Band 6

BIOLOGY, M5 2025 HSC 33

The following diagram shows the cell division processes occurring in two related individuals.

Compare the cell division processes carried out by cells $R$ and $S$ in Individual 1. (3 marks)

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Explain the relationship between Individuals 1 and 2. (2 marks)
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$A$ and $B$ are two separate mutations. Analyse how mutations $A$ and $B$ affect the genetic information present in cells $U$, $V$, $W$ and $X$. (4 marks)
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a. Similarities:

Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
Cell S undergoes meiosis producing four genetically different haploid gametes.
Mitosis in R maintains chromosome number for growth and repair.
Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b. Relationship between Individuals 1 and 2

Individual 2 is the offspring of Individual 1.
This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

c. Mutation’s affect on genetic information

Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
The significance is that only mutation A can be inherited by offspring, while mutation B cannot.

Show Worked Solution

a. Similarities:

Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
Cell S undergoes meiosis producing four genetically different haploid gametes.
Mitosis in R maintains chromosome number for growth and repair.
Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b. Relationship between Individuals 1 and 2

Individual 2 is the offspring of Individual 1.
This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

c. Mutation’s affect on genetic information

Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
The significance is that only mutation A can be inherited by offspring, while mutation B cannot.

CHEMISTRY, M6 2025 HSC 19 MC

0.1 mol of solid sodium acetate is dissolved in 500 mL of 0.1 mol L$^{-1}$ $\ce{HCl}$ in a beaker. This solution has a pH of 4.8 .

500 mL of distilled water is then added to the beaker.

What is the pH of the final solution?

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$C$

Show Worked Solution

The initial reaction between sodium acetate and hydrochloric acid runs to completion as $\ce{HCl}$ is a strong acid:
$\ce{CH3COO-(aq) + HCl(aq) -> CH3COOH(aq) + Cl-(aq)}$
$n(\ce{CH3COO-}) = 0.1\ \text{mol}$
$n(\ce{HCl}) = 0.5\ \text{L} \times 0.1\ \text{mol L}^{-1} = 0.05\ \text{mol}$
As they react in a $1:1$ ratio, $\ce{HCl}$ is the limiting reagent.
$n(\ce{CH3COO-_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{initial}}})-n(\ce{CH3COO_{\text{reacted}}}) = 0.1-0.05 = 0.05\ \text{mol}$
$n(\ce{CH3COOH_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{reacted}}}) = 0.05\ \text{mol}$
The following equilibrium reaction is then established below dilution
$\ce{CH3COOH(aq) + H2O(l) \leftrightharpoons CH3COO-(aq) + H3O+(aq)}$
Therefore the following ice table can be constructed:

\begin{array} {|c|c|c|c|}
\hline & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 & 0.1 & 0 \\
\hline \text{Change} & -10^{-4.8} & +10^{-4.8} & +10^{-4.8} \\
\hline \text{Equilibrium} & 0.1-10^{-4.8} & 0.1+10^{-4.8} & +10^{-4.8} \\
\hline \end{array}

$\therefore K_a = \dfrac{(0.1+10^{-4.8})(10^{-4.8})}{0.1-10^{-4.8}} = 1.585395 \times 10^{-5}$

Then considering the dilution which would shift the equilibrium position to the right.

\begin{array} {|c|c|c|c|}
\hline & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.05-x & 0.05+x & +x \\
\hline \end{array}

As $K_{eq}$ is small, $0.05 -x \approx 0.05$ and $0.05 + x \approx 0.05$.
$\therefore K_{eq} = \dfrac{0.05x}{0.05} = x = 1.585395 \times 10^{-5}$
$\text{pH} = -\log_{10}\ce{[H3O+]} = -\log_{10}(1.585395 \times 10^{-5}) = 4.79962 = 4.8\ \text{(1 d.p.)}$

$\Rightarrow C$

Vectors, EXT2 V1 2025 HSC 16c

Consider the point $B$ with three-dimensional position vector $\underset{\sim}{b}$ and the line $\ell: \underset{\sim}{a}+\lambda \underset{\sim}{d}$, where $\underset{\sim}{a}$ and $\underset{\sim}{d}$ are three-dimensional vectors, $\abs{\underset{\sim}{d}}=1$ and $\lambda$ is a parameter.

Let $f(\lambda)$ be the distance between a point on the line $\ell$ and the point $B$.

Find $\lambda_0$, the value of $\lambda$ that minimises $f$, in terms of $\underset{\sim}{a}, \underset{\sim}{b}$ and $\underset{\sim}{d}$. (2 marks)

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Let $P$ be the point with position vector $\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}$.
Show that $PB$ is perpendicular to the direction of the line $\ell$. (1 mark)

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Hence, or otherwise, find the shortest distance between the line $\ell$ and the sphere of radius 1 unit, centred at the origin $O$, in terms of $\underset{\sim}{d}$ and $\underset{\sim}{a}$.
You may assume that if $B$ is the point on the sphere closest to $\ell$, then $O B P$ is a straight line. (3 marks)

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i. $\lambda_0=\underset{\sim}{d}(\underset{\sim}{b}-\underset{\sim}{a})$

ii. $\text{See Worked Solutions.}$

iii. $d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}$

Show Worked Solution

i. $\ell=\underset{\sim}{a}+\lambda \underset{\sim}{d}, \quad\abs{\underset{\sim}{d}}=1$

$\text{Vector from point \(B$ to a point on $\ell$}:\ \underset{\sim}{a}+\lambda \underset{\sim}{d}-\underset{\sim}{b}\)

$f(\lambda)=\text{distance between \(\ell$ and $B$}\)

$f(\lambda)=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}$

$\text{At} \ \ \lambda_0, f(\lambda) \ \ \text{is a min}\ \Rightarrow \ f(\lambda)^2 \ \ \text {is also a min}$

$f(\lambda)^2$	$=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2$
	$=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})$
	$=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot \underset{\sim}{d}$
	$=\lambda^2\|\underset{\sim}{d}\|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2$
	$=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2$

$f(\lambda)^2 \ \ \text{is a concave up quadratic.}$

$f(\lambda)_{\text {min}}^2 \ \ \text{occurs at the vertex.}$

$\lambda_0=-\dfrac{b}{2 a}=-\dfrac{2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b})}{2}=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})$

ii. $P \ \text{has position vector} \ \ \underset{\sim}{a}+\lambda_0 \underset{\sim}{d}$

$\text{Show} \ \ \overrightarrow{PB} \perp \ell:$

$\overrightarrow{PB}=\underset{\sim}{b}-\underset{\sim}{p}=\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}$

$\overrightarrow{P B} \cdot \underset{\sim}{d}$	$=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}$
	$=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}$
	$=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2$
	$=0$

$\therefore \overrightarrow{PB}\ \text{is perpendicular to the direction of the line}\ \ell. $

iii. $\text{Shortest distance between} \ \ell \ \text{and sphere (radius\(=1$)}\)

$=\ \text{(shortest distance \(\ell$ to $O$)}-1\)

$f\left(\lambda_0\right)=\text{shortest distance \(\ell$ to point $B$}\)

$\text{Set} \ \ \underset{\sim}{b}=0 \ \Rightarrow \ f\left(\lambda_0\right)=\text{shortest distance \(\ell$ to $0$}\)

$\Rightarrow \lambda_0=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})=-\underset{\sim}{d} \cdot \underset{\sim}{a}$

$f\left(\lambda_0\right)$	$=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}$
$f\left(\lambda_0\right)^2$	$=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2$
	$=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2$
$f\left(\lambda_0\right)$	$=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}$

$\text {Shortest distance of \(\ell$ to sphere $\left(d_{\min }\right)$:}\)

$d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}$

BIOLOGY, M5 2025 HSC 18 MC

The fruit fly, Drosophila melanogaster, normally has large round eyes. A mutation of the eye shape gene, found on the X-chromosome, causes the eye to be a narrow slit (bar-eye), which is a dominant allele for eye shape.

Offspring were produced when a bar-eyed male was crossed with a normal female. Males are XY and females are XX.

Which row in the table shows the correct percentage of male and female offspring produced with bar-eye?

	$\quad\quad\quad\textit{Percentage of offspring with bar-eyes}\quad\quad\quad$
	$\quad\textit{Males}\quad$	$\quad\textit{Females}\quad$
A.	$100$	$100$
B.	$0$	$100$
C.	$100$	$0$
D.	$50$	$0$

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$B$

Show Worked Solution

B is correct: Males inherit Y from father; all females inherit X with bar-eye.

Other Options:

A is incorrect: Males get Y chromosome from father, not X with mutation.
C is incorrect: Males cannot inherit bar-eye from father; females must inherit it.
D is incorrect: All females receive father’s X chromosome with bar-eye allele.

Mechanics, EXT2 M1 2025 HSC 16b

A particle of mass 1 kg is projected from the origin with a speed of 50 ms$^{-1}$, at an angle of $\theta$ below the horizontal into a resistive medium.

The position of the particle $t$ seconds after projection is $(x, y)$, and the velocity of the particle at that time is $\underset{\sim}{v}=\displaystyle \binom{\dot{x}}{\dot{y}}$.

The resistive force, $\underset{\sim}{R}$, is proportional to the velocity of the particle, so that $\underset{\sim}{R}=-k \underset{\sim}{v}$, where $k$ is a positive constant.

Taking the acceleration due to gravity to be 10 ms$^{-2}$, and the upwards vertical direction to be positive, the acceleration of the particle at time $t$ is given by:

$\underset{\sim}{a}=\displaystyle \binom{-k \dot{x}}{-k \dot{y}-10}$. (Do NOT prove this.)

Derive the Cartesian equation of the motion of the particle, given $\sin \theta=\dfrac{3}{5}$. (5 marks)

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Show Worked Solution

$\sin \theta=\dfrac{3}{5} \ \Rightarrow \ \cos \theta=\dfrac{4}{5}$

$\text{Components of initial velocity:}$

$\dot{x}(0)=50\, \cos \theta=50 \times \dfrac{4}{5}=40 \ \text{ms}^{-1}$

$\dot{y}(0)=50\, \sin \theta=50 \times \dfrac{3}{5}=-30\ \text{ms}^{-1}$

$\text{Horizontal motion:}$

$\dfrac{d \dot{x}}{dt}$	$=-k \dot{x} \ \ \text{(given)}$
$\dfrac{dt}{d \dot{x}}$	$=-\dfrac{1}{k \dot{x}}$
$\displaystyle \int dt$	$=-\dfrac{1}{k} \int \dfrac{1}{\dot{x}}\, d x$
$t$	$=-\dfrac{1}{k} \ln \dot{x}+c$

$\text{When} \ \ t=0, \ \dot{x}=40 \ \Rightarrow \ c=\dfrac{1}{k} \ln 40$

$t$	$=\dfrac{1}{k} \ln 40-\dfrac{1}{k} \ln \abs{\dot{x}}=\dfrac{1}{k} \ln \abs{\dfrac{40}{\dot{x}}}$
$k t$	$=\ln \abs{\dfrac{40}{\dot{x}}}$
$e^{k t}$	$=\dfrac{40}{\dot{x}}$
$\dot{x}$	$=40 e^{-k t}$
$x$	$\displaystyle=\int 40 e^{-k t}\, d t$
	$=-\dfrac{40}{k} \times e^{-k t}+c$

$\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{40}{k}$

$x=\dfrac{40}{k}-\dfrac{40}{k} e^{-k t}=\dfrac{40}{k}\left(1-e^{-k t}\right)\ \ldots\ (1)$

$\text{Vertical Motion }$

$\dfrac{d \dot{y}}{dt}$	$=-k \dot{y}-10 \quad \text{(given)}$
$\dfrac{d t}{d \dot{y}}$	$=-\dfrac{1}{k} \times \dfrac{1}{\dot{y}+\frac{10}{k}}$
$t$	$=-\dfrac{1}{k} \displaystyle \int \dfrac{1}{\dot{y}+\frac{10}{k}} \, d \dot{y}$
	$=-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}+c$

$\text{When} \ \ t=0, \, \dot{y}=-30 \ \ \Rightarrow\ \ c=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}$

$t$	$=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}$
	$=\dfrac{1}{k} \ln \abs{\frac{-30+\frac{10}{k}}{\dot{y}+\frac{10}{k}}}$
$e^{k t}$	$=\abs{\dfrac{-30+\frac{10}{k}}{y+\frac{10}{k}}}$
$\dot{y}$	$=\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10}{k}$
$y$	$=\displaystyle \left(-30+\dfrac{10}{k}\right) \int e^{-kt}\, d t-\int \dfrac{10}{k}\, dt$
	$=-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10 t}{k}+c$

$\text{When} \ \ t=0, y=0 \ \Rightarrow \ c=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)$

$y$	$=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10t}{k}$
	$=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right)\left(1-e^{-k t}\right)-\dfrac{10 t}{k}\ \ldots\ (2)$

$\text {Cartesian equation (using (1) above):}$

$x$	$=\dfrac{40}{k}\left(1-e^{-k t}\right)$
$\dfrac{k x}{40}$	$=1-e^{-k t}$
$e^{-k t}$	$=1-\dfrac{k x}{40}$
$-k t$	$=\ln \abs{1-\dfrac{k x}{40}}$
$t$	$=-\dfrac{1}{k} \ln \abs{1-\dfrac{kx}{40}}$

$y$	$=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right) \times \dfrac{k x}{40}+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}$
	$=\left(\dfrac{1-3 k}{4 k}\right) x+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}$

Complex Numbers, EXT2 N2 2025 HSC 15c

Show that
$\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.$ (3 marks)

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Use De Moivre's theorem to show that the sixth roots of $-1$ are given by
$\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)$ for $k=0,1,2,3,4,5$. (2 marks)

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Hence, or otherwise, show the solutions to $\left(\dfrac{z-1}{z+1}\right)^6=-1$ are
$z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)$, and $i \cot \left(\dfrac{11 \pi}{12}\right)$. (2 marks)

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i. $\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)$

$\text{LHS}$	$=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}$
	$=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}$
	$=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}$
	$=i \cot \left(\frac{\theta}{2}\right)$

ii. $z=\cos \theta+i \sin \theta$

$\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}$

$z^6=\cos (6 \theta)+i \sin (6 \theta)=-1$

$\cos (6 \theta)$	$=-1 \ \text{and} \ \ \sin (6 \theta)=0$
$6 \theta$	$=\pi, 3 \pi, 5 \pi, \ldots$
$\theta$	$=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5$

$\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for} \ \ k=0,1, \ldots, 5$

iii. $\left(\dfrac{z-1}{z+1}\right)^6=-1$

$\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1$

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

$\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)$

$\alpha=\dfrac{z-1}{z+1}\ \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}$

$\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}$

$z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}$

$\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)$

$\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)$

Show Worked Solution

i. $\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)$

$\text{LHS}$	$=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}$
	$=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}$
	$=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}$
	$=i \cot \left(\frac{\theta}{2}\right)$

ii. $z=\cos \theta+i \sin \theta$

$\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}$

$z^6=\cos (6 \theta)+i \sin (6 \theta)=-1$

$\cos (6 \theta)$	$=-1 \ \text{and} \ \ \sin (6 \theta)=0$
$6 \theta$	$=\pi, 3 \pi, 5 \pi, \ldots$
$\theta$	$=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5$

$\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for} \ \ k=0,1, \ldots, 5$

iii. $\left(\dfrac{z-1}{z+1}\right)^6=-1$

$\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1$

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

$\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)$

$\alpha=\dfrac{z-1}{z+1}\ \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}$

$\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}$

$z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}$

$\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)$

$\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)$

BIOLOGY, M8 2025 HSC 32

A population lives across three regions, $A,\ B$ and $C$.

People in community $B$ developed an environmental disease. An epidemiological study was carried out to determine the risk of developing the disease due to age at exposure. The results of this study are shown in the graph.

Design an epidemiological study that could be used to produce the results shown in the graph. Justify the features of your design. (7 marks)

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Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Participants: Recruit individuals from community B across three age groups: 10-year-olds, 20-year-olds and 30-year-olds at the time of exposure to the environmental factor. This is justified because the graph displays separate curves for exposure at these three ages.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

Longitudinal Follow-up: Monitor all participants annually for disease development over 60 years. This is justified because the graph tracks disease risk across this timeframe and shows when risk peaks and declines.

Data Collection: Document whether each participant develops the disease and calculate the percentage of each age cohort affected at yearly intervals. This is justified because the y-axis shows risk as a percentage.

Control Variables: Ensure all participants experience similar levels of environmental exposure in community B. This is justified because the study isolates age at exposure as the only variable affecting disease risk.

Statistical Analysis: Calculate risk percentages for each time point after exposure for each age group. This is justified because it produces the three distinct curves showing risk declining differently based on initial exposure age.

Show Worked Solution

Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

BIOLOGY, M6 2025 HSC 30

PAI-1 protein is encoded by the SERPINE 1 gene in humans. Anopheles mosquitoes have been genetically modified to express PAI-1, which blocks the entry of the malarial Plasmodium into the mosquito gut. This disrupts the Plasmodium life cycle, resulting in reduced transmission of malaria.

Describe a process that could be used to produce mosquitoes which express PAI-1. (4 marks)

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'Genetic technologies are beneficial for society.'
Evaluate this statement. (7 marks)

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a. Mosquito Production

The SERPINE 1 gene is isolated from human DNA using restriction enzymes that cut at specific recognition sites.
The same restriction enzymes are used to cut mosquito DNA, creating complementary sticky ends.
The human PAI-1 gene is inserted into the mosquito DNA using DNA ligase to form recombinant DNA.
The recombinant DNA is introduced into mosquito eggs or embryos using a vector or microinjection technique.
Modified mosquitoes are screened to confirm PAI-1 gene expression and successful integration into the genome.
Transgenic mosquitoes produce PAI-1 protein that blocks Plasmodium entry into the gut, disrupting malaria transmission.

b. Evaluation Statement

Genetic technologies are highly beneficial for society when evaluated against health improvements and food security criteria. Despite some ethical and environmental concerns requiring careful management, the overall benefits are substantial.

Health Benefits
Evidence supporting includes:

Genetically modified mosquitoes expressing PAI-1 significantly reduce malaria transmission, potentially saving millions of lives annually.
Recombinant DNA technology produces insulin and vaccines, improving accessibility to life-saving treatments for diabetes and infectious diseases.
Gene therapy offers potential cures for inherited genetic disorders, dramatically improving quality of life for affected individuals.

The health criterion strongly meets beneficial status because these technologies address major global health challenges.

Food Security and Agricultural Benefits
Evidence supporting includes:

Genetically modified crops like Bt cotton and Golden Rice increase crop yields and nutritional content, addressing food scarcity.
Drought-resistant GM crops enable farming in challenging environments, supporting population growth and farmer livelihoods.

However, concerns exist about reduced genetic diversity and corporate control over seeds, creating inequalities in access.

Final Evaluation

Weighing these factors shows genetic technologies are substantially beneficial for society. The health improvements and food security gains outweigh the manageable ethical concerns. While challenges like biodiversity impacts and equitable access require ongoing attention, the overall societal benefit remains considerable through life-saving medical applications and enhanced food production.

Show Worked Solution

a. Mosquito Production

The SERPINE 1 gene is isolated from human DNA using restriction enzymes that cut at specific recognition sites.
The same restriction enzymes are used to cut mosquito DNA, creating complementary sticky ends.
The human PAI-1 gene is inserted into the mosquito DNA using DNA ligase to form recombinant DNA.
The recombinant DNA is introduced into mosquito eggs or embryos using a vector or microinjection technique.
Modified mosquitoes are screened to confirm PAI-1 gene expression and successful integration into the genome.
Transgenic mosquitoes produce PAI-1 protein that blocks Plasmodium entry into the gut, disrupting malaria transmission.

b. Evaluation Statement

Health Benefits
Evidence supporting includes:

Genetically modified mosquitoes expressing PAI-1 significantly reduce malaria transmission, potentially saving millions of lives annually.
Recombinant DNA technology produces insulin and vaccines, improving accessibility to life-saving treatments for diabetes and infectious diseases.
Gene therapy offers potential cures for inherited genetic disorders, dramatically improving quality of life for affected individuals.

The health criterion strongly meets beneficial status because these technologies address major global health challenges.

Food Security and Agricultural Benefits
Evidence supporting includes:

Genetically modified crops like Bt cotton and Golden Rice increase crop yields and nutritional content, addressing food scarcity.
Drought-resistant GM crops enable farming in challenging environments, supporting population growth and farmer livelihoods.

However, concerns exist about reduced genetic diversity and corporate control over seeds, creating inequalities in access.

Final Evaluation

BIOLOGY, M6 2025 HSC 20 MC

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

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

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

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

Show Answers Only

$D$

Show Worked Solution

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

Other Options:

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

BIOLOGY, M5 2025 HSC 19 MC

The grasshopper has an average genome size of 120 million base pairs. 42% of these base pairs contain cytosine and guanine.

What is the number of DNA nucleotides with a thymine base found in each cell?

25.2 million
34.8 million
69.6 million
120 million

Show Answers Only

$C$

Show Worked Solution

C is correct: If 42% are C+G, then 58% are A+T; thymine is 29% of 240 million nucleotides.

Other Options:

A is incorrect: This calculates as 21% instead of 29% of total nucleotides.
B is incorrect: This calculates as 29% of base pairs, not total nucleotides.
D is incorrect: This is the total number of base pairs, not thymine nucleotides.

Measurement, STD2 M6 2025 HSC 37

The diagram shows a park consisting of two equilateral triangles. The shaded triangle is a grassed section. All measurements on the diagram are in metres.

How long will it take to mow the grassed section if it takes 5 minutes to mow 20 m² ? Give your answer to the nearest minute. (4 marks)

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$7 \ \text{minutes}$

Show Worked Solution

$\text{Large triangle is equilateral (all sides = 12 m)}$

$\text{Area of large} \ \triangle$	$=\dfrac{1}{2}ab \, \sin C$
	$=\dfrac{1}{2} \times 12 \times 12 \times \sin 60^{\circ}$
	$=36 \sqrt{3}$

$\text{Area of} \ \ \triangle_1=\dfrac{1}{2} \times 3 \times 9 \times \sin 60^{\circ}=\dfrac{27 \sqrt{3}}{4}$

$\text{Grassed area}=36 \sqrt{3}-3 \times \dfrac{27 \sqrt{3}}{4}=27.2798 \ldots \ \text{m}^2$

$\text{Time to mow}$	$=\dfrac{27.2798 \ldots}{20} \times 5$
	$=6.81 \ldots$
	$=7 \ \text{minutes (nearest min)}$

Measurement, STD2 M6 2025 HSC 35

The triangle $PTA$ is shown. The length of $PA$ is 75 m and the length of $PT$ is 51 m.

The angle of depression from $T$ to $A$ is 36°, and the angle $PTA$ is obtuse.

Find the length of $TA$. Give your answer correct to 2 decimal places. (3 marks)

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$TA=35.03 \ \text{m}$

Show Worked Solution

$\angle TAP=36^{\circ} \ \text {(alternate)}$

$\text{Using sine rule in} \ \triangle TAP:$

$\dfrac{\sin \angle PTA}{75}$	$=\dfrac{\sin 36^{\circ}}{51}$
$\sin \angle PTA$	$=75 \times \dfrac{\sin 36^{\circ}}{54}=0.864 \ldots$
$\angle PTA$	$=\sin ^{-1}(0.864 \ldots)=180-59.81=120.19^{\circ}\ \ \text{(obtuse)}$

$\angle PTX=120.19-54=66.19^{\circ}$

$\angle TPA=90-66.19=23.81^{\circ}\ \left(180^{\circ}\ \text{in}\ \triangle \right)$

$\text{Using sine rule in} \ \triangle TAP:$

$\dfrac{TA}{\sin 23.81^{\circ}}$	$=\dfrac{51}{\sin 36^{\circ}}$
$TA$	$=\dfrac{51 \times \sin 23.81^{\circ}}{\sin 36^{\circ}}$
	$=35.03 \ \text{m (2 d.p.)}$

Measurement, STD1 M4 2025 HSC 28

The table provides information about a $2 coin and a $5 note.

\begin{array} {|c|c|c|}
\hline \text{Coin/note} & \text{Quantity needed to} & \text{Mass of this number} \\ & \text{make \$1000} & \text{of notes/coins} \\& & \text{(kg)} \\
\hline \$2 & 500 & 3.3 \\
\hline \$5 & 200 & 0.157 \\
\hline \end{array}

Calculate the mass of a $2 coin in grams, correct to 1 decimal point? (2 marks)
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Suppose the $2 coin is to be replaced with a note that has the same mass as a $5 note.
What is the mass of $1000 in $2 notes in grams? Give your answer correct to the nearest gram. (2 marks)
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a. $ 6.6\ \text{g}$

b. $ 393\ \text{g (nearest gram)}$

Show Worked Solution

a. $\text{Using 1 kg = 1000 grams:}$

$\text{Mass of \$2 coin}=\dfrac{3.3\times 1000}{500}=6.6\ \text{g}$

b. $ \text{Number of \$2 notes}=\dfrac{1000}{2}=500$

$\text{Mass of one \$5 note} = \dfrac{0.157}{200}=0.000785\ \text{kg}$

$\text{Since \$2 note weighs the same as \$5 note:}$

$\text{Mass of 500 \$2 notes}$	$=500\times 0.000785$
	$=0.3925\ \text{kg}$
	$=393\ \text{g (nearest gram)}$

Financial Maths, STD1 F3 2025 HSC 27

The graph shows the salvage value of a car over 5 years.

The salvage values are based on the declining-balance method.

By what amount will the car’s value depreciate during the 10th year? (4 marks)

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$$1476.40$

Show Worked Solution

$\text{Find}\ r:$

$\text{When}\ \ n=1, \ S=$44\ 000\ \ \text{(see graph)}$

$S$	$=V_0(1-r)^n$
$44\ 000$	$=55\ 000(1-r)^1$
$\dfrac{44\ 000}{55\ 000}$	$=1-r$
$1-r$	$=0.8$
$r$	$=1-0.8=0.20$

$\text{Find \(S$ when}\ \ n=9\ \ \text{and}\ \ n=10:\)

$S_9=55\ 000(1-0.20)^{9}=$7381.97504$

$S_{10}=55\ 000(1-0.20)^{10}=$5905.5800$

$S_9-S_{10}=$7381.9750-$5905.580=$1476.40\ \text{(nearest cent)}$

$\therefore\ \text{The car’s value will depreciate by \$1476.40 in the 10th year.}$

Financial Maths, STD1 F3 2025 HSC 21

A credit card has an interest-free period of 45 days from and including the date of purchase. Interest is charged on purchases made, compounding daily at a rate of 13.74% per annum, from and including the day following the interest-free period.

Concert tickets were purchased for a total of $392 using this credit card.

Full payment was made on the 68th day from the date of purchase. There were no other purchases on this credit card.

What was the total interest charged when the account was paid in full? (3 marks)

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$\text{Interest charged}\ =\$ 3.41$

Show Worked Solution

$\text{Day 1-45: no interest is charged}$

$\text{Day 46-68: interest charged (23 days)}$

$\text{Daily interest rate}=\dfrac{13.74}{365} \%=\dfrac{13.74}{365 \times 100}$

$\text{Amount owing}=392\left(1+\dfrac{13.74}{365 \times 100}\right)^{23}=\$ 395.41$

$\text{Interest charged}=395.41-392=\$ 3.41$

Calculus, EXT1 C2 2025 HSC 10 MC

For the function $f(x)$, it is known that $f(3)=1, f^{\prime}(3)=2$ and $f^{\prime \prime}(3)=4$.

Let $g(x)=f^{-1}(x)$.

What is the value of $g^{\prime \prime}(1)$ ?

$\dfrac{1}{4}$
$-\dfrac{1}{4}$
$-\dfrac{1}{2}$
$-1$

Show Answers Only

$C$

Show Worked Solution

$f(3)=1, f^{\prime}(3)=2, f^{\prime \prime}(3)=4$

$\text{Given} \ \ g(x)=f^{-1}(x):$

$f(g(x))=x \ \ \text{(Definition of an inverse fn)}$

$\text{Differentiate both sides:}$

$g^{\prime}(x) \cdot f^{\prime}(g(x))=1 \ \ \Rightarrow \ \ g^{\prime}(x)=\dfrac{1}{f^{\prime}(g(x))}$

$g^{\prime \prime}(x)=\dfrac{d}{d x}\left(\dfrac{1}{f^{\prime}(g(x))}\right)=-\dfrac{f^{\prime \prime}(g(x)) \cdot g^{\prime}(x)}{\left[f^{\prime}(g(x))\right]^2}$

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

$g^{\prime}(1)$	$=\dfrac{1}{f^{\prime}(g(1))}=\dfrac{1}{f^{\prime}(3)}=\dfrac{1}{2}$
$g^{\prime \prime}(1)$	$=-\dfrac{f^{\prime \prime}(g(1)) \cdot g^{\prime}(1)}{\left[f^{\prime}(g(1))\right]^2}=-\dfrac{f^{\prime \prime}(3) \cdot \dfrac{1}{2}}{\left[f^{\prime}(3)\right]^2}=-\dfrac{4 \times \dfrac{1}{2}}{2^2}=-\dfrac{1}{2}$

$\Rightarrow C$

Functions, EXT1 F2 2025 HSC 14e

It is given that $\tan \alpha, \tan \beta$ and $\tan \gamma$ are the three real roots of the polynomial equation $x^3+b x^2+c x-1+b+c=0$, where $b$ and $c$ are real numbers and $c \neq 1$.

Find the smallest positive value of $\alpha+\beta+\gamma$. (3 marks)

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$\alpha+\beta+\gamma=\dfrac{3 \pi}{4}$

Show Worked Solution

$x^3+b x^2+c x-1+b+c=0$

$\text{Roots:} \ \tan \alpha, \tan \beta, \tan \gamma$

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

$\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma=c$

$\tan \alpha \cdot \tan \beta \cdot \tan \gamma=-\dfrac{d}{a}=1-b-c$

$\text { Find smallest +ve value of} \ \ \alpha+\beta+\gamma:$

$\tan (\alpha+\beta+\gamma)$	$=\dfrac{\tan (\alpha+\beta)+\tan \gamma}{1-\tan (\alpha+\beta) \tan \gamma}$
	$=\dfrac{\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}+\tan \gamma}{1-\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta} \times \tan \gamma}$
	$=\dfrac{\tan \alpha+\tan \beta+\tan \gamma(1-\tan \alpha \cdot \tan \beta)}{1-\tan \alpha \cdot \tan \beta-(\tan \alpha+\tan \beta) \tan \gamma}$
	$=\dfrac{\tan \alpha+\tan \beta+\tan \gamma-\tan \alpha \cdot \tan \beta \cdot \tan \gamma}{1-(\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma)}$
	$=\dfrac{-b-(1-b-c)}{1-c}$
	$=\dfrac{-1+c}{1-c}$
	$=-1$

$\therefore \ \text{Smallest +ve value of} \ \ \alpha+\beta+\gamma=\dfrac{3 \pi}{4}$

Calculus, EXT1 C2 2025 HSC 14d

The function $f(x)$ is defined by $f(x)=\cos ^{-1}(\sin x)$ in the domain $(0, \pi)$.

Find $f^{\prime}(x)$ for those values of $x$ where it is defined. (3 marks)

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$f^{\prime}(x)=-1 \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)$

$f^{\prime}(x)=1 \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)$

Show Worked Solution

$f(x)$	$=\cos ^{-1}(\sin x)$
$f^{\prime}(x)$	$=-\dfrac{\cos x}{\sqrt{1-\sin ^2 x}}$
	$=-\dfrac{\cos x}{\abs{\cos x}}$
	$= \pm 1$

$\text{Consider limitations:}$

$1-\sin ^2 x \neq 0 \ \Rightarrow \ \sin x \neq \pm 1 \ \Rightarrow \ x \neq \dfrac{\pi}{2}$

$\text{In lst quadrant:} \ \ -\dfrac{\cos x}{\abs{\cos x}}=-1$

$\text{In 2nd quadrant:}\ \ -\dfrac{\cos x}{\abs{\cos x}}=1$

$f^{\prime}(x)=-1 \ \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)$

$f^{\prime}(x)=1 \ \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)$

Measurement, STD1 M5 2025 HSC 26

A scale of $1 : 50$ is used to draw a rectangular area on a 2 mm grid as shown. The actual rectangular area is to be tiled.

The tiles cost $150 per square metre and the tiler orders 15% extra tiles to allow for cutting and breakage.

The tiler charges $90 per hour and will take 20 hours to complete the tiling.

Calculate the total cost of the tiles and tiling. Give your answer to the nearest dollar. (4 marks)

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$$6105.60$

Show Worked Solution

$\text{Using scale where each grid is 2mm × 2 mm:}$

$2\ \text{mm}\ =50 \times 2 = 100\ \text{mm in actual length}$

$\text{100 mm = 10 cm}$

$\Rightarrow \ \text{i.e. each grid represents 10 cm in actual length.}$

$\text{Width}\ = 52 \times 10 = 520\ \text{cm}\ = 5.2\ \text{m}$

$\text{Height}\ = 48 \times 10 = 480\ \text{cm}\ = 4.8\ \text{m}$

$\text{Area}=5.2 \times 4.8=24.96\ \text{m}^2$

$\text{Cost of tiles}=24.96\times $150\times 1.15=$4305.60$

$\text{Labour Cost}=90\times 20=$1800$

$\therefore\ \text{Total Cost}=$4305.60+$1800=$6105.60$

Calculus, 2ADV C3 2025 HSC 10 MC

The graph of $y=f(x)$, with all its stationary points, is shown.

How many stationary points does the graph of $y=f\left(e^x\right)$ have?

Show Answers Only

$C$

Show Worked Solution

$y=f(e^{x})\ \ \Rightarrow\ \ y^{\prime}=e^{x} \times f(e^{x}) $

$\text{Find number of \(x$ values where}\ \ y^{\prime}=0.\)

$\text{Since}\ e^{x} \in (0, \infty)\ \text{for all}\ x: $

$\text{Stationary points of \(f(e^x)$ = 2 (SP’s of $f(x)$ for}\ x \in (0, \infty)).\)

$\Rightarrow C$

Statistics, STD2 S4 2025 HSC 25

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. $\text{Form: Linear. Direction: Negative}$

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

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

Trigonometry, 2ADV T2 2025 HSC 31

The equation $\cos \, p x=\dfrac{1}{2}$ has 2 solutions where $0 \leq x \leq 2 \pi$ and $p>0$.

Find all possible values of $p$. (3 marks)

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$\dfrac{5}{6} \leqslant p<\dfrac{7}{6}$

Show Worked Solution

$\cos (p x)=\dfrac{1}{2}\ \ \Rightarrow\ \ px=\cos ^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{3}, \dfrac{5 \pi}{3}, \dfrac{7 \pi}{3}$

$\text{Since there are 2 solutions in range} \ \ 0 \leq x<2 \pi:$

$p x=\dfrac{5 \pi}{3} \ \Rightarrow \ x=\dfrac{5 \pi}{3 p}$

$\dfrac{5 \pi}{3 p}$	$\leqslant 2 \pi$
$3 p$	$\geqslant \dfrac{5}{2}$
$p$	$\geqslant \dfrac{5}{6}$

$\text{Since}\ \ p x=\dfrac{7 \pi}{3} \ \ \text{cannot be a solution in the range} \ \ 0 \leq x \leq 2 \pi:$

$p x=\dfrac{7 \pi}{3} \ \Rightarrow \ x=\dfrac{7 \pi}{3 p}$

$\dfrac{7 \pi}{3 p}$	$>2 \pi$
$p$	$<\dfrac{7}{6}$

$\therefore \dfrac{5}{6} \leqslant p<\dfrac{7}{6}$

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 \(O$ from $F$}=021^{\circ} \)

Trigonometry, 2ADV T1 2025 HSC 28

A farmer wants to use a straight fence to divide a circular paddock of radius 10 metres into two segments. The smaller segment is $\dfrac{1}{4}$ of the paddock and is shaded in the diagram. The fence subtends an angle of $\theta$ radians at the centre of the circle as shown.

Show that $\theta=\sin \theta+\dfrac{\pi}{2}$. (2 marks)

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The graph of $y=\sin \theta+\dfrac{\pi}{2}$ is shown.
Use the graph and the result in part (a) to estimate the arc length of the smaller segment to the nearest metre. (2 marks)

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a. $\text{See Worked Solutions}$

b. $\text{Arc length} \ \approx 23 \ \text{metres.}$

Show Worked Solution

a. $\text{Area of paddock} =\pi \times 10^2=100 \pi$

$\text{Area of segment} =\dfrac{1}{4} \times 100 \pi =25 \pi$

$\text{Area of sector} =\dfrac{\theta}{2 \pi} \times 100 \pi=50 \theta$

$\text{Area of triangle} =\dfrac{1}{2} ab \, \sin C=50 \, \sin \theta$

$\text{Equating sector areas:}$

$25 \pi$	$=50 \theta-50\, \sin \theta$
$50 \theta$	$=25 \pi+50 \, \sin \theta$
$\theta$	$=\dfrac{\pi}{2}+\sin \theta$

b. $\text{Find where} \ \ \theta=\sin \theta+\dfrac{\pi}{2}$

$\text {Intersection occurs where: }$

$y=\theta \ \ \text{intersects with} \ \ y=\sin \theta+\dfrac{\pi}{2}$

$\text{At intersection (from graph):} \ \ \theta \approx 2.3 \ \text{radians}$

$\text{Arc length} \ \approx 10 \times 2.3 \approx 23 \ \text{metres.}$

Calculus, 2ADV C3 2025 HSC 26

A piece of wire is 100 cm long. Some of the wire is to be used to make a circle of radius $r$ cm. The remainder of the wire is used to make an equilateral triangle of side length $x$ cm.

Show that the combined area of the circle and equilateral triangle is given by
$A(x)=\dfrac{1}{4}\left(\sqrt{3} x^2+\dfrac{(100-3 x)^2}{\pi}\right)$. (2 marks)

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By considering the quadratic function in part (a), show that the maximum value of $A(x)$ occurs when all the wire is used for the circle. (3 marks)
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a. $\text{See Worked Solutions}$

b. $\text{See Worked Solutions}$

Show Worked Solution

a. $\text{Area of equilateral triangle:}$

$A_{\Delta}=\dfrac{1}{2} a b\, \sin C=\dfrac{1}{2} \times x^2 \times \sin 60^{\circ}=\dfrac{\sqrt{3} x^2}{4}$

$\text{Wire remaining to make circle}=100-3 x$

$\text {Find radius of circle:}$

$2 \pi r=100-3 x \ \Rightarrow \ r=\dfrac{100-3 x}{2 \pi}$

$\text{Area of circle: }$

$A_{\text {circ }}=\pi r^2=\pi \times\left(\dfrac{100-3 x}{2 \pi}\right)^2=\dfrac{(100-3 x)^2}{4 \pi}$

$\text{Total Area}$	$=\dfrac{\sqrt{3} x^2}{4}+\dfrac{(100-3 x)^2}{4 \pi}$
	$=\dfrac{1}{4}\left(\sqrt{3} x^2+\dfrac{(100-3 x)^2}{\pi}\right)$

b. $\text{Note strategy clue in question: “By considering the quadratic..”}$

$\text {Expanding} \ A(x):$

$A(x)$	$=\dfrac{1}{4}\left(\sqrt{3} x^2+\dfrac{10\,000}{\pi}-\dfrac{600 x}{\pi}+\dfrac{9 x^2}{\pi}\right)$
	$=\dfrac{1}{4}\left(\sqrt{3}+\dfrac{9}{\pi}\right) x^2-\dfrac{150}{\pi} x+\dfrac{2500}{\pi}$

$\text{Consider limits on} \ x:$

$3 x \leqslant 100 \ \Rightarrow \ x \in\left[0,33 \frac{1}{3}\right]$

$\text{Consider vertex of concave up parabola}\ A(x):$

$x=-\dfrac{b}{2 a}=\dfrac{150}{\pi} \ ÷ \ \dfrac{1}{2}\left(\sqrt{3}+\dfrac{9}{\pi}\right) \approx 20.8$

$\text{By symmetry of the quadratic for} \ x \in\left[0,33 \dfrac{1}{3}\right],$

$A(x)_{\text{max}} \ \text{occurs at} \ \ x=0.$

$\text{i.e. when the wire is all used in the circle.}$

Calculus, 2ADV C4 2025 HSC 25

Show that $\dfrac{d}{d x}(\sin x-x\, \cos x)=x\, \sin x$. (2 marks)

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Hence, find the value of $\displaystyle\int_0^{2025 \pi} x\, \sin x \, dx$. (2 marks)

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The regions bounded by the $x$-axis and the graph of $y=x\, \sin x$ for $x \geq 0$ are shown.

Let $A_n=\displaystyle \int_{(n-1) \pi}^{n \pi} x\, \sin x \,dx$, where $n$ is a positive integer.
It can be shown that $\left|A_n\right|=(2 n-1) \pi$. (Do NOT prove this.)
Find the exact total area of the regions bounded by the curve $y=x \sin x$, and the $x$-axis between $x=0$ and $x=2025 \pi$. (2 marks)
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Show Answers Only

a. $\text{See Worked Solutions}$

b. $2025 \pi$

c. $4\,100\,625 \pi \ \text{units}^2$

Show Worked Solution

a.	$\dfrac{d}{dx}(\sin x-x\, \cos x)$	$=\dfrac{d}{dx} \sin x-\dfrac{d}{dx} x\, \cos x$
		$=\cos x+x\, \sin x-\cos x$
		$=x\, \sin x$

b.	$\displaystyle\int_0^{2025 \pi} x\, \sin x$	$=\Big[\sin x-x\, \cos x\Big]_0^{2025 \pi}$
		$=\Big[(\sin (2025\pi)-2025 \pi \times \cos (2025 \pi))-0\Big]$
		$=0-2025 \pi \times -1$
		$=2025 \pi$

c. $\text{Area}=\displaystyle \int_0^\pi x\, \sin x \, dx+\left|\int_\pi^{2 \pi} x\, \sin x \, dx\right|+\cdots+\int_{2024 \pi}^{2025 \pi} x\, \sin x \, dx$

$\text{Using}\ \ \left|A_n\right|=(2n-1) \pi:$

$A_1=\pi, A_2=3 \pi, A_3=5 \pi, \ldots, A_{2025}=4049 \pi$

\begin{aligned}
\rule{0pt}{2.5ex} \text{Area}& =\underbrace{\pi+3 \pi+5 \pi+\ldots+4049 \pi}_{\text {AP where } a=\pi, \ l=4049 \pi, \ n=2025} \\
\rule{0pt}{4.5ex} & =\frac{2025}{2}(\pi+4049 \pi) \\
\rule{0pt}{3.5ex} & =4\,100\,625 \pi \ \text{units }^2
\end{aligned}

Statistics, STD2 S5 2025 HSC 40

In a flock of 12 600 sheep, the ratio of males to females is $1:20$.
The weights of the male sheep are normally distributed with a mean of 76.2 kg and a standard deviation of 6.8 kg.
In the flock, 15 of the male sheep each weigh more than $x$ kg.
Find the value of $x$. (4 marks)

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The weights of the female sheep are also normally distributed but have a smaller mean and smaller standard deviation than the weights of male sheeр.
Explain whether it could be expected that 300 of the females from the flock each weigh more than $x$ kg, where $x$ is the value found in part (a). (1 mark)

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Show Answers Only

a. $x=89.8 \ \text{kg}$

b. $\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}$

$\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m$

$\text{Consider the value of}\ x_f\ \text{when \(z$-score = 2}:\)

$\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m$

$\therefore \ \text{It is not expected that 300 females weigh > 89.8 kg.}$

Show Worked Solution

a. $12\,600 \ \text{sheep} \ \Rightarrow \ \text{male : female}=1:20$

$\text{Number of sheep in “1 part”} = \dfrac{12\,600}{21}=600$

$\Rightarrow \ \text{male : female}=600:12\,000$

$\text{male sheep} \ \%=\dfrac{15}{600}=0.025\%$

$z \text{-score }(0.025 \%)=2$

$\text{Using } \ z=\dfrac{x-\bar{x}}{s}, \ \text{find} \ \ x_m:$

$2$	$=\dfrac{x_m-76.2}{6.8}$
$x_m$	$=76.2+2 \times 6.8=89.8 \ \text{kg}$

b. $\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}$

$\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m$

$\text{Consider the value of}\ x_f\ \text{when \(z$-score = 2}:\)

$\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m$

$\therefore \ \text{It is not expected that 300 females weigh > 89.8 kg.}$

Statistics, STD2 S4 2025 25

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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Show Answers Only

a. $\text{Form: Linear. Direction: Negative}$

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

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

BIOLOGY, M7 2021 VCE 11

Two students designed an experiment to investigate antibiotic resistance in Escherichia coli bacteria. They began with an E. coli culture. The following procedure was conducted in a filtered air chamber using aseptic techniques:

On Day 0, spread 1 mL of E. coli culture onto a nutrient agar plate containing $0 \ \mu \text{g} / \text{mL}$ (micrograms per millilitre) of the antibiotic ampicillin. Spread $1 \ \text{mL}$ of the $E. coli $ culture onto a separate nutrient agar plate containing $1 \ \mu \text{g} / \mathrm{mL}$ of ampicillin. Cover each plate with an airtight lid.
On Day 1, transfer a sample of bacteria from one of the Day 0 plates to one of the Day 1 plates containing $2 \ \mu \text{g} / \text{mL}$ of ampicillin. Transfer a sample of bacteria from the other Day 0 plate to the other Day 1 plate, which also contains $2 \ \mu \text{g} / \text{mL}$ of ampicillin. Cover as before.
On Day 2, transfer a sample of bacteria from one of the Day 1 plates to one of the Day 2 plates containing $4 \ \mu \text{g} / \text{mL}$ of ampicillin. Transfer a sample of bacteria from the other Day 1 plate to the other Day 2 plate, which also contains $4 \ \mu \text{g} / \text{mL}$ of ampicillin. Cover as before.
On Day 3, transfer a sample of bacteria from one of the Day 2 plates to one of the Day 3 plates containing $6 \ \mu \text{g} / \text{mL}$ of ampicillin. Transfer a sample of bacteria from the other Day 2 plate to the other Day 3 plate, which also contains $6 \ \mu \text{g} / \text{mL}$ of ampicillin. Cover and seal the plates.

All plates were incubated at 37 °C for each 24-hour period. Used plates were refrigerated until the end of the experiment. They were then photographed to compare the amount of bacterial growth and disposed of safely.

The students drew a diagram (Figure 1) to help explain the experimental design and to show their predicted results in each condition at the end of each day.

Identify any two controlled variables for this experiment. (2 marks)

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Write a suitable hypothesis for this experiment. (2 marks)

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The refrigerated plates kept from Days 0,1,2 and 3 of the experiment were photographed. The diagrams in Figure 2 represent the bacterial growth seen in the photographs.

i. Analyse the results of the experiment shown in Figure 2. (3 marks)
ii. Explain whether the results of the students' experiment shown in Figure 2 support the predicted results shown in Figure 1. (2 marks)

Show Answers Only

a. Examples of suitable responses included two of the following:

volume of E. coli first applied to plates
type of nutrient agar used
size of agar plates used
batch or source of ampicillin
duration of incubation
temperature of incubation
exposure time to each concentration of ampicillin.

b. Hypothesis:

If E. coli is exposed to low OR 1 µg/ml concentration of ampicillin, it will exhibit increased growth when exposed to higher levels of ampicillin.

c.i. Experiment analysis:

No ampicillin: More E. coli growth on Day 0.
2 µg/mL ampicillin: No effect on bacterial growth.
Days 2/3: Fewer colonies for both groups.
Higher ampicillin (4 µg/mL, 6 µg/mL): Significant effect, killed many E. coli.
More growth than predicted on all plates.
By end: More ampicillin-resistant bacteria on experimental plate.
1 µg/mL ampicillin: Initial exposure led to greater number of resistant E. coli.

c.ii. If the experimental results did not support the predicted results:

E. coli was still present on both plates on Day 3.
This shows exposing E. coli to 1 µg/ml ampicillin does not increase its resistance to ampicillin.
Resistant E. coli grew regardless of initial exposure to ampicillin.

Any two of the following, if the experimental results did support the predicted results:

The number of colonies in both groups decreased over time.
More E. coli developed ampicillin resistance in the experimental group.
There were more colonies in the experimental group than in the control group on Day 3.
There were more colonies than expected as ampicillin concentration increased.
No growth was expected at the end of Day 3 for the control group.

Show Worked Solution

a. Examples of suitable responses included two of the following:

volume of E. coli first applied to plates
type of nutrient agar used
size of agar plates used
batch or source of ampicillin
duration of incubation
temperature of incubation
exposure time to each concentration of ampicillin.

b. Hypothesis:

If E. coli is exposed to low OR 1 µg/ml concentration of ampicillin, it will exhibit increased growth when exposed to higher levels of ampicillin.

♦♦♦ Mean mark (b) 40%.

c.i. Experiment analysis:

No ampicillin: More E. coli growth on Day 0.
2 µg/mL ampicillin: No effect on bacterial growth.
Days 2/3: Fewer colonies for both groups.
Higher ampicillin (4 µg/mL, 6 µg/mL): Significant effect, killed many E. coli.
More growth than predicted on all plates.
By end: More ampicillin-resistant bacteria on experimental plate.
1 µg/mL ampicillin: Initial exposure led to greater number of resistant E. coli.

♦♦♦ Mean mark (c)(i) 6%.

c.ii. If the experimental results did not support the predicted results:

E. coli was still present on both plates on Day 3.
This shows exposing E. coli to 1 µg/ml ampicillin does not increase its resistance to ampicillin.
Resistant E. coli grew regardless of initial exposure to ampicillin.

Any two of the following, if the experimental results did support the predicted results:

The number of colonies in both groups decreased over time.
More E. coli developed ampicillin resistance in the experimental group.
There were more colonies in the experimental group than in the control group on Day 3.
There were more colonies than expected as ampicillin concentration increased.
No growth was expected at the end of Day 3 for the control group.

♦♦ Mean mark (c)(ii) 50%.

CHEMISTRY, M1 EQ-Bank 19

Silicon $\ce{(Si)}$ is in Group 14 and Period 3 of the periodic table. Sodium $\ce{(Na)}$ is in Group 1 and Period 3. (6 marks)

Compare the properties of silicon and sodium with reference to their:

- Metallic character
- Electrical conductivity
- Ion formation

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Show Answers Only

Metallic character:

Sodium has much greater metallic character than silicon. Sodium is an alkali metal (Group 1) with typical metallic properties such as being shiny, malleable, and ductile.
Silicon is a metalloid with properties intermediate between metals and non-metals.

Electrical conductivity:

Sodium is an excellent electrical conductor, while silicon is a semiconductor. As a Group 1 metal, sodium has one valence electron that is delocalised in a metallic lattice, allowing it to move freely and conduct electricity very effectively.
Silicon, as a metalloid, has moderate electrical conductivity that increases with temperature – its conductivity is much lower than sodium’s but higher than non-metals.

Ion formation:

Sodium readily forms positive ions (cations) with a 1+ charge ($\ce{Na^+}$) by losing its single valence electron to achieve a stable electron configuration similar to neon. Silicon typically does not form simple ions due to its position as a metalloid.
Instead, silicon forms covalent bonds by sharing its four valence electrons with other atoms. While silicon can theoretically form $\ce{Si^4+}$ or $\ce{Si^4-}$ ions, the energy required to remove or add four electrons is too high.

Show Worked Solution

Metallic character:

Sodium has much greater metallic character than silicon. Sodium is an alkali metal (Group 1) with typical metallic properties such as being shiny, malleable, and ductile.
Silicon is a metalloid with properties intermediate between metals and non-metals.

Electrical conductivity:

Sodium is an excellent electrical conductor, while silicon is a semiconductor. As a Group 1 metal, sodium has one valence electron that is delocalised in a metallic lattice, allowing it to move freely and conduct electricity very effectively.
Silicon, as a metalloid, has moderate electrical conductivity that increases with temperature – its conductivity is much lower than sodium’s but higher than non-metals.

Ion formation:

Sodium readily forms positive ions (cations) with a 1+ charge ($\ce{Na^+}$) by losing its single valence electron to achieve a stable electron configuration similar to neon. Silicon typically does not form simple ions due to its position as a metalloid.
Instead, silicon forms covalent bonds by sharing its four valence electrons with other atoms. While silicon can theoretically form $\ce{Si^4+}$ or $\ce{Si^4-}$ ions, the energy required to remove or add four electrons is too high.

CHEMISTRY, M1 EQ-Bank 16

Hexane and water are liquids that are immiscible with each other. Some of their properties are shown in the table.

\begin{array} {|c|c|c|}
\hline & \text{Boiling point } (^{\circ}\text{C}) & \text{Density } (\text{g mL}^{-1})\\
\hline \text{Hexane} & 68.7 & 0.66 \\
\hline \text{Water} & 100 & 1.00 \\
\hline \end{array}

A chemist finds a bottle containing hexane and water and needs to determine whether she should use a separating funnel or distillation to separate the liquids.

Assess the effectiveness of each technique when separating hexane and water. (4 marks)

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Show Answers Only

Hexane and water are immiscible, so they do not mix and, instead, form separate layers.
As a result, they can be separated effectively with a separating funnel. The density of hexane is lower than the density of water, so the hexane will be the upper layer and the water will be the lower layer.
The liquids could also be separated effectively through distillation.
They have a large difference in boiling points. Hexane has a lower boiling point than water (68.7°C). As such, it will be the first fraction to be collected during distillation.

Show Worked Solution

Hexane and water are immiscible, so they do not mix and, instead, form separate layers.
As a result, they can be separated effectively with a separating funnel. The density of hexane is lower than the density of water, so the hexane will be the upper layer and the water will be the lower layer.
The liquids could also be separated effectively through distillation.
They have a large difference in boiling points. Hexane has a lower boiling point than water (68.7°C). As such, it will be the first fraction to be collected during distillation.

CHEMISTRY, M1 EQ-Bank 10 MC

Element $\ce{Y}$ forms the following ionic compound:

What is the correct spdf notation for $\ce{Y}$ in this compound?

$1s^2\,2s^2\,2p^6$
$1s^2\,2s^2\,2p^6\,3s^2$
$1s^2\,2s^2\,2p^6\,3s^2\,3p^1$
$1s^2\,2s^2\,2p^6\,3s^2\,3p^2$

Show Answers Only

$A$

Show Worked Solution

In $\ce{YCl2}$, each chlorine becomes $\ce{Cl-}$, therefore $\ce{Y}$ must form a $\ce{Y^2+}$.
The neutral atom that forms a 2+ ion is typically in group 2 (the alkaline earth metals), which have a valence configuration of $ns^2$.
The neutral atom’s configuration is $1s^2\,2s^2\,2p^6\,3s^2$. When it loses two eelctrons to form $\ce{Y^2+}$, those are removed from the outer $3s$ shell leaving: $\ce{Y^2+}: 1s^2\,2s^2\,2p^6$

$\Rightarrow A$

CHEMISTRY, M1 EQ-Bank 8 MC

Which of the alternatives below identifies the electron configuration of the cation and anion present in the compound magnesium oxide?

\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{Cation}\rule[-1ex]{0pt}{0pt}& \text{Anion} \\
\hline
\rule{0pt}{2.5ex} 1s^2\,2s^2\,2p^6\,3s^1\rule[-1ex]{0pt}{0pt}& 1s^2\,2s^2\,2p^6\\
\hline
\rule{0pt}{2.5ex} 1s^2\,2s^2\,2p^6\rule[-1ex]{0pt}{0pt}& 1s^2\,2s^2\,2p^6\\
\hline
\rule{0pt}{2.5ex} 1s^2\,2s^2\,2p^6\,3s^2\rule[-1ex]{0pt}{0pt}& 1s^2\,2s^2\,2p^4 \\
\hline
\rule{0pt}{2.5ex} 1s^2\,2s^2\,2p^6\rule[-1ex]{0pt}{0pt}& 1s^2\,2s^2\,2p^6\,3s^2\,3p^6 \\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

In its neutral state $\ce{Mg}$ has an electron configuration of $1s^2\,2s^2\,2p^6\,3s^2$.
In $\ce{MgO}$, magnesium forms $\ce{Mg^2+}$ by losing its two $3s$ electrons. Hence the magnesium ion has the electron configuration of $1s^2\,2s^2\,2p^6$.
In its neutral state $\ce{O}$ has an electron configuration of $1s^2\,2s^2\,2p^4$.
In $\ce{MgO}$, oxygen forms $\ce{O^2-}$ by gaining two electrons to complete the $2p$ subshell. Hence the oxygen ion has the electron configuration of $1s^2\,2s^2\,2p^6$.

$\Rightarrow B$

CHEMISTRY, M1 EQ-Bank 6

Describe the process by which emission line spectra are formed. (4 marks)

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Electrons in atoms exist in discrete energy levels.
When an atom absorbs energy (e.g. from heat or electricity), an electron is excited to a higher energy level.
The excited electron is unstable and will eventually fall back to a lower energy level.
As it does so, it releases energy in the form of a photon of light.
The energy of the photon corresponds to the difference between the two energy levels, so only specific wavelengths of light are emitted.
Passing this light through a spectroscope produces a series of discrete coloured lines known as the emission spectrum.

Show Worked Solution

Electrons in atoms exist in discrete energy levels.
When an atom absorbs energy (e.g. from heat or electricity), an electron is excited to a higher energy level.
The excited electron is unstable and will eventually fall back to a lower energy level.
As it does so, it releases energy in the form of a photon of light.
The energy of the photon corresponds to the difference between the two energy levels, so only specific wavelengths of light are emitted.
Passing this light through a spectroscope produces a series of discrete coloured lines known as the emission spectrum.

CHEMISTRY, M1 EQ-Bank 7 MC

What is the electron configuration of the element with atomic number 34?

$1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^4$
$1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^9\, 4p^5$
$1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^6\, 4p^2$
$1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 4p^{4}\, 4d^{10}$

Show Answers Only

$A$

Show Worked Solution

Electrons occupy orbitals in increasing energy order: `1s` → `2s` → `2p` → `3s` → `3p` → `4s` → `3d` → `4p`.
`s` orbitals can hold two electrons, `p` can hold 6 electrons and `d` orbitals can hold 10 electrons.
Filling the 34 electrons into these orbitals gives, $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^4$.

$\Rightarrow A$

CHEMISTRY, M1 EQ-Bank 14

Compare and explain the reactivity of Group 1 (alkali metals) and Group 2 (alkaline earth metals) with water. In your answer, link your explanation to electron configuration, atomic radius, and ionisation energy. (6 marks)

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Show Answers Only

Group 1 metals (e.g. $\ce{Li, Na, K}$) react vigorously with water to form a hydroxide and hydrogen gas.
Group 2 metals (e.g. $\ce{Mg, Ca}$) also react with water but much less vigorously, especially at the top of the group. For example, magnesium reacts only slowly with cold water.
Electron configuration: Group 1 metals have one valence electron, while Group 2 metals have two valence electrons. Losing one electron requires less energy than losing two, making Group 1 metals more reactive.
Atomic radius and ionisation energy: Down both groups, the atomic radius increases, shielding increases, and ionisation energy decreases. This means reactivity with water increases down the group.
Therefore: Reactivity increases down both groups, but Group 1 metals show higher reactivity with water compared with Group 2 metals in the same period.

Show Worked Solution

Group 1 metals (e.g. $\ce{Li, Na, K}$) react vigorously with water to form a hydroxide and hydrogen gas.
Group 2 metals (e.g. $\ce{Mg, Ca}$) also react with water but much less vigorously, especially at the top of the group. For example, magnesium reacts only slowly with cold water.
Electron configuration: Group 1 metals have one valence electron, while Group 2 metals have two valence electrons. Losing one electron requires less energy than losing two, making Group 1 metals more reactive.
Atomic radius and ionisation energy: Down both groups, the atomic radius increases, shielding increases, and ionisation energy decreases. This means reactivity with water increases down the group.
Therefore: Reactivity increases down both groups, but Group 1 metals show higher reactivity with water compared with Group 2 metals in the same period.

v1 Algebra, STD2 A4 2021 HSC 35

A toy store releases a limited edition LEGO set for $20 each. At this price, 3000 LEGO sets are sold each week and the revenue is `3000 xx 20=$60\ 000`.

The toy store considers increasing the price. For every dollar price increase, 15 fewer LEGO sets will be sold.

If the toy store charges `(20+x)` dollars for each LEGO set, a quadratic model for the revenue raised, `R`, from selling them is

`R=-15x^2+2700x+60\ 000`

What price should be charged per LEGO set to maximise the revenue? (2 marks)

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How many LEGO sets are sold when the revenue is maximised? (2 marks)

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Find the value of the intercept of the parabola with the vertical axis. (1 mark)

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Show Answers Only

a. `$110`

b. `1650`

c. `$60\ 000`

Show Worked Solution

a. `text{Highest revenue}\ (R_text{max})\ text(occurs halfway between)\ \ x= -20 and x=200.`

`text{Midpoint}\ =(-20 + 200)/2 = 90`

`:.\ text(Price of LEGO set for)\ R_text(max)`

`=90 + 20`

`=$110`

b. `text{LEGO sets sold when}\ R_{max}`

`=3000-(90 xx 15)`

`=1650`

c. `ytext(-intercept → find)\ R\ text(when)\ \ x=0:`

`R`	`= -15(0)^2 + 2700(0) + 60\ 000`
	`=$60\ 000`

CHEMISTRY, M1 EQ-Bank 14

Copper and copper $\text{(II)}$ oxide both conduct electricity in the molten state. However, copper also conducts electricity in the solid state, whereas copper $\text{(II)}$ oxide does not.

Explain the electrical conductivity of copper and copper $\text{(II)}$ oxide in terms of their structure and bonding. (4 marks)

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Show Answers Only

Copper $\text{(II)}$:

Copper is a metal with a giant metallic lattice structure. In both the solid and molten states, copper atoms are arranged in a lattice surrounded by a “sea of delocalised valence electrons”.
These delocalised electrons are free to move and carry charge, so copper conducts electricity in the solid state as well as when molten.

Copper $\text{(II)}$ oxide :

Copper $\text{(II)}$ oxide is an ionic compound composed of $\ce{Cu^2+}$ cations and $\ce{O^2-}$ anions arranged in a giant ionic lattice.
In the solid state, the ions are held in fixed positions by strong electrostatic forces and cannot move, so $\ce{CuO}$ does not conduct electricity.
In the molten state, the ionic lattice breaks apart and the ions become mobile, allowing them to act as charge carriers, so molten $\ce{CuO}$ does conduct electricity.

Show Worked Solution

Copper $\text{(II)}$:

Copper is a metal with a giant metallic lattice structure.
In both the solid and molten states, copper atoms are arranged in a lattice surrounded by a “sea of delocalised valence electrons”.
These delocalised electrons are free to move and carry charge, so copper conducts electricity in the solid state as well as when molten.

Copper $\text{(II)}$ oxide :

Copper $\text{(II)}$ oxide is an ionic compound composed of $\ce{Cu^2+}$ cations and $\ce{O^2-}$ anions arranged in a giant ionic lattice.
In the solid state, the ions are held in fixed positions by strong electrostatic forces and cannot move, so $\ce{CuO}$ does not conduct electricity.
In the molten state, the ionic lattice breaks apart and the ions become mobile, allowing them to act as charge carriers, so molten $\ce{CuO}$ does conduct electricity.

v1 Algebra, STD2 A4 2013 HSC 22 MC

Jevin wants to build a rectangular chicken pen. He has 32 metres of fencing and will use a barn wall as one side of the pen. The width of the pen is $d$ metres.

Which equation gives the area, $P$, of the chicken pen?

$P = 16d-d^2$
$P = 32d-d^2$
$P = 16d-\dfrac{d^2}{2}$
$P = 16d-2d^2$

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$C$

Show Worked Solution

♦♦♦ Mean mark 24% (lowest mean of any MC question in 2013 exam)

$\text{Length of pen}\ = \dfrac{1}{2}(32-d)$

$\text{Area}\ =d \times \dfrac{1}{2}(32-d)=16d-\dfrac{d^2}{2}$

$\Rightarrow C$

CHEMISTRY, M1 EQ-Bank 12

Draw a Lewis electron dot diagram for each of the following compounds:
i. $\ce{CO2}$ (1 mark)
ii. $\ce{MgCl2}$ (1 mark)

Using the two substances in (a) as examples, compare the bonding in ionic compounds with the bonding in covalent molecular compounds. (3 marks)

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The boiling point of $\ce{CO2}$ is –78°C, while the boiling point of $\ce{MgCl2}$ is 1412$^{\circ}$C. Account for the difference in boiling points between these two substances. (3 marks)

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a.i.

a.ii.

b. Bonding in ionic compounds vs covalent molecular compounds:

Ionic compounds are solids at STP and consist of positively and negatively charged ions held together in a lattice by strong electrostatic forces.
In magnesium chloride, doubly charged magnesium cations are surrounded by chloride anions in a ratio of 1:2.
Covalent molecular compounds are groups of atoms in which one or more pairs of electrons are shared between atoms.
In carbon dioxide, carbon forms double covalent bonds with two oxygen atoms, sharing its valence electrons. Carbon dioxide molecules are bound together only by relatively weak dispersion forces and therefore exist as a gas at STP.

c. Differences in boiling points:

$$\ce{CO2}$$ molecules are held together by weak dispersion forces between neutral molecules. These forces require little energy to overcome, resulting in a very low boiling point –78°C.
$\ce{MgCl2}$ has a giant ionic lattice. Strong electrostatic forces between $\ce{Mg^2+}$ and $\ce{Cl^-}$ ions extend throughout the solid. Large amounts of energy are required to break these ionic bonds, giving it a very high boiling point at 1412°C.
Therefore, the huge difference in boiling points is due to weak intermolecular forces in covalent molecular substances compared with strong ionic bonds in ionic compounds.

Show Worked Solution

a.i.

a.ii.

b. Bonding in ionic compounds vs covalent molecular compounds:

Ionic compounds are solids at STP and consist of positively and negatively charged ions held together in a lattice by strong electrostatic forces.
In magnesium chloride, doubly charged magnesium cations are surrounded by chloride anions in a ratio of 1:2.
Covalent molecular compounds are groups of atoms in which one or more pairs of electrons are shared between atoms.
In carbon dioxide, carbon forms double covalent bonds with two oxygen atoms, sharing its valence electrons. Carbon dioxide molecules are bound together only by relatively weak dispersion forces and therefore exist as a gas at STP.

c. Differences in boiling points:

$$\ce{CO2}$$ molecules are held together by weak dispersion forces between neutral molecules. These forces require little energy to overcome, resulting in a very low boiling point –78°C.
$\ce{MgCl2}$ has a giant ionic lattice. Strong electrostatic forces between $\ce{Mg^2+}$ and $\ce{Cl^-}$ ions extend throughout the solid. Large amounts of energy are required to break these ionic bonds, giving it a very high boiling point at 1412°C.
Therefore, the huge difference in boiling points is due to weak intermolecular forces in covalent molecular substances compared with strong ionic bonds in ionic compounds.

HMS, TIP EQ-Bank 419

Compare how program customisation approaches differ between recreational participants and elite athletes. (6 marks)

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Benefits

[P] Assessment-driven customisation significantly improves program effectiveness by targeting individual needs precisely and efficiently.
[E] Personalised programs address specific strengths and weaknesses identified through systematic testing rather than generic approaches that may miss individual requirements.
[Ev] A recreational client with poor cardiovascular fitness but strong upper body receives aerobic-focused training whilst maintaining strength levels through targeted exercises.
[L] This demonstrates how customisation maximises training efficiency and individual progress rates through focused interventions.
[P] Customised programs enhance participant motivation and adherence through relevant and achievable goal setting based on personal capabilities.
[E] Individual assessment data enables realistic target establishment that maintains engagement throughout extended training periods and prevents unrealistic expectations.
[Ev] Basketball players setting 5cm vertical jump improvements based on testing results show greater commitment than arbitrary goals without data foundation.
[L] Therefore, data-driven customisation creates meaningful progress markers that sustain long-term participation and program compliance.

Challenges

[P] Assessment-based customisation requires significant professional expertise and time investment from qualified fitness practitioners and specialists.
[E] Comprehensive testing, data analysis, and individualised program design demand extensive knowledge and planning resources that increase program costs.
[Ev] Detailed performance analysis for elite athletes involves multiple testing sessions and complex program modifications requiring specialised equipment and facilities.
[L] This highlights how customisation complexity can limit accessibility for some fitness professionals and clients with restricted budgets.

Show Worked Solution

Benefits

[P] Assessment-driven customisation significantly improves program effectiveness by targeting individual needs precisely and efficiently.
[E] Personalised programs address specific strengths and weaknesses identified through systematic testing rather than generic approaches that may miss individual requirements.
[Ev] A recreational client with poor cardiovascular fitness but strong upper body receives aerobic-focused training whilst maintaining strength levels through targeted exercises.
[L] This demonstrates how customisation maximises training efficiency and individual progress rates through focused interventions.
[P] Customised programs enhance participant motivation and adherence through relevant and achievable goal setting based on personal capabilities.
[E] Individual assessment data enables realistic target establishment that maintains engagement throughout extended training periods and prevents unrealistic expectations.
[Ev] Basketball players setting 5cm vertical jump improvements based on testing results show greater commitment than arbitrary goals without data foundation.
[L] Therefore, data-driven customisation creates meaningful progress markers that sustain long-term participation and program compliance.

Challenges

[P] Assessment-based customisation requires significant professional expertise and time investment from qualified fitness practitioners and specialists.
[E] Comprehensive testing, data analysis, and individualised program design demand extensive knowledge and planning resources that increase program costs.
[Ev] Detailed performance analysis for elite athletes involves multiple testing sessions and complex program modifications requiring specialised equipment and facilities.
[L] This highlights how customisation complexity can limit accessibility for some fitness professionals and clients with restricted budgets.

HMS, TIP EQ-Bank 413

Analyse the relationship between sport-specific assessment and training program effectiveness for elite athletes. (8 marks)

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Overview Statement

Sport-specific assessment and training program effectiveness demonstrate strong interconnected relationships in elite athlete development.
Assessment precision directly influences program targeting and subsequent performance improvements.

Component Relationship 1

Sport-specific assessments identify key performance indicators that directly relate to competitive success requirements in individual sports.
Testing protocols measure capabilities most relevant to performance outcomes rather than general fitness parameters.
This relationship becomes critical when kickboxers undergo leg strength testing to reveal bilateral imbalances affecting competition performance.
Assessment specificity connects to program effectiveness as targeted training addresses exact performance limitations.
Therefore, sport-specific testing determines the precision of subsequent training interventions and performance improvement potential.

Component Relationship 2

Training program effectiveness depends on assessment accuracy in identifying genuine performance-limiting factors versus minor weaknesses.
Precise measurement influences program focus allocation and resource distribution across different training components.
This relationship demonstrates how sprinters with identified acceleration deficits receive explosive training emphasis rather than generic speed development.
Program success interacts with assessment quality as accurate testing leads to targeted interventions.
Consequently, sport-specific assessment precision enables maximum training effectiveness through focused performance improvement strategies.

Implications and Synthesis

The relationship between sport-specific assessment and program effectiveness reveals that testing precision directly determines training success rates.
Assessment accuracy establishes program targeting quality while training specificity maximises performance improvement efficiency.
This pattern shows that elite athlete development requires integrated assessment and training approaches where testing precision guides intervention effectiveness.

Show Worked Solution

Overview Statement

Sport-specific assessment and training program effectiveness demonstrate strong interconnected relationships in elite athlete development.
Assessment precision directly influences program targeting and subsequent performance improvements.

Component Relationship 1

Sport-specific assessments identify key performance indicators that directly relate to competitive success requirements in individual sports.
Testing protocols measure capabilities most relevant to performance outcomes rather than general fitness parameters.
This relationship becomes critical when kickboxers undergo leg strength testing to reveal bilateral imbalances affecting competition performance.
Assessment specificity connects to program effectiveness as targeted training addresses exact performance limitations.
Therefore, sport-specific testing determines the precision of subsequent training interventions and performance improvement potential.

Component Relationship 2

Training program effectiveness depends on assessment accuracy in identifying genuine performance-limiting factors versus minor weaknesses.
Precise measurement influences program focus allocation and resource distribution across different training components.
This relationship demonstrates how sprinters with identified acceleration deficits receive explosive training emphasis rather than generic speed development.
Program success interacts with assessment quality as accurate testing leads to targeted interventions.
Consequently, sport-specific assessment precision enables maximum training effectiveness through focused performance improvement strategies.

Implications and Synthesis

The relationship between sport-specific assessment and program effectiveness reveals that testing precision directly determines training success rates.
Assessment accuracy establishes program targeting quality while training specificity maximises performance improvement efficiency.
This pattern shows that elite athlete development requires integrated assessment and training approaches where testing precision guides intervention effectiveness.

HMS, TIP EQ-Bank 410 MC

A 10-kilometre runner completes monthly time trials throughout their training program. After three months of consistent training, the results show no improvement in performance times. This scenario best demonstrates which aspect of assessment-based program development for elite athletes?

The athlete requires more intensive training volume to achieve performance breakthroughs in endurance events
Monthly testing frequency is insufficient for tracking meaningful performance changes in distance running
Continuous monitoring enables identification of training ineffectiveness and indicates need for program adjustments
Performance plateaus are natural indicators that athletes have reached their genetic potential limitations

Show Answers Only

$C$

Show Worked Solution

C is correct: Continuous performance monitoring reveals when current training approaches aren’t producing desired results, requiring systematic program modifications.

Other Options:

A is incorrect: While volume may need adjustment, the key principle is using monitoring data to identify ineffective approaches.
B is incorrect: Monthly frequency is typically adequate for endurance performance tracking rather than being the primary limitation.
D is incorrect: Plateaus often indicate training issues rather than genetic limitations, especially over short three-month periods.

HMS, TIP EQ-Bank 407

How does regular progress monitoring through reassessment enhance program effectiveness for recreational participants? (5 marks)

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Regular reassessment provides objective data about fitness improvements and program effectiveness over extended time periods. This enables trainers to identify successful training elements and strategies. Therefore, proven approaches can be maintained whilst ineffective methods are systematically modified.
Progress tracking demonstrates tangible results that motivate recreational participants to continue their fitness journey enthusiastically. Visible improvements create positive reinforcement for sustained participation and engagement. Consequently, participants develop stronger commitment to long-term fitness goals.
Monitoring data reveals when training adaptations require program adjustments to prevent plateaus and maintain steady progression. Assessment results indicate optimal timing for intensity increases or exercise modifications. This leads to continuous improvement rather than stagnated performance outcomes.
Reassessment identifies potential overtraining or excessive fatigue requiring immediate program modifications for participant safety. Early detection prevents injury development and burnout issues effectively. As a result, participants maintain consistent training without setbacks or extended recovery periods.

Show Worked Solution

Regular reassessment provides objective data about fitness improvements and program effectiveness over extended time periods. This enables trainers to identify successful training elements and strategies. Therefore, proven approaches can be maintained whilst ineffective methods are systematically modified.
Progress tracking demonstrates tangible results that motivate recreational participants to continue their fitness journey enthusiastically. Visible improvements create positive reinforcement for sustained participation and engagement. Consequently, participants develop stronger commitment to long-term fitness goals.
Monitoring data reveals when training adaptations require program adjustments to prevent plateaus and maintain steady progression. Assessment results indicate optimal timing for intensity increases or exercise modifications. This leads to continuous improvement rather than stagnated performance outcomes.
Reassessment identifies potential overtraining or excessive fatigue requiring immediate program modifications for participant safety. Early detection prevents injury development and burnout issues effectively. As a result, participants maintain consistent training without setbacks or extended recovery periods.

HMS, TIP EQ-Bank 097

To what extent has GPS tracking technology revolutionised tactical analysis and performance improvement in team sports? (6 marks)

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Judgment Statement

GPS tracking technology has significantly revolutionised tactical analysis and performance improvement in team sports.
This is due to its role in workload monitoring, tactical decision-making and injury prevention.

Workload and Tactical Insights

GPS provides data on distance covered, speed, acceleration and positional heat maps.
Evidence supporting this includes the growing use of GPS technology by rugby coaches to monitor player fatigue and adjust substitutions accordingly.
Soccer teams use GPS to map player positioning, ensuring tactical discipline and improving defensive structures.
This shows GPS has transformed coaching decisions by replacing guesswork with objective data.
As a result, both tactical outcomes and player performance are improved.

Limitations in Real-Time Application

However, GPS is less useful in real-time decision-making during matches.
Data often needs post-game analysis which limits immediate tactical adjustments.
Cost can also restrict access for smaller teams.
Despite this, GPS remains a key driver of performance improvement because post-game insights refine future tactics and reduce injury risk.

Reaffirmation

In summary, GPS has largely revolutionised tactical analysis and performance improvement.
Its data-driven insights create stronger tactical planning and safer workload management.
While there are practical limits, its impact on modern team sports, particularly at the well resourced elite level, is profound.

Show Worked Solution

Judgment Statement

GPS tracking technology has significantly revolutionised tactical analysis and performance improvement in team sports.
This is due to its role in workload monitoring, tactical decision-making and injury prevention.

Workload and Tactical Insights

GPS provides data on distance covered, speed, acceleration and positional heat maps.
Evidence supporting this includes the growing use of GPS technology by rugby coaches to monitor player fatigue and adjust substitutions accordingly.
Soccer teams use GPS to map player positioning, ensuring tactical discipline and improving defensive structures.
This shows GPS has transformed coaching decisions by replacing guesswork with objective data.
As a result, both tactical outcomes and player performance are improved.

Limitations in Real-Time Application

However, GPS is less useful in real-time decision-making during matches.
Data often needs post-game analysis which limits immediate tactical adjustments.
Cost can also restrict access for smaller teams.
Despite this, GPS remains a key driver of performance improvement because post-game insights refine future tactics and reduce injury risk.

Reaffirmation

In summary, GPS has largely revolutionised tactical analysis and performance improvement.
Its data-driven insights create stronger tactical planning and safer workload management.
While there are practical limits, its impact on modern team sports, particularly at the well resourced elite level, is profound.

HMS, TIP EQ-Bank 401

Analyse how the relationship between health screening and performance testing contributes to effective exercise program development. (8 marks)

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Overview Statement

Health screening and performance testing create complementary assessment components that provide comprehensive information for safe and effective program development.
Their relationship ensures both safety considerations and performance objectives are addressed.

Component Relationship 1

Health screening identifies potential medical risks that must be considered before performance testing protocols.
Risk factor identification influences test selection as medical conditions require modified assessment approaches.
This relationship becomes critical when cardiovascular concerns require submaximal rather than maximal testing.
Pre-exercise questionnaires connect to safe testing by revealing conditions like hypertension.
Therefore, health screening determines appropriate performance testing boundaries for individual clients.

Component Relationship 2

Performance testing results interact with health screening data to create comprehensive fitness profiles for program design.
Baseline measurements depend on health status interpretation to ensure appropriate training intensity prescription.
This relationship demonstrates how previous knee injury would receive modified strength assessments leading to rehabilitation exercises.
Health limitations influence performance expectations while fitness capabilities inform training loads.
Consequently, combined data enables personalised program development addressing health maintenance and performance improvement.

Implications and Synthesis

The relationship between health screening and performance testing reveals that effective program development requires both safety-focused and performance-oriented approaches.
Health screening establishes safe parameters while performance testing provides capability measurements.
This pattern shows that comprehensive assessment integrates medical safety with fitness evaluation for individually tailored exercise programs.

Show Worked Solution

Overview Statement

Health screening and performance testing create complementary assessment components that provide comprehensive information for safe and effective program development.
Their relationship ensures both safety considerations and performance objectives are addressed.

Component Relationship 1

Health screening identifies potential medical risks that must be considered before performance testing protocols.
Risk factor identification influences test selection as medical conditions require modified assessment approaches.
This relationship becomes critical when cardiovascular concerns require submaximal rather than maximal testing.
Pre-exercise questionnaires connect to safe testing by revealing conditions like hypertension.
Therefore, health screening determines appropriate performance testing boundaries for individual clients.

Component Relationship 2

Performance testing results interact with health screening data to create comprehensive fitness profiles for program design.
Baseline measurements depend on health status interpretation to ensure appropriate training intensity prescription.
This relationship demonstrates how previous knee injury would receive modified strength assessments leading to rehabilitation exercises.
Health limitations influence performance expectations while fitness capabilities inform training loads.
Consequently, combined data enables personalised program development addressing health maintenance and performance improvement.

Implications and Synthesis

The relationship between health screening and performance testing reveals that effective program development requires both safety-focused and performance-oriented approaches.
Health screening establishes safe parameters while performance testing provides capability measurements.
This pattern shows that comprehensive assessment integrates medical safety with fitness evaluation for individually tailored exercise programs.

HMS, TIP EQ-Bank 090

To what extent are psychological recovery strategies as important as physiological strategies in achieving optimal athlete recovery? (6 marks)

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Judgment Statement

Psychological strategies are important to a large extent, though slightly secondary to physiological methods in short-term recovery.
Justification rests on three factors: sustaining long-term performance, preventing burnout and complementing physical repair.

Long-Term Psychological Value

Evidence supporting this includes the impact of strategies like mindfulness, meditation and progressive muscle relaxation.
These methods reduce cortisol, enhance emotional regulation and improve sleep quality.
Elite athletes such as Simone Biles and Naomi Osaka highlight the need for mental recovery breaks.
Sustained psychological wellbeing ensures resilience, sharper focus and consistent performance over an entire season.
This shows psychological recovery underpins both performance readiness and career longevity.

Immediate Physiological Necessity

However, physiological strategies often deliver the most immediate benefits.
Cold water immersion reduces inflammation and DOMS, while cool-downs aid waste removal and circulation.
Without these processes, athletes would experience stiffness, cramps and compromised training capacity.
Despite this, physiological recovery alone cannot maintain long-term focus, confidence or resilience, reinforcing the value of psychological methods.

Reaffirmation

Psychological recovery is important to a large extent, complementing but not replacing physiological recovery.
Athletes achieve optimal outcomes when both strategies are integrated.
The implication is that effective recovery programs must deliberately balance body repair with psychological resilience to sustain peak performance.

Show Worked Solution

Judgment Statement

Psychological strategies are important to a large extent, though slightly secondary to physiological methods in short-term recovery.
Justification rests on three factors: sustaining long-term performance, preventing burnout and complementing physical repair.

Long-Term Psychological Value

Evidence supporting this includes the impact of strategies like mindfulness, meditation and progressive muscle relaxation.
These methods reduce cortisol, enhance emotional regulation and improve sleep quality.
Elite athletes such as Simone Biles and Naomi Osaka highlight the need for mental recovery breaks.
Sustained psychological wellbeing ensures resilience, sharper focus and consistent performance over an entire season.
This shows psychological recovery underpins both performance readiness and career longevity.

Immediate Physiological Necessity

However, physiological strategies often deliver the most immediate benefits.
Cold water immersion reduces inflammation and DOMS, while cool-downs aid waste removal and circulation.
Without these processes, athletes would experience stiffness, cramps and compromised training capacity.
Despite this, physiological recovery alone cannot maintain long-term focus, confidence or resilience, reinforcing the value of psychological methods.

Reaffirmation

Psychological recovery is important to a large extent, complementing but not replacing physiological recovery.
Athletes achieve optimal outcomes when both strategies are integrated.
The implication is that effective recovery programs must deliberately balance body repair with psychological resilience to sustain peak performance.

HMS, TIP EQ-Bank 395

Analyse how effective strategy implementation depends on the relationship between individual player roles and overall team objectives in group sports. (8 marks)

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Overview Statement

Effective strategy implementation in group sports requires clear relationships between individual player responsibilities and collective team goals.
Success depends on how well individual roles integrate with broader strategic objectives during competition.

Component Relationship 1

Individual role clarity connects to team strategic success when players understand their specific jobs and responsibilities clearly.
Each player must be aware of their specific function within the overall tactical framework, creating systematic performance execution.
This relationship becomes critical when basketball players execute set plays requiring coordinated movements and timing precision.
Position-specific training influences strategic effectiveness as players develop specialised skills that contribute to team objectives.
Therefore, individual competence determines collective strategic implementation success through coordinated role execution.

Component Relationship 2

Team communication systems interact with individual decision-making processes to create unified strategic responses during competition pressure.
Strategic objectives depend on individual players adapting their roles based on changing game circumstances while maintaining team coherence.
This relationship demonstrates how football teams modify defensive formations where individual positional adjustments serve collective tactical purposes.
Player understanding of role flexibility enables strategic adaptation without losing team coordination.
Consequently, effective implementation requires balance between individual autonomy and collective strategic discipline.

Implications and Synthesis

The relationship between individual roles and team objectives reveals that strategic success depends on systematic integration rather than isolated performance.
Individual excellence must align with collective goals to achieve effective implementation.
This pattern shows that successful group sport strategies require both personal accountability and team coordination, demonstrating how individual and collective elements work together for optimal strategic execution.

Show Worked Solution

Overview Statement

Effective strategy implementation in group sports requires clear relationships between individual player responsibilities and collective team goals.
Success depends on how well individual roles integrate with broader strategic objectives during competition.

Component Relationship 1

Individual role clarity connects to team strategic success when players understand their specific jobs and responsibilities clearly.
Each player must be aware of their specific function within the overall tactical framework, creating systematic performance execution.
This relationship becomes critical when basketball players execute set plays requiring coordinated movements and timing precision.
Position-specific training influences strategic effectiveness as players develop specialised skills that contribute to team objectives.
Therefore, individual competence determines collective strategic implementation success through coordinated role execution.

Component Relationship 2

Team communication systems interact with individual decision-making processes to create unified strategic responses during competition pressure.
Strategic objectives depend on individual players adapting their roles based on changing game circumstances while maintaining team coherence.
This relationship demonstrates how football teams modify defensive formations where individual positional adjustments serve collective tactical purposes.
Player understanding of role flexibility enables strategic adaptation without losing team coordination.
Consequently, effective implementation requires balance between individual autonomy and collective strategic discipline.

Implications and Synthesis

The relationship between individual roles and team objectives reveals that strategic success depends on systematic integration rather than isolated performance.
Individual excellence must align with collective goals to achieve effective implementation.
This pattern shows that successful group sport strategies require both personal accountability and team coordination, demonstrating how individual and collective elements work together for optimal strategic execution.

HMS, TIP EQ-Bank 391

Evaluate the effectiveness of small-sided games versus traditional drills for developing tactical awareness in team sports. (6 marks)

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Evaluation Statement

Small-sided games are more effective than traditional drills for developing tactical awareness due to their realistic pressure situations and enhanced decision-making requirements.

Game-like Conditions

Small-sided games provide performance-like pressure that closely mimics actual competition environments.
Players experience realistic timing constraints and spatial challenges that develop genuine tactical awareness.
Evidence shows athletes transfer skills more effectively from small-sided games to full competition compared to isolated drill practice.

Decision-making Development

Traditional drills focus on repetitive technical execution without tactical pressure or decision-making requirements.
Small-sided games force continuous tactical decisions under competitive stress, developing adaptive thinking skills.
However, traditional drills provide greater repetition volume for technical skill refinement without tactical complexity.

Final Evaluation

While traditional drills offer technical skill development benefits, small-sided games demonstrate superior effectiveness for tactical awareness development.
The realistic pressure and decision-making requirements create more transferable tactical skills for competitive performance.
Small-sided games should form the primary tactical development method with traditional drills supporting technical skill refinement.

Show Worked Solution

Evaluation Statement

Small-sided games are more effective than traditional drills for developing tactical awareness due to their realistic pressure situations and enhanced decision-making requirements.

Game-like Conditions

Small-sided games provide performance-like pressure that closely mimics actual competition environments.
Players experience realistic timing constraints and spatial challenges that develop genuine tactical awareness.
Evidence shows athletes transfer skills more effectively from small-sided games to full competition compared to isolated drill practice.

Decision-making Development

Traditional drills focus on repetitive technical execution without tactical pressure or decision-making requirements.
Small-sided games force continuous tactical decisions under competitive stress, developing adaptive thinking skills.
However, traditional drills provide greater repetition volume for technical skill refinement without tactical complexity.

Final Evaluation

While traditional drills offer technical skill development benefits, small-sided games demonstrate superior effectiveness for tactical awareness development.
The realistic pressure and decision-making requirements create more transferable tactical skills for competitive performance.
Small-sided games should form the primary tactical development method with traditional drills supporting technical skill refinement.

HMS, TIP EQ-Bank 379

Analyse how environmental conditions influence strategic decision-making differently in individual and group sports. (8 marks)

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Overview Statement

Environmental conditions create distinct strategic challenges for individual and group sports due to different support systems and coordination requirements.
Decision-making processes vary significantly based on athlete autonomy levels and team dynamics during environmental adaptation.

Component Relationship 1

Individual sports emphasise complete personal responsibility for environmental adaptation without external guidance during competition.
Weather conditions directly influence athlete decision-making as they must assess and respond independently to changing circumstances.
This relationship becomes critical when cyclists adjust cornering speeds in rain or runners modify pacing strategies in wind.
Personal experience and preparation connect to real-time environmental assessment, enabling immediate tactical modifications.
Therefore, individual athletes develop comprehensive environmental awareness skills that determine their competitive success in adverse conditions.

Component Relationship 2

Group sports integrate collective environmental responses through coordinated team communication and shared decision-making processes.
Environmental factors interact with team dynamics as coaches and players collaborate to modify strategies during competition.
This influences tactical implementation as rugby teams collectively reduce passing games in wet conditions while maintaining forward dominance.
Team coordination depends on unified environmental assessment and synchronised tactical adjustments across all players.
Consequently, group sports require systematic communication protocols that enable rapid strategic modifications while maintaining team cohesion under environmental pressure.

Implications and Synthesis

The relationship between environmental adaptation and sport structure reveals that individual sports prioritise personal autonomy while group sports emphasise collective coordination.
Individual athletes develop complete self-reliance for environmental decision-making whereas team sports create shared responsibility systems for strategic adaptation.
Thus demonstrating how sport structure fundamentally shapes environmental response strategies.

Show Worked Solution

Overview Statement

Environmental conditions create distinct strategic challenges for individual and group sports due to different support systems and coordination requirements.
Decision-making processes vary significantly based on athlete autonomy levels and team dynamics during environmental adaptation.

Component Relationship 1

Individual sports emphasise complete personal responsibility for environmental adaptation without external guidance during competition.
Weather conditions directly influence athlete decision-making as they must assess and respond independently to changing circumstances.
This relationship becomes critical when cyclists adjust cornering speeds in rain or runners modify pacing strategies in wind.
Personal experience and preparation connect to real-time environmental assessment, enabling immediate tactical modifications.
Therefore, individual athletes develop comprehensive environmental awareness skills that determine their competitive success in adverse conditions.

Component Relationship 2

Group sports integrate collective environmental responses through coordinated team communication and shared decision-making processes.
Environmental factors interact with team dynamics as coaches and players collaborate to modify strategies during competition.
This influences tactical implementation as rugby teams collectively reduce passing games in wet conditions while maintaining forward dominance.
Team coordination depends on unified environmental assessment and synchronised tactical adjustments across all players.
Consequently, group sports require systematic communication protocols that enable rapid strategic modifications while maintaining team cohesion under environmental pressure.

Implications and Synthesis

The relationship between environmental adaptation and sport structure reveals that individual sports prioritise personal autonomy while group sports emphasise collective coordination.
Individual athletes develop complete self-reliance for environmental decision-making whereas team sports create shared responsibility systems for strategic adaptation.
Thus demonstrating how sport structure fundamentally shapes environmental response strategies.

HMS, TIP EQ-Bank 373

Analyse the relationship between psychological factors and athletic performance in different competitive environments. (8 marks)

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Overview Statement

Psychological factors demonstrate complex relationships with athletic performance that vary significantly across different competitive environments.
Performance outcomes depend on the interaction between mental preparation, environmental pressures and individual athlete characteristics.

Component Relationship 1

Concentration skills interact with competitive environment demands to determine performance quality during crucial moments.
High-pressure situations influence attention control as athletes must focus on task execution rather than external distractions.
This relationship becomes critical in individual sports where athletes compete without teammate support or coaching guidance.
Tennis players demonstrate how concentration training connects to improved performance under crowd pressure and media attention.
Therefore, mental focus strategies enable athletes to maintain technical proficiency despite environmental stressors and competitive intensity.

Component Relationship 2

Arousal management relates to environmental factors as different competitive settings require varying optimal performance states.
Team sports create collective psychological dynamics that influence individual athlete arousal through shared energy and communication patterns.
This interacts with performance outcomes as basketball teams benefit from coordinated mental preparation before high-stakes matches.
Confidence levels connect to past competitive experiences in similar environments, affecting current performance expectations.
Consequently, successful psychological adaptation depends on matching mental strategies to specific competitive contexts and environmental demands.

Implications and Synthesis

The relationship between psychological preparation and competitive environment determines optimal performance strategy selection for different athletic contexts.
Individual sports require complete mental self-reliance while team environments enable collective psychological support systems.
This pattern reveals that effective performance psychology must consider both internal mental factors and external competitive pressures for maximum athletic success.

Show Worked Solution

Overview Statement

Psychological factors demonstrate complex relationships with athletic performance that vary significantly across different competitive environments.
Performance outcomes depend on the interaction between mental preparation, environmental pressures and individual athlete characteristics.

Component Relationship 1

Concentration skills interact with competitive environment demands to determine performance quality during crucial moments.
High-pressure situations influence attention control as athletes must focus on task execution rather than external distractions.
This relationship becomes critical in individual sports where athletes compete without teammate support or coaching guidance.
Tennis players demonstrate how concentration training connects to improved performance under crowd pressure and media attention.
Therefore, mental focus strategies enable athletes to maintain technical proficiency despite environmental stressors and competitive intensity.

Component Relationship 2

Arousal management relates to environmental factors as different competitive settings require varying optimal performance states.
Team sports create collective psychological dynamics that influence individual athlete arousal through shared energy and communication patterns.
This interacts with performance outcomes as basketball teams benefit from coordinated mental preparation before high-stakes matches.
Confidence levels connect to past competitive experiences in similar environments, affecting current performance expectations.
Consequently, successful psychological adaptation depends on matching mental strategies to specific competitive contexts and environmental demands.

Implications and Synthesis

The relationship between psychological preparation and competitive environment determines optimal performance strategy selection for different athletic contexts.
Individual sports require complete mental self-reliance while team environments enable collective psychological support systems.
This pattern reveals that effective performance psychology must consider both internal mental factors and external competitive pressures for maximum athletic success.

HMS, TIP EQ-Bank 369

To what extent can anxiety management techniques help athletes overcome performance challenges in competitive situations? (5 marks)

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Judgment Statement

Anxiety management techniques can significantly help athletes overcome performance challenges, though effectiveness varies based on individual factors and technique selection.

Evidence Supporting Effectiveness

Relaxation techniques demonstrate strong success in controlling physical anxiety symptoms including muscle tension and rapid heart rate. Mental rehearsal shows proven benefits through repeated visualisation of successful competitive outcomes.
Athletes using these techniques report improved confidence and reduced performance anxiety during high-pressure situations.

Limitations and Qualifications

Individual differences affect technique effectiveness as some athletes respond better to physical relaxation while others benefit from mental approaches. Severe anxiety disorders may require professional intervention beyond basic sport psychology techniques.
Technique effectiveness depends on consistent practice and proper implementation during training phases.

Overall Assessment

Evidence indicates anxiety management techniques provide considerable benefits for most athletes in overcoming competitive performance challenges.
While not universally effective for all individuals or anxiety levels, properly applied techniques significantly improve performance outcomes.

Show Worked Solution

Judgment Statement

Anxiety management techniques can significantly help athletes overcome performance challenges, though effectiveness varies based on individual factors and technique selection.

Evidence Supporting Effectiveness

Relaxation techniques demonstrate strong success in controlling physical anxiety symptoms including muscle tension and rapid heart rate. Mental rehearsal shows proven benefits through repeated visualisation of successful competitive outcomes.
Athletes using these techniques report improved confidence and reduced performance anxiety during high-pressure situations.

Limitations and Qualifications

Individual differences affect technique effectiveness as some athletes respond better to physical relaxation while others benefit from mental approaches. Severe anxiety disorders may require professional intervention beyond basic sport psychology techniques.
Technique effectiveness depends on consistent practice and proper implementation during training phases.

Overall Assessment

Evidence indicates anxiety management techniques provide considerable benefits for most athletes in overcoming competitive performance challenges.
While not universally effective for all individuals or anxiety levels, properly applied techniques significantly improve performance outcomes.

HMS, TIP EQ-Bank 365

Analyse how stress management techniques can be applied effectively in individual and group sports to improve performance. (8 marks)

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Overview Statement

Stress management techniques require different applications in individual and group sports due to varying support systems and performance contexts.
Effective implementation depends on understanding sport-specific stressor patterns and athlete needs.

Component Relationship 1

Individual sports emphasise personal stress management strategies that athletes can implement independently during competition.
Mental rehearsal connects to performance improvement by allowing athletes to practice stress responses in visualised scenarios.
This relationship enables tennis players and swimmers to develop personalised coping mechanisms for high-pressure moments.
Progressive muscle relaxation interacts with individual preparation routines, helping athletes control physical tension before events.
Therefore, self-directed techniques become essential for managing isolation stress inherent in individual competition formats.

Component Relationship 2

Group sports integrate collective stress management approaches with individual athlete needs to create team cohesion.
Team meditation sessions influence group arousal levels by synchronising mental preparation across all players.
This connects to improved performance through unified focus and reduced collective anxiety during crucial match moments.
Goal setting strategies interact with both individual and team objectives, creating shared stress management frameworks.
Consequently, basketball and football teams benefit from coordinated approaches that address both personal and collective pressure sources.

Implications and Synthesis

The relationship between individual autonomy and team support determines optimal stress management strategy selection in different sporting contexts.
Individual sports enable complete personalisation of techniques while group sports require balance between personal needs and collective team dynamics for maximum effectiveness.

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Overview Statement

Stress management techniques require different applications in individual and group sports due to varying support systems and performance contexts.
Effective implementation depends on understanding sport-specific stressor patterns and athlete needs.

Component Relationship 1

Individual sports emphasise personal stress management strategies that athletes can implement independently during competition.
Mental rehearsal connects to performance improvement by allowing athletes to practice stress responses in visualised scenarios.
This relationship enables tennis players and swimmers to develop personalised coping mechanisms for high-pressure moments.
Progressive muscle relaxation interacts with individual preparation routines, helping athletes control physical tension before events.
Therefore, self-directed techniques become essential for managing isolation stress inherent in individual competition formats.

Component Relationship 2

Group sports integrate collective stress management approaches with individual athlete needs to create team cohesion.
Team meditation sessions influence group arousal levels by synchronising mental preparation across all players.
This connects to improved performance through unified focus and reduced collective anxiety during crucial match moments.
Goal setting strategies interact with both individual and team objectives, creating shared stress management frameworks.
Consequently, basketball and football teams benefit from coordinated approaches that address both personal and collective pressure sources.

Implications and Synthesis

The relationship between individual autonomy and team support determines optimal stress management strategy selection in different sporting contexts.
Individual sports enable complete personalisation of techniques while group sports require balance between personal needs and collective team dynamics for maximum effectiveness.

HMS, TIP EQ-Bank 359

Discuss the benefits and challenges of optimising arousal levels for athletes in individual and group sports. (5 marks)

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Benefits

[P] Optimal arousal levels enhance athletic performance through improved focus and concentration.
[E] Athletes who achieve their ideal arousal zone demonstrate better skill execution and decision-making abilities.
[Ev] Tennis players maintaining moderate arousal levels show increased serve accuracy compared to over-aroused competitors.
[L] This demonstrates how proper arousal management directly improves competitive outcomes.
[P] Team arousal strategies help all players reach the same mental readiness level.
[E] Group sports work better when players prepare mentally together to match team objectives.
[Ev] Basketball teams using shared warm-up routines achieve better on-court communication and tactical execution.
[L] Therefore, arousal optimisation enhances both individual and team performance simultaneously.

Challenges

[P] Individual differences make arousal optimisation complex to achieve consistently across all athletes.
[E] Personal factors like experience, personality and skill level affect optimal arousal requirements significantly.
[Ev] Novice athletes often struggle with over-arousal while experienced competitors may need higher stimulation levels.
[L] This highlights the difficulty of developing universal arousal management strategies for diverse athletic populations.

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Benefits

[P] Optimal arousal levels enhance athletic performance through improved focus and concentration.
[E] Athletes who achieve their ideal arousal zone demonstrate better skill execution and decision-making abilities.
[Ev] Tennis players maintaining moderate arousal levels show increased serve accuracy compared to over-aroused competitors.
[L] This demonstrates how proper arousal management directly improves competitive outcomes.
[P] Team arousal strategies help all players reach the same mental readiness level.
[E] Group sports work better when players prepare mentally together to match team objectives.
[Ev] Basketball teams using shared warm-up routines achieve better on-court communication and tactical execution.
[L] Therefore, arousal optimisation enhances both individual and team performance simultaneously.

Challenges

[P] Individual differences make arousal optimisation complex to achieve consistently across all athletes.
[E] Personal factors like experience, personality and skill level affect optimal arousal requirements significantly.
[Ev] Novice athletes often struggle with over-arousal while experienced competitors may need higher stimulation levels.
[L] This highlights the difficulty of developing universal arousal management strategies for diverse athletic populations.

HMS, TIP EQ-Bank 355

Analyse how psychological strategies can be applied differently in individual and group sports to improve performance. (8 marks)

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Overview Statement

Individual and group sports require different psychological strategy applications due to different performance environments and athlete responsibilities.
Competitive pressures affect mental preparation differently in these sporting contexts.

Component Relationship 1

Individual sports emphasise personal mental preparation and complete self-reliance strategies during competition situations.
Athletes utilise mental rehearsal for technical skill visualisation and personalised goal setting approaches tailored to their specific needs.
This connects to performance improvement through enhanced confidence levels and reduced pre-competition anxiety.
Tennis players develop concentration skills to maintain focus during lengthy matches without external coaching support or team encouragement.
Marathon runners apply self-hypnosis techniques to manage pain and maintain motivation during challenging race sections when competing alone.

Component Relationship 2

Group sports integrate team-based psychological approaches with individual mental preparation needs and collective team goals.
Teams apply group relaxation sessions before major competitions to build team unity and shared focus among all players.
This interacts with individual strategies as players must balance personal mental state management with team dynamics and communication needs.
Basketball teams use group goal setting to align individual objectives with team performance targets.
Football teams implement team meditation sessions to coordinate mental preparation and reduce group anxiety levels before important matches.

Implications and Synthesis

The relationship between individual focus and team dynamics determines the best strategy selection in different sporting contexts.
Individual sports enable complete personalisation of psychological techniques allowing athletes to address specific mental challenges.
Group sports require coordination between personal and team mental preparation approaches for optimal team performance outcomes and effective teamwork.

Show Worked Solution

Overview Statement

Individual and group sports require different psychological strategy applications due to different performance environments and athlete responsibilities.
Competitive pressures affect mental preparation differently in these sporting contexts.

Component Relationship 1

Individual sports emphasise personal mental preparation and complete self-reliance strategies during competition situations.
Athletes utilise mental rehearsal for technical skill visualisation and personalised goal setting approaches tailored to their specific needs.
This connects to performance improvement through enhanced confidence levels and reduced pre-competition anxiety.
Tennis players develop concentration skills to maintain focus during lengthy matches without external coaching support or team encouragement.
Marathon runners apply self-hypnosis techniques to manage pain and maintain motivation during challenging race sections when competing alone.

Component Relationship 2

Group sports integrate team-based psychological approaches with individual mental preparation needs and collective team goals.
Teams apply group relaxation sessions before major competitions to build team unity and shared focus among all players.
This interacts with individual strategies as players must balance personal mental state management with team dynamics and communication needs.
Basketball teams use group goal setting to align individual objectives with team performance targets.
Football teams implement team meditation sessions to coordinate mental preparation and reduce group anxiety levels before important matches.

Implications and Synthesis

The relationship between individual focus and team dynamics determines the best strategy selection in different sporting contexts.
Individual sports enable complete personalisation of psychological techniques allowing athletes to address specific mental challenges.
Group sports require coordination between personal and team mental preparation approaches for optimal team performance outcomes and effective teamwork.

HMS, TIP EQ-Bank 350

Compare the sport-specific fitness components and skill requirements for a marathon runner and a basketball player. (6 marks)

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Similarities

Both athletes require cardiovascular fitness for sustained performance during their respective competitions. Marathon runners and basketball players need efficient heart and lung function.
Mental resilience and concentration skills are essential for both sports under competitive pressure. Athletes must maintain focus during physical stress and make strategic decisions.
Both require sport-specific technical skills that must be practised repeatedly for mastery. Proper biomechanics and movement efficiency are crucial for optimal performance outcomes.
Training periodisation is important for both athletes to peak at competition times. Both sports require structured preparation phases and recovery periods for adaptation.

Differences

Marathon runners require exceptional aerobic capacity for prolonged 42-kilometre efforts over two hours. Basketball players need anaerobic power for explosive jumping and rapid directional changes.
Distance running emphasises slow-twitch muscle fibres for oxygen efficiency and sustained endurance. Basketball demands fast-twitch fibres for acceleration, agility and vertical leap ability.
Marathon technique focuses on biomechanical efficiency and rhythm maintenance throughout the race distance. Basketball skills include ball handling, shooting accuracy, passing precision and defensive footwork.
Running training prioritises high-volume progression and aerobic base development over extended periods. Basketball balances individual skill refinement with team coordination, tactical awareness and positional play development.

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Similarities

Both athletes require cardiovascular fitness for sustained performance during their respective competitions. Marathon runners and basketball players need efficient heart and lung function.
Mental resilience and concentration skills are essential for both sports under competitive pressure. Athletes must maintain focus during physical stress and make strategic decisions.
Both require sport-specific technical skills that must be practised repeatedly for mastery. Proper biomechanics and movement efficiency are crucial for optimal performance outcomes.
Training periodisation is important for both athletes to peak at competition times. Both sports require structured preparation phases and recovery periods for adaptation.

Differences

Marathon runners require exceptional aerobic capacity for prolonged 42-kilometre efforts over two hours. Basketball players need anaerobic power for explosive jumping and rapid directional changes.
Distance running emphasises slow-twitch muscle fibres for oxygen efficiency and sustained endurance. Basketball demands fast-twitch fibres for acceleration, agility and vertical leap ability.
Marathon technique focuses on biomechanical efficiency and rhythm maintenance throughout the race distance. Basketball skills include ball handling, shooting accuracy, passing precision and defensive footwork.
Running training prioritises high-volume progression and aerobic base development over extended periods. Basketball balances individual skill refinement with team coordination, tactical awareness and positional play development.

HMS, TIP EQ-Bank 341 MC

A tennis coach designs a 6-week mesocycle to specifically target strength development before the competition season begins. The coach includes volume increases in weeks 1-3, followed by intensity focus in weeks 4-6.

This approach demonstrates which key principle of effective subphase planning?

Inadequate planning as strength should only be developed during off-season periods
Poor periodisation because mesocycles should focus on skill development exclusively
Effective targeted preparation allowing specific adaptation within manageable timeframes
Incorrect timing as competition preparation requires only technical skill refinement

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$C$

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C is correct: Mesocycles (4-8 weeks) allow coaches to target specific areas needing development within manageable blocks, with the progressive volume-to-intensity approach showing sound periodisation principles.

Other Options:

A is incorrect: Strength development occurs throughout various training phases, not exclusively in off-season periods.
B is incorrect: Mesocycles can target any training component including fitness, skills or tactical elements.
D is incorrect: Competition preparation requires comprehensive development including physical attributes, not just technical skills.

HMS, TIP EQ-Bank 297

Evaluate the effectiveness of supplement use for high-performance athletes compared to meeting nutritional requirements through whole food sources alone. (8 marks)

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Evaluation Statement

Supplement use demonstrates moderate effectiveness for high-performance athletes.
Whole food sources on the other hand show high effectiveness for meeting nutritional requirements in most athletic contexts.

Supplement Use Effectiveness

Supplements prove highly effective for athletes with demanding training schedules who cannot consume adequate whole foods for their nutritional needs.
Evidence supporting this includes situations where food availability is limited during travel or competition, making supplementation necessary for performance maintenance.
Research demonstrates supplements effectively address specific deficiencies in athletes with food allergies, intolerances or dietary preferences like vegetarian diets.
Studies show supplements provide convenient nutrient delivery when athletes require precise dosages or cannot meet increased requirements through food alone.
However, supplements only partially fulfil optimal nutrition as they may lack the complex nutrient interactions and bioavailability found in whole foods.

Whole Food Effectiveness

Whole food sources demonstrate superior effectiveness as most athletes can meet dietary requirements through consuming varied, balanced diets.
Research reveals whole foods provide essential nutrients in natural combinations that enhance absorption and utilisation compared to isolated supplements.
Evidence shows whole foods offer additional beneficial compounds including antioxidants and phytonutrients that supplements typically cannot replicate.
Studies indicate whole foods pose minimal risk of overconsumption and associated side effects compared to supplement use.

Final Evaluation

Assessment reveals whole food sources provide superior effectiveness for meeting nutritional requirements in most athletic situations.
While supplements offer targeted benefits for specific circumstances, whole foods remain the foundation for optimal athletic nutrition and performance.

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Evaluation Statement

Supplement use demonstrates moderate effectiveness for high-performance athletes.
Whole food sources on the other hand show high effectiveness for meeting nutritional requirements in most athletic contexts.

Supplement Use Effectiveness

Supplements prove highly effective for athletes with demanding training schedules who cannot consume adequate whole foods for their nutritional needs.
Evidence supporting this includes situations where food availability is limited during travel or competition, making supplementation necessary for performance maintenance.
Research demonstrates supplements effectively address specific deficiencies in athletes with food allergies, intolerances or dietary preferences like vegetarian diets.
Studies show supplements provide convenient nutrient delivery when athletes require precise dosages or cannot meet increased requirements through food alone.
However, supplements only partially fulfil optimal nutrition as they may lack the complex nutrient interactions and bioavailability found in whole foods.

Whole Food Effectiveness

Whole food sources demonstrate superior effectiveness as most athletes can meet dietary requirements through consuming varied, balanced diets.
Research reveals whole foods provide essential nutrients in natural combinations that enhance absorption and utilisation compared to isolated supplements.
Evidence shows whole foods offer additional beneficial compounds including antioxidants and phytonutrients that supplements typically cannot replicate.
Studies indicate whole foods pose minimal risk of overconsumption and associated side effects compared to supplement use.

Final Evaluation

Assessment reveals whole food sources provide superior effectiveness for meeting nutritional requirements in most athletic situations.
While supplements offer targeted benefits for specific circumstances, whole foods remain the foundation for optimal athletic nutrition and performance.

HMS, TIP EQ-Bank 278

Evaluate the effectiveness of pre-performance, during-performance and post-performance hydration routines for maintaining optimal movement quality and preventing performance decline in endurance athletes. (8 marks)

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Evaluation Statement

Pre-performance hydration routines demonstrate high effectiveness while during-performance routines show moderate effectiveness and post-performance routines prove highly effective for endurance athletes.

Pre-Performance Hydration Effectiveness

Pre-performance hydration routines strongly meet optimal movement requirements through systematic fluid preparation before exercise begins.
Evidence supporting this includes recommendations to drink 400-600mL water 2-3 hours before exercise and 200-300mL thirty minutes before starting.
Research demonstrates that beginning exercise well-hydrated reduces dehydration risk and maintains blood volume for optimal oxygen delivery.
Studies show proper pre-hydration prevents early onset fatigue and maintains coordination essential for sustained endurance performance.
However, pre-performance routines only partially address total fluid needs as they cannot account for individual sweat rates during exercise.

During and Post-Performance Hydration Effectiveness

During-performance hydration shows limited effectiveness due to absorption delays and potential gastrointestinal discomfort during intense exercise.
Research reveals that fluid replacement during exercise often lags behind fluid loss rates in endurance events.
Post-performance hydration demonstrates high effectiveness through systematic replacement of 125-150 percent of fluids lost over 4-6 hours.
Evidence indicates post-exercise hydration with electrolyte replacement effectively restores fluid balance and prepares athletes for subsequent training.

Final Evaluation

Assessment reveals that pre-performance and post-performance routines provide superior effectiveness compared to during-performance strategies.
While during-exercise hydration supports performance, pre and post routines prove more critical for maintaining movement quality and preventing decline.
Therefore structured hydration routines demonstrate greatest effectiveness when emphasising preparation and recovery phases.

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Evaluation Statement

Pre-performance hydration routines demonstrate high effectiveness while during-performance routines show moderate effectiveness and post-performance routines prove highly effective for endurance athletes.

Pre-Performance Hydration Effectiveness

Pre-performance hydration routines strongly meet optimal movement requirements through systematic fluid preparation before exercise begins.
Evidence supporting this includes recommendations to drink 400-600mL water 2-3 hours before exercise and 200-300mL thirty minutes before starting.
Research demonstrates that beginning exercise well-hydrated reduces dehydration risk and maintains blood volume for optimal oxygen delivery.
Studies show proper pre-hydration prevents early onset fatigue and maintains coordination essential for sustained endurance performance.
However, pre-performance routines only partially address total fluid needs as they cannot account for individual sweat rates during exercise.

During and Post-Performance Hydration Effectiveness

During-performance hydration shows limited effectiveness due to absorption delays and potential gastrointestinal discomfort during intense exercise.
Research reveals that fluid replacement during exercise often lags behind fluid loss rates in endurance events.
Post-performance hydration demonstrates high effectiveness through systematic replacement of 125-150 percent of fluids lost over 4-6 hours.
Evidence indicates post-exercise hydration with electrolyte replacement effectively restores fluid balance and prepares athletes for subsequent training.

Final Evaluation

Assessment reveals that pre-performance and post-performance routines provide superior effectiveness compared to during-performance strategies.
While during-exercise hydration supports performance, pre and post routines prove more critical for maintaining movement quality and preventing decline.
Therefore structured hydration routines demonstrate greatest effectiveness when emphasising preparation and recovery phases.

HMS, TIP EQ-Bank 266

Evaluate the effectiveness of different pre-performance meal timing strategies for endurance athletes competing in events lasting over 90 minutes, considering performance optimisation and digestive comfort. (8 marks)

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Evaluation Statement

Earlier meal timing strategies demonstrate high effectiveness for performance optimisation while later timing shows moderate effectiveness but increased digestive risk.

3-4 Hour Pre-Performance Timing Effectiveness

Consuming meals 3-4 hours before competition proves highly effective for complete digestion and nutrient absorption processes.
Evidence demonstrates this timing allows carbohydrates to be fully processed and stored as glycogen without remaining in the digestive system.
Research shows athletes can consume mixed macronutrient meals including protein, carbohydrates and moderate fat content without performance interference.
Studies indicate this strategy effectively prevents hunger during competition while ensuring adequate fuel availability for working muscles.
However, this timing may be impractical for early morning events or require athletes to wake very early for proper fuelling.

1-2 Hour Pre-Performance Timing Effectiveness

Consuming snacks 1-2 hours before competition shows moderate effectiveness for final glycogen topping while increasing digestive risk.
Evidence reveals this timing works well for light carbohydrate snacks but becomes problematic with larger meal consumption.
Research demonstrates increased risk of gastrointestinal discomfort when consuming substantial food quantities close to competition time.
Studies show this strategy can provide last-minute fuel but may compromise performance if digestive issues occur.

Final Evaluation

Assessment reveals earlier meal timing provides superior effectiveness for performance optimisation with minimal digestive complications.
While later timing offers convenience advantages, the increased risk of digestive issues makes earlier strategies more suitable for competitive athletes.
Therefore 3-4 hour pre-performance meal timing proves most effective for balancing fuel availability with digestive comfort during competition.

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Evaluation Statement

Earlier meal timing strategies demonstrate high effectiveness for performance optimisation while later timing shows moderate effectiveness but increased digestive risk.

3-4 Hour Pre-Performance Timing Effectiveness

Consuming meals 3-4 hours before competition proves highly effective for complete digestion and nutrient absorption processes.
Evidence demonstrates this timing allows carbohydrates to be fully processed and stored as glycogen without remaining in the digestive system.
Research shows athletes can consume mixed macronutrient meals including protein, carbohydrates and moderate fat content without performance interference.
Studies indicate this strategy effectively prevents hunger during competition while ensuring adequate fuel availability for working muscles.
However, this timing may be impractical for early morning events or require athletes to wake very early for proper fuelling.

1-2 Hour Pre-Performance Timing Effectiveness

Consuming snacks 1-2 hours before competition shows moderate effectiveness for final glycogen topping while increasing digestive risk.
Evidence reveals this timing works well for light carbohydrate snacks but becomes problematic with larger meal consumption.
Research demonstrates increased risk of gastrointestinal discomfort when consuming substantial food quantities close to competition time.
Studies show this strategy can provide last-minute fuel but may compromise performance if digestive issues occur.

Final Evaluation

Assessment reveals earlier meal timing provides superior effectiveness for performance optimisation with minimal digestive complications.
While later timing offers convenience advantages, the increased risk of digestive issues makes earlier strategies more suitable for competitive athletes.
Therefore 3-4 hour pre-performance meal timing proves most effective for balancing fuel availability with digestive comfort during competition.

\(f(\lambda)^2\)	\(=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2\)
	\(=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})\)
	\(=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot \underset{\sim}{d}\)
	\(=\lambda^2\|\underset{\sim}{d}\|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)
	\(=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)

\(\overrightarrow{P B} \cdot \underset{\sim}{d}\)	\(=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}\)
	\(=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}\)
	\(=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2\)
	\(=0\)

\(f\left(\lambda_0\right)\)	\(=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}\)
\(f\left(\lambda_0\right)^2\)	\(=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2\)
	\(=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2\)
\(f\left(\lambda_0\right)\)	\(=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}\)

	\(\quad\quad\quad\textit{Percentage of offspring with bar-eyes}\quad\quad\quad\)
	\(\quad\textit{Males}\quad\)	\(\quad\textit{Females}\quad\)
A.	\(100\)	\(100\)
B.	\(0\)	\(100\)
C.	\(100\)	\(0\)
D.	\(50\)	\(0\)

\(\dfrac{d \dot{x}}{dt}\)	\(=-k \dot{x} \ \ \text{(given)}\)
\(\dfrac{dt}{d \dot{x}}\)	\(=-\dfrac{1}{k \dot{x}}\)
\(\displaystyle \int dt\)	\(=-\dfrac{1}{k} \int \dfrac{1}{\dot{x}}\, d x\)
\(t\)	\(=-\dfrac{1}{k} \ln \dot{x}+c\)

\(t\)	\(=\dfrac{1}{k} \ln 40-\dfrac{1}{k} \ln \abs{\dot{x}}=\dfrac{1}{k} \ln \abs{\dfrac{40}{\dot{x}}}\)
\(k t\)	\(=\ln \abs{\dfrac{40}{\dot{x}}}\)
\(e^{k t}\)	\(=\dfrac{40}{\dot{x}}\)
\(\dot{x}\)	\(=40 e^{-k t}\)
\(x\)	\(\displaystyle=\int 40 e^{-k t}\, d t\)
	\(=-\dfrac{40}{k} \times e^{-k t}+c\)

\(t\)	\(=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}\)
	\(=\dfrac{1}{k} \ln \abs{\frac{-30+\frac{10}{k}}{\dot{y}+\frac{10}{k}}}\)
\(e^{k t}\)	\(=\abs{\dfrac{-30+\frac{10}{k}}{y+\frac{10}{k}}}\)
\(\dot{y}\)	\(=\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10}{k}\)
\(y\)	\(=\displaystyle \left(-30+\dfrac{10}{k}\right) \int e^{-kt}\, d t-\int \dfrac{10}{k}\, dt\)
	\(=-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10 t}{k}+c\)

\(y\)	\(=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10t}{k}\)
	\(=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right)\left(1-e^{-k t}\right)-\dfrac{10 t}{k}\ \ldots\ (2)\)

\(x\)	\(=\dfrac{40}{k}\left(1-e^{-k t}\right)\)
\(\dfrac{k x}{40}\)	\(=1-e^{-k t}\)
\(e^{-k t}\)	\(=1-\dfrac{k x}{40}\)
\(-k t\)	\(=\ln \abs{1-\dfrac{k x}{40}}\)
\(t\)	\(=-\dfrac{1}{k} \ln \abs{1-\dfrac{kx}{40}}\)