Band 3

Proof, EXT2 P1 2025 HSC 5 MC

Consider the statement:

If $x^2-2 x \geq 0$, then $x \leq 0$.

Which of the following is the contrapositive of the statement?

If $x>0$, then $x^2-2 x<0$.
If $x \leq 0$, then $x^2-2 x \geq 0$.
If $x^2-2 x<0$, then $x<0$.
If $x^2-2 x \leq 0$, then $x>0$.

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$A$

Show Worked Solution

$\text{Statement: If}\ \ p\ \Rightarrow \ q$

$\text{Contrapositive statement: If}\ \ \neg\ q\ \Rightarrow \ \neg\ p$

$\therefore\ \text{If}\ \ x>0\ \ \Rightarrow\ \ x^2-2 x<0$.

$\Rightarrow A$

BIOLOGY, M5 2025 HSC 29

The Varroa mite is an external parasite of European honey bees and considered to be the most serious pest of honey bees worldwide.

Why is the Varroa mite infection considered to be an infectious disease. (2 mark2)

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In June 2022, the Varroa mite was detected for the first time in Australia at the Port of Newcastle. It then spread to surrounding areas.
Explain TWO procedures that could have been employed to prevent the spread of the Varroa mite in honey bees. (4 marks)

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a. Varroa mite infection classification

Varroa mite infection is classified as an infectious disease because the mite acts as a pathogen.
The mite transmits from one honey bee host to another through direct contact.
This enables the infection to spread between individual bees and across different hive populations.

b. Procedure 1: Quarantine

Quarantine works by isolating infected hives at the Port of Newcastle to prevent mite movement.
This stops infected bees from contacting healthy colonies in surrounding areas.
Movement restrictions on beekeeping equipment reduce the risk of accidentally transporting mites to new locations

Procedure 2: Surveillance and Destruction

Regular inspection of hives allows early detection of Varroa mite presence before widespread establishment.
Destruction of heavily infested hives eliminates the source of infection and prevents further transmission.
This creates a containment zone that limits geographic spread of the parasite.

Show Worked Solution

a. Varroa mite infection classification

Varroa mite infection is classified as an infectious disease because the mite acts as a pathogen.
The mite transmits from one honey bee host to another through direct contact.
This enables the infection to spread between individual bees and across different hive populations.

b. Procedure 1: Quarantine

Quarantine works by isolating infected hives at the Port of Newcastle to prevent mite movement.
This stops infected bees from contacting healthy colonies in surrounding areas.
Movement restrictions on beekeeping equipment reduce the risk of accidentally transporting mites to new locations

Procedure 2: Surveillance and Destruction

Regular inspection of hives allows early detection of Varroa mite presence before widespread establishment.
Destruction of heavily infested hives eliminates the source of infection and prevents further transmission.
This creates a containment zone that limits geographic spread of the parasite.

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)

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

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

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

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

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

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

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

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

BIOLOGY, M5 2025 HSC 22a

Outline a process used by fungi for reproduction. (2 marks)

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Spore production (asexual reproduction):

Fungi produce spores within specialised structures called sporangia.
Spores are released into the environment and dispersed by wind or water.
When conditions are favourable, spores germinate to form new fungal hyphae.
This produces genetically identical offspring.

Alternative acceptable response – Budding:

A small outgrowth (bud) forms on the parent fungal cell.
The bud grows and receives a copy of the nucleus.
The bud detaches to become an independent organism.
This is common in yeasts.

Show Worked Solution

Spore production (asexual reproduction):

Fungi produce spores within specialised structures called sporangia.
Spores are released into the environment and dispersed by wind or water.
When conditions are favourable, spores germinate to form new fungal hyphae.
This produces genetically identical offspring.

Alternative acceptable response – Budding:

A small outgrowth (bud) forms on the parent fungal cell.
The bud grows and receives a copy of the nucleus.
The bud detaches to become an independent organism.
This is common in yeasts.

BIOLOGY, M7 2025 HSC 22b

Outline an adaptation in a pathogen that facilitates transmission between hosts. (2 marks)

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Answers could include ONE of the following:

Influenza virus:

Influenza produces viral particles that are expelled in respiratory droplets when infected individuals cough or sneeze.
These droplets can be inhaled by nearby individuals, facilitating airborne transmission.

Plasmodium (malaria parasite):

Plasmodium has a complex life cycle adapted to mosquito vectors.
The parasite develops specific stages (sporozoites) in mosquito salivary glands.
This enables injection into new human hosts during blood feeding.

Bacterial spores:

Some bacteria form resistant endospores that survive harsh environmental conditions.
Spores can persist on surfaces or in soil for extended periods.
This increases opportunity for transmission to new hosts.

Show Worked Solution

Answers could include ONE of the following:

Influenza virus

Influenza produces viral particles that are expelled in respiratory droplets when infected individuals cough or sneeze.
These droplets can be inhaled by nearby individuals, facilitating airborne transmission.

Plasmodium (malaria parasite):

Plasmodium has a complex life cycle adapted to mosquito vectors.
The parasite develops specific stages (sporozoites) in mosquito salivary glands.
This enables injection into new human hosts during blood feeding.

Bacterial spores:

Some bacteria form resistant endospores that survive harsh environmental conditions.
Spores can persist on surfaces or in soil for extended periods.
This increases opportunity for transmission to new hosts.

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)

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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)

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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)

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

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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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?

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$B$

Show Worked Solution

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

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

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

$\Rightarrow B$

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

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

$z^2=3-4 i$

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

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

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

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

$\text{By inspection:}$

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

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

$\Rightarrow C$

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, M6 2025 HSC 2 MC

What type of mutagen is UV light?

Biochemical
Biological
Chemical
Electromagnetic

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$D$

Show Worked Solution

D is correct: UV light is electromagnetic radiation that causes DNA mutations.

Other Options:

A is incorrect: Biochemical mutagens are biological molecules like enzymes or toxins.
B is incorrect: Biological mutagens include viruses or transposons that alter DNA.
C is incorrect: Chemical mutagens are substances like alkylating agents or base analogs.

BIOLOGY, M5 2025 HSC 1 MC

Amoeba reproduce by the process shown.

Which of the following is a characteristic of the daughter cells?

Genetically different to parent amoeba
Genetically identical to parent amoeba
More organelles due to the cell division
Fewer chromosomes than the parent amoeba

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$B$

Show Worked Solution

B is correct: Asexual reproduction produces genetically identical offspring.

Other Options:

A is incorrect: Sexual reproduction produces genetic variation, not asexual.
C is incorrect: Cell division distributes existing organelles between daughter cells.
D is incorrect: Mitosis maintains the same chromosome number in offspring.

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

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

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

$\Rightarrow B$

Statistics, STD1 S3 2025 HSC 15

A researcher is using the statistical investigation process to investigate a possible relationship between average number of minutes per day a person spends watching television, and the average number of minutes per day the person spends exercising.

State the statistical question being posed. (1 mark)
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Participants were asked to record the number of minutes they spent watching television each day and the number of minutes they spent exercising each day. The averages for each participant were recorded and graphed, and a line of best fit was included.

From the graph, identify the dependent variable. (1 mark)

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Describe the bivariate dataset in terms of its form and direction. (2 marks)

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The points $(0, 70)$ and $(60, 10)$ lie on the line of best fit. By first plotting these points on the graph, find the gradient and the $y$-intercept of the line of best fit. (3 marks)
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Explain why it is NOT appropriate to extrapolate the line of best fit to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day. (1 mark)
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a. $\text{How does the average daily time spent watching television}$

$\text{relate to the average daily time spent exercising?}$

b. $\text{Dependent variable: Average minutes per day exercising, or }y.$

c. $\text{Form: Linear}$

$\text{Direction: Negative}$

d. $\text{Gradient}=-1$

e. $\text{The extrapolation of the graph past 70 minutes produces}$

$\text{negative average minutes per day exercising (impossible).}$

Show Worked Solution

a. $\text{How does the average daily time spent watching television}$

$\text{relate to the average daily time spent exercising?}$

b. $\text{Dependent variable:}$

$\text{Average minutes per day exercising, or }y.$

c. $\text{Form: Linear}$

$\text{Direction: Negative}$

$y-\text{intercept = 70}$

$\text{Gradient}=\dfrac{\text{rise}}{\text{run}}=\dfrac{-60}{60}=-1$

e. $\text{The extrapolation of the graph past 70 minutes produces}$

$\text{negative average minutes per day exercising (impossible).}$

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)

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

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

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

Probability, STD1 S2 2025 HSC 7 MC

A biased die is made from this net.

The die is rolled once.

What is the probability of rolling a 2?

$\dfrac{1}{6}$
$\dfrac{1}{4}$
$\dfrac{1}{3}$
$\dfrac{1}{2}$

Show Answers Only

$C$

Show Worked Solution

$P(2)=\dfrac{2}{6}=\dfrac{1}{3}$

$\Rightarrow C$

Measurement, STD1 M1 2025 HSC 4 MC

What is $6\ 280\ 000$ in standard form?

$628\times 10^4$
$62.8\times 10^5$
$6.28\times 10^6$
$0.628\times 10^7$

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$C$

Show Worked Solution

$6\ 280\ 000=6.28\times 10^6$

$\Rightarrow C$

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

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

$\Rightarrow B$

Statistics, STD1 S1 2025 HSC 1 MC

Which of the following could be classified as discrete data?

Colour of a car
Time taken to swim 200 m
Temperature of an ice block
Number of children in a class

Show Answers Only

$D$

Show Worked Solution

$\text{Discrete data refers to individual and countable items that can be listed.}$

$\Rightarrow D$

Measurement, STD2 M1 2025 HSC 26

A toy has a curved surface on the top which has been shaded as shown. The toy has a uniform cross-section and a rectangular base.

Use two applications of the trapezoidal rule to find an approximate area of the cross-section of the toy. (2 marks)

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The total surface area of the plastic toy is 1300 cm².
What is the approximate area of the curved surface? (2 marks)

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The measurements shown on the diagram are given to the nearest millimetre.
What is the percentage error of the measurement of 10.2 cm? Give your answer correct to 3 significant figures. (2 marks)

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a. $34.68 \ \text{cm}^2$

b. $582.64 \ \text{cm}^2$

c. $0.490 \%$

Show Worked Solution

a. $\begin{array}{|c|c|c|c|}
\hline\rule{0pt}{2.5ex} \quad x \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 5.1 \quad & \quad 10.2 \quad\\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 6 & 3.8 & 0 \\
\hline
\end{array}$

$A$	$\approx \dfrac{h}{2}\left(y_0+2y_1+y_2\right)$
	$\approx \dfrac{5.1}{2}\left(6+2 \times 3.8+0\right)$
	$\approx 34.68 \ \text{cm}^2$

b. $\text{Toy has 5 sides.}$

$\text{Area of base}=10.2 \times 40=408 \ \text{cm}^2$

$\text{Area of rectangle}=6.0 \times 40=240 \ \text{cm}^2$

$\text{Approximated areas}=2 \times 34.68=69.36 \ \text{cm}^2$

$\therefore \ \text{Area of curved surface}$	$=1300-(408+240+69.36)$
	$=582.64 \ \text{cm}^2$

c. $\text{Convert cm to mm:}$

$10.2 \ \text{cm}\times 10=102 \ \text{mm}$

$\text{Absolute error}=\dfrac{1}{2} \times \text{precision}=0.5 \ \text{mm}$

$\% \ \text{error}=\dfrac{0.5}{102} \times 100=0.490 \%$

Combinatorics, EXT1 A1 2025 HSC 13e

The Pascal's triangle relation can be expressed as
1. $\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}.$ (Do NOT prove this.)
Show that $\displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}$. (1 mark)

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Hence, or otherwise, prove that
$\displaystyle\binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}$. (2 marks)
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i. $\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)$

$\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}$

$\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}$

$\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}$

$\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}$

ii. $\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}$

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

$\operatorname{LHS}$	$=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots$
	$\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]$
	$=1-1+\displaystyle \binom{2051}{2001}$
	$=\displaystyle \binom{2051}{2001}$

Show Worked Solution

i. $\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)$

$\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}$

$\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}$

$\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}$

$\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}$

ii. $\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}$

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

$\operatorname{LHS}$	$=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots$
	$\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]$
	$=1-1+\displaystyle \binom{2051}{2001}$
	$=\displaystyle \binom{2051}{2001}$

Functions, EXT1 F2 2025 HSC 11f

The roots of $2 x^3+6 x^2+x-1=0$ are $\alpha, \beta$ and $\gamma$.

What is the value of $\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}$ ? (2 marks)

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$\dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=-6$

Show Worked Solution

$2 x^3+6 x^2+x-1=0$

$\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{\alpha+\beta+\gamma}{\alpha \beta \gamma}$

$\alpha+\beta+\gamma=-\dfrac{b}{a}=-3$

$\alpha \beta \gamma=-\dfrac{d}{a}=\dfrac{1}{2}$

$\therefore \dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{-3}{\frac{1}{2}}=-6$

Vectors, EXT1 V1 2025 HSC 11e

For what value of $m$ is the vector $\displaystyle \binom{1}{m}$ parallel to the vector $\displaystyle \binom{2}{6}$? (1 mark)

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$m=3$

Show Worked Solution

$\text{If vectors are parallel:}$

$\displaystyle \binom{2}{6}=k\binom{1}{m} \ \Rightarrow \ k=2$

$2m$	$=6$
$m$	$=3$

Trigonometry, EXT1 T2 2025 HSC 11b

Solve $\sin 2 \theta-\sin \theta=0$ for $0 \leq \theta \leq \pi$. (3 marks)

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$\theta=0, \dfrac{\pi}{3}\ \text{or}\ \pi$

Show Worked Solution

$\sin\,2\theta-\sin\,\theta$	$=0$
$2\,\sin\,\theta\,\cos\,\theta-\sin\,\theta$	$=0$
$\sin\,\theta(2\,\cos\,\theta-1)$	$=0$

$\sin\,\theta=0\ \ \Rightarrow\ \ \theta=0, \pi$

$2\,\cos\,\theta-1=0\ \ \Rightarrow\ \ \cos\,\theta=\dfrac{1}{2}\ \ \Rightarrow\ \ \theta=\dfrac{\pi}{3}$

$\therefore \theta=0, \dfrac{\pi}{3}\ \text{or}\ \pi.$

Functions, EXT1 F1 2025 HSC 11a

Find the inverse function, $f^{-1}(x)$, of the function $f(x)=1-\dfrac{1}{x-2}$. (2 marks)

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

Show Worked Solution

$y=1-\dfrac{1}{x-2}$

$\text{Inverse: swap}\ x ↔ y$

$x$	$=1-\dfrac{1}{y-2}$
$\dfrac{1}{y-2}$	$=1-x$
$y-2$	$=\dfrac{1}{1-x}$
$y$	$=2+\dfrac{1}{1-x}$

Vectors, EXT1 V1 2025 HSC 2 MC

The projection of $\underset{\sim}{u}$ onto $\underset{\sim}{v}$ is given by $\left(\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{v}|^2}\right) \underset{\sim}{v}$.

What is the projection of $\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}$ onto $\underset{\sim}{v}=2 \underset{\sim}{i}-3 \underset{\sim}{j}$ ?

$-\dfrac{4}{5}(\underset{\sim}{i}+2 \underset{\sim}{j})$
$-\dfrac{4}{13}(2 \underset{\sim}{i}-3 \underset{\sim}{j})$
$-\dfrac{4}{\sqrt{5}}(\underset{\sim}{i}+2 \underset{\sim}{j})$
$-\dfrac{4}{\sqrt{13}}(2 \underset{\sim}{i}-3 \underset{\sim}{j})$

Show Answers Only

$B$

Show Worked Solution

$\underset{\sim}{u}=\displaystyle\binom{1}{2},|\underset{\sim}{u}|=\sqrt{1^2+2^2}=\sqrt{5}$

$\underset{\sim}{v}=\displaystyle \binom{2}{-3},|\underset{\sim}{v}|=\sqrt{2^2+(-3)^2}=\sqrt{13}$

$\operatorname{proj}_{\underset{\sim}{v}}{\underset{\sim}{u}}$	$=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{\|\underset{\sim}{v}\|^2} \times \underset{\sim}{v}$
	$=\dfrac{2-6}{13}(\underset{\sim}{2i}-3\underset{\sim}{j})$
	$=-\dfrac{4}{13}(\underset{\sim}{2i}-3\underset{\sim}{j})$

$\Rightarrow B$

Functions, EXT1 F1 2025 HSC 1 MC

What is the solution to $\abs{2 x+3}<5$ ?

$-4<x<1$
$x<-4$ or $x>1$
$-1<x<4$
$x<-1$ or $x>4$

Show Answers Only

$A$

Show Worked Solution

$\abs{2 x+3}<5$

$-5 < 2x+3 <5$

$-8< 2x < 2$

$-4<x<1$

$\Rightarrow A$

Measurement, STD2 M7 2025 HSC 21

A house has a reverse-cycle air conditioner which uses 2.5 kW of power for cooling and 3.2 kW of power for heating. The cost of electricity is 29 cents per kWh .

Find the cost, in dollars and cents, of cooling the house for 6 hours. (1 mark)

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The cost of operating the air conditioner to heat the house during winter last year was $640. There are 92 days in winter.
Find the number of hours, to 1 decimal place, that the air conditioner was used on average per day. (2 marks)

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a. $\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35$

b. $7.5\ \text{hours}$

Show Worked Solution

a. $\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35$

b. $\text{Let \(h$ = hours used per day}\)

$\text{Cost per day}\ = h \times 3.2 \times 0.29$

$\text{Cost (92 days )}\ = 92 \times h \times 3.2 \times 0.29$

$\text{Find \(h$ when cost = \$640:}\)

$640$	$=92 \times h \times 3.2 \times 0.29$
$h$	$=\dfrac{640}{92 \times 3.2 \times 0.29}=7.5\ \text{hours (1 d.p.)}$

Networks, STD2 N3 2025 HSC 19

The activities and corresponding durations in days for a project are shown in the network diagram.

Complete the table showing the immediate prerequisites for each activity. Indicate with an $\text{X}$ any activities without any immediate prerequisites. (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

Find the critical path for this project AND state the minimum duration for the project. (2 marks)

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The duration of activity $ A $ is increased by 2. Does this affect the critical path for the project? Give a reason for your answer. (1 mark)

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\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \text{X} \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & C,D \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & E \\
\hline
\end{array}

b. $\text{Critical Path:}\ BDEFH$

$\text{Minimum Duration}\ =4+5+5+7+5=26\ \text{days}$

c. $\text{If duration of activity \(A$ is increased by 2:}\)

$\text{The critical path remains unchanged (EST of activity \(E$ remains = 9)}\)

Show Worked Solution

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \text{X} \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & C,D \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & E \\
\hline
\end{array}

$\text{Critical Path:}\ BDEFH$

$\text{Minimum Duration}\ =4+5+5+7+5=26\ \text{days}$

c. $\text{If duration of activity \(A$ is increased by 2:}\)

$\text{The critical path remains unchanged (EST of activity \(E$ remains = 9)}\)

Measurement, STD2 M7 2025 HSC 16

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{Savings}\ = $160$

Show Worked Solution

$\text{Cost of 4 litre cans:}$

$ C_1=\dfrac{80}{4} \times 90=$1800 $

$\text{Cost of 10 litre cans:}$

$ C_2=\dfrac{80}{10} \times 205=$1640$

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

Measurement, STD2 M1 2025 HSC 5 MC

Which of the following is arranged from largest to smallest?

$ 6.2 \times 10^{-3}, 4.5 \times 10^{-4}, 3.2 \times 10^{-1} $
$ 4.5 \times 10^{-4}, 6.2 \times 10^{-3}, 3.2 \times 10^{-1} $
$ 3.2 \times 10^{-1}, 4.5 \times 10^{-4}, 6.2 \times 10^{-3} $
$ 3.2 \times 10^{-1}, 6.2 \times 10^{-3}, 4.5 \times 10^{-4}$

Show Answers Only

$D$

Show Worked Solution

$\text{Largest to smallest:}$

$ 3.2 \times 10^{-1}, 6.2 \times 10^{-3}, 4.5 \times 10^{-4}$

$\Rightarrow D$

Financial Maths, STD2 F1 2025 HSC 4 MC

Frankie takes four weeks of annual leave. His weekly pay is $350 and his annual leave loading is $ 17 \dfrac{1}{2} $% of four weeks pay.

What is Frankie's total pay for this period of annual leave?

$245.00
$411.25
$1461.25
$1645.00

Show Answers Only

$D$

Show Worked Solution

$\text{4 weeks pay}\ = 4 \times 350 = $1400$

$\text{Annual leave loading}\ =17.5\% \times 1400 = $245$

$\text{Total pay}\ = 1400+245=$1645$

$\Rightarrow D$

Algebra, STD2 A4 2025 HSC 2 MC

Which graph could represent $y=4^x$ ?

Show Answers Only

$A$

Show Worked Solution

$\Rightarrow A$

Functions, 2ADV F2 2025 HSC 2 MC

Which graph could represent $y=4^x$ ?

Show Answers Only

$A$

Show Worked Solution

$\Rightarrow A$

Networks, STD2 N2 2025 HSC 3 MC

The network shows the distances, in kilometres, along a series of roads that connect towns.

What is the value of the largest weighted edge included in the minimum spanning tree for this network?

Show Answers Only

$C$

Show Worked Solution

$\text{Minimum spanning tree:}$

$\text{Using Kruskal’s algorithm:}$

$\text{Edge 1 = 4, Edge 2/3 = 4, Edge 4 = 6, Edge 5 = 9}$

$\therefore\ \text{The largest weighted edge in the MST = 9.}$

$\Rightarrow C$

Calculus, 2ADV C4 2024 HSC 5 MC v1

What is $ {\displaystyle \int x(3 x^2+1)^4 d x} $ ?

$ \dfrac{1}{30}(3x^2+1)^5+C $
$ \dfrac{1}{5}(3x^2+1)^5+C $
$ \dfrac{5}{6}(3x^2+1)^5+C $
$ \dfrac{6}{5}(3x^2+1)^5+C $

Show Answers Only

$ A $

Show Worked Solution

\[ \int x(3x^2+1)^4 dx\]	$=\dfrac{1}{5} \cdot \dfrac{1}{6x}x(3x^2+1)^5+C$
	$=\dfrac{1}{30}(3x^2+1)^5+C$

$ \Rightarrow A $

NOTE: Integrating by the reverse chain rule only works if some form of the derivative is already present outside of the brackets.

Calculus, 2ADV C1 EO-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = t^3-4t^2 +5t + 6` when `t >= 0`

Calculate the velocity when `t = 3`. (1 mark)

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When is the particle stationary? (2 marks)

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Show Answers Only

`8\ text(ms)^(−1)`
`1, 5/3\ text(s)`

Show Worked Solution

i. `x =t^3 -4t^2 +5t + 6`

`v = (dx)/(dt) = 3t^2 – 8t +5`

`text(When)\ t = 3,`

`v`	`= 3 xx 3^2 -8 · 3 +5`
	`= 8\ text(ms)^(−1)`

ii. `text(Particle is stationary when)\ \ v = 0`

`3t^2 -8t +5`	`= 0`
`3t^2-3t-5t+5`	`= 0`
`3t(t-1)-5(t-1)`	`= 0`
`(t-1)(3t-5)`	`= 0`
`t`	`= 1, 5/3\ text(s)`

Trigonometry, 2ADV T1 2025 HSC 29

The point $T$ is the peak of a mountain and the point $O$ is directly below the mountain's peak. The point $Y$ is due east of $O$ and the angle of elevation of $T$ from $Y$ is 60°. The point $F$ is 4 km south-west of $Y$. The points $O, Y$ and $F$ are on level ground. The angle of elevation of $T$ from $F$ is 45°.

Let the height of the mountain be $h$.
Show that $O Y=\dfrac{h}{\sqrt{3}}$. (1 mark)

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Hence, or otherwise, find the value of $h$, correct to 2 decimal places. (3 marks)

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Find the bearing of point $O$ from point $F$, correct to the nearest degree. (3 marks)

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a. $\text{See Worked Solutions}$

b. $h=3.03 \ \text{km}$

c. $\text {Bearing of \(O$ from $F$}=021^{\circ} \)

Show Worked Solution

a. $\text{In}\ \triangle TOY:$

$\tan 60^{\circ}$	$=\dfrac{h}{OY}$
$OY$	$=\dfrac{h}{\tan 60^{\circ}}$	$=\dfrac{h}{\sqrt{3}}$

b. $\text{In}\ \triangle TOF:$

$\tan 45^{\circ}=\dfrac{h}{OF} \ \Rightarrow \ OF=h$

$\text{Since \(Y$ is due east of $O$ and $F$ is south-west of $Y$:}\)

$\angle OYF =45^{\circ}$

$\text{Using cosine rule in} \ \triangle OYF:$

$OY^2+YF^2-2 \times OY \times YF\, \cos 45^{\circ}$	$=OF^2$
$\left(\dfrac{h}{\sqrt{3}}\right)^2+4^2-2 \times \dfrac{h}{\sqrt{3}} \times 4 \times \dfrac{1}{\sqrt{2}}$	$=h^2$
$\dfrac{h^2}{3}+16-\dfrac{8}{\sqrt{6}} h$	$=h^2$
$\dfrac{2}{3} h^2+\dfrac{8}{\sqrt{6}} h-16$	$=0$

$h$

$=\dfrac{\dfrac{-8}{\sqrt{6}}+\sqrt{\frac{64}{6}+4 \times \frac{2}{3} \times 16}}{2 \times \frac{2}{3}}\ \ \ (h>0)$

$=3.0277 \ldots$

$=3.03 \ \text{km (to 2 d.p.)}$

c. $\text {Using sine rule in} \ \ \triangle OYF:$

$\dfrac{\sin\angle FOY}{4}$	$=\dfrac{\sin 45^{\circ}}{3.03}$
$\sin \angle F O Y$	$=\dfrac{4 \times \sin 45^{\circ}}{3.03}=0.93347$
$\angle FOY$	$=180-\sin^{-1}(0.93347)=180-68.98 \ldots = 111^{\circ} \ \text{(angle is obtuse)}$

$\angle FOS^{\prime}=111-90=21^{\circ}$

$\angle \text{N}^{\prime}FO=21^{\circ}(\text {alternate})$

$\therefore \ \text {Bearing of \(O$ from $F$}=021^{\circ} \)

Calculus, 2ADV C4 2025 HSC 27

The shaded region is bounded by the graph $y=\left(\dfrac{1}{2}\right)^x$, the coordinate axes and $x=2$.

Use two applications of the trapezoidal rule to estimate the area of the shaded region. (2 marks)

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Show that the exact area of the shaded region is $\dfrac{3}{4 \ln 2}$. (2 marks)

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Using your answers from part (a) and part (b), deduce $e<2 \sqrt{2}$. (2 marks)

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a. $A\approx \dfrac{9}{8}\ \text{units}^2$

b.	$\text{Area}$	$=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x$
		$=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2$
		$=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}$
		$=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}$
		$=\dfrac{3}{4 \ln 2}$

c. $\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}$

$\text{between (0,1) and \((2,\dfrac{1}{4})$}\)

$\Rightarrow \ \text{Area using trap rule > Actual area}$

$\dfrac{9}{8}$	$>\dfrac{3}{4 \ln 2}$
$36\, \ln 2$	$>24$
$\ln 2$	$>\dfrac{2}{3}$
$e^{\small \dfrac{2}{3}}$	$>2$
$e$	$>2^{\small \dfrac{3}{2}}$
$e$	$>2 \sqrt{2}$

Show Worked Solution

\begin{array}{|c|c|c|c|}
\hline \ \ x \ \ & \ \ 0 \ \ & \ \ 1 \ \ & \ \ 2 \ \ \\
\hline y & 1 & \dfrac{1}{2} & \dfrac{1}{4} \\
\hline
\end{array}

$A$	$\approx \dfrac{h}{2}\left[1 \times 1+2 \times \dfrac{1}{2}+1 \times \dfrac{1}{4}\right]$
	$\approx \dfrac{1}{2}\left(\dfrac{9}{4}\right)$
	$\approx \dfrac{9}{8}\ \text{units}^2$

b.	$\text{Area}$	$=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x$
		$=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2$
		$=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}$
		$=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}$
		$=\dfrac{3}{4 \ln 2}$

c. $\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}$

$\text{between (0,1) and \((2,\dfrac{1}{4})$}\)

$\Rightarrow \ \text{Area using trap rule > Actual area}$

$\dfrac{9}{8}$	$>\dfrac{3}{4 \ln 2}$
$36\, \ln 2$	$>24$
$\ln 2$	$>\dfrac{2}{3}$
$e^{\small \dfrac{2}{3}}$	$>2$
$e$	$>2^{\small \dfrac{3}{2}}$
$e$	$>2 \sqrt{2}$

Statistics, 2ADV S2 2025 HSC 14

In a research study, participants were asked to record the number of minutes they spent watching television and the number of minutes they spent exercising each day over a period of 3 months. The averages for each participant were recorded and graphed.

Describe the bivariate dataset in terms of its form and direction. (2 marks)
Form: ..................................................................
Direction: ............................................................
--- 0 WORK AREA LINES (style=lined) ---

The equation of the least-squares regression line for this dataset is

$y=64.3-0.7 x$

Interpret the values of the slope and $y$-intercept of the regression line in the context of this dataset. (2 marks)

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Jo spends an average of 42 minutes per day watching television.
Use the equation of the regression line to determine how many minutes on average Jo is expected to exercise each day. (1 mark)

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Explain why it is NOT appropriate to extrapolate the regression line to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day. (1 mark)

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

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a. $\text{Form: Linear}$

$\text{Direction: Negative}$

b. $\text{Slope}=-0.7$

$\text{This means that for each added minute of watching television per day, a participant, on average,}$

$\text{will exercise for 0.7 minutes less.}$

$y \text{-intercept}=64.3$

$\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}$

c. $\text{34.9 minutes}$

d. $\text{At} \ \ x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7$

$\text{The model predicts a negative value for time spent exercising, which is not possible.}$

Show Worked Solution

a. $\text{Form: Linear}$

$\text{Direction: Negative}$

b. $\text{Slope}=-0.7$

$\text{This means that for each added minute of watching television per day, a participant, on average,}$

$\text{will exercise for 0.7 minutes less.}$

$y \text{-intercept}=64.3$

$\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}$

c. $\text{At} \ \ x=42:$

$y=64.3-0.7 \times 42=34.9$

$\therefore \ \text{Jo is expected to exercise for 34.9 minutes}$

d. $\text{At} \ \ x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7$

$\text{The model predicts a negative value for time spent exercising, which is not possible.}$

Financial Maths, 2ADV M1 2025 HSC 13

The numbers, 75, $p$, $q$, 2025, form a geometric sequence.

Find the values of $p$ and $q$. (2 marks)

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$p=225, \ q=675$

Show Worked Solution

$a=75, \ 75r=p, \ 75r^2=q, \ 75r^3=2025$

$\text{Using}\ \ 75r^3=2025:$

$r=\sqrt[3]{\dfrac{2025}{75}}=3$

$p=75 \times 3 = 225$

$q=75 \times 3^{2}=675$

Calculus, 2ADV C3 2025 HSC 12

Find the equation of the tangent to $y=5 x^3-\dfrac{2}{x^2}-9$ at the point $(1,-6)$. (3 marks)

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$y=19 x-25$

Show Worked Solution

$y=5 x^3-2 x^{-2}-9$

$y^{\prime}=15 x^2+4 x^{-3} $

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

$y^{\prime}=15+4=19$

$\text{Equation of line} \ \ m=19 \ \ \text {through}\ \ (1,-6): $

$y+6$	$=19(x-1)$
$y+6$	$=19 x-19$
$y$	$=19 x-25$

Algebra, STD2 A4 2025 HSC 20

The graph of a quadratic function represented by the equation $h=t^2-8 t+12$ is shown.

Find the values of $t$ and $h$ at the turning point of the graph. (2 marks)

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The graph shows $h=12$ when $t=0$.
What is the other value of $t$ for which $h=12$? (1 mark)

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Show Answers Only

a. $\text{Turning point at} \ \ (4,-4)$

b. $t=8$

Show Worked Solution

a. $\text{Axis of quadratic occurs when}\ \ t= \dfrac{2+6}{2} = 4$

$\text{At} \ \ t=4:$

$h=4^2-8 \times 4+12=-4$

$\therefore \ \text{Turning point at} \ \ (4,-4)$

b. $\text {When} \ \ h=12:$

$t^2-8 t+12$	$=12$
$t(t-8)$	$=0$

$\therefore \ \text{Other value:} \ \ t=8$

Functions, 2ADV F1 2025 HSC 11

The graph of a quadratic function represented by the equation $h=t^2-8 t+12$ is shown.

Find the values of $t$ and $h$ at the turning point of the graph. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
The graph shows $h=12$ when $t=0$.
What is the other value of $t$ for which $h=12$? (1 mark)

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

Show Answers Only

a. $\text{Turning point at} \ \ (4,-4)$

b. $t=8$

Show Worked Solution

a. $\text{Strategy 1 (no calculus)}$

$\text{Axis of quadratic occurs when}\ \ t= \dfrac{2+6}{2} = 4$

$\text{At} \ \ t=4:$

$h=4^2-8 \times 4+12=-4$

$\therefore \ \text{Turning point at} \ \ (4,-4)$

$\text{Strategy 2 (using calculus)}$

$h=t^2-8 t+12$

$h^{\prime}=2 t-8$

$\text{Find \(t$ when} \ \ h^{\prime}=0:\)

$2 t-8=0 \ \Rightarrow \ t=4$

b. $\text {When} \ \ h=12:$

$t^2-8 t+12$	$=12$
$t(t-8)$	$=0$

$\therefore \ \text{Other value:} \ \ t=8$

Calculus, 2ADV C4 2025 HSC 5 MC

What is $\displaystyle \int \frac{1}{\sqrt{x+5}} d x$ ?

$\dfrac{1}{2} \sqrt{x+5}+C$
$2 \sqrt{x+5}+C$
$-\dfrac{1}{2} \sqrt{x+5}+C$
$-2 \sqrt{x+5}+C$

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$B$

Show Worked Solution

$\displaystyle \int \frac{1}{\sqrt{x+5}}\ d x$	$=\displaystyle \int (x+5)^{-\frac{1}{2}}\ d x$
	$= 2 \sqrt{x+5}+C$

$\Rightarrow B$

Functions, 2ADV F1 2025 HSC 3 MC

What is the domain of the function $y=\sqrt{6-x^2}$ ?

$\left(0, \sqrt{6}\right)$
$\left[0, \sqrt{6}\right]$
$\left(-\sqrt{6}, \sqrt{6}\right)$
$\left[-\sqrt{6}, \sqrt{6}\right]$

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$D$

Show Worked Solution

$6-x^2 \geq 6\ \ \Rightarrow\ \ x^2 \leq 6$

$-\sqrt{6} \leq x \leq \sqrt{6} $

$\Rightarrow D$

Probability, 2ADV S1 2025 HSC 1 MC

The probability distribution table for a discrete random variable $X$ is shown.

\begin{array}{|c|c|}
\hline \quad \quad x \quad \quad & \quad P(X=x) \quad\\
\hline 1 & 0.4 \\
\hline 2 & 0.2 \\
\hline 3 & \\
\hline
\end{array}

What is the value of $P(X=3)$ ?

$0.2$
$0.4$
$1.2$
$2.0$

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$B$

Show Worked Solution

$\text{Probabilities sum to 1:}$

$0.4+0.2+P(X=3) = 1\ \ \Rightarrow\ \ P(X=3) = 0.4$

$\Rightarrow B$

Calculus, 2ADV C1 2018 HSC 12d v1

The displacement of a particle moving along the `x`-axis is given by

`x = 1/4t^4 -t^3 -1/2t^2 +3t,`

where `x` is the displacement from the origin in metres and `t` is the time in seconds, for `t >= 0`.

What is the initial velocity of the particle? (1 mark)

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At which times is the particle stationary? (2 marks)

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Find the position of the particle when the acceleration is `4/3`. (2 marks)

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`3\ text(ms)^(-1)`
`t = 1 or 3\ text(seconds)`
`-329/324\ text(m)`

Show Worked Solution

i. `x = 1/4t^4 -t^3 -1/2t^2 +3t`

`v = (dx)/(dt) = t^3-3t^2-t+3`

`text(Find)\ \ v\ \ text(when)\ \ t = 0:`

`v`	`= 0 -3(0) -0+ 3`
	`= 3\ text(ms)^(-1)`

ii. `text(Particle is stationary when)\ \ v = 0`

`t^3-3t^2-t+3`	`=0`
`t^2(t-3)-1(t-3)`	`=0`
`(t-3)(t^2-1)`	`=0`
`(t-3)(t-1)(t+1)`	`=0`

`t = 1 or 3\ text(seconds), t >= 0`

iii. `a = (dv)/(dt) = 3t^2-6t-1`

`text(Find)\ \ t\ \ text(when)\ \ a = 4/3`

`3t^2-6t-1`	`= 4/3`
`3t^2-6t-7/3`	`= 0`
`9t^2-18t-7`	`=0`
`(3t-7)(3t+1)`	`=0`
`t`	`=7/3`, `t >= 0`

`x(7/3)`	`= 1/4(7/3)^4 -(7/3)^3 -1/2(7/3)^2 +3(7/3)`
	`= -329/324\ text(m)`

Calculus, 2ADV C1 EO-Bank 2

Find the equations of the tangents to the curve `y = x^2-5x+6` at the points where the curve cuts the `x`-axis. (2 marks)

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Where do the tangents intersect? (2 marks)

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`y = −x+2`
`y = x-3`
`(5/2, −1/2)`

Show Worked Solution

a. `y`	`= x^2-5x+6`
	`= (x-2)(x-3)`

`text(Cuts)\ xtext(-axis at)\ \ x = 2\ \ text(or)\ \ x = 3`

`(dy)/(dx) = 2x-5`

`text(At)\ \ x = 2 \ => \ (dy)/(dx) = -1`

`T_1\ text(has)\ \ m = −1,\ text{through (2, 0)}`

`y -0`	`= -1(x-2)`
`y`	`= -x+2`

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 1`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y -0`	`= 1(x-3)`
`y`	`= x -3`

b. `text(Intersection occurs when:)`

`-x+2`	`= x-3`
`2x`	`= 5`
`x`	`= 5/2`

`y = 5/2 – 3 = −1/2`

`:.\ text(Intersection at)\ \ (5/2, −1/2)`

Calculus, 2ADV C1 2023 HSC 14 v1

Find the equation of the tangent to the curve `y=x(3x+2)^2` at the point `(1,25)`. (3 marks)

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`y=55x-30`

Show Worked Solution

`y`	`=x(3x+2)^2`
`dy/dx`	`=6x(3x+2) + (3x+2)^2`
	`=(3x+2)(9x+2)`

`text{At}\ x=1\ \ =>\ \ dy/dx=(3(1)+2)(9(1)+2)=55`

`text{Find equation of line}\ \ m=55,\ text{through}\ (1,25)`

`y-y_1`	`=m(x-x_1)`
`y-25`	`=55(x-1)`
`y`	`=55x-30`

Calculus, 2ADV C1 2009 HSC 1d v1

Find the gradient of the tangent to the curve `y = 2x^3 - 5x^2 + 4` at the point `(2, 0)`. (2 marks)

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`text(Gradient = 2.)`

Show Worked Solution

`y`	`= 2x^3-5x^2 + 4`
`dy/dx`	`= 6x^2 – 10x`

`text(At)\ x = 2`

`dy/dx`	`= 6(2)^2 – 10(2)`
	`=24-20`
	`=4`

`:.\ text(Gradient of tangent at)\ (2, 0) = 4.`

Calculus, 2ADV C1 2019 HSC 11c v1

Differentiate `(4x + 3)/(3x-4)`. (2 marks)

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`-25/(3x-4)^2`

Show Worked Solution

`text(Using quotient rule:)`

`u=4x+3,`	`v=3x-4`
`u^{′} = 4,`	`v^{′} = 3`

`y^{′}`	`= (u^{′} v-v^{′} u)/v^2`
	`= (4(3x-4)-3(4x+3))/(3x-4)^2`
	`= (12x-16-12x-9)/(3x-4)^2`
	`= -25/(3x-4)^2`

Calculus, 2ADV C1 2015 HSC 12c v1

Find `f^{′}(x)`, where `f(x) = (2x^2-3x)/(2-x).` (2 marks)

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`((x-3) (x + 1))/(x-1)^2`

Show Worked Solution

`f(x) = (2x^2-3x)/(2-x)`

`text(Using the quotient rule:)`

`u`	`= 2x^2-3x`	`\ \ \ \ \ \ v`	`= 2-x`
`u^{′}`	`= 4x-3`	`\ \ \ \ \ \ v^{′}`	`= -1`

`f^{′}(x)`	`= (u^{′} v-uv^{′})/v^2`
	`= ((4x-3)(2-x)-(2x^2-3x) xx -1)/(2-x)^2`
	`= (-2x^2 + 8x-6)/(x-2)^2`
	`= (-2(x^2-4x+3)/(x-2)^2`
	`= (-2(x-3) (x-1))/(x-2)^2`

Calculus, 2ADV C1 EQ-Bank 3

Use differentiation by first principles to find $y^{′}$, given $y = 2x^2 + 5x$. (2 marks)

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Find the equation of the tangent to the curve when $x = 1$. (1 mark)

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`y^{′} = 4x+5`
`y = 9x-2`

Show Worked Solution

i.	`f(x)`	`= 2x^2 + 5x`
	`f^{′}(x)`	`= lim_(h->0) (f(x + h)-f(x))/h`
		`= lim_(h->0) ((2(x + h)^2 +5(x + h))-(2x^2+5x))/h`
		`= lim_(h->0)(2x^2 + 4xh + 2h^2+5x+5h-2x^2-5x)/h`
		`= lim_(h->0)(4xh + 2h^2+5h)/h`
		`= lim_(h->0)(h(4x+5 +2h))/h`

`:.\ y^{′} = 4x+5`

ii. `text(When)\ \ x = 1, y = 7`

`y^{′} = 4+5 = 9`

`:. y-7`	`= 9(x-1)`
`y`	`= 9x-2`

CHEMISTRY, M1 EQ-Bank 18

A geochemist analyzes a rock sample and finds it contains 15 600 kg of ore. Chemical analysis shows that the ore contains 3.8% w/w of chromium metal.

Calculate the mass of chromium in the rock sample. Express your answer in scientific notation. (2 marks)

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$5.93 \times 10^2\ \text{kg}$

Show Worked Solution

Mass of chromium	$= 3.8\% \times 15\,600\ \text{kg}$
	$=0.038 \times 15\,600\ \text{kg}$
	$=5.93 \times 10^2\ \text{kg}$

CHEMISTRY, M1 EQ-Bank 11 MC

Which statement about the periodic table is correct?

Non-metals are found on the left side of the periodic table and are generally good conductors of electricity.
Metalloids have properties intermediate between metals and non-metals and are found along the staircase line.
Metals generally have high electronegativity and form negative ions.
Noble gases are highly reactive due to their incomplete valence shells.

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$B$

Show Worked Solution

Metalloids (such as silicon, germanium, and arsenic) are located along the staircase line separating metals and non-metals on the periodic table.
They exhibit properties of both metals and non-metals, such as being semiconductors of electricity.

$\Rightarrow B$

CHEMISTRY, M1 EQ-Bank 9 MC

The formula for the sulfate ion is $\ce{SO4^2-}$. What is the formula for aluminium sulfate?

$\ce{AlSO4}$
$\ce{Al2SO4}$
$\ce{Al(SO4)3}$
$\ce{Al2(SO4)3}$

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$D$

Show Worked Solution

Aluminum has a charge of $3+$ and sulfate has a charge of $2-$. To form a neutral compound, the charges must balance.
Using the criss-cross method, we need 2 aluminum ions (total charge: $2 \times 3 = 6+$) and 3 sulfate ions (total charge: $3 \times 2- = 6-$) to achieve a neutral compound.
Therefore, the formula is $\ce{Al2(SO4)3}$.

$\Rightarrow D$

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

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

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

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

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

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

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

\(A\)	\(\approx \dfrac{h}{2}\left(y_0+2y_1+y_2\right)\)
	\(\approx \dfrac{5.1}{2}\left(6+2 \times 3.8+0\right)\)
	\(\approx 34.68 \ \text{cm}^2\)

\(\therefore \ \text{Area of curved surface}\)	\(=1300-(408+240+69.36)\)
	\(=582.64 \ \text{cm}^2\)