $2\, \sin (x-\alpha)$, where $0<\alpha<\dfrac{\pi}{2}$. (1 mark)

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Hence, or otherwise, solve $\sqrt{3}\, \sin x=\cos x+1$, where $0 \leq x \leq 2 \pi$. (2 marks)

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i. $\text{See Worked Solutions}$

ii. $x=\dfrac{\pi}{3}, \pi$

Show Worked Solution

i. $2 \sin (x-\alpha)=\sqrt{3} \sin x-\cos x$

$2 \sin x \, \cos \alpha-2 \cos x \, \sin \alpha=\sqrt{3} \sin x-\cos x$

$\text{Equating coefficients:}$

$2 \cos \alpha=\sqrt{3}, \ \ 2 \sin \alpha=1$

$\text{Since} \ \ \cos \alpha>0 \ \ \text{and} \ \ \sin \alpha>0$

$\Rightarrow \alpha \ \text{is in 1st quadrant.}$

$\tan \alpha=\dfrac{1}{\sqrt{3}} \ \Rightarrow \ \alpha=\dfrac{\pi}{6}$

$\therefore \sqrt{3}\, \sin x-\cos x=2\, \sin \left(x-\dfrac{\pi}{6}\right)$

ii.	$\sqrt{3} \sin x$	$=\cos x+1$
	$\sqrt{3} \sin x-\cos x$	$=1$
	$2 \sin \left(x-\dfrac{\pi}{6}\right)$	$=1$
	$\sin \left(x-\dfrac{\pi}{6}\right)$	$=\dfrac{1}{2}$
	$x-\dfrac{\pi}{6}$	$=\dfrac{\pi}{6}, \dfrac{5 \pi}{6}$
	$x$	$=\dfrac{\pi}{3}, \pi \quad(0 \leqslant x \leqslant 2 \pi)$

\(\dfrac{mv_B^2}{r}\)	\(=T+mg\)
\(v_B^2\)	\(=rg \ \ (T=0)\)
\(v_B\)	\(=\sqrt{r g}\)

\(\text {Total} \ ME\)	\(=\dfrac{1}{2} m v_B^2+mg(2r)\)
	\(=\dfrac{1}{2} m \times r g+2 mrg\)
	\(=\dfrac{5}{2} m r g\)

\(\dfrac{5}{2} mrg\)	\(=\dfrac{1}{2} m v_A^2+m r g\)
\(\dfrac{1}{2} mv_A^2\)	\(=\dfrac{3}{2} m r g\)
\(v_A^2\)	\(=3 rg\)
\(v_A\)	\(=\sqrt{3rg}=\sqrt{3} \times v_B\)

\(\dfrac{mv_B^2}{r}\)	\(=T+mg\)
\(v_B^2\)	\(=rg \ \ (T=0)\)
\(v_B\)	\(=\sqrt{r g}\)

\(\text {Total} \ ME\)	\(=\dfrac{1}{2} m v_B^2+mg(2r)\)
	\(=\dfrac{1}{2} m \times r g+2 mrg\)
	\(=\dfrac{5}{2} m r g\)

\(x\)	\(=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2a}\)
	\(=\dfrac{-2 \pm \sqrt{2^2-4 \times 3 \times-4}}{2 \times 3}\)
	\(=\dfrac{-2 \pm \sqrt{52}}{6}\)
	\(=\dfrac{-1 \pm \sqrt{13}}{3}\)

a.	\(y\)	\(=x^2-4 x+6\)
		\(=x^2-4 x+4+2\)
		\(=(x-2)^2+2\)

\(-2\)	\(=-\dfrac{3}{5} \times 7+\dfrac{11}{5}\)
\(-2\)	\(=-\dfrac{21}{5}+\dfrac{11}{5}\)
\(-2\)	\(=-2 \ \text{(correct)}\)

\(y-1\)	\(=-\dfrac{3}{5}(x-2)\)
\(y\)	\(=-\dfrac{3}{5} x+\dfrac{11}{5}\)