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Mechanics, EXT2 M1 2025 HSC 8 MC

The graph shows the velocity of a particle as a function of its displacement.
 

  

Which of the following graphs best shows the acceleration of the particle as a function of its displacement?
 

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\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(v(x)\ \Rightarrow\ \text{degree 3 (see graph)},\ \ \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 2}\)

\(a=v \cdot \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 5 (eliminate C and D)}\)

\(\text{At}\ \ x=0,\ \ v(x)>0\ \ \text{and}\ \ \dfrac{dv}{dx}<0\ \ \Rightarrow\ \ v \cdot \dfrac{dv}{dx} \neq 0\ \text{(eliminate B)}\)

\(\Rightarrow A\)

Complex Numbers, EXT2 N2 2025 HSC 15c

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

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

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

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

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

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

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

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

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

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

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

Mechanics, EXT2 M1 2025 HSC 15b

A particle moves in simple harmonic motion about the origin with amplitude \(A\), and it completes two cycles per second. When it is \(\dfrac{1}{4}\) metres from the origin, its speed is half its maximum speed.

Find the maximum positive acceleration of the particle during its motion.   (4 marks)

Show Answers Only

\(a_{\text{max}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Show Worked Solution

\(v^2=-n^2\left(x^2-A^2\right)\)

\(\operatorname{Period}\ (T)=\dfrac{1}{2} \ \Rightarrow \ \dfrac{2 \pi}{n}=\dfrac{1}{2} \ \Rightarrow \ n=4 \pi\)

\(\text{Max velocity occurs at}\ \  x=0:\)

   \(v_{\text{max}}^2=-(4 \pi)^2\left(0-A^2\right)=16 \pi^2 A^2\)

   \(v_{\text{max}}=\sqrt{16 \pi^2 A^2}=4 \pi A\)
 

\(\text{At} \ \ x=\dfrac{1}{4}, \ v=\dfrac{1}{2} \times 4 \pi A=2 \pi A\)

\((2 \pi A)^2\) \(=-(4 \pi)^2\left(\dfrac{1}{16}-A^2\right)\)
\(\dfrac{A^2}{4}\) \(=A^2-\dfrac{1}{16}\)
\(\dfrac{3 A^2}{4}\) \(=\dfrac{1}{16}\)
\(A^2\) \(=\dfrac{1}{12}\)
\(A\) \(=\dfrac{1}{\sqrt{12}}\)

 

\(\text{Max positive acceleration occurs at} \ \ x=-\dfrac{1}{\sqrt{12}}:\)

\(a_{\text{max}}=-n^2 x=-(4 \pi)^2 \times-\dfrac{1}{\sqrt{12}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Proof, EXT2 P1 2025 HSC 14d

Positive real numbers \(a, b, c\) and \(d\) are chosen such that  \(\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}\)  and  \(\dfrac{1}{d}\)  are consecutive terms in an arithmetic sequence with common difference \(k\), where  \(k \in \mathbb{R} , k>0\).

Show that  \(b+c<a+d\).   (3 marks)

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\(\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)\)

\(\text{Show} \ \ b+c<a+d\)

\(\dfrac{1}{a}<\dfrac{1}{a}+k(k>0)\)

\(\dfrac{1}{a}<\dfrac{1}{b}\)

\(\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\  \ldots\  \text{such that}\)

\(\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)\)
 

\(\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab\)

\(\text {Similarly:}\)

\(\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd\)
 

\(\text{Using \(\ a>b>c>d\ \)  (see (1) above):}\)

\(a-b>c-d\)

\(b+c<a+d\)

Show Worked Solution

\(\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)\)

\(\text{Show} \ \ b+c<a+d\)

\(\dfrac{1}{a}<\dfrac{1}{a}+k\ \ (k>0)\)

\(\dfrac{1}{a}<\dfrac{1}{b}\)

\(\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\  \ldots\  \text{such that}\)

\(\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)\)
 

\(\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab\)

\(\text {Similarly:}\)

\(\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd\)
 

\(\text{Using \(\ a>b>c>d\ \)  (see (1) above):}\)

\(a-b>c-d\)

\(b+c<a+d\)

Complex Numbers, EXT2 N2 2025 HSC 14c

Let \(w\) be a complex number such that  \(1+w+w^2+\cdots+w^6=0\).

  1. Show that \(w\) is a 7th root of unity.   (1 mark)

The complex number  \(\alpha=w+w^2+w^4\)  is a root of the equation  \(x^2+b x+c=0\), where \(b\) and \(c\) are real and \(\alpha\) is not real.

  1. Find the other root of  \(x^2+b x+c=0\)  in terms of positive powers of \(w\).  (2 marks)

  2. Find the numerical value of \(c\).  (1 mark)

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i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \text{(given)}\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7=1\) \(=1\)

ii.   \(w^6+w^5+w^3\)

iii.  \(2\)

Show Worked Solution

i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \ \text{(given,}\ w\neq 1)\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7\) \(=1\)

 
ii.    \(\text {Find the other root of:} \ \ x^2+b x+c=0\)

\(\text{Since \(b, c\) are real (given),}\)

\(\text{Using conjugate root theory, other root}\ =\bar{\alpha}\)

\(\bar{\alpha}\) \(=\overline{w+w^2+w^4}\)
  \(=\overline{w}+\overline{w^2}+\overline{w^4}\)
  \(=\dfrac{1}{w}+\dfrac{1}{w^2}+\dfrac{1}{w^4} \quad\left( \bar{w}=\dfrac{1}{w} \ \text{since} \ \ \abs{w}=1\right)\)
  \(=\dfrac{w^7}{w}+\dfrac{w^7}{w^2}+\dfrac{w^7}{w^4}\)
  \(=w^6+w^5+w^3\)

 

iii.    \(\text{Product of roots}=\dfrac{c}{a}=c\)

\(c\) \(=\left(w+w^2+w^4\right)\left(w^6+w^5+w^3\right)\)
  \(=w^7+w^6+w^4+w^8+w^7+w^5+w^{10}+w^9+w^7\)
  \(=1+w^6+w^4+\left(w^7 \cdot w\right)+1+w^5+\left(w^7 \cdot w^3\right)+\left(w^7 \cdot w^2\right)+1\)
  \(=2+\underbrace{1+w+w^2+w^3+w^4+w^5+w^6}_{=0}\)
  \(=2\)

Mechanics, EXT2 M1 2025 HSC 14b

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

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

  2. Find the time taken for the particle to travel 3 metres from the origin.   (2 marks)

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i.    \(\text{See Worked Solutions}\)

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Calculus, EXT2 C1 2025 HSC 14a

Let  \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta\)  for integers  \(n \geq 0\).

  1. Show that  \(I_n=\dfrac{1}{2 n-1}-I_{n-1}\) for \(n>0\),  given that  \(\dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\).   (3 marks)

  2. Hence, or otherwise, calculate \(I_2\).   (1 mark)

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i.    \(\text{See Worked Solutions}\)

ii.   \(\dfrac{\pi}{4}-\dfrac{2}{3}\)

Show Worked Solution

i.    \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta \ \ \Rightarrow \ \ \)\(I_{n-1}=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)

\(\text{Show} \ \ I_n=\dfrac{1}{2n-1}-I_{n-1} \ \ \text{for} \ \ n>0\)

\(\text{Given} \ \ \dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\ \ldots\ (1)\)

\(I_n\) \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \cot ^2 \theta \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta\left(\operatorname{cosec}^2 \theta-1\right) \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \operatorname{cosec}^2 \theta \, d \theta-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)
  \(=-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \dfrac{d}{d \theta}(\cot \theta) d \theta-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} \theta}{2 n-1}\right]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} (\frac{\pi}{2})}{2 n-1}-\dfrac{\cot ^{2 n-1} (\frac{\pi}{4})}{2 n-1}\right]-I_{n-1}\)
  \(=-\left[0-\dfrac{1}{2 n-1}\right]-I_{n-1}\)
  \(=\dfrac{1}{2n-1}-I_{n-1}\)

 

ii.     \(I_2\) \(=\dfrac{1}{3}-I_1\)
    \(=\dfrac{1}{3}-\left[ \dfrac{1}{2-1}-\displaystyle \int_{\tfrac{\pi}{2}}^{\tfrac{\pi}{2}} 1 \, d\theta \right]\)
    \(=\dfrac{1}{3}-\left[1-\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)\right]\)
    \(=\dfrac{\pi}{4}-\dfrac{2}{3}\)

BIOLOGY, M5 2025 HSC 29

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

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

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

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a.    Varroa mite infection classification

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

b.    Procedure 1: Quarantine

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

Procedure 2: Surveillance and Destruction

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

a.    Varroa mite infection classification

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

b.    Procedure 1: Quarantine

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

Procedure 2: Surveillance and Destruction

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

Proof, EXT2 P1 2025 HSC 13c

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

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

Show Answers Only

 

Show Worked Solution

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

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

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

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

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

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

 

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

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

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

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

BIOLOGY, M8 2025 HSC 25

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

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

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

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Evaluation Judgment:

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

Supporting Evidence:

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

Conclusion:

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

Show Worked Solution

Evaluation Judgment:

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

Supporting Evidence:

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

Conclusion:

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

BIOLOGY, M6 2025 HSC 23

Compare the processes of artificial insemination and artificial pollination.   (3 marks)

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Similarities:

  • Both involve controlled transfer of reproductive cells to produce offspring with desired traits.

Differences:

  • Artificial insemination transfers sperm into female animal reproductive tracts using veterinary equipment.
  • Artificial pollination manually transfers pollen between plant flowers using brushes.
  • Insemination requires timing with oestrus cycles whilst pollination uses controlled plant environments.
  • Insemination improves livestock genetics whereas pollination enhances crop breeding programs.

Show Worked Solution

Similarities:

  • Both involve controlled transfer of reproductive cells to produce offspring with desired traits.

Differences:

  • Artificial insemination transfers sperm into female animal reproductive tracts using veterinary equipment.
  • Artificial pollination manually transfers pollen between plant flowers using brushes.
  • Insemination requires timing with oestrus cycles whilst pollination uses controlled plant environments.
  • Insemination improves livestock genetics whereas pollination enhances crop breeding programs.

BIOLOGY, M5 2025 HSC 22a

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

Show Answers Only

Spore production (asexual reproduction):

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

Alternative acceptable response – Budding:

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

Show Worked Solution

Spore production (asexual reproduction):

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

Alternative acceptable response – Budding:

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

BIOLOGY, M7 2025 HSC 22b

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

Show Answers Only

Answers could include ONE of the following:

Influenza virus:

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

Plasmodium (malaria parasite):

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

Bacterial spores:

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

Show Worked Solution

Answers could include ONE of the following:

Influenza virus

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

Plasmodium (malaria parasite):

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

Bacterial spores:

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

Vectors, EXT2 V1 2025 HSC 13b

Let  \(\underset{\sim}{c}=x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}\)  be a unit vector that is perpendicular to both  \(\underset{\sim}{a}=2 \underset{\sim}{i}+4 \underset{\sim}{j}-3 \underset{\sim}{k}\)  and  \(\underset{\sim}{b}=-4 \underset{\sim}{i}-5 \underset{\sim}{j}+3 \underset{\sim}{k}\).

Find all possible vectors \(\underset{\sim}{c}\).   (4 marks)

Show Answers Only

\(\underset{\sim}{c}= \pm \dfrac{1}{3}(\underset{\sim}{i}-2\underset{\sim}{j}-2 \underset{\sim}{k})\)

Show Worked Solution

\(\underset{\sim}{c} \cdot \underset{\sim}{a}=\left(\begin{array}{l}x \\ y \\ z\end{array}\right)\left(\begin{array}{c}2 \\ 4 \\ -3\end{array}\right)=2 x+4 y-3 z=0\ \ldots\ (1)\)

\(\underset{\sim}{c} \cdot \underset{\sim}{b}=\left(\begin{array}{l}x \\ y \\ z\end{array}\right)\left(\begin{array}{c}-4 \\ -5 \\ 3\end{array}\right)=-4 x-5 y+3 z=0\ \ldots\ (2)\)

\(\text{Given} \ \ \abs{\underset{\sim}{c}}=1:\)

\(x^2+y^2+z^2=1\ \ldots\ (3)\)
 

\(\text{Add}\ \ (1)+(2):\)

\(-2 x-y=0 \ \ \Rightarrow\ \ y=-2 x\)
 

\(\text{Substitute}\ \ y=-2 x \ \ \text{into (1):}\)

\(-6 x-3 z=0 \ \ \Rightarrow\ \ z=-2 x\)
 

\(\text{Substituting into (3):}\)

\(x^2+4 x^2+4 x^2=1\ \ \Rightarrow\ \ x^2=\dfrac{1}{9}\ \ \Rightarrow\ \ x= \pm \dfrac{1}{3}\)

\(\underset{\sim}{c}\) \(=\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}\)
  \(= \pm \dfrac{1}{3}(\underset{\sim}{i}-2\underset{\sim}{j}-2 \underset{\sim}{k})\)

Mechanics, EXT2 M1 2025 HSC 12e

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

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

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

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

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

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

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

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

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

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

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

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

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

Calculus, EXT2 C1 2025 HSC 12d

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

Show Answers Only

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

Show Worked Solution

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

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

\(\text{If}\ \  x=4:\)

\(17 A=16-8+9=17 \ \ \Rightarrow\ \ A=1\)

\(\text{If}\ \  x=0:\)

\(1+c \times 4=9 \ \ \Rightarrow\ \ C=2\)

\(\text{If}\ \  x=1:\)

\(2+3 B+6=8 \ \ \Rightarrow\ \ B=0\)

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

Complex Numbers, EXT2 N2 2025 HSC 12c

Sketch the region of the complex plane defined by  \(\abs{z+5-i}>\abs{z-3+3 i}\).   (3 marks)

Show Answers Only

Show Worked Solution

\(\abs{z+5-i}=\abs{z-(-5+i)}\)

\(\abs{z-3+3 i}=\abs{z-(3-3 i)}\)

\(\text{Find equation of} \ \perp \ \text{bisector between}\ \ (-5,1) \ \ \text{and}\ \ (3,-3):\)

\(\text{Midpoint} \equiv \left(\dfrac{-5+3}{2}, \dfrac{1-3}{2}\right) \equiv (-1,-1)\)

\(m=\dfrac{-3-1}{3+5}=-\dfrac{1}{2} \ \Rightarrow \ m_{\perp}=2\)
 

\(\text{Equation of line} \ \ m=2 \ \ \text{through}\ \ (-1,-1):\)

\(y+1\) \(=2(x+1)\)
\(y\) \(=2 x+1\)

 
\(\text{Sketch:}\ \abs{z+5-i}>\abs{z-3+3 i}\)

Proof, EXT2 P1 2025 HSC 11e

Prove by contradiction that  \(\sqrt{3}+\sqrt{5}>\sqrt{11}\).   (2 marks)

Show Answers Only

\(\text{Prove} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

\(\text{Assume}\ \ \sqrt{3}+\sqrt{5}\) \(\leqslant\sqrt{11}\)
\((\sqrt{3}+\sqrt{5})^2\) \(\leqslant 11\)
\(3+2 \sqrt{15}+5\) \(\leqslant 11\)
\(2 \sqrt{15}\) \(\leqslant 3\)
\(60\) \(\leqslant 9 \ \text{(incorrect)}\)

 

\(\therefore \ \text{By contradiction} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

Show Worked Solution

\(\text{Prove} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

\(\text{Assume}\ \ \sqrt{3}+\sqrt{5}\) \(\leqslant\sqrt{11}\)
\((\sqrt{3}+\sqrt{5})^2\) \(\leqslant 11\)
\(3+2 \sqrt{15}+5\) \(\leqslant 11\)
\(2 \sqrt{15}\) \(\leqslant 3\)
\(60\) \(\leqslant 9 \ \text{(incorrect)}\)

 

\(\therefore \ \text{By contradiction} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

Vectors, EXT2 V1 2025 HSC 11d

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

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

  6. Calculate  \({\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}\), where \({\underset{\sim}{F}}_3\) is the resultant force from part (ii) and  \(\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}\).   (1 mark)

Show Answers Only

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

Proof, EXT2 P2 2025 HSC 12b

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

Show Answers Only

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

Show Worked Solution

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

\(\text{If } \ \ n=1:\)

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

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

\(\therefore \ \text{True for} \ \ n=1\)
 

\(\text{Assume true for}\ \  n=k:\)

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

\(\text{Prove true for}\ \  n=k+1:\)

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

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

 

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

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

Probability, STD2 S2 2025 HSC 29

A bag contains 7 blue lollies and 9 yellow lollies. One lolly is selected at random and eaten. A second lolly is then selected from the remaining lollies in the bag.

Find the probability that the two lollies selected are the same colour.   (2 marks)

Show Answers Only

\(P\text{(same colour)}=\dfrac{19}{40}\)

Show Worked Solution

\(P\text{(same colour)}\) \(=P(BB)+P(YY)\)
  \(=\dfrac{7}{16} \times \dfrac{6}{15}+\dfrac{9}{16} \times \dfrac{8}{15}\)
  \(=\dfrac{19}{40}\)

BIOLOGY, M5 2025 HSC 17 MC

Some phases of cell division in an organism are shown.
  

What is the correct sequence of the process?

  1. Q, S, R, T
  2. Q, R, S, T
  3. S, T, R, Q
  4. S, R, T, Q
Show Answers Only

\(A\)

Show Worked Solution
  • A is correct: Interphase (Q) → Metaphase (S) → Anaphase (R) → Telophase (T).

Other Options:

  • B is incorrect: Chromosomes must align at metaphase before separating in anaphase.
  • C is incorrect: Cell division starts from interphase, not metaphase.
  • D is incorrect: Anaphase occurs after metaphase, not before.

BIOLOGY, M8 2025 HSC 15 MC

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

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

Which is the correct order for this process?

  1. 1, 4, 3, 2, 5
  2. 1, 3, 2, 4, 5
  3. 3, 1, 5, 2, 4
  4. 3, 1, 2, 5, 4
Show Answers Only

\(C\)

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

Other Options:

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

BIOLOGY, M7 2025 HSC 14 MC

A pathogen that produces an immune response is shown.

Which antibody will be produced as a response to the pathogen?

Show Answers Only

\(B\)

Show Worked Solution
  • B is correct: Antibodies are specific to antigens; matches rectangular antigens on pathogen.

Other Options:

  • A is incorrect: Square binding sites don’t match rectangular antigens on pathogen.
  • C is incorrect: While rectangular, the binding site shape doesn’t complement the antigen.
  • D is incorrect: Square binding sites don’t match the pathogen’s rectangular antigens.

BIOLOGY, M8 2025 HSC 9 MC

Dialysis is used to assist people with loss of kidney function.

Which row of the table correctly describes the composition of blood before and after
dialysis?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}{Blood \ before \ dialysis}\rule[-1ex]{0pt}{0pt}& {Blood \ after \ dialysis} \\
\hline
\rule{0pt}{2.5ex}\text{High urea and high glucose}\rule[-1ex]{0pt}{0pt}&\text{Low urea and high glucose }\\
\hline
\rule{0pt}{2.5ex}\text{High urea and high glucose}\rule[-1ex]{0pt}{0pt}& \text{High urea and low glucose }\\
\hline
\rule{0pt}{2.5ex}\text{Low urea and low glucose }\rule[-1ex]{0pt}{0pt}& \text{Low urea and high glucose } \\
\hline
\rule{0pt}{2.5ex}\text{Low urea and high glucose}\rule[-1ex]{0pt}{0pt}& \text{Low urea and low glucose} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • A is correct: Dialysis removes urea waste but maintains glucose levels in blood.

Other Options:

  • B is incorrect: Dialysis removes urea; it doesn’t stay high after treatment.
  • C is incorrect: Before dialysis, urea is high due to kidney failure.
  • D is incorrect: Before dialysis, blood has high urea from waste accumulation.

BIOLOGY, M7 2025 HSC 6 MC

Rabies is a viral disease spread by infected animals. If bitten by an infected animal, a person can be treated by receiving an injection of antibodies.

What type of immunity will this person have following the injection?

  1. Innate active
  2. Natural passive
  3. Acquired active
  4. Acquired passive
Show Answers Only

\(D\)

Show Worked Solution
  • D is correct: Receiving pre-made antibodies from external source is acquired passive immunity.

Other Options:

  • A is incorrect: Innate immunity is non-specific and present from birth.
  • B is incorrect: Natural passive involves antibodies from mother through placenta or milk.
  • C is incorrect: Acquired active requires person’s immune system to produce own antibodies.

BIOLOGY, M7 2025 HSC 5 MC

An infectious disease is spread through direct contact between hosts.

Under which conditions will this disease spread most rapidly?

  1. Low population density with infected individuals rarely dying from the disease
  2. High population density with infected individuals rarely dying from the disease
  3. Low population density with infected individuals quickly dying from the disease
  4. High population density with infected individuals quickly dying from the disease
Show Answers Only

\(B\)

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

Other Options:

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

BIOLOGY, M7 2025 HSC 3 MC

Which row in the table correctly identifies the role of phagocytes and lymphocytes?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}{Phagocytes}\rule[-1ex]{0pt}{0pt}& {Lymphocytes} \\
\hline
\rule{0pt}{2.5ex}\text{Engulf bacteria}\rule[-1ex]{0pt}{0pt}&\text{Produce antibodies }\\
\hline
\rule{0pt}{2.5ex}\text{Engulf bacteria }\rule[-1ex]{0pt}{0pt}& \text{Produce antigens}\\
\hline
\rule{0pt}{2.5ex}\text{Produce antibodies}\rule[-1ex]{0pt}{0pt}& \text{Engulf bacteria} \\
\hline
\rule{0pt}{2.5ex}\text{Produce antigens}\rule[-1ex]{0pt}{0pt}& \text{Produce antibodies} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • A is correct: Phagocytes engulf bacteria through phagocytosis and lymphocytes produce antibodies.

Other Options:

  • B is incorrect: Lymphocytes produce antibodies, not antigens.
  • C is incorrect: Roles are reversed; phagocytes engulf, lymphocytes produce antibodies.
  • D is incorrect: Phagocytes engulf bacteria, they don’t produce antigens.

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}

  1. Calculate the mass of a $2 coin in grams, correct to 1 decimal point?   (2 marks)
  3. Suppose the $2 coin is to be replaced with a note that has the same mass as a $5 note.
  4. What is the mass of $1000 in $2 notes in grams? Give your answer correct to the nearest gram.   (2 marks)
Show Answers Only

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 24

A used car has a sale price of $24 200. In addition to the sale price, the following costs are charged:

  • transfer of registration $50
  • stamp duty which is calculated at $3 for every $100, or part thereof, of the sale price.

Kat borrows the total amount to be paid for the car, including transfer of registration and stamp duty. Simple interest at the rate of 6.8% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 3 years.

Calculate Kat’s monthly repayment.   (5 marks)

Show Answers Only

\($835.31\)

Show Worked Solution

\(\text{Stamp Duty} =\dfrac{24\ 200}{100}\times 3=$726\)

\(\text{Total Cost}\ \) \(\text{= Price + Transfer + Stamp Duty}\)
  \(\text{= }24\ 200+50+726\)
  \(\text{= }$24\ 976\)

 
\(\text{Interest}=Prn=24\,976\times 0.068\times 3=$5095.104\)

\(\text{Loan amount}\ \) \(\text{= Total Cost + Interest}\)
\(\text{= }24\ 976+5095.104\)
\(\text{= }$30\ 071.104\)

 
\(\text{3 years}= 3 \times 12=36\ \text{months}\)

\(\text{Monthly repayment}\) \(=\dfrac{30\ 071.104}{36}\)
  \(=$835.308444\)
  \(\approx $835.31\)

Measurement, STD1 M3 2025 HSC 22

An isosceles triangle is drawn inside a circle as shown. The base of the triangle is 4.8 cm long, the length of other sides is 4 cm and the height is \(h\) cm.
 

  1. Calculate the height, \(h\), of triangle \(ABC\).   (2 marks)
  1. The area of triangle \(ABC\) is 7.68 cm².
  2. The radius of the circle is 2.5 cm.
  3. Express the area of triangle \(ABC\) as a percentage of the area of the circle, correct to 1 decimal place.   (2 marks)
Show Answers Only

a.    \(h=3.2\ \text{cm}\)

b.    \(39.1\%\)

Show Worked Solution

a.   \(\text{Since}\ \Delta ABC\ \text{is isosceles:}\)

\(BM\ \text{bisects}\ AC\ \ \Rightarrow\ \ AM=MC=2.4\)
 

\(\text{Using Pythagoras:}\)

\(h^2=4^2-2.4^2=16-5.76=10.24\)

\(h = \sqrt{10.24} = 3.2\ \text{cm}\)
 

b.  \(\text{Area of circle}=\pi\times 2.5^2\)

\(\text{Percentage}\) \(=\dfrac{\text{Area of triangle}}{\text{Area of circle}}\times 100\%\)
  \(=\dfrac{7.68}{\pi\times 2.5^2}\times 100\%\)
  \(=39.11…\%\)
  \(\approx 39.1\%\ \text{(1 decimal place)}\)

Measurement, STD1 M3 2025 HSC 20

A map of a park containing a duck pond is shown.

A fence is built passing through the points \(A\), \(B\) and \(C\) around the duck pond.
 

  1. Using the scale provided on the map, calculate the length of the fence \(AB\).   (2 marks)
  1. The length of \(AB\) is equal to the length of \(BC\).
  2. Use Pythagoras’ theorem to calculate the length of \(AC\) in metres. Give your answer correct to 3 significant figures.   (3 marks)
  3. What is the true bearing of point \(A\) from point \(C\) ?   (2 marks)
Show Answers Only

a.    \(75\ \text{metres}\)

b.    \(106\ \text{metres}\)

c.    \(225^\circ\)

Show Worked Solution

a.   \(\text{Scale: 1 grid width = 5 metres}\)

\(AB = 15 \times 5 = 75\ \text{metres}\)
 

b.   \(\text{Using Pythagoras:}\)

\(AC^2=AB^2+BC^2\)

\(AC^2=75^2+75^2=11250\)

  \(\therefore\ AC\) \(=\sqrt{11250}\)
    \(=106.066…\)
    \(\approx 106\ \text{m (3 sig fig)}\)

  
c.    \(\text{Since }AB=BC:\)

\(\angle BAC=\angle BCA=45^\circ\)

\(\text{Bearing of \(A\) from \(C\)}\ =180+45=225^\circ\)

Financial Maths, STD1 F1 2025 HSC 19

At the end of the 2024−2025 financial year, Alex’s taxable income was $148 600.

  1. The table shows the income tax rate for Australian residents for the 2024−2025 financial year.  

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{    Taxable income}\rule[-1ex]{0pt}{0pt} & \textit{    Tax payable}\\
\hline
\rule{0pt}{2.5ex}\text{\$0 – \$18 200}\rule[-1ex]{0pt}{0pt} & \text{Nil}\\
\hline
\rule{0pt}{2.5ex}\text{\$18 201 – \$45 000}\rule[-1ex]{0pt}{0pt} & \text{16 cents for each \$1 over \$18 200}\\
\hline
\rule{0pt}{2.5ex}\text{\$45 001 – \$135 000}\rule[-1ex]{0pt}{0pt} & \text{\$4288 plus 30 cents for each \$1 over \$45 000}\\
\hline
\rule{0pt}{2.5ex}\text{\$135 001 – \$190 000}\rule[-1ex]{0pt}{0pt} & \text{\$31 288 plus 37 cents for each \$1 over \$135 000}\\
\hline
\rule{0pt}{2.5ex}\text{\$190 001 and over}\rule[-1ex]{0pt}{0pt} & \text{\$51 638 plus 45 cents for each \$1 over \$190 000}\\
\hline
\end{array}

  1. Using the table, calculate Alex’s tax payable.   (3 marks)

  2. The Medicare levy is 2% of taxable income.
  3. Calculate the Medicare levy payable by Alex.   (1 mark)

Show Answers Only

a.    \($36\ 320\)

b.    \($2972\)

Show Worked Solution
a.     \(\text{Tax payable}\) \(=31\ 288+0.37\times(148\ 600-135\ 000)\)
    \(=31\ 288+0.37\times 13\ 600\)
    \(=31288+5032\)
    \(=$36\ 320\)

 

b.    \(\text{Medicare}\) \(=0.02\times 148\ 600\)
    \(=$2\ 972\)

Measurement, STD1 M4 2025 HSC 18

Ramon took 1.25 hours to travel the length of a freeway. The length of the freeway is 100 km.

  1. Ramon’s total travel time was made up of:
    • 20 minutes to travel to the start of the freeway
    • the time taken to travel the length of the freeway
    • 35 minutes after exiting the freeway to get to his destination.
  1. What was the total time Ramon spent travelling? Give your answer in hours and minutes.   (2 marks)
  1. What was Ramon’s average speed on the freeway?   (1 mark)
Show Answers Only

a.    \(2\ \text{hours}\ 10\ \text{minutes}\)

b.    \(80\ \text{km/h}\)

Show Worked Solution
a.     \(\text{Total time}\) \(=20+75+35\)
    \(=130\ \text{minutes}\)
    \(=2\text{ h }10\text{ min}\)

 

b.     \(\text{Average speed}\) \(=\dfrac{\text{distance}}{\text{time}}\)
    \(=\dfrac{100}{1.25}\)
    \(=80\ \text{km/h}\)

Measurement, STD1 M4 2025 HSC 17

Paint is sold in two sizes at a local shop.

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

Mina needs to purchase 80 litres of paint.

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

Show Answers Only

\(\text{Saving}=$160\)

Show Worked Solution

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

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

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

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

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

Statistics, STD1 S3 2025 HSC 15

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

  1. State the statistical question being posed.   (1 mark)

Participants were asked to record the number of minutes they spent watching television each day and the number of minutes they spent exercising each day. The averages for each participant were recorded and graphed, and a line of best fit was included.
 

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

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

  3. The points \((0, 70)\) and \((60, 10)\) lie on the line of best fit. By first plotting these points on the graph, find the gradient and the \(y\)-intercept of the line of best fit.   (3 marks)
  4. Explain why it is NOT appropriate to extrapolate the line of best fit to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)
Show Answers Only

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

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

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

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d.   \(\text{Gradient}=-1\)
 

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

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

Show Worked Solution

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

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

b.    \(\text{Dependent variable:}\)

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

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d. 


 

\(y-\text{intercept = 70}\)

\(\text{Calculating gradient using}\ (0, 70)\ \text{and}\ (60, 10):\)

\(\text{Gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\Big(\dfrac{10-70}{60-0}\Big)=-1\)
 

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

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

Networks, STD1 N1 2025 HSC 14

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

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

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

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

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

     

  2. Explain whether the path in part (a) is still the shortest path from \(A\) to \(F\) after the new roads are added.   (1 mark)

Show Answers Only

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

\(\text{Path: }A → B → C → D → E → F\)
 

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

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

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

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

Show Worked Solution

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

\(\text{Path: }A → B → C → D → E → F\)
 

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

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

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

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

Measurement, STD1 M1 2025 HSC 13

A trapezium is shown.
 

  1. Write an expression for the area of this trapezium.   (1 mark)

  2. Find the value of \(a\), given that the area of the trapezium is 330 cm².   (2 marks)
Show Answers Only

a.   \(\text{A}=6(a+30)\ \text{or}\ A=6a+180\)

b.   \(a=25\)

Show Worked Solution
a.     \(A\) \(=\dfrac{h}{2}\Big(a+b\Big)\)
    \(=\dfrac{12}{2}\Big(a+30\Big)\)
    \(=6a+180\)

 

b.    \(\text {When} \ A=330:\)

\(6a+180\) \(=330\)
\(6a\) \(=330-180\)
\(6a\) \(=150\)
\(a\) \(=25\)

Measurement, STD1 M4 2025 HSC 12

Vijay’s heart rate before and after his morning run is shown.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \text{Before run} \rule[-1ex]{0pt}{0pt} & \text{72 beats per minute} \\
\hline
\rule{0pt}{2.5ex} \text{After run} \rule[-1ex]{0pt}{0pt} & \text{126 beats per minute} \\
\hline
\end{array}

What is the percentage increase in Vijay’s heart rate?    (2 marks)

Show Answers Only

\(75\%\)

Show Worked Solution
\(\text{% increase}\) \(=\Big[\dfrac{\text{Change in HR}}{\text{Original HR}}\Big]\times 100\%\)
  \(=\Big[\dfrac{126-72}{72}\Big]\times 100\%\)
  \(=75\%\)

Measurement, STD1 M1 2025 HSC 11

Consider the composite shape shown.
 

What is the area of the shape in cm² ?   (3 marks)

Show Answers Only

\(21\ \text{cm}^2\)

Show Worked Solution

\(\text{Height of triangle}=3+2=5\ \text{cm}\)

\(\text{Area}\) \(=\text{Area of triangle}+\text{Area of rectangle}\)
  \(=\dfrac{1}{2}\times 6\times 5+3\times 2\)
  \(=15+6\)
  \(=21\ \text{cm}^2\)

Financial Maths, STD1 F2 2025 HSC 9 MC

An amount of $90 000 is invested at 4% per annum, compounded quarterly.

Which expression gives the value of this investment, in dollars, after 6 years?

  1. \(90\ 000(1+0.04)^6\)
  2. \(90\ 000(1+0.04)^{24}\)
  3. \(90\ 000(1+0.01)^6\)
  4. \(90\ 000(1+0.01)^{24}\)
Show Answers Only

\(D\)

Show Worked Solution

\(PV=90\ 000,\  r=\dfrac{4\%}{4}=1\%=0.01,\  n=4\times 6=24\)

\(FV\) \(=PV(1+r)^n\)
  \(=90\ 000(1+0.01)^{24}\)

  
\(\Rightarrow D\)

Financial Maths, STD2 F4 2025 HSC 33

A used car has a sale price of $24 200. In addition to the sale price, the following costs are charged:

  • transfer of registration $50
  • stamp duty which is calculated at $3 for every $100, or part thereof, of the sale price.

Kat borrows the total amount to be paid for the car, including transfer of registration and stamp duty. Simple interest at the rate of 6.8% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 3 years.

Calculate Kat’s monthly repayment.   (5 marks)

Show Answers Only

\($835.31\)

Show Worked Solution

\(\text{Stamp Duty} =\dfrac{24\ 200}{100}\times 3=$726\)

\(\text{Total Cost}\ \) \(=\ \text{Price + Transfer + Stamp Duty}\)
  \(=24\ 200+50+726\)
  \(=$24\ 976\)

 
\(\text{Interest}=Prn=24\,976\times 0.068\times 3=$5095.104\)

\(\text{Loan amount}\ \) \(=\ \text{Total Cost + Interest}\)
  \(=24\ 976+5095.104\)
  \(=$30\ 071.104\)

 
\(\text{3 years}= 3 \times 12=36\ \text{months}\)

\(\text{Monthly repayment}\) \(=\dfrac{30\ 071.104}{36}\)
  \(=$835.308444\)
  \(\approx $835.31\)

Measurement, STD2 M1 2025 HSC 26

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

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

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

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

Show Answers Only

a.   \(34.68 \ \text{cm}^2\)

b.   \(582.64 \ \text{cm}^2\)

c.   \(0.490 \%\)

Show Worked Solution

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

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

 

b.    \(\text{Toy has 5 sides.}\)

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

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

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

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

 

c.   \(\text{Convert cm to mm:}\)

\(10.2 \ \text{cm}\times 10=102 \ \text{mm}\)

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

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

Combinatorics, EXT1 A1 2025 HSC 13e

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

  3. Hence, or otherwise, prove that
  4. \(\displaystyle\binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\).   (2 marks)
Show Answers Only

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

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

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

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)
 

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

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

\(\operatorname{LHS}\) \(=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots\)
  \(\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]\)
  \(=1-1+\displaystyle \binom{2051}{2001}\)
  \(=\displaystyle \binom{2051}{2001}\)
Show Worked Solution

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

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

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

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)
 

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

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

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

Vectors, EXT1 V1 2025 HSC 13c

At time \(t\), a particle has position vector  \(\underset{\sim}{r}(t)=t \underset{\sim}{i}+\dfrac{t^2}{9} \underset{\sim}{j}\), velocity vector \(\underset{\sim}{v}(t)\) and acceleration vector \(\underset{\sim}{a}(t)\).

Find the time when the angle between \(\underset{\sim}{v}(t)\) and \(\underset{\sim}{a}(t)\) is \(\dfrac{\pi}{4}\).   (4 marks)

Show Answers Only

\(t=\dfrac{9}{2}\)

Show Worked Solution

\(\underset{\sim}{r}(t)=t \underset{\sim}{i}+\dfrac{t^2}{9} \underset{\sim}{j}, \ \ \underset{\sim}{v}(t)=\underset{\sim}{i}+\dfrac{2}{9} t \underset{\sim}{j}, \ \ \underset{\sim}{a}(t)=\dfrac{2}{9} \underset{\sim}{j}\)

\(\underset{\sim}{v}=\displaystyle \binom{1}{\frac{2}{9} t}, \quad \underset{\sim}{a}=\displaystyle \binom{0}{\frac{2}{9}}\)

\(\abs{\underset{\sim}{v}}=\sqrt{1^2+\left(\dfrac{2}{9} t\right)^2}=\sqrt{1+\dfrac{4}{81} t^2}\)

\(\abs{\underset{\sim}{a}}=\sqrt{\left(\dfrac{2}{9}\right)^2}=\dfrac{2}{9}\)
 

\(\text{Angle between vectors}=\dfrac{\pi}{4}:\)

\(\cos \dfrac{\pi}{4}\) \(=\dfrac{1 \times 0+\dfrac{2}{9} t \times \dfrac{2}{9}}{\dfrac{2}{9} \times \sqrt{1+\dfrac{4}{81} t^2}}\)
\(\dfrac{1}{\sqrt{2}}\) \(=\dfrac{\dfrac{2}{9} t}{\sqrt{1+\dfrac{4}{81} t^2}}\)
\(\sqrt{1+\dfrac{4}{81} t^2}\) \(=\dfrac{2 \sqrt{2}}{9} t\)
\(1+\dfrac{4}{81} t^2\) \(=\dfrac{8}{81} t^2\)
\(81+4 t^2\) \(=8 t^2\)
\(4 t^2\) \(=81\)
\(t^2\) \(=\dfrac{81}{4}\)
\(t\) \(=\dfrac{9}{2} \quad(t>0)\)

Calculus, EXT1 C3 2025 HSC 13a

It is given that  \(\dfrac{d y}{d x}=\dfrac{5}{y}\)  and  \(y=-4\)  when  \(x=0\).

Find \(y\) as a function of \(x\).   (3 marks)

Show Answers Only

\(y=-\sqrt{10 x+16}\)

Show Worked Solution
\(\dfrac{dy}{dx}\) \(=\dfrac{5}{y}\)
\(\displaystyle \int y\,dy\) \(=\displaystyle \int 5 \,d x\)
\(\dfrac{y^2}{2}\) \(=5 x+c\)

 

\(\text{Given} \ \ y=-4 \ \ \text{when} \ \ x=0:\)

\(\dfrac{(-4)^2}{2}\) \(=0+c \ \Rightarrow \ c=8\)
\(\dfrac{y^2}{2}\) \(=5 x+8\)
\(y^2\) \(=10 x+16\)
\(y\) \(=-\sqrt{10 x+16} \quad \text{(Since \((0,-4)\) lies on graph)}\)

Trigonometry, EXT1 T3 2025 HSC 12e

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

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

Show Answers Only

i.    \(\text{See Worked Solutions}\)

ii.   \(x=\dfrac{\pi}{3}, \pi\)

Show Worked Solution

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

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

\(\text{Equating coefficients:}\)

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

\(\text{Since} \ \ \cos \alpha>0 \ \ \text{and} \ \ \sin \alpha>0\)

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

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

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

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

Calculus, EXT1 C3 2025 HSC 12d

Find the solution of  \(\dfrac{dy}{dx}=\sqrt{(2-y)(2+y)}\), given that  \(y=1\)  when  \(x=0\).   (3 marks)

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\(y=2\, \sin \left(x+\dfrac{\pi}{6}\right)\)

Show Worked Solution
\(\dfrac{d y}{d x}\) \(=\sqrt{(2-y)(2+y)}\) \(=\sqrt{4-y^2}\)
\(\dfrac{d x}{d y}\) \(=\dfrac{1}{\sqrt{4-y^2}}\)
\(\displaystyle \int d x\) \(=\displaystyle \int \dfrac{1}{\sqrt{4-y^2}} d y\)
\(x\) \(=\sin ^{-1}\left(\dfrac{y}{2}\right)+c\)

 

\(\text{When} \ \ x=0, y=1:\)

\(0=\sin ^{-1}\left(\dfrac{1}{2}\right)+c \ \ \Rightarrow \ \ c=-\dfrac{\pi}{6}\)

\(x\) \(=\sin ^{-1}\left(\dfrac{y}{2}\right)-\dfrac{\pi}{6}\)
\(\sin ^{-1}\left(\dfrac{y}{2}\right)\) \(=x+\dfrac{\pi}{6}\)
\(\dfrac{y}{2}\) \(=\sin \left(x+\dfrac{\pi}{6}\right)\)
\(y\) \(=2\, \sin \left(x+\dfrac{\pi}{6}\right)\)

Proof, EXT1 P1 2025 HSC 12c

Prove by mathematical induction that

\(1 \times(1!)+2 \times(2!)+\cdots+n \times(n!)=(n+1)!-1\)

for integers  \(n \geq 1\).   (3 marks)

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\(\text{See Worked Solution}\)

Show Worked Solution

\(1 \times 1!+2 \times 2!+\ldots+n \times n!=(n+1)!-1\)

\(\text{Prove true for}\ \ n=1:\)

\(\text{LHS}=1 \times 1!=1\)

\(\text{RHS}=(2)!-1=1=\text{LHS }\)

\(\therefore\ \text{True for}\ \ n=1.\)
 

\(\text{Assume true for} \ \ n=k:\)

\(1 \times 1!+2 \times 2!+\ldots+k \times k!=(k+1)!-1\)
 

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

\(\text{i.e.} \ \ 1 \times 1!+2 \times 2!+\ldots +k \times k!+(k+1) \times(k+1)!=(k+2)!-1\)

\(\text{LHS}\) \(=1\times 1!+\ldots+k \times k!+(k+1) \times(k+1)!\)
  \(=(k+1)!-1+(k+1) \times (k+1)!\)
  \(=(k+1)!(1+k+1)-1\)
  \(=(k+1)!(k+2)-1\)
  \(=(k+2)!-1\)
  \(= \operatorname{RHS}\)

 

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

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

Calculus, EXT1 C3 2025 HSC 12b

Consider the region bounded by the hyperbola  \(y=\dfrac{1}{x}\),  the \(y\)-axis and the lines  \(y=1\)  and  \(y=a\)  for  \(a>1\).

Find the volume of the solid of revolution formed when the region is rotated about the \(y\)-axis.   (2 marks)

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\(\pi\left(1-\dfrac{1}{a}\right)\ \text{u}^3\)

Show Worked Solution

\(y=\dfrac{1}{x} \ \Rightarrow \ x^2=\dfrac{1}{y^2}\)

\(V\) \(=\pi \displaystyle \int_1^a x^2\, dy\)
  \(=\pi \displaystyle \int_1^a y^{-2}\, d y\)
  \(=-\pi\left[y^{-1}\right]_1^a\)
  \(=-\pi\left(\dfrac{1}{a}-1\right)\)
  \(=\pi\left(1-\dfrac{1}{a}\right)\ \text{u}^3\)

Calculus, EXT1 C1 2025 HSC 12a

The radius, \(r\) cm, and angle, \(\theta\) radians, of a sector vary in such a way that its area remains a constant 10 cm².
 

The angle \(\theta\) is increasing at a constant rate of 2 radians per second.

Find the rate at which the radius is changing when the radius is 4 cm.   (3 marks)

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\(\text{Radius is decreasing at} \ \dfrac{16}{5} \ \text{cm per second.}\)

Show Worked Solution

\(\dfrac{d \theta}{d t}=2\)

\(\dfrac{dv}{d t}=\dfrac{dv}{d \theta} \times \dfrac{d \theta}{d t}\)

\(A\) \(=\dfrac{\theta}{2 \pi} \times \pi r^2\)  
\(10\) \(=\dfrac{\theta}{2} \times r^2\)  
\(\theta\) \(=20\, r^{-2}\)  

 
\(\dfrac{d \theta}{dv}=-40 r^{-3} \)

\(\dfrac{dv}{d \theta}=-\dfrac{r^3}{40}\)
 

\(\text{ Find \(\dfrac{dv}{dt}\)  when  \(r=4\):}\)

\(\dfrac{dv}{dt}=-\dfrac{4^3}{40} \times 2=-\dfrac{16}{5}\)

\(\therefore \ \text{Radius is decreasing at} \ \dfrac{16}{5} \ \text{cm per second.}\)

Calculus, EXT1 C2 2025 HSC 11g

Evaluate \(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3 x) d x\).   (3 marks)

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\(\dfrac{\pi}{12}\)

Show Worked Solution
\(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3x) dx\) \(=\displaystyle \dfrac{1}{2} \int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} (\cos6x+1) dx \)
  \(=\dfrac{1}{2}\left[\dfrac{1}{6} \sin 6 x+x\right]_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}}\)
  \(=\dfrac{1}{2}\left[\left(\dfrac{1}{6} \sin 2 \pi+\dfrac{\pi}{3}\right)-\left(\dfrac{1}{6} \sin \pi+\dfrac{\pi}{6}\right)\right]\)
  \(=\dfrac{1}{2}\left(\dfrac{\pi}{3}-\dfrac{\pi}{6}\right)\)
  \(=\dfrac{\pi}{12}\)

Trigonometry, EXT1 T1 2025 HSC 11d

Sketch the graph of  \(y=\dfrac{1}{3} \cos ^{-1}(2 x)\).   (2 marks)

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Show Worked Solution

\(y=\dfrac{1}{3} \cos ^{-1}(2 x)\)

\(\text{Domain:} \ -1 \leqslant 2 x \leqslant 1 \ \Rightarrow \ -\dfrac{1}{2} \leqslant x \leqslant \dfrac{1}{2}\)

\(\text{Range:} \ \cos (2 x) \in[0, \pi] \  \Rightarrow \  \dfrac{1}{3} \cos (2x) \in\left[0, \dfrac{\pi}{3}\right]\)
 

Calculus, EXT1 C2 2025 HSC 11c

Find \(\displaystyle \int \sin 3x \, \cos x \, dx\).   (2 marks)

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\(-\dfrac{1}{8} \cos 4 x-\dfrac{1}{4} \cos 2 x+c\)

Show Worked Solution
\(\displaystyle\int \sin3x \, \cos x \, dx\) \(=\displaystyle\frac{1}{2} \int \sin 4 x+\sin2x \,dx\)
  \(=-\dfrac{1}{8} \cos 4 x-\dfrac{1}{4} \cos 2 x+c\)

Calculus, EXT1 C2 2025 HSC 3 MC

Consider the integral  \(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\).

The substitution  \(x=5 \sin \theta\)  is applied.

Which of the following is obtained?

  1. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec} \theta \, d \theta\)
  2. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \,  d \theta\)
  3. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec}^2 \theta \, d \theta\)
  4. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec ^2 \theta \, d \theta\)
Show Answers Only

\(\Rightarrow B\)

Show Worked Solution

\(x=5\, \sin \theta\)

\(\dfrac{dx}{d \theta}=5\, \cos \theta \ \Rightarrow \ dx=5\, \cos \theta \, d \theta\)

\(\text{When} \ \ x=\dfrac{5}{2} \ \Rightarrow \ \sin\, \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)

\(\text{When} \ \ x=-\dfrac{5}{2} \ \Rightarrow \ \sin \theta=-\dfrac{1}{2} \ \Rightarrow \ \theta=-\dfrac{\pi}{6}\)

\(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\) \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\left(\dfrac{1}{25-25\, \sin ^2 \theta}\right) \cdot 5\  \cos \theta \, d \theta\)
  \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{5\, \cos \theta}{25\, \cos ^2 \theta} \,  d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{1}{\cos \theta} \, d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \, d \theta\)

\(\Rightarrow B\)

Calculus, EXT1 C3 2025 HSC 7 MC

A slope field is shown.
 

Which of the following could be the differential equation represented by the slope field?

  1. \(\dfrac{d y}{d x}=x^2\)
  2. \(\dfrac{d y}{d x}=x^2+C, C \neq 0\)
  3. \(\dfrac{d y}{d x}=x^3\)
  4. \(\dfrac{d y}{d x}=x^3+C, C \neq 0\)
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\(A\)

Show Worked Solution

\(\text{For all \(x<0\), gradients are positive (from graph):}\)

\(\text{Eliminate C and D.}\)

\(\text{At \(x=0\), gradient = 0 (from graph):}\)

\(\text{Eliminate B.}\)

\(\Rightarrow A\)

Calculus, EXT1 C3 2025 HSC 6 MC

Given that \(a\) is a non-zero constant, which of the following integrals is equal to zero?

  1. \(\displaystyle \int_{-a}^a x\, \cos ^{-1}(x) d x\)
  2. \(\displaystyle\int_{-a}^a x^2\, \cos ^{-1}(x) d x\)
  3. \(\displaystyle\int_{-a}^a x\, \tan ^{-1}(x) d x\)
  4. \(\displaystyle\int_{-a}^a x^2\, \tan ^{-1}(x) d x\)
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\(D\)

Show Worked Solution

\(\text{Since the limits are symmetrical about 0:}\)

\(\text{Integral will equal zero if function is odd.}\)

\(\text{Consider option D:}\)

\(f(x)=x^2\, \tan^{-1}(x)\)

\(f(-x)=(-x)^2\, \tan^{-1}(-x)=-x^2\, \tan ^{-1}(x)=-f(x)\ \text{(odd)}\)

\(\Rightarrow D\)

Statistics, EXT1 S1 2025 HSC 4 MC

A Bernoulli random variable \(X\) has probability distribution

\(P(x)=\dfrac{x+1}{3}\)  for  \(x=0,1\).

What are the mean and variance of \(X\) ?

  1. \(E(X)=\dfrac{1}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{9}\)
  2. \(E(X)=\dfrac{1}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{3}\)
  3. \(E(X)=\dfrac{2}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{9}\)
  4. \(E(X)=\dfrac{2}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{3}\)
Show Answers Only

\(C\)

Show Worked Solution

\(P(0)=\dfrac{1}{3}, \ P(1)=\dfrac{2}{3} \)

\(E(X) = \dfrac{1}{3} \times 0 + \dfrac{2}{3} \times 1 = \dfrac{2}{3}\)

\(E(X^2) = \dfrac{1}{3} \times 0^2 + \dfrac{2}{3} \times 1^2 = \dfrac{2}{3} \)

\(\text{Var}(X) = E(X^2)-E(X)^2 = \dfrac{2}{3}-\dfrac{4}{9}=\dfrac{2}{9} \)

\(\Rightarrow C\)

Trigonometry, 2ADV T3 2025 HSC 15

A sound wave can be modelled using a function  \(P(t)=k\, \sin a t\), where \(P\) is air pressure in Pascals, \(t\) is time in milliseconds (ms) and \(k\) and \(a\) are constants.

  1. Write the equation for a sound wave \(P_1(t)\) that has an amplitude of 2 Pascals and a period of 5 ms.   (2 marks)

  2. The graph of \(P_1(t)\) from part (a) is shown.
  3. On the diagram, sketch the graph of  \(P_2(t)=4 \sin \left(\dfrac{\pi}{10} t\right)\)  for  \(0 \leq t \leq 10\).   (2 marks)
     

  1. Hence, find the values of \(t\), where  \(0<t<10\),  for which functions \(P_1(t)\) and \(P_2(t)\) are BOTH decreasing.   (2 marks)

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a.  \(P_1(t)=2\, \sin \left(\dfrac{2 \pi t}{5}\right)\)

b.   
       

c.  \(\text{Both decreasing for} \ \ 6.25<t<8.75\)

Show Worked Solution

a.    \(\text{Amplitude}=2 \Rightarrow k=2\)

\(\text{Period}=5\)

\(\dfrac{2 \pi}{a}\) \(=5\)
\(5a\) \(=2 \pi\)
\(a\) \(=\dfrac{2 \pi}{5}\)

 
\(\therefore P_1(t)=2\, \sin \left(\dfrac{2 \pi t}{5}\right)\)
 

b.   
     

\(P_2(t)=4\, \sin \left(\dfrac{\pi}{10} t\right)\)

\(\text{Amplitude}=4\)

\(\text{Period}=\dfrac{2 \pi}{\frac{\pi}{10}}=20\)
 

c.    \(\text {By inspection of graph:}\)

\(P_2(t) \ \text {is decreasing for} \ \ 5<t \leq10\)

\(P_1(t) \text { is decreasing for} \ \ 1.25<t<3.75 \ \ \text{and}\ \ 6.25<t<8.75\)

\(\therefore \ \text{Both decreasing for} \ \ 6.25<t<8.75\)

Functions, 2ADV F2 2025 HSC 6 MC

The graph of  \(y=f(x)\) is shown.  
 

Which of the following is the graph of   \(y=-f(-x)\) ?
 

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\(C\)

Show Worked Solution

\(y=-f(x)\ \ \Rightarrow\ \ \text{Reflect \(f(x)\) in the \(x\)-axis.}\)

\(y=-f(-x)\ \ \Rightarrow\ \ \text{Reflect \(-f(x)\) in the \(y\)-axis.}\)

\(\text{In this combination of translations, the order is not important.}\)

\(\Rightarrow C\)