HSC Maths Exam Paper Solutions

Embrace HSC Exams with confidence

We provide solutions to popular math exams and prepare you to nail your exam
PLATO

2023 HSC Mathsc Extension 2 (4U) Exam Paper Solutions

Section I :

Question 1.

Which of the following is equal to $(a+i b)^3$ ?
  1. $\left(a^3-3 a b^2\right)+i\left(3 a^2 b+b^3\right)$
  2. $\left(a^3+3 a b^2\right)+i\left(3 a^2 b+b^3\right)$
  3. $\left(a^3-3 a b^2\right)+i\left(3 a^2 b-b^3\right)$
  4. $\left(a^3+3 a b^2\right)+i\left(3 a^2 b-b^3\right)$
Answer Solution
C $$ \begin{aligned} (a+i b)^3 & =a^3+3\left(a^2\right)(i b)+3(a)(i b)^2+(i b)^3 \\ & =\left(a^3-3 a b^2\right)+i\left(3 a^2 b-b^3\right) \end{aligned} $$

Question 2.

Consider the following statement. If an animal is a herbivore, then it does not eat meat.
  1. If an animal is a herbivore, then it eats meat.
  2. If an animal is not a herbivore, then it eats meat.
  3. If an animal eats meat, then it is not a herbivore.
  4. If an animal does not eat meat, then it is a herbivore.
Answer Solution
D statement: $P \rightarrow Q$.
converse: $Q \rightarrow P$ : if it does not eat meat, then it is a herbivore.

Question 3.

A complex number $z$ lies on the unit circle in the complex plane, as shown in the diagram.
Which of the following complex numbers is equal to $\bar{z}$ ?
  1. $-z$
  2. $z^2$
  3. $-z^3$
  4. $z^4$
Answer Solution
B $\arg (z)=\pi+\frac{\pi}{3}=\frac{4 \pi}{3} \quad|z|=1 \quad \therefore \text { it lies on unit circle } \\$ $\arg (\bar{z})=-\arg (z)=-\frac{4 \pi}{3}=\frac{2 \pi}{3} \\$ $$ \begin{aligned} & \text { A: } \arg (-z)=\pi+\arg (z)=\pi+\frac{4 \pi}{3}=\frac{7 \pi}{3}=\frac{\pi}{3} & {\color{red}\mathit{X}} \\ & \text { B: } \arg \left(z^2\right)=2 \arg (z)=2 \times \frac{4 \pi}{3}=\frac{8 \pi}{3}=\frac{2 \pi}{3} & {\color{green}\checkmark} \\ & \text { C: } \arg \left(-z^3\right)=\pi+3 \arg (z)=\pi+3 \times \frac{4 \pi}{3}=5 \pi & {\color{red}\mathit{X}} \\ & \text { D: } \arg \left(z^4\right)=4 \arg (z)=4 \times \frac{4 \pi}{3}=\frac{16 \pi}{3}=\frac{4 \pi}{3} & {\color{red}\mathit{X}} \end{aligned} $$

Question 4.

Consider the following statement about real numbers. Whichever positive number $r$ you pick, it is possible to find a number $x$ greater than $1$ such that $\frac{\ln x}{x^3}\lt r$. When this statement is written in the formal language of proof, which of the following is obtained?
  1. $\forall x>1 \quad \exists r>0 \quad \frac{\ln x}{x^3}\lt r$
  2. $\exists x>1 \quad \forall r>0 \quad \frac{\ln x}{x^3}\lt r$
  3. $\forall r>0 \quad \exists x>1 \quad \frac{\ln x}{x^3}\lt r$
  4. $\exists r>0 \quad \forall x>1 \quad \frac{\ln x}{x^3}\lt r$
Answer Solution
C $$ \left.\begin{array}{lcl} \text{"Whichever positive number } r \text{ you pick, } & = & \forall r>0,\\ \text{it is possible to find a number } x \text{ greater than } 1 & = &\exists x>1, \\ \text{such that } \frac{\ln x}{x^3} \lt r \text{ "} & = & \frac{\ln x}{x^3} \lt r. \end{array}\right\} $$

Question 5.

Which of the following is a true statement about the lines $\ell_1=\left(\begin{array}{c}-1 \\ 2 \\ 5\end{array}\right)+\lambda\left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right)$ and $ \ell_2=\left(\begin{array}{c} 3 \\ -10 \\ 1 \end{array}\right)+\mu\left(\begin{array}{c} 1 \\ -3 \\ -1 \end{array}\right)$ ?
  1. $\ell_1$ and $\ell_2$ are the same line.
  2. $\ell_1$ and $\ell_2$ are not parallel and they intersect.
  3. $\ell_1$ and $\ell_2$ are parallel and they do not intersect.
  4. $\ell_1$ and $\ell_2$ are not parallel and they do not intersect.
Answer Solution
A Consider $l_1=\left(\begin{array}{c}-1 \\ 2 \\ 5\end{array}\right)+\lambda\left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right)$, if $\lambda=-4,\ l_1=\left(\begin{array}{c}3 \\ -10 \\ 1\end{array}\right)$, which is the position vector of $l_2$.
Consider $l_2=\left(\begin{array}{c}3 \\ -10 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}1 \\ -3 \\ -1\end{array}\right)$, if $\mu=-4, l_2=\left(\begin{array}{c}-1 \\ 2 \\ 5\end{array}\right)$, which is the position vector of $l_1$. Also, the direction vector of $l_1=-1\ \times$ direction vector of $l_2 \Rightarrow$ same line.

Question 6.

Which of the following functions does NOT describe simple harmonic motion?
  1. $x=\cos ^2 t-\sin 2 t$
  2. $x=\sin 4 t+4 \cos 2 t$
  3. $x=2 \sin 3 t-4 \cos 3 t+5$
  4. $x=4 \cos \left(2 t+\frac{\pi}{2}\right)+5 \sin \left(2 t-\frac{\pi}{4}\right)$
Answer Solution
B A function describes SHM if $\ddot{x}=-n^2(x-c)$ $$ \begin{array}{l} \text { A: } & x & =\cos ^2 t-\sin 2 t & \\ &\dot{x} & =-2 \cos t \sin t-2 \cos 2 t=-\sin 2 t-2 \cos 2 t & \\ &\ddot{x} & =-2 \cos 2 t+4 \sin 2 t \\ &\ddot{x} & =-2\left(2 \cos ^2 t-1\right)+4 \sin 2 t & \\ &\ddot{x} & =-4\left(\cos ^2 t-\sin ^2 t-1\right) & \\ &\therefore\ \ddot{x} & =-4(x-1) & \therefore \text { SHH } \\ \\ \text { B: } & x & =\sin 4 t+4 \cos 2 t & \\ &\dot{x} & =4 \cos 4 t-8 \sin 2 t & \\ &\ddot{x} & =-16 \sin 4 t-16 \cos 2 t & \\ &\ddot{x} & =-16(\sin 4 t-\cos 2 t) & \therefore \text { NOT SHH } \\ \\ \text { C: } & x & =2 \sin 3 t-4 \cos 3 t+5 & \\ &\dot{x} & =6 \cos 3 t+12 \sin 3 t & \\ &\ddot{x} & =-18 \sin 3 t+36 \cos 3 t & \\ &\ddot{x} & =-9(2 \sin 3 t-4 \cos 3 t) & \\ &\therefore\ \ddot{x} & =-9(x-5) & \therefore \text { SHH } \\ \\ \text { D: } & x & =4 \cos \left(2 t+\frac{\pi}{2}\right)+5 \sin \left(2 t-\frac{\pi}{4}\right) & \\ &\dot{x} & =-8 \sin \left(2 t+\frac{\pi}{2}\right)+10 \cos \left(2 t-\frac{\pi}{4}\right) & \\ &\ddot{x} & =-16 \cos \left(2 t+\frac{\pi}{2}\right)-20 \sin \left(2 t-\frac{\pi}{4}\right) & \\ &\ddot{x} & =-4\left(4 \cos \left(2 t+\frac{\pi}{2}\right)+5 \sin \left(2 t-\frac{\pi}{4}\right)\right) & \\ &\therefore\ \ddot{x} & =-4 x & \therefore \text { SHH } \end{array} $$

Question 7.

Which of the following statements about complex numbers is true?
  1. For all real numbers $x, y, \theta$ with $x \neq 0$, $$ \tan \theta=\frac{y}{x} \Rightarrow x+i y=r e^{i \theta}, $$ for some real number $r$.
  2. For all non-zero complex numbers $z_1$ and $z_2$, $$ \operatorname{Arg}\left(z_1\right)=\theta_1 \text { and } \operatorname{Arg}\left(z_2\right)=\theta_2 \Rightarrow \operatorname{Arg}\left(z_1 z_2\right)=\theta_1+\theta_2, $$ where $\operatorname{Arg}$ denotes the principal argument.$
  3. For all real numbers $r_1, r_2, \theta_1, \theta_2$ with $r_1, r_2>0$, $$ r_1 e^{i \theta_1}=r_2 e^{i \theta_2} \Rightarrow r_1=r_2 \text { and } \theta_1=\theta_2 \text {. } $$
  4. For all real numbers $x, y, r, \theta$ with $r>0$ and $x \neq 0$, $$ x+i y=r e^{i \theta} \Rightarrow \theta=\arctan \left(\frac{y}{x}\right) \text {. } $$
Answer Solution
A $$ \begin{array}{llc} \text{A: }\quad & \forall x, y, \theta \in \mathbb{R}, x \neq 0, \quad \tan \theta \cdot \frac{y}{x} \Rightarrow x+i y=r e^{i \theta} & \\ & \text{for some real } r & \text{(A)} \\ \\ \text{B: }\quad & \text{for } z_1 \notin z_2 \in \mathbb{C} \quad\text{ s.t }\quad z_1 \neq 0, z_2 \neq 0, & \\ &\operatorname{Arg}\left(z_1\right)=\theta_1 \neq \operatorname{Arg}\left(z_2\right)=\theta_2 \Rightarrow \operatorname{Arg}\left(z_1 z_2\right)=\theta_1+\theta_2 & \\ & \text{consider } z_1=e^{i \pi} \text{ and } z_2=e^{i \pi} \Rightarrow z_1 z_2=e^{i 2 \pi}=1 & \\ & \operatorname{Arg}\left(z_1 z_2\right) \text{ in this case is $0$, which is not } \theta_1+\theta_2 & \text{ NOT (B) } \\ \\ \text{C: }\quad & \forall\ r_1,\ r_2,\ \theta_1,\ \theta_2 \in \mathbb{R},\ r_1,\ r_2>0, &\\ & r_1 e^{i \theta_1}=r_2 e^{i \theta_2} \Rightarrow r_1=r_2 \text{ and } \theta_1=\theta_2 &\\ & \text{consider } e^{i \pi}=e^{i 3 \pi}, \text{ but } \pi \neq 3 \pi. \text{ not always true }& \text{NOT (C)} \\ \\ \text{D: }\quad & \forall\ x,\ y,\ r,\ \theta,\ \text{ with } r>0 \text{ and } x \neq 0, &\\ & x+i y=r e^{i \theta} \Rightarrow \theta=\arctan \left(\frac{y}{x}\right) & \\ & \text{consider } -1+i=\sqrt{2} e^{i\left(\frac{3 \pi}{4}\right)} \text{ but } \arctan \left(\frac{1}{-1}\right)=-\frac{\pi}{4} \neq \frac{3 \pi}{4} & \text{NOT (D)} \end{array} $$

Question 8.

A shaded region on a complex plane is shown.
Which relation best describes the region shaded on the complex plane?
  1. $|z-i|>2|z-1|$
  2. $|z-i|\lt 2|z-1|$
  3. $|z-1|>2|z-i|$
  4. $|z-1|\lt 2|z-i|$
Answer Solution
D $\begin{array}{|ll|l|l|c|} \hline \\ & & \text{test } (0,1) & \text{test } (0,-1) & \\ & & \text{this should make} & \text{this should make} & \\ & & \text{the statement false} & \text{the statement false} & \\ \\ \hline \\ \text{A:} & |z-i|>2|z-1| & |0|>2|-1+i| \quad {\color{red} \times} & |-2 i|>2|-1-i| \quad {\color{red} \times} & \text{NOT (A)} \\ \\ \hline \\ \text{B:} & |z-i| \lt 2|z-1| & |0| \lt 2|-1+i| \quad {\color{green} \checkmark} & |-2 i| \lt 2|-1-i| \quad {\color{green} \checkmark} & \text{NOT (B)} \\ \\ \hline \\ \text{C:} & |z-i|>2|z-1| & |-1+i|>2|0| \quad {\color{green} \checkmark} & |-1-i|>2|-2 i| \quad {\color{red} \times} & \text{NOT (C)} \\ \\ \hline \\ \text{D:} & |z-i|>2|z-1| & |-1+i| \lt 2|0| \quad {\color{green} \times} & |-1-i| \lt 2|-2 i| \quad {\color{red} \checkmark} & \text{(D)} \\ \\ \hline \end{array}$

Question 9.

A particle travels along a curve from $O$ to $E$ in the $xy$-plane, as shown in the diagram.
The position vector of the particle is $\boldsymbol{r}$, its velocity is $\boldsymbol{v}$, and its acceleration is $\boldsymbol{a}$.
While travelling from $O$ to $E$, the particle is always slowing down.
Which of the following is consistent with the motion of the particle?
  1. $\boldsymbol{r} \cdot \boldsymbol{v} \leq 0$ and $\boldsymbol{a} \cdot \boldsymbol{v} \geq 0$
  2. $\boldsymbol{r} \cdot \boldsymbol{v} \leq 0$ and $\boldsymbol{a} \cdot \boldsymbol{v} \leq 0$
  3. $\boldsymbol{r} \cdot \boldsymbol{v} \geq 0$ and $\boldsymbol{a} \cdot \boldsymbol{v} \geq 0$
  4. $\boldsymbol{r} \cdot \boldsymbol{v} \geq 0$ and $\boldsymbol{a} \cdot \boldsymbol{v} \leq 0$
Answer Solution
D
$\because$ the particle is moving away from $0$
$\therefore$ $\underset{\sim}{r} \cdot \underset{\sim}{v} \geq 0$ ie. $\underset{\sim}{r}$ and $\underset{\sim}{v} $ are in the same direction.
$\because$ the particle is slowing down,
$\therefore$ $\underset{\sim}{a} \cdot \underset{\sim}{v} \leq 0$ ie. $\underset{\sim}{a}$ and $\underset{\sim}{v} $ are working against each other.

Question 10.

Consider any three-dimensional vectors $\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}$ and $\underset{\sim}{c}=\overrightarrow{O C}$ that satisfy these three conditions $$ \begin{aligned} & \underset{\sim}{a} \cdot \underset{\sim}{b}=1 \\ & \underset{\sim}{b} \cdot \underset{\sim}{c}=2 \\ & \underset{\sim}{c} \cdot \underset{\sim}{a}=3 . \end{aligned} $$ Which of the following statements about the vectors is true?
  1. Two of $\underset{\sim}{a}, \underset{\sim}{b}$ and $\underset{\sim}{c}$ could be unit vectors.
  2. The points $A, B$ and $C$ could lie on a sphere centred at $O$.
  3. For any three-dimensional vector $\underset{\sim}{a}$, vectors $\underset{\sim}{b}$ and $\underset{\sim}{c}$ can be found so that $\underset{\sim}{a}, \underset{\sim}{b}$ and $\underset{\sim}{c}$ satisfy these three conditions.
  4. $\forall \underset{\sim}{a}, \underset{\sim}{b}$ and $\underset{\sim}{c}$ satisfying the conditions, $\exists r, s$ and $t$ such that $r, s$ and $t$ are positive real numbers and $r \underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}$.
Answer Solution
B $$ \begin{aligned} & \underset{\sim}{a} \cdot \underset{\sim}{b}=1 \\ & \underset{b}{c} \cdot \underset{\sim}{c}=2 \\ & \underset{\sim}{c} \cdot \underset{\sim}{a}=3 \end{aligned} $$ $ \begin{array}{llc} \text{A: } & \text{ if } \underset{\sim}{a} \text{ and } \underset{\sim}{b} \text{ are the } 2 \text{unit vectors, then } \underset{\sim}{a}-\underset{\sim}{b}=1 \text{ implies } \underset{\sim}{a}=\underset{\sim}{b}, & \\ & \text{ if } \underset{\sim}{a}=\underset{\sim}{b}, \text{ then } \underset{\sim}{b} \cdot \underset{\sim}{c}=\underset{\sim}{a} \cdot \underset{\sim}{c} \text{ but } 2 \neq 3 & \\ & \text{ This will also occur if } \underset{\sim}{a} \neq \underset{\sim}{c} \text{ or } \underset{\sim}{b} \neq \underset{\sim}{c} \text{ are the unit vectors} & \text{ NOT (A)} \\ \\ \text{B: } & \text{ if } \underset{\sim}{a}, \underset{\sim}{b} \text{ and } \underset{\sim}{c} \text{ lied on a sphere centred at } O \text{ with radius } R, & \\ & \Rightarrow|\underset{\sim}{a}|=|\underset{\sim}{b}|=|\underset{\sim}{c}|=R & \\ & \because \text{ there is no limitations on the megnatudes of the vector } \underset{\sim}{a} \cdot \underset{\sim}{b}\neq\underset{\sim}{c} & \\ & \therefore \text{ B is true} & \text{ (B)} \end{array} $
$\begin{aligned} \begin{array}{llcl} & \underset{\sim}{a} \cdot \underset{\sim}{b} & = & \left|\underset{\sim}{a}\right|\left|\underset{\sim}{b}\right| \cos \theta_1 \\ & 1 & = & R^2\cos\theta \\ & \cos\theta_1 & = & \frac1{R^2} \end{array} & \quad \begin{array}{llcl} & \underset{\sim}{b} \cdot \underset{\sim}{c} & = & \left|\underset{\sim}{b}\right|\left|\underset{\sim}{c}\right| \cos \theta_2 \\ & 2 & = & R^2\cos\theta_2 \\ & \cos\theta_2 & = & \frac2{R^2} \end{array} & \quad \begin{array}{llcl} & \underset{\sim}{a} \cdot \underset{\sim}{c} & = & \left|\underset{\sim}{a}\right|\left|\underset{\sim}{c}\right| \cos \theta_3 & \\ & 3 & = & R^2\cos\theta_3 & \\ & \cos\theta_3 & = & \frac1{R_2} & \end{array} \end{aligned}$

ie. it is possible that $A, B \notin C$ cold lie on a sphere.

$ \begin{array}{llc} \text{C: } & \text{ consider if } \underset{\sim}{a} \text{ or } \underset{\sim}{b} \text{ or } \underset{\sim}{c} \text{ were the zero vector}, & \\ & \text{ the conditions would not be satisfied} & \text{ NOT (C)} \\ \\ \text{D: } & \text{consider if } \underset{\sim}{a}=(1,0,3), \quad \underset{\sim}{b}=(1,2,0) \text{ and } \underset{\sim}{c}=(0,1,1), & \\ & \text{ the 3 conditions would be satisfied but there exists No positive real values of } r, s, t & \\ & \text{ such that } r \underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=0 & \text{ NOT (D)} \end{array} $

Section II:

Question 11-a.

Solve the quadratic equation $$ z^2-3 z+4=0 $$ where $z$ is a complex number. Give your answers in Cartesian form.
Solution
$z^2-3 z+4=0 \quad$ where $z=x+i y \quad \forall x, y \in \mathbb{R}$, $$ \begin{array} & z=\frac{3 \pm \sqrt{9-16}}{2}=\frac{3 \pm \sqrt{-7}}{2}=\frac{3 \pm i \sqrt{7}}{2} \\ \boxed{\therefore z=\frac{3}{2}+i \frac{\sqrt{7}}{2} \text { or } z=\frac{3}{2}-i \frac{\sqrt{7}}{2}} \end{array} $$

Question 11-b.

Find the angle between the vectors $$ \begin{aligned} & \underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k} \\ & \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}, \end{aligned} $$ giving your answer to the nearest degree.
Solution
$ \begin{array}{ll} \underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k} & |\underset{\sim}{a}|=\sqrt{1+4+9}=\sqrt{14} \\ \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k} & |\underset{\sim}{b}|=\sqrt{1+16+4}=\sqrt{21} \end{array} $
$ \cos \theta=\frac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{|a||b|}=\frac{1}{\sqrt{14} \times \sqrt{21}} \Rightarrow \theta=87^{\circ} \quad \text { (nearest degree) } $

Question 11-c.

Find a vector equation of the line through the points $A(-3,1,5)$ and $B(0,2,3)$.
Solution
$ \begin{array}{lcl} \text{let } \overrightarrow{O A} \text{ be position vector } & \rightarrow & \overrightarrow{O A}=\left(\begin{array}{c}-3 \\ 1 \\ 5\end{array}\right) \\ \text{let } \overrightarrow{A B} \text{ be direction vector } & \rightarrow & \overrightarrow{A B}=\left(\begin{array}{l}0--3 \\ 2-1 \\ 3-5\end{array}\right)=\left(\begin{array}{c}3 \\ -1 \\ -2\end{array}\right) \\ \end{array} $
$ \therefore \underset{\sim}{r}=\left(\begin{array}{c} -3 \\ 1 \\ 5 \end{array}\right)+\lambda\left(\begin{array}{c} 3 \\ -1 \\ -2 \end{array}\right) \quad, \forall \lambda \in \mathbb{R} $

Question 11-d.

The quadrilaterals $A B C D$ and $A B E F$ are parallelograms.
By considering $\overrightarrow{A B}$, show that $C D F E$ is also a parallelogram.
Solution
Considering $\overrightarrow{A B}$,
$\overrightarrow{A B}=\overrightarrow{D C} \quad \because A B C D$ is a parallelogram,
$\overrightarrow{A B}=\overrightarrow{F E} \quad \because A B E F$ is a parallelogram,
$\therefore \overrightarrow{D C}=\overrightarrow{F E} \Rightarrow \overrightarrow{C D}=\overrightarrow{E F},
\therefore C D F E$ is a parallelogram.

Question 11-e.

A particle moves in simple harmonic motion described by the equation $$ \ddot{x}=-9(x-4) . $$ Find the period and the central point of motion.
Solution
$\ddot{x}=-9(x-4)$ which is in the form $x=-n^2(x-c)$
$ \begin{aligned} & n^2=9 & & \\ & \therefore n=3 & \rightarrow & \text { period }=\frac{2 \pi}{3} \\ & c=4 & \rightarrow & \text { centre: } x=4 \end{aligned} $

Question 11-f.

Find $\int_0^2 \frac{5 x-3}{(x+1)(x-3)} d x$.
Find the period and the central point of motion.
Solution
let $ \frac{5 x-3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}$
$5 x-3=A(x-3)+B(x+1)$
let $ x=3, \quad 12=4 B \rightarrow B=3$
let $x=-1,-8=-4 A \rightarrow A=2$
$ \begin{aligned} \int_0^2 \frac{5 x-3}{(x+1)(x-3)} d x & =\int_0^2 \frac{2}{x+1}+\frac{3}{x-3} d x \\ &=[2 \ln |x+1|+3 \ln |x-3|]_0^2 \\ &=2 \ln 3+\cancel{3 \ln} |\cancel{-2 \ln} |-3 \ln 3 \\ &= -\ln 3 \end{aligned} $

Question 12-a.

Prove that $\sqrt{23}$ is irrational.
Solution
Suppose $\sqrt{23}$ is rational $\Rightarrow \sqrt{23}=\frac{p}{q}$ for $p, q \in \mathbb{Z}, q \neq 0$ where $p$ and $q$ are coprime.

$ \begin{aligned} & \sqrt{23}=\frac{p}{q} \\ & 23=\frac{p^2}{q^2} \\ & p^2=23 q^2 \Rightarrow p^2 \text { is a mutriple of } 23 \\ & \Rightarrow p \text { is a multiple of } 23 \\ & \therefore \text { let } p=23 k, k \in \mathbb{Z} \\ & \end{aligned} $
$ \begin{aligned} (23 k)^2 & =23 q^2 \\ 23^2 k^2 & =23 q^2 \\ q^2 & =23 k^2 \\ & \Rightarrow q^2 \text { is a multiple of } 23 \\ & \Rightarrow q \text { is a multiple ot } 23 \end{aligned} $
but $p$ and $q$ must not share any common factors,
$\therefore$ this is a contradiction
$\therefore \sqrt{23}$ is irrational

Question 12-b.

Prove that for all real numbers $x$ and $y$, where $x^2+y^2 \neq 0$, $$ \frac{(x+y)^2}{x^2+y^2} \leq 2 \text {. } $$.
Solution
$ $$ \begin{aligned}{} \forall x, y \in \mathbb{R},\ (x-y)^2 & \geqslant 0 \\ x^2-2 x y+y^2 & \geqslant 0 \\ x^2+y^2 & \geqslant 2 x y \\ \because x^2+y^2 \neq 0 \quad \therefore \quad \frac{2 x y}{x^2+y^2} & \leqslant 1 & \text { (*) } \end{aligned} $$

Now, consider :

$$ \begin{aligned} \frac{(x + y)^2}{x^2 + y^2} & = \frac{x^2 + 2xy + y^2}{} \\ & = 1 + \frac{}{} \\ & \leqslant 1 + 1 & \text{using (*)} \\ & = 2 \\ \therefore \frac{}{} & \leqslant 2\ \forall x, y \in \mathbb{R} \end{aligned} $$

Question 12-c.

An object with mass $m$ kilograms slides down a smooth inclined plane with velocity $\underset{\sim}{v}(t)$, where $t$ is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle $\theta$ with the horizontal, as shown in the diagram. The normal reaction force is $\underset{\sim}{R}$. The acceleration due to gravity is $\underset{\sim}{g}$ and has magnitude $g$. No other forces act on the object.

The vectors $\underset{\sim}{i}$ and $\underset{\sim}{j}$ are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram.

Question 12-c-(i).

Show that the resultant force on the object is $\underset{\sim}{F} = -(m g \sin \theta)\underset{\sim}{i}$.
Solution

Question 12-d.

Find the cube roots of $2-2i$. Give your answer in exponential form.
Solution
Let $z=r e^{i \theta}$ such that $z=\sqrt[3]{2-2 i}$, where $2-2 i=2 \sqrt{2} e^{\frac{-i \pi}{4}}$
$ \begin{aligned} \therefore z^3=\left(r e^{i \theta}\right)^3 & =2 \sqrt{2} e^{-\frac{i \pi}{4}} \\ \Rightarrow r^3 e^{i(3 \theta)} & =2 \sqrt{2} e^{-i \pi / 4} \\ r^3=2 \sqrt{2} \Rightarrow r & =\sqrt{2} \\ 3 \theta & =-\frac{\pi}{4}, -\frac{\pi}{4} + 2\pi = \frac{7 \pi}{4}, -\frac{\pi}{4} - 2\pi = -\frac{9 \pi}{4}\\ \Rightarrow \theta & =-\frac{\pi}{12}, \frac{7 \pi}{12},-\frac{3 \pi}{4} \\ &\boxed{\therefore z = \sqrt{2}e^{\frac{-i\pi}{12}}, \sqrt{2}e^{\frac{i7\pi}{12}}, \sqrt{2}e^{\frac{-i3\pi}{12}}} \end{aligned} $

Question 12-e.

The complex number $2+i$ is a zero of the polynomial $P(z) = z^4 - 3z^3 + cz^2 + dz - 30$ where $c$ and $d$ are real numbers.

Question 12-e-(i).

Explain why $2-i$ is also a zero of the polynomial $P(z)$.
Solution
Since all coefficients of $P(z)$are real,
$\therefore,$ complex roots occur in conjugate pairs.
ie. if $2+i$ is a zero, then $2-i$ is also a zero.

Question 12-e-(ii).

Find the remaining zeros of the polynomial $P(z)$.
Solution
Let roots be $2+i$, $2-i$, $\alpha$, $\beta$.
$ \begin{array} \text{SUM: } & 4 + \alpha + \beta & = & 3 & \\ & \alpha + \beta & = & -1 & \text{(1)} \\ \text{PRODUCT: } & (2+i)(2-i)\alpha\beta & = & -30 & \\ & 5\alpha\beta & = & -30 & \\ & \therefore \alpha\beta & = & -6 & \text{(2)} \\ \end{array} $
$\alpha$ and $\beta$ are roots of the equation $x^2 + x - 6 = 0$
$ \begin{array} \ x^2 + x - 6 & = 0 \\ (x+3)(x-2) & = 0 \\ x = -3,\ x = 2 & \end{array} $
$\therefore$ zeros of $P(z)$ are $2+i$, $2-i$, $-3$ and $2$

Question 13-a.

Find $\int \frac{1-x}{\sqrt{5-4x - x^2}}dx$.
Solution
$ \begin{aligned} \int \frac{1-x}{\sqrt{5-4x - x^2}}dx & = \int \frac{1-x}{\sqrt{5-(x^2+4x+4) + 4}}dx \\ & = \int \frac{1-x}{\sqrt{9-(x+2)^2}}dx \\ & = \int \frac{1}{\sqrt{9-(x+2)^2}}dx + \int \frac{-x}{\sqrt{9-(x+2)^2}}dx \\ \\ & \text{let } x + 2 = 3\sin\theta \Rightarrow dx = 3\cos\theta d\theta \\ \\ \int \frac{1-x}{\sqrt{5-4x - x^2}}dx & =\sin ^{-1}\left(\frac{x+2}{3}\right)+\int \frac{(2-3 \sin \theta) \times 3 \cos \theta}{\sqrt{9-9 \sin ^2 \theta}} d \theta \\ & =\sin ^{-1}\left(\frac{x+2}{3}\right)+\int \frac{(2-3 \sin \theta) \times \cancel{3} \cancel{\cos \theta}}{\cancel{3} \cancel{\sqrt{1-\sin ^2 \theta}}} d \theta \qquad \because \sqrt{1-\sin\theta} = \sqrt{cos^2\theta} = \cos\theta \\ & =\sin ^{-1}\left(\frac{x+2}{3}\right)+\int 2-3 \sin \theta d \theta \\ & =\sin ^{-1}\left(\frac{x+2}{3}\right)+2 \theta+3 \cos \theta+C \quad \because x+2=3 \sin \theta,\ \therefore \theta=\sin ^{-1}\left(\frac{x+2}{3}\right) \\ & =\sin ^{-1}\left(\frac{x+2}{3}\right)+ \sqrt{5 - 4x - x^2 }+C \end{aligned} $

Question 13-b-(i).

Show that $k^2 - 2k - 3 \geq 0$ for $k \geq 3$.
Solution
Consider $k^2 - 2k - 3 = (k-3)(k+1)$
for $k\geq 0$, $(k-3) \geq 0$
for $k\geq 0$, $(k+1) \geq 0$
$\therefore$ $(k-3)(k+1) \geq 0 \Leftrightarrow k^2 - 2k-3 \geq 0$ for $k \geq 0$

Question 13-b-(ii).

Hence, or otherwise, use mathematical induction that $s(n): 2^n \geq n^2-2$ for all integers $n \geq 3$.
Solution
$ \begin{aligned} \text { S(3): } \text { LHS }&=2^3 & \text { RHS }=&3^2-2 \\ & =8 &=&9-2 \\ & & =&7 & \\ & &\leqslant &\text { LHS } \quad \therefore \text { true for } n=3 \\ & \end{aligned} $
Assume true for $n=k$, ie. $s(k): 2^k \geqslant k^2-2 \quad$ for $k \geqslant 3$
Prove true for $n=k+1$,
RTP: $\quad s(k+1): \quad 2^{k+1} \geqslant(k+1)^2-2$
$ \begin{aligned} \text { LHS } & =2^{k+1} \\ & =2 \times 2^k \\ & \geqslant 2\left(k^2-2\right) \quad \text { (by assumption) } \\ & =2 k^2-4 \\ & =k^2+\left(k^2-2 k-3\right)-1+2 k \quad \text { (by part I) } \\ & \geqslant k^2+2 k-1 \\ & =\left(k^2+2 k+1\right)-2 \\ & =(k+1)^2-2 \\ & = \text{ RHS } \quad \therefore S(k+1) \text{ is true if } s(k) \text{ is true } \end{aligned} $
By mathematical induction, $2^n \geqslant n^2-2$ for $n \geqslant 3$

Question 13-c.

A particle of mass $1 \mathrm{~kg}$ is projected from the origin with speed $40 \mathrm{~m} \mathrm{~s}^{-1}$ at an angle $30^{\circ}$ to the horizontal plane.
Solution

Question 13-c-(i).

Use the information above to show that the initial velocity of the particle is $\mathbf{v}(0)=\left(\begin{array}{c}20 \sqrt{3} \\ 20\end{array}\right)$.
Solution
$\dot{x} = 40\cos 30 = 20\sqrt{3}$
$\dot{y} = 40\sin 30 = 20$
$\therefore \underset{\sim}{V}(0) = \left(\begin{array}{c}20\sqrt{3}\\ 20\end{array}\right)$

Question 13-c-(ii).

The forces acting on the particle are gravity and air resistance. The air resistance is proportional to the velocity vector with a constant of proportionality 4 . Let the acceleration due to gravity be $10 \mathrm{~m} \mathrm{~s}^{-2}$.
The position vector of the particle, at time $t$ seconds after the particle is projected, is $\mathbf{r}(t)$ and the velocity vector is $\mathbf{v}(t)$.

Show that $\mathbf{v}(t)=\left(\begin{array}{c}20 \sqrt{3} e^{-4 t} \\ \frac{45}{2} e^{-4 t}-\frac{5}{2}\end{array}\right)$.
Solution
Forces $(\underset{\sim}{F})$ acting on particle:
gravity: $m \underset{\sim}{g}=m\left(\begin{array}{c}0 \\ -10\end{array}\right)$
air resistance: $m\underset{\sim}{k}=-4 m \underset{\sim}{v}=-4 m\left(\begin{array}{l}\dot{x} \\ \dot{y}\end{array}\right)$
$$ \begin{aligned} \underset{\sim}{F} & =\underset{\sim}{g}+\underset{\sim}{k} \\ & =\underset{\sim}{\tilde{g}}-4 \underset{\sim}{v}\\ \because \underset{\sim}{F}&=\operatorname{m}\underset{\sim}{a} & \therefore\ \underset{\sim}{a}=\left(\begin{array}{c} 0 \\ -10 \end{array} \right)-4\left(\begin{array}{l} \dot{x} \\ \dot{y} \end{array}\right)=\left(\begin{array}{c} -4 \dot{x} \\ -10-4 \dot{y} \end{array}\right) \\ \end{aligned} $$ $$ \begin{array}{l} \frac{d \dot{x}}{d t}=-4 \dot{x} & &\frac{d \dot{y}}{d t}=-10-4 \dot{y} \\ \int \frac{d \dot{x}}{\dot{x}}=\int-4 d t & &\int \frac{d y}{\frac{5}{2}+\dot{y}}=\int-4 d t \\ \ln |\dot{x}|=-4 t+C_1 & & \ln \left|\frac{5}{2}+\dot{y}\right|=-4 t+C_2 \\ \text{ went } t=0,\ \dot{x}=20 \sqrt{3},\ \therefore C_1=\ln 20 \sqrt{3} & & \text{ when } t=0,\ \dot{y}=20,\ \therefore C_2=\ln \frac{45}{2} \\ \ln |\dot{x}| =-4 t+\ln 20 \sqrt{3} & & \ln \left|\frac{5}{2}+\dot{y}\right| =-4 t+\ln \frac{45}{2} \\ \dot{x} =e^{-4 t+\ln 20 \sqrt{3}} & & \dot{y} =e^{-4 t+\ln \frac{45}{2}}-\frac{5}{2} \\ \therefore \dot{x} =20 \sqrt{3} e^{-4 t} & & \therefore \dot{y} =\frac{45}{2} e^{-4 t}-\frac{5}{2} \\ \end{array} $$ $$ \Rightarrow \underset{\sim}{V}(t)=\left(\begin{array}{c} 20 \sqrt{3} e^{-4 t} \\ \frac{45}{2} e^{-4 t}-\frac{5}{2} \end{array}\right) $$

Question 13-c-(iii).

Show that $\mathbf{r}(t)=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \frac{45}{8}\left(1-e^{-4 t}\right)-\frac{5}{2} t\end{array}\right)$.
Solution
$ \underset{\sim}{V}(t) = \left(\begin{array}{c}\dfrac{dx}{dt}\\ \dfrac{dy}{dt}\end{array}\right)$, $$ \begin{aligned}{} \frac{dx}{dt} & = 20\sqrt{3}e^{-4t} & \qquad & \frac{dy}{dt} & =& \frac{45}{2}e^{-4t}-\frac{5}{2} \\ \int dx & = \int 20\sqrt{3}e^{-4t} dt & & \int dy & =& \int \frac{45}{2}e^{-4t}-\frac{5}{2} dt \\ x &= -5\sqrt{3}e^{-4t} + C_3 & & y & = & -\frac{45}{2}e^{-4t}-\frac{5}{2}t + C_4 \\ &\text{when } t = 0,\ x=0,\ \therefore C_3 = 5\sqrt{3} & & & &\text{when } t = 0,\ y=0,\ \therefore C_4 = \frac{45}8\\ \therefore x &= -5\sqrt{3}e^{-4t} + 5\sqrt{3} & & \therefore y & = & -\frac{45}{8}e^{-4t}-\frac{5}{2}t + \frac{45}8 \\ x = & 5\sqrt{3}( 1 -e^{-4t}) & & y & = & \frac{45}{8}(1-e^{-4t}) - \frac{5}{2}t \\ \end{aligned} $$ $$ \therefore \underset{\sim}{r}(t) = \left(\begin{array}{c}5\sqrt{3}\left( 1 -e^{-4t}\right)\\ \\ \frac{45}{8}(1-e^{-4t}) - \frac{5}{2}t\end{array}\right) $$

Question 13-c-(iv).

The graphs $y=1-e^{-4 x}$ and $y=\frac{4 x}{9}$ are given in the diagram below.
Using the diagram, find the horizontal range of the particule, giving your answer rounded to one decimal place.
Solution
For the range of the projectile, $y=0$
$$ \begin{array}{rcl} \frac{45}{8}\left(1-e^{-4t}\right) - \frac{5}{2}t & = & 0 \\ \frac{45}{8}\left(1-e^{-4t}\right) & = & \frac{5}{2}t \\ 1-e^{-4t} & = & \frac{5}{2}t \times\frac{8}{45} \\ \therefore 1-e^{-4t} & = & \frac{4}{9}t \end{array} $$ From the diagram, the solution to $1-e^{-4t} = \frac{4}{9}t$ is $t=2.25$ when $t=2.25$, $$ \begin{array}{rcl} x & = & 5\sqrt{3}\left(1-e^{-4(2.25)}\right) \\ & = & 8.659 \dots \\ & \approx & 8.7 \mathrm{m} \text{ (1dp)} \end{array} $$ $\therefore$ range of projectile is $8.7 \mathrm{m}$.

Question 14-a.

Let $z$ be the complex number $z=e^{\frac{i \pi}{6}}$ and $w$ be the complex number $w=e^{\frac{3 i \pi}{4}}$.

Question 14-a-(i).

By first writing $z$ and $w$ in Cartesian form, or otherwise, show that $$ |z+w|^2=\frac{4-\sqrt{6}+\sqrt{2}}{2} \text {. } $$
Solution
$$ z=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}+i \frac{1}{2} $$ $$ \begin{aligned} w & =\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2} \\ z+w & =\frac{\sqrt{3}-\sqrt{2}}{2}+i\left(\frac{1+\sqrt{2}}{2}\right) \\ |z+w|^2 & =\left(\frac{\sqrt{3}-\sqrt{2}}{2}\right)^2+\left(\frac{1+\sqrt{2}}{2}\right)^2 \\ & =\frac{3-2 \sqrt{6}+2+1+2 \sqrt{2}+2}{4} \\ & =\frac{8-2 \sqrt{6}+2 \sqrt{2}}{4} \end{aligned} $$ $$ |z+w|^2=\frac{4-\sqrt{6}+\sqrt{2}}{2} $$

Question 14-a-(ii).

The complex numbers $z, w$ and $z+w$ are represented in the complex plane by the vectors $\overrightarrow{O A}, \overrightarrow{O B}$ and $\overrightarrow{O C}$ respectively, where $O$ is the origin. Show that $\angle A O C=\frac{7 \pi}{24}$.
Solution
let $\angle A O C=\theta$
$\theta = \frac{\mathrm{Arg}(w)-\mathrm{Arg}(z)}{2}$
$\because$ diagonals disect opp. angles in Thombus $OACB$ $\left(|z| = |w|=1\right)$
$\theta = \frac{\frac{3\pi}{4} - \frac{\pi}{6}}2$
$\therefore \angle AOC = \frac{7\pi}{24}$

Question 14-a-(iii).

Deduce that $\cos \frac{7 \pi}{24}=\frac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}$.
Solution
using cosine rule in $\triangle O A C$, $$ \begin{aligned} \cos \frac{7 \pi}{24} & =\frac{|z+\omega|^2+|z|^2-|w|^2}{2 \times|z+w| \times|z|} \\ & =\frac{\frac{4-\sqrt{6}+\sqrt{2}}{2}+\cancel{1} - \cancel{1}}{2 \times 1 \times \frac{\sqrt{4-\sqrt{6}+\sqrt{2}}}{\sqrt{2}}} \\ & =\frac{4-\sqrt{6}+\sqrt{2}}{2} \times \frac{\sqrt{2}}{2 \sqrt{4-\sqrt{6}+\sqrt{2}}} \\ & =\frac{\sqrt{2} \times \sqrt{4-\sqrt{6}+\sqrt{2}}}{4} \\ \therefore \cos \frac{7 \pi}{24} & =\frac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4} \end{aligned} $$

Question 14-b.

The point $P$ is 4 metres to the right of the origin $O$ on a straight line.
A particle is released from rest at $P$ and moves along the straight line in simple harmonic motion about $O$, with period $8 \pi$ seconds.
After $2 \pi$ seconds, another particle is released from rest at $P$ and also moves along this straight line in simple harmonic motion about $O$, with period $8 \pi$ seconds.
Find when and where the two particles first collide.
Solution
Period:
$ \begin{array} \ \frac{2\pi}{n} & = & 8\pi \\ \therefore n & = & \frac{1}4 \end{array} $
Particle 1 released from $P(x=4)$, centred at $0(x=0)$ from rest $(V=0)$ with period $8 \pi\left(n=\frac{1}{4}\right)$. Since the particle starts at max $p t$, it can be modelled by $x_1=4 \cos \left(\frac{1}{4} t\right)$ for $t \geqslant 0$.
Particle 2 is release $2 \pi$ seconds of tor Particle 1 under the same conditions, $\therefore$ Particle 2 's motion can be modelled by $x_2=4 \cos \left(\frac{1}{4}(t-2 \pi)\right)$ for $t \geqslant 2 \pi$ $$ x_2=4 \cos \left(\frac{1}{4}(t-2 \pi)\right)=4 \cos \left(\frac{1}{4} t-\frac{\pi}{2}\right)=4 \sin \left(\frac{1}{4} t\right) $$ (a cosine curve shifted by $\frac{\pi}{2}$ is the sire curve)

Particle 1 and 2 collide when $x_1=x_2$ $$ \begin{aligned} 4 \cos \left(\frac{1}{4} t\right) & =4 \sin \left(\frac{1}{4} t\right) \\ \tan \left(\frac{1}{4} t\right) & =1 \\ \frac{1}{4} t & =\frac{5 \pi}{4} \quad \frac{1}{4} t \geqslant \frac{\pi}{2} \quad \text { for } x_2 \\ \therefore t & =5 \pi \\ \text { at } t=5 \pi, x_1=4 \cos \left(\frac{5 \pi}{4}\right) & =\frac{-4 \sqrt{2}}{2}=-2 \sqrt{2} \mathrm{~m} \end{aligned} $$ $\therefore$ the 2 particles collide $2 \sqrt{2}$ metres to the LEFT of the origin at $5 \pi$ seconds

Question 14-c.

A projectile of mass $M \mathrm{~kg}$ is launched vertically upwards from the origin with an initial speed $v_0 \mathrm{~m} \mathrm{~s}^{-1}$. The acceleration due to gravity is $\mathrm{g} \mathrm{m} \mathrm{s}^{-2}$.
The projectile experiences a resistive force of magnitude $k M v^2$ newtons, where $k$ is a positive constant and $v$ is the speed of the projectile at time $t$ seconds. After $2 \pi$ seconds, another particle is released from rest at $P$ and also moves along this straight line in simple harmonic motion about $O$, with period $8 \pi$ seconds.
Find when and where the two particles first collide.

Question 14-c-(i).

The maximum height reached by the particle is $H$ metres.
Show that $H=\frac{1}{2 k} \ln \left(\frac{k v_0{ }^2+g}{g}\right)$.
Solution
$$ \begin{array}{rl} F=&M \ddot{x}=-M g-k M v^2 \\ \ddot{x}=&-g-k v^2\\ v \frac{d v}{d x}=&-g-k v^2 \\ \end{array} $$ $$ \begin{aligned} & \frac{1}{2 k} \int \frac{2 k v}{g+k v^2} d v=\int-d x \\ & \frac{1}{2 k} \ln \left|g+k v^2\right|=-x+C_1 \\ & \text { when } x=0, \quad V=V_0,\quad \therefore C_1=\frac{1}{2 k} \ln \left|g+k v_0^2\right| \\ & \frac{1}{2 k} \ln \left|g+k v^2\right|=-x+\frac{1}{2 k} \ln \left|g+k v_0^2\right| \end{aligned} $$ when $v=0, x=H$, $$ \begin{aligned} & \frac{1}{2 k} \ln |g|=-H+\frac{1}{2 k} \ln \left|g+k v_0^2\right| \\ H & =\frac{1}{2 k}\left(\ln \left|g+k v_0^2\right|-\ln |g|\right) \\ \therefore H & =\frac{1}{2 k} \ln \left(\frac{k v_0^2+g}{g}\right) \end{aligned} $$

Question 14-c-(ii).

When the projectile lands on the ground, its speed is $v_1 \mathrm{~m} \mathrm{~s}^{-1}$, where $v_1$ is less than the magnitude of the terminal velocity.
Show that $g\left(v_0^2-v_1^2\right)=k v_0^2 v_1^2$.
Solution
$$ \begin{array}{rl} F=&M \ddot{x}=-M g-k M v^2 \\ \ddot{x}=&-g-k v^2\\ v \frac{d v}{d x}=&-g-k v^2 \\ \end{array} $$ $$ \begin{aligned} & -\frac{1}{2 k} \int \frac{-2 k v}{g-k v^2} d v=\int d x \\ & -\frac{1}{2 k} \ln \left|g-k v^2\right|=x+c_2 \end{aligned} $$ when $x=0 \quad$ v $0,\quad \therefore C_2=-\frac{1}{2 k}\ln| g \mid$ $$ \begin{aligned} & -\frac{1}{2 k} \ln \left|g-k v^2\right|=x-\frac{1}{2 k} \ln |g| \\ & \therefore x=\frac{1}{2 k} \ln \left|\frac{g}{g-k v^2}\right| \end{aligned} $$ When $x=H_1 \quad V=V_1$ $$ \begin{gathered} H=\frac{1}{2 k} \ln \left|\frac{g}{g-k v_1^2}\right| \\ \frac{1}{2 k} \ln \left|\frac{g+k v_0^2}{g}\right|=\frac{1}{2 k} \ln \left|\frac{g}{g-k v_1^2}\right| \\ \frac{g+k v_0^2}{g}=\frac{g}{g-k v_1^2} \\ \left(g+k v_0^2\right)\left(g-k v_1^2\right)=g^2 \\ \cancel{g^2}-k g v_1^2+k g v_0^2-k^2 v_0^2 v_1^2=\cancel{g^2} \\ k g\left(v_0^2-v_1^2\right)=k^2 v_0^2 v_1^2 \\ \therefore g\left(v_0^2-v_1^2\right)=k v_0^2 v_1^2 \end{gathered} $$

Question 15-a-(i).

Let $J_n = \int^{\frac\pi2}_0\sin^n\theta d\theta$ where $n\geq0$ is an integer.
Show that $J_n=\frac{n-1}nJ_{n-2}$. Show that $g\left(v_0^2-v_1^2\right)=k v_0^2 v_1^2$.
Solution
$ \begin{array} \ J_n=\int_0^{\frac{\pi}{2}} \sin ^n \theta d \theta\quad \text{ for } \quad n \geqslant 0 \\ =\int_0^{\frac{\pi}{2}} \sin \theta \cdot \sin ^{n-1} \theta d \theta \end{array} $
using IBP: $\quad u=-\cos \theta \quad v=\sin ^{n-1} \theta$ $\int u^{\prime} v d \theta=u v-\int u v^{\prime} d \theta \quad u^{\prime}=\sin \theta \quad v^{\prime}=(n-1) \cos \theta \sin ^{n-2} \theta$ $$ \begin{aligned} & \begin{aligned} J_n & =\left[-\cos \theta \sin ^{n-1} \theta\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}(n-1) \cos ^2 \theta \sin ^{n-2} \theta d \theta \\ & =(n-1) \int_0^{\frac{\pi}{2}}\left(1-\sin ^2 \theta\right) \sin ^{n-2} \theta d \theta \\ & =(n-1) \int_0^{\frac{\pi}{2}} \sin ^{n-2} \theta-\sin ^n \theta d \theta \end{aligned} \\ & \begin{aligned} J_n & =(n-1)\left(J_{n-2}-J_n\right) \end{aligned} \\ & J_n=(n-1) J_{n-2}-(n-1) J_n \\ & J_n+(n-1) J_n=(n-1) J_{n-2} \\ & J_n(n)=(n-1) J_{n-2} \\ & \therefore J_n=\frac{n-1}{n} J_{n-2} \quad \text { for } n \geqslant 2 \end{aligned} $$

Question 15-a-(ii).

Let $I_n = \int^1_0x^n(1-x)^ndx$ where $n$ is a positive integer.
By using the substitution $x=\sin^2\theta$, or otherwise, show that $I_n \frac1{2^{2n}}\int_0^{\frac\pi2}\sin^{2n+1}\theta d\theta$.
Solution
$$ \begin{array}{rll} I_n &=\int_0^1 x^n(1-x)^n d x & \\ \text { let } x=\sin ^2 \theta & \text { when } x=0, \theta=0 \\ d x=2 \sin \theta \cos \theta d \theta & x=1, \theta=\frac{\pi}{2} \\ I_n & =\int_0^{\frac{\pi}{2}} \sin ^{2 n} \theta\left(1-\sin ^2 \theta\right)^n \times 2 \sin \theta \cos \theta d \theta \\ & =\int_0^{\frac{\pi}{2}} 2 \sin ^{2 n+1} \theta \cos ^{2 n+1} \theta d \theta \\ & =\frac{1}{2^{2 n}} \int_0^{\frac{\pi}{2}} 2^{2 n+1} \sin ^{2 n+1} \theta \cos ^{2 n+1} \theta d \theta \\ & =\frac{1}{2^{2 n}} \int_0^{\frac{\pi}{2}}(\sin 2 \theta)^{2 n+1} d \theta\\ \text{ let } u=2 \theta & \text{ when } \theta=0, u=0\\ du = 2d\theta & \theta = \frac\pi2,\ u = \pi\\ I_n & =\frac{1}{2^{2 n}} \int_0^\pi \sin ^{2 n+1} u \frac{d u}{2} \\ & =\frac{1}{2^{2 n+1}} \int_0^\pi \sin ^{2 n+1} u d u \\ & \because \sin ^{2 n+1} u \text { is symmetrical about } u=\frac{\pi}{2}, \\ & \Rightarrow \frac{1}{2^{2 n+1}} \int_0^\pi \sin ^{2 n+1} u d u=2 \times \frac{1}{2^{n+1}} \int_0^{\frac{\pi}{2}} \sin ^{2 n+1} u d u \\ \therefore I_n &=\frac{1}{2^{2 n}} \int_0^{\frac{\pi}{2}} \sin ^{2 n+1} u d u \\ & \text{ let } n \text{ be a dummy variable for } \theta \\ \therefore I_n &=\frac{1}{2^{2 n}} \int_0^{\frac{\pi}{2}} \sin ^{2 n+1} \theta d \theta \end{array} $$

Question 15-a-(iii).

Hence, or otherwise, show that $I_n = \frac n{4n+2}I_{n-1}$, for all integers $n\geq1$.
Solution
$$ \begin{aligned} I_n & =\frac{1}{2^{2 n}} \int_0^{\frac{\pi}{2}} \sin ^{2 n+1} \theta d \theta \\ & =\frac{1}{2^{2 n}} \cdot J_{2 n+1} \\ & =\frac{1}{2^{2 n}} \cdot \frac{2 n}{2 n+1} \cdot J_{2 n-1} \qquad\qquad\qquad \text { (using result of part I) } \\ & =\frac{2 \times 2 n}{2^{2 n} \times 2(2 n+1)} \int_0^{\frac{\pi}{2}} \sin ^{2 n-1} \theta d \theta \\ & =\frac{2^2 n}{2^{2 n}(4 n+2)} \int_0^{\frac{\pi}{2}} \sin ^{2 n-1} \theta d \theta \\ & =\frac{n}{2^{2 n-2}(4 n+2)} \int_0^{\frac{\pi}{2}} \sin ^{2 n-1} \theta d \theta \\ & =\frac{n}{4 n+2} \times \underset{I_{n-1}}{\underline{\frac{1}{2^{2(n-1)}} \int_0^{\frac{\pi}{2}} \sin ^{2(n-1)+1} \theta d \theta}} \\ \therefore I_n & =\frac{n}{4 n+2} I_{n-1} \end{aligned} $$

Question 15-b.

On the triangular pyramid $ABCD$, $L$ is the midpoint of $AB$, $M$ is the midpoint of $AD$, $P$ is the midpoint of $CD$, $Q$ is the midpoint of $BD$ and $R$ is the midpoint of $BC$.
Let $\underset{\sim}{b} = \overrightarrow{AB}$, $\underset{\sim}{c} = \overrightarrow{AC}$ and $\underset{\sim}{d} = \overrightarrow{AD}$.

Question 15-b-(i).

Show that $\overrightarrow{L P} = \frac{1}{2} ( -\underset{\sim}{b}+ \underset{\sim}{c}+ \underset{\sim}{d})$.
Solution
$$\begin{aligned} \overrightarrow{L P} & =\overrightarrow{L B}+\overrightarrow{B C}+\overrightarrow{C P} \\ & =\frac{1}{2} \overrightarrow{A B}+(\overrightarrow{B A}+\overrightarrow{A C})+\frac{1}{2} \overrightarrow{C D} \\ & =\frac{1}{2} \overrightarrow{A B}+(\overrightarrow{B A}+\overrightarrow{A C})+\frac{1}{2}(\overrightarrow{C A}+\overrightarrow{A D}) \\ & =\frac{1}{2} \underset{\sim}{b}+(-\underset{\sim}{b}+\underset{\sim}{c})+\frac{1}{2}(-\underset{\sim}{c}+\underset{\sim}{d}) \\ & =-\frac{1}{2} \underset{\sim}{b}+\frac{1}{2} \underset{\sim}{c}+\frac{1}{2} \underset{\sim}{d}\\ \therefore \overrightarrow{L P} & = \frac{1}{2} ( -\underset{\sim}{b}+ \underset{\sim}{c}+ \underset{\sim}{d})\\ \end{aligned}$$

Question 15-b-(ii).

It can be shown that $\overrightarrow{M Q}=\frac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})$ and \overrightarrow{N R}=\frac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}). (Do NOT prove these.)
Prove that $$|\overrightarrow{A B}|^2+|\overrightarrow{A C}|^2+|\overrightarrow{A D}|^2+|\overrightarrow{B C}|^2+|\overrightarrow{B D}|^2+|\overrightarrow{C D}|^2 = 4\left(|\overrightarrow{L P}|^2+|\overrightarrow{M Q}|^2+|\overrightarrow{N R}|^2\right).$$
Solution
$ \begin{aligned} &&\text { NOTE LHS VECTORS: } \\ \text { given } & \overrightarrow{LP}=\frac{1}{2}(-\underset{\sim}{b}+\underset{\sim}{c}+\underset{\sim}{d}) & \overrightarrow{A B}=\underset{\sim}{b} \quad \overrightarrow{B C}=-\underset{\sim}{b}+\underset{\sim}{c} \\ & \overrightarrow{M Q}=\frac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) & \overrightarrow{A C}=\underset{\sim}{c} \quad \overrightarrow{B D}=-\underset{\sim}{b}+\stackrel{d}{d} \\ & \overrightarrow{N R}=\frac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}) & \overrightarrow{A D}=\underset{\sim}{d} \quad \overrightarrow{C D}=-\underset{\sim}{c}+\underset{\sim}{d} \end{aligned} $ $\begin{aligned} \text{RHS} & =4\left(|\overrightarrow{L P}|^2+|\overrightarrow{M Q}|^2+|\overrightarrow{N R}|^2\right) \\ & = 4(\overrightarrow{L P} \cdot \overrightarrow{L P}+\overrightarrow{M Q} \cdot \overrightarrow{M Q}+\overrightarrow{N R} \cdot \overrightarrow{N R}) \\ & =4\left(\frac{1}{2}(-\underset{\sim}{b}+\underset{\sim}{c}+\underset{\sim}{d}) \cdot \frac{1}{2}(-\underset{\sim}{b}+\underset{\sim}{c}+\underset{\sim}{d})+\frac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) \cdot \frac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})+\frac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\frac{d}{\sim}) \cdot \frac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d})\right) \\ & =(\overrightarrow{B C}+\overrightarrow{A D}) \cdot(\overrightarrow{B C}+\overrightarrow{A D})+(\overrightarrow{A B}+\overrightarrow{C D}) \cdot(\overrightarrow{A B}+\overrightarrow{C D})+(\overrightarrow{A C}-\overrightarrow{B D}) \cdot(\overrightarrow{A C}-\overrightarrow{B D}) \\ & =|\overrightarrow{B C}|^2+2 \overrightarrow{B C} \cdot \overrightarrow{A D}+|\overrightarrow{A D}|^2+|\overrightarrow{A B}|^2+2 \overrightarrow{A B} \cdot \overrightarrow{C D}+|\overrightarrow{C D}|^2+|\overrightarrow{A C}|^2-2 \overrightarrow{A C} \cdot \overrightarrow{B D}+|\overrightarrow{B D}| \\ & =|\overrightarrow{A B}|^2+|\overrightarrow{A C}|^2+|\overrightarrow{A D}|^2+|\overrightarrow{B C}|^2+|\overrightarrow{B D}|^2+|\overrightarrow{C D}|^2+2(\overrightarrow{B C} \cdot \overrightarrow{A D}+\overrightarrow{A B} \cdot \overrightarrow{C D}-\overrightarrow{A C} \cdot \overrightarrow{B D}) \\ & \text{consider } \overrightarrow{B C} \cdot \overrightarrow{A D}+\overrightarrow{A B} \cdot \overrightarrow{C D}-\overrightarrow{A C} \cdot \overrightarrow{B D} = (-\underset{\sim}{b}+\underset{\sim}{c})\cdot \underset{\sim}{d} + \underset{\sim}{b} \cdot (-\underset{\sim}{c}+\underset{\sim}{d}) - \underset{\sim}{c} \cdot (-\underset{\sim}{b}+\underset{\sim}{d}) \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad = -\underset{\sim}{b} \cdot \underset{\sim}{d} +\underset{\sim}{c} \cdot \underset{\sim}{d} -\underset{\sim}{b} \cdot \underset{\sim}{c} +\underset{\sim}{b} \cdot \underset{\sim}{d} +\underset{\sim}{b} \cdot \underset{\sim}{c} -\underset{\sim}{c} \cdot \underset{\sim}{d} \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad = 0 \\ & =|\overrightarrow{A B}|^2+|\overrightarrow{A C}|^2+|\overrightarrow{A D}|^2+|\overrightarrow{B C}|^2+|\overrightarrow{B D}|^2+|\overrightarrow{C D}|^2 \\ & = \text{LHS} \end{aligned} $

Question 15-c.

A curve $\mathcal{C}$ spirals 3 times around the sphere centred at the origin and with radius 3 , as shown. A particle is initially at the point $(0,0,-3)$ and moves along the curve $\mathcal{C}$ on the surface of the sphere, ending at the point $(0,0,3)$.
By using the diagram below, which shows the graphs of the functions $f(x)=\cos (\pi x)$ and $g(x)=\sqrt{9-x^2}$, and considering the graph $y=f(x) g(x)$, give a possible set of parametric equations that describe the curve $\mathcal{C}$.
Solution
given $(0,0,-3)$ and $(0,0,3)$
let the parametric eq$^n$ for $x(t)=f(t) g(t)$
$\therefore x(t)=\cos (\pi t) \sqrt{9-t^2}$
let the parametric eq$^n$ for $z(t)=t$

Question 16-a.

Let $w$ be the complex number $w=e^\frac{2i\pi}3$.

Question 16-a-i.

Show that $1+w+w^2=0$.
Solution
$$ \text { LHS: } \begin{aligned} 1+w+w^2 & =1+e^{\frac{2 i \pi}{3}}+e^{\frac{4 i \pi}{3}} \\ & =1\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)+\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right) \\ & =0 \\ & =\text { RHS } \end{aligned}

Question 16-a-ii.

The vertives of the triangle can be labelled $A$, $B$ and $C$ in anticlockwise or clockwise direction, as shown.
Three complex numbers $a$, $b$, and $c$ are represented in the complex plan by points $A$, $B$ and $C$ respectively.

Show that if triangle $ABC$ is anticlockwise and equilateral, then $a + bw + cw^2 = 0$.
Solution
$B C$ is an anticlockwise rotation by $\frac{2 \pi}{3}$ of $A B$,
$ \begin{array}{rll} \Rightarrow \quad(b-a) w & =c-b \\ b w-a w & =c-b \\ b w+b & =a w+c \\ b(w+1) & =a w+c \\ b\left(-w^2\right) & =a w+c & w+1=-w^2 \text { from (I) } \\ b w^2+a w+c & =0 \\ \text{multiply by } w^2: b w^4+a w^3 + c w^2 & =0 \\ \therefore a+b w+c w^2 & =0 & \because w^3=1 \end{array} $

Question 16-a-iii.

It can be shown that if triangle $ABC$ is clockwise and equilateral, then $a + bw^2 + cw = 0$. (Do NOT prove this.)

Show that if $ABC$ is equilateral triangle, then $a^2 + b^2 + c^2 = ab + bc + ca$.
Solution
given $a + bw + cw^2 = 0 \quad \cdots (1)$ for $\triangle ABC$ anticlockwise \phantom{given }$a + bw^2 + cw = 0 \quad \cdots (2)$ for $\triangle ABC$ clockwise
$\begin{array}{ll} (1) \times (2):& \left(a+b w+c w^2\right)\left(a+b w^2+c w\right)=0 \\ & a^2+a b w^2+a c w+a b w+b^2 w^3+b c w^2+a c w^2+b c w^4+c^2 w^3=0 \\ & a^2+b^2+c^2+a b\left(w^2+w\right)+a c\left(w^2+w\right)+b c\left(w^2+w\right)=0 \\ & a^2+b^2+c^2+a b(-1)+a c(-1)+b c(-1)=0 \qquad w^2+w=-1 \text { from (i) } \\ & \therefore a^2+b^2+c^2=a b+a c+b c \end{array}$

Question 16-b-i.

Prove that $x \gt \ln x$, for $x \gt 0$.
Solution
Which shows that $f(x)$ lies above $g(x)$. However, we must consider their gradient to confirm that $g(x)$ will not overtake $f(x)$ as $x\rightarrow \infty$
Consider their gradients for $x \gt 0$, $f'(x) = 1$ and $g'(x)= \frac1x$
$ \begin{array}{rl} \ \therefore 1>\frac{1}{x} for x>0 & \Rightarrow f(x) \text{ increases at a greater rate than }\\ & g(x) \text{ for } x>0 \\ & \Rightarrow\quad \therefore x >\ln x \text { for } x>0 \end{array} $

Question 16-b-ii.

Using part (i), or otherwise, prove that for all positive integers $n$, $e^{n^2+n} \gt \left(n!\right)$
Solution
.

Question 16-c.

The complex numbers $w$ and $z$ both have modulus 1 , and $\frac{\pi}{2}<\operatorname{Arg}\left(\frac{z}{w}\right)<\pi$, where Arg denotes the principal argument.

For real numbers $x$ and $y$, consider the complex number $\frac{x z+y w}{z}$.

On an $x y$-plane, clearly sketch the region that contains all points $(x, y)$ for which $\frac{\pi}{2}<\operatorname{Arg}\left(\frac{x z+y w}{z}\right)<\pi$
Solution
$\left.\begin{array}{rl}1 & >\ln 1 \\ 2 & >\ln 2 \\ 3 & >\ln 3 \\ \vdots \\ n-1 & >\ln (n-1) \\ n & >\ln (n)\end{array}\right\} \begin{array}{l} \text{adding these results would yield:} \\ 1+2+3+\cdots+(n-1)+n>\ln 1+\ln 2+\ln 3+\cdots+\ln (n-1)+\ln (n) \\ \text { AP: } a=1, l=n,\ n \text { terms } \end{array} $
$\begin{array}{rll} &\frac{n}{2}(1+n) & \gt \ln (1 \times 2 \times 3 \times \cdots \times n-1 \times n) \\ &\frac{n}{2}(n+1) & >\ln (n !) \\ &n^2+n & \gt 2 \ln (n !) \\ &n^2+n & \gt \ln \left((n !)^2\right) \\ \Rightarrow & e^{n^2+n}& \gt (n !)^2 \end{array}$