Band 3 – Page 2

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}$

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$D$

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 $y=4^x$ ?

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$A$

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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i. `8\ text(ms)^(−1)`

ii. `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: ............................................................
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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)

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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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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)

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

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

Show Answers Only

$B$

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 $y=\sqrt{6-x^2}$ ?

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

Show Answers Only

$D$

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 $P(X=3)$ ?

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

Show Answers Only

$B$

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)

Show Answers Only

`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)

--- 6 WORK AREA LINES (style=lined) ---
Find the equation of the tangent to the curve when $x = 1$. (1 mark)

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

Show Answers Only

`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.

Show Answers Only

$B$

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?

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

Show Answers Only

$D$

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 $\ce{Al2(SO4)3}$.

$\Rightarrow D$

CHEMISTRY, M1 EQ-Bank 7 MC

Which row of the table correctly identifies a typical physical property of a metal element and a typical physical property of a non-metal element?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Physical property of metal}\rule[-1ex]{0pt}{0pt}& \text{Physical property of non-metal} \\
\hline
\rule{0pt}{2.5ex}\text{Good conductor of electricity}\rule[-1ex]{0pt}{0pt}&\text{Brittle}\\
\hline
\rule{0pt}{2.5ex}\text{Poor conductor of electricity}\rule[-1ex]{0pt}{0pt}& \text{Malleable}\\
\hline
\rule{0pt}{2.5ex}\text{Dull appearance}\rule[-1ex]{0pt}{0pt}& \text{High density} \\
\hline
\rule{0pt}{2.5ex}\text{Low melting point}\rule[-1ex]{0pt}{0pt}& \text{Lustrous} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$A$

Show Worked Solution

Metals: conduct heat and electricity well, are malleable and ductile, and often have high densities.
Non-metals: poor conductors (insulators), brittle when solid, and often dull in appearance.
Only option A shows a correct pairing of metal and non-metal physical properties.

$\Rightarrow A$

CHEMISTRY, M1 EQ-Bank 5 MC

Which of the following techniques would be used to separate a mixture of oil and water?

Filtration
Chromatography
Using a separating funnel
Evaporation

Show Answers Only

$C$

Show Worked Solution

Oil and water are immiscible — they form two distinct layers due to different densities.
A separating funnel allows the denser liquid (water) to be drained off first, leaving the oil layer behind.
Filtration only separates solids from liquids.
Chromatography separates substances based on solubility and adsorption, not density.
Evaporation removes a solvent, not separate two immiscible liquids.

$\Rightarrow C$

CHEMISTRY, M1 EQ-Bank 2 MC

Which of the following lists contains only metals?

Calcium, copper, zinc, potassium
Magnesium, aluminium, silicon, iron
Sodium, tin, iodine, lead
Lithium, carbon, nickel, bromine

Show Answers Only

$A$

Show Worked Solution

In the incorrect answers silicon is a metalloid while iodine, carbon and bromine are all non-metals.

$\Rightarrow A$

CHEMISTRY, M1 EQ-Bank 1 MC

What is the formula of magnesium nitride?

$\ce{MgN}$
$\ce{Mg2N3}$
$\ce{Mg3N2}$
$\ce{MgN2}$

Show Answers Only

$C$

Show Worked Solution

The mageniusm forms a $\ce{Mg^2+}$ ion and nitrogen forms a $\ce{N^3-}$.
Using the swap and drop method, the $2$ from the magnesium becomes the subscript for the nitrogen and vise versa.
Hence the correct formula is $\ce{Mg3N2}$.

$\Rightarrow C$

CHEMISTRY, M1 EQ-Bank 6 MC

Radon-222 has a half-life of 3.8 days. How long will it take for a 64.0 g sample to decay to 4.00 g?

7.6 days
11.4 days
15.2 days
26.6 days

Show Answers Only

$C$

Show Worked Solution

When the sample of Radon-222 has decayed to $4.00\ \text{g}$ there is $\dfrac{4.00}{64.0} = \dfrac{1}{16}$ left of the original sample.
As $\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$, the sample has had 4 half-lives pass.
Hence $4 \times 3.8 = 15.2\ \text{days}$ have passed.

$\Rightarrow C$

CHEMISTRY, M1 EQ-Bank 4 MC

Which of the following pairs represents isotopes of the same element?

$\ce{^12_6C}$ and $\ce{^13_6C}$
$\ce{^14_6C}$ and $\ce{^14_7N}$
$\ce{^23_11Na}$ and $\ce{^24_12Mg}$
$\ce{^35_17C}$ and $\ce{^35_18Ar}$

Show Answers Only

$A$

Show Worked Solution

Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons.
Option A: Both are carbon and so have 6 protons but differ in the number of neutrons they have hence leading to different mass numbers.

$\Rightarrow A$

CHEMISTRY, M1 EQ-Bank 3

Thorium exists in several isotopic forms. The existence of these isotopes can be shown by placing a thorium sample in a mass spectrometer, in which atoms are vaporised, electrically charged, and the ratio of the mass/charge for each is compared.

A mass spectrogram of thorium is shown below. The mass number is displayed on the x-axis and the % abundance on the y-axis.

What is the $Z$ for thorium? (1 mark)

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Write the three isotopes of thorium. (1 mark)

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Using the mass spectrogram, calculate the relative atomic weight of natural thorium. Show your working. (2 marks)

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Thorium-232 decays to radium-228. Write the nuclear equation to show the radioactive decay of thorium-232. (2 marks)

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

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a. $Z=90$

b. $\ce{^228_90Th, ^230_90Th, ^232_90Th}$.

c. $232.0$

d. $\ce{^232_90Th -> ^228_88Ra + ^4_2He}$

Show Worked Solution

a. $Z = 90$ (atomic number)

b. $\ce{^228_90Th, ^230_90Th, ^232_90Th}$.

c.	$M$	$=\dfrac{(228 \times 0.02) + (230 \times 0.8) + (232 \times 99.18)}{100}$
		$=231.98$
		$=232.0\ \text{(4 sig.fig)}$

d. Thorium-232 undergoes alpha decay:

$\ce{^232_90Th -> ^228_88Ra + ^4_2He}$

v1 Measurement, STD2 M7 2016 HSC 11 MC

The concentration of an iron supplement is 250 mg / 5 mL. A patient is prescribed 1500 mg of the supplement.

How much medication should be given to the patient?

12.5 mL
20 mL
30 mL
300 mL

Show Answers Only

`C`

Show Worked Solution

`text(Volume required)`

`= 1500/250 xx 5`

`= 30\ text(mL)`

`=> C`

CHEMISTRY, M1 EQ-Bank 1 MC

Consider the elements in Group 1 of the periodic table.

What is the general trend in the first ionisation energy of these elements as you move down the group?

Increases from lithium to caesium
Decreases from lithium to caesium
Increases from lithium to potassium then decreases to caesium
Remains constant from lithium to caesium

Show Answers Only

$B$

Show Worked Solution

The first ionisation energy decreases down the group, this is because the outermost electron is further from the nucleus and more shielded by inner electron shells, making it easier to remove.

$\Rightarrow B$

v1 Measurement, STD2 M7 2019 HSC 18

Dana, Eden and Flynn are the first three players in the school basketball team. In a recent game, Dana scored 24 points, Eden scored 18 points and Flynn scored 30 points.

What is the ratio of Dana's to Eden's to Flynn's points scored, in simplest form? (2 marks)

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In this game, the ratio of the total number of points scored by Dana, Eden and Flynn to the total number of points scored by the whole team is `9:20`.
How many points were scored by the whole team? (2 marks)

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`4:3:5`
`160\ text(points)`

Show Worked Solution

a.	`D:E:F`	`= 24:18:30`
		`= 4:3:5`

b.	`text(Total points by)\ D,E,F`	`= 24 + 18 + 30`
		`= 72`

`text(Let)\ P`	`=\ text(team points)`
`P/72`	`= 20/9`
`:. P`	`= (20 xx 72)/9`
	`= 160\ text(points)`

v1 Algebra, STD2 A4 2021 HSC 24

A population of Tasmanian devils, `D`, is to be modelled using the function `D = 650 (0.8)^t`, where `t` is the time in years.

What is the initial population? (1 mark)

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Find the population after 2 years. (1 mark)

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On the axes below, draw the graph of the population against time, in the period `t = 0` to `t = 6`. (2 marks)

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

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a. `650`

b. `416`

c. `text{See Worked Solutions}`

Show Worked Solution

a. `text{Initial population occurs when}\ \ t = 0:`

`D=650(0.8)^0=650 xx 1= 650`

b. `text{Find} \ D \ text{when} \ \ t = 5: `

`D= 650 (0.8)^2= 416`

♦ Mean mark (c) 48%.

c. `\text{At}\ t=6:`

`D=650(0.8)^6=170.39…`

CHEMISTRY, M1 EQ-Bank 2 MC

The table below contains data with reference to four common metals.

\begin{array} {|l|c|c|c|c|}
\hline \text{Metal} & \ce{Al} & \ce{Zn} & \ce{Sn} & \ce{Pb} \\
\hline \text{Z} & 13 & 30 & 50 & 82\\
\hline \text{A} & 26.98 & 65.38 & 118.7 & 207.2 \\
\hline \text{Melting point } (^{\circ}\text{C}) & 660 & 420 & 232 & 327 \\
\hline \text{Boiling point } (^{\circ}\text{C}) & 2470 & 907 & 2602 & 1749 \\
\hline
\end{array}

Which element remains a liquid over the greatest temperature range?

$\ce{Al}$
$\ce{Zn}$
$\ce{Sn}$
$\ce{Pb} $

Show Answers Only

$C$

Show Worked Solution

Range $=$ Boiling point $-$ Melting point

$\ce{Al}: \ 2470-660= 1810^{\circ}\text{C}$

$\ce{Zn}: \ 907-420= 487^{\circ}\text{C}$

$\ce{Sn}: \ 2602-232= 2370^{\circ}\text{C}$

$\ce{Pb}: \ 1749-327= 1422^{\circ}\text{C}$

$\Rightarrow C$

v1 Measurement, STD2 M7 2019 HSC 2 MC

Rice is sold in four different sized packets. Which is the best buy?

250 g for $1.10
500 g for $2.00
1 kg for $3.80
2 kg for $7.40

Show Answers Only

`D`

Show Worked Solution

$\text{Price per kilogram}$

$\text{250 g:}\ 1.10 \times 4 = $4.40 $

$\text{500 g:}\ 2.00 \times 2 = $4.00 $

$\text{1 kg:}\ $3.80 $

$\text{2 kg:}\ 7.40\ ÷\ 2 = $3.70 $

$\text{Best buy is 2 kg at \$3.70/kg.}$

$\Rightarrow D$

v2 Measurement, STD2 M7 2016 HSC 9 MC

An old air conditioner uses 2.8 kW of electricity per hour. A new air conditioner uses 1.2 kW per hour. How much electricity is saved each year if the air conditioner runs for 5 hours per day for 200 days using the new model?

1200 kWh
1600 kWh
2400 kWh
2800 kWh

Show Answers Only

$B$

Show Worked Solution

$\text{Electricity used by old air conditioner}\ = 2.8 \times 5 \times 200 = 2800\ \text{kWh}$

$\text{Electricity used by new air conditioner}\ = 1.2 \times 5 \times 200 = 1200\ \text{kWh}$

$\therefore\ \text{Electricity saved} = 2800-1200 = 1600\ \text{kWh}$

$\Rightarrow B$

v1 Measurement, STD2 M7 2016 HSC 9 MC

An old dishwasher uses 45 L of water per cycle. A new dishwasher uses 15 L per cycle. How much water is saved each year if three cycles are run each week using the new dishwasher?

2340 L
3640 L
4680 L
7020 L

Show Answers Only

$C$

Show Worked Solution

$\text{Water used by old dishwasher}\ = 45 \times 3 \times 52 = 7020\ \text{L}$

$\text{Water used by new dishwasher}\ = 15 \times 3 \times 52 = 2340\ \text{L}$

$\text{Water saved}\ = 7020-2340 = 4680\ \text{L}$

$\Rightarrow C$

HMS, TIP EQ-Bank 414 MC

A fitness professional designs different programs for two clients with similar fitness goals but different assessment results. One client shows strong upper body strength but poor cardiovascular endurance, while the other demonstrates good aerobic capacity but limited muscular strength. This approach primarily demonstrates which principle of assessment-based program development?

Standardised program design ensures consistent outcomes across all client populations regardless of individual differences
Assessment data enables targeted program customisation that addresses individual strengths and weaknesses effectively
Similar goals require identical training approaches to maintain professional standards and program consistency
Program complexity should increase proportionally with client fitness levels to ensure appropriate challenge progression

Show Answers Only

$B$

Show Worked Solution

B is correct: Assessment data enables targeted customisation by addressing each client’s specific strengths and weaknesses for optimal program effectiveness.

Other Options:

A is incorrect: The scenario demonstrates individualised rather than standardised program design based on assessment differences.
C is incorrect: Similar goals don’t require identical approaches when assessment data reveals different individual capabilities and needs.
D is incorrect: Program complexity relates to individual needs rather than fitness levels, and doesn’t address the customisation principle shown.

HMS, TIP EQ-Bank 405

Why is personalisation essential when developing training programs for recreational participants? (3 marks)

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Personalisation ensures programs match individual fitness levels, preventing excessive challenge that leads to injury risk and dropout.
Individual preferences require specific adaptations to enhance motivation. This results in sustained participation and fitness success.
Assessment data reveals unique strengths and weaknesses needing targeted attention. This causes more effective development.
Personal goals demand tailored program design to achieve specific outcomes. This creates meaningful progress toward aspirations.

Show Worked Solution

Personalisation ensures programs match individual fitness levels, preventing excessive challenge that leads to injury risk and dropout.
Individual preferences require specific adaptations to enhance motivation. This results in sustained participation and fitness success.
Assessment data reveals unique strengths and weaknesses needing targeted attention. This causes more effective development.
Personal goals demand tailored program design to achieve specific outcomes. This creates meaningful progress toward aspirations.

HMS, TIP EQ-Bank 402 MC

A recreational client wants to improve cardiovascular fitness but expresses strong dislike for running. Based on exercise assessment principles, which approach would be most appropriate for program development?

Start with short running intervals to gradually build tolerance and positive associations with the activity
Program alternative activities like cycling and swimming to achieve cardiovascular goals while respecting preferences
Include running as a minor component alongside preferred activities to ensure comprehensive cardiovascular development
Focus on high-intensity interval training using bodyweight exercises to maximise cardiovascular benefits efficiently

Show Answers Only

$B$

Show Worked Solution

B is correct: Personalisation principles involve using preferred activities to achieve fitness goals, enhancing motivation and program adherence.

Other Options:

A is incorrect: While gradual exposure seems logical, forcing disliked activities reduces long-term adherence and motivation in recreational settings.
C is incorrect: Including disliked activities even minimally can negatively impact overall program enjoyment and compliance rates.
D is incorrect: While HIIT is effective, it doesn’t address the personalisation aspect of respecting client preferences for sustainable programming.

HMS, TIP EQ-Bank 094

Describe how equipment innovations in assistive technology enhance performance for athletes with disability. In your answer, give two specific examples. (4 marks)

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Racing wheelchairs: Made from lightweight carbon fibre with angled wheels for stability. Custom-fitted seats match athlete’s body shape for maximum power transfer. These features allow wheelchair racers to reach speeds over 30km/h.
Audible balls in blind cricket: Ball contains ball bearings that rattle when bowled or hit. Players with vision impairment hear the ball’s speed and direction. This enables accurate batting, bowling and fielding responses.
Performance benefits – Both technologies allow athletes to compete at elite levels. They maximise athletes’ abilities rather than focusing on disability.

Show Worked Solution

Racing wheelchairs: Made from lightweight carbon fibre with angled wheels for stability. Custom-fitted seats match athlete’s body shape for maximum power transfer. These features allow wheelchair racers to reach speeds over 30km/h.
Audible balls in blind cricket: Ball contains ball bearings that rattle when bowled or hit. Players with vision impairment hear the ball’s speed and direction. This enables accurate batting, bowling and fielding responses.
Performance benefits – Both technologies allow athletes to compete at elite levels. They maximise athletes’ abilities rather than focusing on disability.

HMS, TIP EQ-Bank 092

Outline how wearable technology enhances performance monitoring for endurance athletes. (3 marks)

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Real-time data tracking – Monitors heart rate, pace and distance during training. Athletes can adjust intensity immediately to stay in target zones.
Recovery monitoring – Tracks sleep quality and heart rate variability. Helps athletes know when to train hard or rest.
Performance trends – Records data over weeks and months. Shows improvements in fitness and identifies when athletes might be overtraining.

Show Worked Solution

Real-time data tracking – Monitors heart rate, pace and distance during training. Athletes can adjust intensity immediately to stay in target zones.
Recovery monitoring – Tracks sleep quality and heart rate variability. Helps athletes know when to train hard or rest.
Performance trends – Records data over weeks and months. Shows improvements in fitness and identifies when athletes might be overtraining.

HMS, TIP EQ-Bank 399

Describe the key components involved in conducting a comprehensive exercise assessment. (3 marks)

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Health screening identifies potential risk factors and medical considerations affecting exercise safety. Pre-exercise questionnaires gather essential health history information.
Performance testing evaluates current physical capabilities across relevant fitness components. Tests measure baseline levels in cardiovascular endurance and muscular strength.
Goal identification clarifies individual objectives such as weight loss or performance improvement. This ensures programs align with personal aspirations.
Data integration combines all assessment information to create individualised training foundations for safe program design.

Show Worked Solution

Health screening identifies potential risk factors and medical considerations affecting exercise safety. Pre-exercise questionnaires gather essential health history information.
Performance testing evaluates current physical capabilities across relevant fitness components. Tests measure baseline levels in cardiovascular endurance and muscular strength.
Goal identification clarifies individual objectives such as weight loss or performance improvement. This ensures programs align with personal aspirations.
Data integration combines all assessment information to create individualised training foundations for safe program design.

HMS, TIP EQ-Bank 396 MC

A fitness professional conducts a Rockport 1.6-kilometre walk test with a sedentary client before designing their exercise program. This assessment primarily serves which fundamental purpose?

Establishing a baseline fitness level to ensure appropriate program intensity and reduce injury risk
Determining the client's maximum exercise capacity for high-intensity training prescription
Identifying specific sport skills that need improvement for competitive performance goals
Measuring advanced fitness parameters required for elite athletic training programs

Show Answers Only

$A$

Show Worked Solution

A is correct: The Rockport walk test establishes baseline aerobic capacity, ensuring programs are appropriately challenging while reducing injury risk.

Other Options:

B is incorrect: The walk test assesses moderate fitness levels rather than determining maximum exercise capacity for high-intensity work.
C is incorrect: This test measures general aerobic fitness rather than identifying specific sport skills for competition.
D is incorrect: The Rockport test is designed for general fitness assessment rather than advanced elite athletic parameters.

HMS, TIP EQ-Bank 104 MC

What is the primary purpose of GPS technology in sport performance?

Reducing shoe weight to enhance speed
Monitoring an athlete's location
Monitoring hydration and nutrition that flags over training
Tracking distance, speed and movement intensity

Show Answers Only

$D$

Show Worked Solution

D is correct: GPS technology tracks distance, speed and movement intensity, giving athletes and coaches data to manage workload, performance and recovery.

Other options:

A is incorrect: Shoe weight reduction relates to equipment innovation, not GPS.
B is incorrect: While GPS involves location data, its purpose in sport is broader performance monitoring.
C is incorrect: Hydration and nutrition monitoring are achieved through apps and biometric sensors, not GPS.

HMS, TIP EQ-Bank 103 MC

Which of the following best describes the role of virtual reality (VR) in athlete training?

Monitors heart rate and breathing patterns during exercise
Creates simulated environments for practising skills and tactics
Records movement data for post-training analysis
Provides feedback on muscle activation patterns

Show Answers Only

$A$

Show Worked Solution

B is correct: VR creates simulated environments where athletes can practise skills and tactics, enhancing decision-making and performance in realistic conditions without physical strain.

Other options:

A is incorrect: Monitoring heart rate and breathing is the role of wearable devices, not VR.
C is incorrect: Recording movement data is the purpose of video or motion analysis tools.
D is incorrect: Muscle activation feedback comes from biomechanics or EMG systems, not VR.

HMS, TIP EQ-Bank 086

A marathon runner completes a 42km race in hot conditions. Outline TWO physiological recovery strategies that would be most beneficial for this athlete immediately post-race. (3 marks)

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Recovery 1: Cool-down (10-15 minutes)

Gradual walking/light jogging reduces heart rate and removes lactic acid preventing muscle stiffness.
Static stretching helps lengthen tight muscles and prevents blood pooling and dizziness.

Cold water immersion (10-15°C for 10-15 minutes)

Reduces blood flow to damaged muscles which decreases inflammation and swelling.
Speeds up recovery from muscle soreness which helps remove metabolic waste products.

Show Worked Solution

Recovery 1: Cool-down (10-15 minutes)

Gradual walking/light jogging reduces heart rate and removes lactic acid preventing muscle stiffness.
Static stretching helps lengthen tight muscles and prevents blood pooling and dizziness.

Cold water immersion (10-15°C for 10-15 minutes)

Reduces blood flow to damaged muscles which decreases inflammation and swelling.
Speeds up recovery from muscle soreness which helps remove metabolic waste products.

HMS, TIP EQ-Bank 389

Describe the key features of a games-centred approach to tactical development in team sports. (3 marks)

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Games-centred approach replicates performance-like situations in training environments. Athletes experience competitive pressure that tests tactical integrity under realistic conditions.
Small-sided games reduce complexity while maintaining essential tactical elements. Examples include 3v3 situations that simulate game pressure with fewer players.
Problem-solving opportunities develop decision-making skills through modified competitive scenarios. Players learn tactical responses through direct experience rather than instruction alone.
Performance pressure preparation allows athletes to test strategies before actual competition. Training situations expose tactical weaknesses requiring improvement.

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Games-centred approach replicates performance-like situations in training environments. Athletes experience competitive pressure that tests tactical integrity under realistic conditions.
Small-sided games reduce complexity while maintaining essential tactical elements. Examples include 3v3 situations that simulate game pressure with fewer players.
Problem-solving opportunities develop decision-making skills through modified competitive scenarios. Players learn tactical responses through direct experience rather than instruction alone.
Performance pressure preparation allows athletes to test strategies before actual competition. Training situations expose tactical weaknesses requiring improvement.

HMS, TIP EQ-Bank 386 MC

A basketball coach uses 3-on-3 half-court games during training to develop tactical awareness. This approach demonstrates which key principle of tactical development?

Replicating full game conditions with complete team structures and official rules
Creating performance-like pressure situations with reduced complexity for skill focus
Eliminating competitive elements to allow pure technique development without pressure
Maximising player rest periods by reducing the number of active participants

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$B$

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B is correct: Small-sided games create performance pressure while reducing complexity, allowing focused tactical development.

Other Options:

A is incorrect: Small-sided games deliberately modify full game conditions rather than replicating them completely.
C is incorrect: The approach maintains competitive elements and pressure rather than eliminating them entirely.
D is incorrect: The focus is tactical development rather than rest management for players.

HMS, TIP EQ-Bank 380 MC

A netball team has several tall players but lacks speed and agility. Which tactical approach would best utilise their strengths while minimising their weaknesses?

Focus on fast break attacks and quick transitions between defence and attack
Emphasise lob passes and high ball strategies while avoiding ground-level play
Use short, quick passes at ground level to exploit speed advantages
Implement full-court pressing defence to utilise agility and mobility

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$B$

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B is correct: Lob passes and high ball strategies utilise height advantage while avoiding ground-level play that requires agility.

Other Options:

A is incorrect: Fast breaks require speed and agility, which are identified weaknesses for this team.
C is incorrect: Quick ground passes don’t utilise height advantages and require agility the team lacks.
D is incorrect: Full-court pressing requires speed and mobility, which are team weaknesses rather than strengths.

HMS, TIP EQ-Bank 377

Describe how wind conditions can affect strategies and tactics in both individual and group sports. (3 marks)

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Wind creates resistance affecting ball flight and athlete energy expenditure during competition. Runners must consider windbreak effects when positioning during races.
Group sports experience modified passing accuracy due to unpredictable ball movement. Teams adjust by using shorter passes and ground-based play.
Athletes adapt timing and technique methods to account for wind assistance or resistance. Strategic positioning becomes crucial for competitive advantage.

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Wind creates resistance affecting ball flight and athlete energy expenditure during competition. Runners must consider windbreak effects when positioning during races.
Group sports experience modified passing accuracy due to unpredictable ball movement. Teams adjust by using shorter passes and ground-based play.
Athletes adapt timing and technique methods to account for wind assistance or resistance. Strategic positioning becomes crucial for competitive advantage.

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

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

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

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

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

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

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

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

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