2019 HSC Maths Ext 2 Exam Paper Solutions

The Matrix 2019 HSC Maths Ext 2 Exam Paper Solutions are here!

2019 HSC Maths Ext 2 Exam Paper Solutions

Have you seen the 2019 HSC Mathematics Extension 2 exam Paper yet?

In this post, we will work our way through the 2019 HSC Maths Extension 2 exam paper and give you the solutions, written by our Head of Mathematics Oak Ukrit and his team.

Read on to see how to answer all of the 2019 questions.

Section 1. Multiple Choice

Section 2 Long Response Questions

Question 11a

Part (i)

\begin{align*}
z + \bar{w} &= (1+3i) + (2 + i)\\
& = 3 + 4i
\end{align*}

Part (ii)

\begin{align*}
\frac{z}{w} &= \frac{1+3i}{2-i}\\
&= \frac{1+3i}{2-i} \cdot \frac{2+i}{2+i}\\
&= \frac{2+i+6i-3}{4+1}\\
&= \frac{-1+7i}{5}\\
&= -\frac{1}{5}+\frac{7}{5}i
\end{align*}

Question 11b

Question 11c

\begin{align*}
\int \frac{1}{x^2 + 10x + 29} dx & = \int \frac{1}{x^2 + 10x + 25 + 4} dx\\
& = \int \frac{1}{(x+5)^2+ 4} dx\\
& = \frac{1}{2}\tan^{-1}\Big(\frac{x+5}{2}\Big)+C
\end{align*}

Question 11d

We shall first split the integrand into its partial fractions. That is, we wish to find constants \(A\) and \(B\) such that \(\frac{6}{x^2-9} = \frac{A}{x+3} + \frac{B}{x-3}.\) That is, \(6 \equiv A(x-3) + B(x+3).\)
Substituting \(x=3\) gives us that \(6B=6\) so \(B=1\). And similarly, substituting \(x=-3\) gives us that \(-6A=6\) so \(A=-1\).

Hence
\begin{align*}
\int \frac{6}{x^2-9} dx & = \int \frac{-1}{x+3} + \frac{1}{x-3} dx\\
& = -\ln|x+3| + \ln|x-3| +C\\
& = \ln\Big|\frac{x-3}{x+3}\Big| +C
\end{align*}

Question 11e

Part (i)

\(|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2\) and \(\tan \theta = \frac{\sqrt{3}}{-1}\) so \(arg(z) = \frac{2\pi}{3}\). Thus, \(z=2cis(\frac{2\pi}{3})\).

Part (ii)

By De Moivre’s Theorem, \(z^3 = 2^3cis(\frac{2\pi}{3}\cdot 3) = 8cis(2\pi) = 8\).

Question 12a

Question 12b

Part (i)

Applying cosine rule on \(\triangle ABP\), we have that \(70^2 = x^2 + y^2-2\cdot x \cdot y \cdot \cos(\frac{2\pi}{3})\). Hence, \(70^2 = x^2 +xy+ y^2\)

Part (ii)

As Train \(B\) is moving at a constant speed of \(4\text{ms}^{-1}\), \(\frac{dy}{dt}=4\).

Implicit differentiation on the equation found in part (i) gives us that
\begin{align*}
2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx}&=0 \\
\frac{dy}{dx}(x+2y)&=-2x-y\\
\frac{dy}{dx}&=\frac{-2x-y}{x+2y}
\end{align*}
Thus, when \(x=30\) and \(y=50\), we have that \(\frac{dy}{dx}=\frac{-2\cdot 30 – 50}{30 + 2(50)} = \frac{-110}{130} = \frac{-11}{13}\).

Finally, we have that \(\frac{dx}{dt} = \frac{dx}{dy}\cdot \frac{dy}{dt} = \frac{13}{-11}\cdot 4 = \frac{-52}{11}\)

Question 12c

Using the method of cylindrical shells, we have that the radius of the shell would be \(x\) and that the height of the shell would be \(Ae^{-kx}\).

Hence, the area of the cylindrical shell at distance \(x\) is given by:
\begin{align*}
A & = 2\pi r h\\
&= 2 \pi x Ae^{-kx}
\end{align*}
Thus the required volume is given by
\begin{align*}
V &= \int_{10}^{40} 2 \pi x Ae^{-kx} dx\\
&= 2\pi A \int_{10}^{40} x e^{-kx} dx\\
&= 2\pi A \Big[ [x\cdot \frac{1}{-k} e^{-kx}]_{10}^{40} -\int_{10}^{40} \frac{1}{-k} e^{-kx} dx\Big] \text{ (using integration by parts)}\\
&= 2\pi A \Big[ \big(-\frac{40e^{-40k}}{k}+\frac{10e^{-10k}}{k}\big) +\frac{1}{k} \int_{10}^{40} e^{-kx} dx\Big] \\
&= \frac{2\pi A}{k} \Big[ 10e^{-10k}-40e^{-40k} +\frac{1}{-k}[e^{-kx}]_{10}^{40} \Big] \\
&= \frac{2\pi A}{k^2} \Big[ 10ke^{-10k}-40ke^{-40k} -e^{-40k}+e^{-10k} \Big]
\end{align*}

Question 12d

Part (i)

Part (ii)

Part (iii)

Question 13a

Part (i)

At \(P(a\sec\theta,b\tan\theta)\), we have that \(\frac{dx}{d\theta}=a\sec\theta\tan\theta\) and \(\frac{dy}{d\theta}=b\sec^2\theta\). Hence,
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{d\theta}\cdot \frac{d\theta}{dx}\\
&=\frac{b\sec^2\theta}{a\sec\theta\tan\theta}\\
&=\frac{b\sec\theta}{a\tan\theta}
\end{align*}
Now using the point gradient formula, we have that the equation of the tangent is:
\begin{align*}
y-y_1 &= m(x-x_1)\\
y – b\tan\theta &= \frac{b\sec\theta}{a\tan\theta} (x-a\sec\theta)\\
ay\tan\theta – ab\tan^2\theta &= bx\sec\theta -ab\sec^2\theta\\
ab\sec^2\theta- ab\tan^2\theta &= bx\sec\theta-ay\tan\theta \\
ab(\sec^2\theta- \tan^2\theta) &= bx\sec\theta-ay\tan\theta \\
bx\sec\theta-ay\tan\theta &= ab
\end{align*}

Part (ii)

Tangent at \(P\):
\begin{equation}
bx\sec\theta-ay\tan\theta = ab \label{eq:1}
\end{equation}
Tangent at \(Q\):
\begin{equation}
bx\sec\phi-ay\tan\phi = ab \label{eq:2}
\end{equation}
Multiply both sides of \((1)\) by \(\tan\phi\):
\begin{equation}
bx\sec\theta\tan\phi-ay\tan\theta\tan\phi = ab\tan\phi \label{eq:3}
\end{equation}
Multiply both sides of \((2)\) by \(\tan\theta\):
\begin{equation}
bx\sec\phi\tan\theta-ay\tan\phi\tan\theta = ab\tan\theta \label{eq:4}
\end{equation}
The point is intersection is found by solving these equations simultaneously. Thus, by subtracting \((3)\) from \((4)\), we get that
\begin{align*}
bx_0\sec\phi\tan\theta- bx_0\sec\theta\tan\phi &= ab\tan\theta – ab\tan\phi \\
bx_0(\sec\phi\tan\theta- \sec\theta\tan\phi)&=ab(\tan\theta – \tan\phi)\\
x_0&=\frac{a(\tan\theta – \tan\phi)}{\sec\phi\tan\theta – \sec\theta\tan\phi}
\end{align*}

Part (iii)

To prove that \(O\), \(M\) and \(T\) are collinear, we will show that \(m_{OM} = m_{OT}\).

First, we find the coordinates of \(M\).
\begin{align*}
M\Big(\frac{a\sec\theta + a\sec\phi}{2},\frac{b\tan\theta + b\tan\phi}{2}\Big)
\end{align*}
Now we find the gradient \(OM\).
\begin{align*}
m_{OM} & = \frac{\frac{b\tan\theta + b\tan\phi}{2}-0}{\frac{a\sec\theta + a\sec\phi}{2}-0}\\
& = \frac{b\tan\theta + b\tan\phi}{a\sec\theta + a\sec\phi}\\
& = \frac{b(\tan\theta + \tan\phi)}{a(\sec\theta + \sec\phi)}
\end{align*}

Similarly, we can find the gradient of \(OT\):
\begin{align*}
m_{OT} & = \frac{\frac{b(\sec\theta – \sec\phi)}{\sec\phi\tan\theta – \sec\theta\tan\phi}-0}{\frac{a(\tan\theta – \tan\phi)}{\sec\phi\tan\theta – \sec\theta\tan\phi}-0}\\
& = \frac{b(\sec\theta – \sec\phi)}{a(\tan\theta – \tan\phi)}
\end{align*}

Now to show that \(m_{OM}=m_{OT}\), observe that
\begin{align*}
\sec^2\theta – \tan^2\theta &= \sec^2\phi – \tan^2\phi\\
\sec^2\theta – \sec^2\phi &= \tan^2\theta – \tan^2\phi\\
(\sec\theta + \sec\phi)(\sec\theta – \sec\phi) &= (\tan\theta + \tan\phi)(\tan\theta – \tan\phi)\\
\frac{\sec\theta – \sec\phi}{\tan\theta – \tan\phi} &= \frac{\tan\theta + \tan\phi}{\sec\theta + \sec\phi}
\end{align*}
This shows that \(M_{OM} = M_{OT}\)

Question 13b

Part (i)

As we are finding the area of a trapezium, we need to find the length of the top, bottom and height.

The height is given by the function \(y=1-x^2\). The using the right-angled isoceles triangle, we see that the length of the base of the trapezium is constant and \(1\) unit long.

Notice that \(\angle CDA=45^\circ\). Let \(E\) be the projection of \(C\) onto \(AD\).

Then we have that \(CE=BA=1-x^2\) and so \(DE=CE=1-x^2\). This means that \(BC = AD – DE = 1 – (1-x^2) = x^2\).

Hence, \(Area_{ABCD} = \frac{1+x^2}{2}\cdot (1-x^2) = \frac{1-x^4}{2}\)

Part (ii)

To calculate the volume of the solid, we integrate the area found in (i) from \(-1\) to \(1\).
\begin{align*}
V &= \int_{-1}^1 \frac{1-x^4}{2} dx\\
&= 2\int_0^1 \frac{1}{2} – \frac{x^4}{2} dx\\
&= 2\Big[\frac{x}{2} – \frac{x^5}{10}\Big]_0^1\\
&= 2\Big[\frac{1}{2} – \frac{1}{10}-0+0\Big]\\
&= \frac{4}{5} u^3
\end{align*}

Question 13c

Part (i)

The equations for motion of Object 1 are
\begin{align*}
x &= 20\cos\Big(\frac{\pi}{3}\Big)t\\
x &= 10t\\
\\
y &= -4.9t^2 + 20\sin\Big(\frac{\pi}{3}\Big)t\\
y &= -4.9t^2 + 10\sqrt{3}t
\end{align*}
Hence, Object 1 hits the ground at \(y=0\), that is, \(-4.9t^2 + 10\sqrt{3}t=0\) or \(t(-4.9t + 10\sqrt{3})=0\). This means that \(t = 0\) or \(t = \frac{10\sqrt{3}}{4.9}\). But \(t=0\) is the start of motion, so Object 1 hits the ground at \(t = \frac{10\sqrt{3}}{4.9}\).

This means that the range is \(x = 10 \cdot \frac{10\sqrt{3}}{4.9} = \frac{100\sqrt{3}}{4.9}\).

Part (ii)

We are now trying to find the values of \(v\) and \(\theta\) that will satisfy the equations of motion for Object 2. Since it is projected two seconds after Object 1, its equations will be, \(x = v\cos\theta(t-2)\) and \(y = -4.9(t-2)^2 + v\sin\theta(t-2)\) at the point where \(x=\frac{100\sqrt{3}}{4.9}\), \(y=0\) and \(t=\frac{10\sqrt{3}}{4.9}\). Hence, we have that:
\begin{align}
\frac{100\sqrt{3}}{4.9} &= v\cos\theta\Big(\frac{10\sqrt{3}}{4.9}-2\Big)\nonumber\\
v\cos\theta &= 23.031\ldots \label{eq:5}\\
\nonumber\\
0 &= -4.9\Big(\frac{10\sqrt{3}}{4.9}-2\Big)^2 + v\sin\theta\Big(\frac{10\sqrt{3}}{4.9}-2\Big)\nonumber\\
v\sin\theta &= 7.52050\ldots \label{eq:6}
\end{align}
To solve these simultaneously, divide \(v \sin \theta \) by \(v \cos \theta\)
\begin{align*}
\tan\theta &= 0.3265381\\
\theta &= 18.1^\circ (1 d.p.)
\end{align*}
Substitute this back to find \(v\):
\begin{align*}
v\cos18.1^\circ &= 23.031\\
v &= 24.2m/s~(1 d.p.)
\end{align*}

Question 14a

Let the normal force be \(N\). Since there are no other forces, resolving the forces gives us:
\(\text{Vertically: } N\cos\theta = mg\)
\(\text{Horizontally: } N\sin\theta = \frac{mv^2}{r}\)

Dividing these equations, we obtain:

\begin{align*}
\frac{N\sin\theta}{N\cos\theta} &= \frac{\frac{mv^2}{r}}{mg}\\
\tan\theta &= \frac{v^2}{rg}\\
v^2 &= gr\tan\theta
\end{align*}

Question 14b

Part (i)

The terminal velocity is found by solving \(\ddot{x}=0\). That is,
\begin{align*}
0 &= g-kw\\
w = \frac{g}{k}
\end{align*}

Part (ii)

We require the parachutist to slow down from \(1.6w\) to \(1.1w\).
\begin{align*}
\ddot{x} &= g-kv\\
\frac{dv}{dt} &= g-kv\\
\frac{dt}{dv} &= \frac{1}{g-kv}\\
t &= \int_{1.6w}^{1.1w} \frac{1}{g-kv} dv\\
t &= -\frac{1}{k}[ln|g-kv|]_{1.6w}^{1.1w}\\
t &= -\frac{1}{k}\Big[ln|g-1.1kw|-ln|g-1.6kw|\Big]\\
t &= -\frac{1}{k}ln\Big|\frac{g-1.1kw}{g-1.6kw}\Big|\\
t &= -\frac{1}{k}ln\Big|\frac{g-1.1k\frac{g}{k}}{g-1.6k\frac{g}{k}}\Big|\\
t &= -\frac{1}{k}ln\Big|\frac{g-1.1g}{g-1.6g}\Big|\\
t &= -\frac{1}{k}ln\Big|\frac{-0.1g}{-0.6g}\Big|\\
t &= -\frac{1}{k}ln\Big|\frac{1}{6}\Big|\\
t &= \frac{1}{k}ln|6|
\end{align*}

Part (iii)

\begin{align*}
\ddot{x} &= g-kv\\
v\frac{dv}{dx} &= g-kv\\
\frac{dv}{dx} &= \frac{g-kv}{v}\\
\frac{dx}{dv} &= \frac{v}{g-kv}\\
D &= \int_{1.6w}^{1.1w} \frac{v}{g-kv} dv\\
D &= -\frac{1}{k}\int_{1.6w}^{1.1w} \frac{-kv}{g-kv} dv\\
D &= -\frac{1}{k}\int_{1.6w}^{1.1w} 1 -\frac{g}{g-kv} dv\\
D &= -\frac{1}{k}\Big[v + \frac{g}{k}ln|g-kv| \Big]_{1.6w}^{1.1w}\\
D &= \frac{g}{k^2}\Big[\frac{vk}{g} + ln|g-kv| \Big]_{1.1w}^{1.6w}\\
D &= \frac{g}{k^2}\Big[\frac{1.6wk}{g} + ln|g-1.6wk| – \frac{1.1wk}{g} – ln|g-1.1wk| \Big]\\
D &= \frac{g}{k^2}\Big[0.5\frac{vk}{g} + ln\Big|g-1.6k\frac{g}{k}\Big| – ln\Big|g-1.1k\frac{g}{k}\Big| \Big]\\
D &= \frac{g}{k^2}\Big[0.5 + ln\Big|g-1.6g\Big| – ln\Big|g-1.1g\Big| \Big]\\
D &= \frac{g}{k^2}\Big[\frac{1}{2} + ln\Big|-0.6g\Big| – ln\Big|-0.1g\Big| \Big]\\
D &= \frac{g}{k^2}\Big[\frac{1}{2} + ln\Big|\frac{-0.6g}{-0.1g}\Big|\Big]\\
D &= \frac{g}{k^2}\Big[\frac{1}{2} + ln|6|\Big]
\end{align*}

Question 14c

Part (i)

\begin{align*}
LHS & = \cot x – \cot 2x\\
& = \frac{\cos x}{\sin x} – \frac{\cos 2x}{\sin 2x}\\
& = \frac{\cos x}{\sin x} – \frac{2\cos^2x-1}{2\sin x\cos x}\\
& = \frac{2\cos^2 x}{2\sin x\cos x} – \frac{2\cos^2x-1}{2\sin x\cos x}\\
& = \frac{1}{2\sin x\cos x}\\
& = \frac{1}{\sin 2x}\\
& = \csc 2x
\end{align*}

Part (ii)

Base case: Prove true for \(n=1\).

\(LHS = \csc 2x\), \(RHS = \cot x – \cot 2x\). By part (ii), we see that the statement is true for \(n=1\).

Inductive step: Assume true for \(n=k\) where \(k\) is some integer greater than or equal to 1. That is, assume that \(\sum_{r=1}^k \csc (2^r x) = \cot x – \cot(2^kx). \)
Required to prove true for \(n=k+1\). That is, we are required to prove that \(\sum_{r=1}^{k+1} \csc (2^r x) = \cot x – \cot(2^{k+1}x). \)
Proof:
\begin{align*}
LHS &= \sum_{r=1}^{k+1} \csc (2^r x)\\
&= \sum_{r=1}^{k} [\csc (2^r x)] + \csc (2^{k+1} x)\\
&= \cot x – \cot(2^kx) + \csc (2^{k+1} x)\\
&= \cot x – \cot (2^{k+1} x) \text{ by substituting } x \text{ for } 2^kx \text{ in part (i)}\\
&= RHS
\end{align*}

Question 15a

Part (i)

\begin{align*}
RHS &= \int_{-a}^a\frac{f(-x)}{f(x) + f(-x)} dx\\
&= \int_{a}^{-a}\frac{f(u)}{f(-u) + f(u)} \cdot -du \text{ substituting } u = -x\\
&= \int_{-a}^{a}\frac{f(u)}{f(-u) + f(u)} \cdot du\\
&= \int_{-a}^{a}\frac{f(x)}{f(-x) + f(x)} \cdot dx \text{ replacing } u \text{ with } x\\
&= LHS
\end{align*}

Part (ii)

Let \(I = \int_{-1}^1\frac{e^{x}}{e^x + e^{-x}} dx.\) Then by part (i) we also have that \(I = \int_{-1}^1\frac{e^{-x}}{e^x + e^{-x}} dx.\) Adding the two equations above, we have that
\begin{align*}
2I &= \int_{-1}^1\frac{e^{x}}{e^x + e^{-x}} +\int_{-1}^1\frac{e^{-x}}{e^x + e^{-x}} dx \\
2I &= \int_{-1}^1\frac{e^{x} + e^{-x}}{e^x + e^{-x}} dx\\
2I &= \int_{-1}^1 1 dx \\
2I &= 2 \\
I &= 1
\end{align*}

Question 15b

Part (i)

In the case where \(w=y\), the chance for either play to win on their turn is the same. Since, player \(A\) goes first, they will have a higher chance of winning.

Part (ii)

Note that as the game progresses, the number of balls in the urn never changes until there is a winner. Thus, the probability that \(A\) will win on any turn is \(\frac{w}{w+y}\) and the probability that \(B\) will win on any turn is \(\frac{y}{w+y}.\)
The cases where \(B\) will win the game are:

• \(A\) loses, \(B\) wins with probability \(\frac{y}{w+y} \cdot \frac{y}{w+y} = \big(\frac{y}{w+y}\big)^2\)
• \(A\) loses, \(B\) loses, \(A\) loses, \(B\) wins with probability \(\frac{y}{w+y} \cdot \frac{w}{w+y} \cdot \frac{y}{w+y} \cdot \frac{y}{w+y} = \big(\frac{w}{w+y}\big)\big(\frac{y}{w+y}\big)^3 \)
• \(A\) loses, \(B\) loses, \(A\) loses, \(B\) loses, \(A\) loses, \(B\) wins with probability \( \big(\frac{w}{w+y}\big)^2\big(\frac{y}{w+y}\big)^4 \)
• And so on…

Summing these together, we find that the probability that \(B\) will win is
\begin{align*}
P(B~wins) &= \big(\frac{y}{w+y}\big)^2 + \big(\frac{w}{w+y}\big)\big(\frac{y}{w+y}\big)^3 + \big(\frac{w}{w+y}\big)^2\big(\frac{y}{w+y}\big)^4 + \cdots \\
P(B~wins) &= \frac{\big(\frac{y}{w+y}\big)^2}{1-\big(\frac{w}{w+y}\big)\big(\frac{y}{w+y}\big)}\\
P(B~wins) &= \frac{y^2}{(w+y)^2 – wy}\\
P(B~wins) &= \frac{y^2}{w^2 + 2wy + y^2 – wy}\\
P(B~wins) &= \frac{y^2}{w^2 + wy + y^2}
\end{align*}
In order for \(B\) to have a greater chance of winning, we need \(P(B~win) > \frac{1}{2}\). That is,
\begin{align*}
\frac{y^2}{w^2 + wy + y^2} > \frac{1}{2}\\
2y^2 > w^2 + wy + y^2\\
y^2 – wy – w^2 > 0\\
\Big(\frac{y}{w}\Big)^2 – \frac{y}{w} – 1 > 0
\end{align*}
Thus, \(\frac{y}{w} > \frac{1 + \sqrt{1+4}}{2} \text{ or } \frac{y}{w} < \frac{1 – \sqrt{1+4}}{2}\)
\(\frac{y}{w} > \frac{1 + \sqrt{5}}{2} \text{ or } \frac{y}{w} < \frac{1 – \sqrt{5}}{2}.\)
But \(w\) and \(y\) are positive, so we must have that \(\frac{y}{w} > \frac{1 + \sqrt{5}}{2}\)

Question 15c

Part (i)

\begin{align*}
\int_0^1 \frac{x}{(x+1)^2} dx &= \int_0^1 \frac{x+1 -1}{(x+1)^2} dx\\
&= \int_0^1 \frac{x+1}{(x+1)^2} – \frac{1}{(x+1)^2} dx\\
&= \int_0^1 \frac{1}{x+1} – \frac{1}{(x+1)^2} dx\\
&= \Big[\ln|x+1| + \frac{1}{x+1}\Big]_{0}^1\\
&= \Big[\ln|2| + \frac{1}{2}\Big] – \Big[\ln|1| + \frac{1}{1}\Big]\\
& = \ln2 – \frac{1}{2}
\end{align*}

Part (ii)

\begin{align*}
I_n &= \int_0^1\frac{x^n}{(x+1)^2}dx\\
&=\int_0^1\frac{x^{n-1}((x+1)-1)}{(x+1)^2}dx\\
&=\int_0^1\frac{x^{n-1}}{x+1}dx – I_{n-1}
\end{align*}

\begin{align*}
∴ I_n &= \left[\frac{x^n}{x(1+x)}\right]_0^1 + \frac{1}{n}\int_0^1\frac{x^n}{(1+x)^2}dx – I_{n-1}\\
&= \frac{1}{2n} + \frac{1}{n}I_n – I_{n-1}\\
∴ I_n&(1-\frac{1}{n}) = \frac{1}{2n}-I_{n-1}\\
I_n&(\frac{n-1}{n}) = \frac{1}{2n} – I_{n-1}\\
I_n &= \frac{1}{2(n-1)} – \frac{n}{n-1}I_{n-1} \,\, \textrm{ as required}
\end{align*}

Part (iii)

\begin{align*}
I_3 &= \frac{1}{2(2)} – \frac{3}{2}I_2\\
&= \frac{1}{4} – \frac{3}{2}(\frac{1}{2}-2I_1)\\
&= \frac{1}{4} – \frac{3}{4} + 3(\ln 2 – \frac{1}{2})\\
&= -\frac{1}{2} + 3\ln 2 – \frac{3}{2}\\
&= 3\ln 2 – 2
\end{align*}

Question 16a

Part (i)

\begin{align*}
x^3+px+q=0, \,\,p,q\in\mathbb{R}\,\,p>0\\
\textrm{Letting }r=\sqrt{\frac{4p}{3}} \textrm{ and } \cos 3\theta = -\frac{4q}{r^3}:\\
\textrm{Now }-\frac{4q}{r^3} = 4\cos^3\theta – 3\cos\theta\\
∴ r^3\cos^3\theta – \frac{3r^2}{4}(r\cos\theta)+q = 0\\
\textrm{Since } r=\sqrt{\frac{4p}{3}},\,\,p = \frac{3r^2}{4}\\
\end{align*}

Therefore the above is a polynomial in \(r\cos\theta\) in the form \(x^3+px+q = 0\) and hence \(r\cos\theta\) is a root.

Part (ii)

\begin{align*}
x^3+9x^2+15x-17&=0\\
\textrm{Let }u=\alpha+3, \textrm{ so }\alpha = u&-3\\
\textrm{Then } (u-3)^3 + 9(u-3)^2 + 15(u-3) – 17 &= 0\\
u^3-9u^2+27u-27 + 9u^2 – 54u + 81 + 15u – 45 – 17 &= 0\\
u^3 – 12u – 8 &= 0\\
\textrm{Hence } \alpha+3,\beta+3,\gamma+3 \textrm{ are roots of } x^3-12x-8&=0
\end{align*}

Part (iii)

Solve \(x^3-12x-8=0\) for \(\alpha+3,\beta+3,\gamma+3\) by recognising that it is of the form in part (i) where \(p=12\) and \(q=-8\).
ie. \(r=\sqrt{\frac{4\times12}{3}} = 4\) and \(\cos3\theta = -\frac{4(-8)}{4^3}=\frac{1}{2}\)
The roots will be of the form \(4\cos\theta\) where \(\cos3\theta = \frac{1}{2}\)
\(\Rightarrow3\theta = \frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3}\Rightarrow \theta = \frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9}\)
\(∴ \alpha+3 = 4\cos\frac{\pi}{9}, \beta+3 = \frac{5\pi}{9}, \gamma+3 = \frac{7\pi}{9}\)
and thus the solutions of \(x^3+9x^2+15x-17=0\) are \(4\cos\frac{\pi}{9} – 3, 4\cos\frac{5\pi}{9}-3, 4\cos\frac{7\pi}{9}-3\)

Question 16b

Part (i)

\begin{align*}
\textrm{Let } P(z) = z^4 – 2kz^3 + 2k^2z^2 + mz + 1\\
\textrm{where } k,m\in\mathbb{R}.\textrm{ Roots are } \alpha,\bar{\alpha},\beta,\bar{\beta}.\,\,|\alpha|=|\beta|=1\\
\end{align*}

We use the fact that:

\begin{align*}
(Re(\alpha))^2 + (Re(\beta))^2 &= (Re(\alpha) + Re(\beta))^2 – 2Re(\alpha)Re(\beta)\\
\textrm{where } Re(\alpha)+ Re(\beta) &= \frac{\alpha+\bar{\alpha}+\beta+\bar{\beta}}{2} = \frac{2k}{2} = k\\
\textrm{and } 2Re(\alpha)Re(\beta) &= 2\left(\frac{\alpha+\bar{\alpha}}{2}\right)\left(\frac{\beta+\bar{\beta}}{2}\right)\\
&=\frac{1}{2}(\alpha\beta + \alpha\bar{\beta} + \bar{\alpha}\beta + \bar{\alpha}\bar{\beta})\\
&=\frac{1}{2}[(\alpha\beta + \alpha\bar{\beta} + \bar{\alpha}\beta + \bar{\alpha}\bar{\beta} + \alpha\bar{\alpha} + \beta\bar{\beta}) – (\alpha\bar{\alpha} + \beta\bar{\beta})]\\
&=\frac{1}{2}(2k^2-2)\\
&=k^2-1\\
\textrm{Thus } (Re(\alpha))^2 + (Re(\beta))^2 &= k^2-(k^2-1) = 1 \textrm{ as required}\\
\end{align*}

Part (ii)

From part (i) we have that \(Re(\alpha)^2 + Re(\beta)^2 = 1\). But since \(|a|=1\) we have that \(Re(\alpha)^2 + Im(\alpha)^2 = 1\). Hence, we must have \(Re(\beta)^2 = Im(\alpha)^2\) so that \(Re(\beta)^2 = \pm Im(\alpha)\). Since \(|\beta|=1\) we have that \(\beta\) also lies on the unit circle. Hence, plotting the possible positions of \(\beta\), we have that

Question 16c

\begin{align*}
\angle ACD &= \angle ABD \textrm{ (angles standing on common arcs)}\\
\angle BAC &= \angle BDC \textrm{ (angles standing on common arcs)}\\
\angle ACD &= \pi – \angle AED \textrm{ (opposite angles on cyclic quad)}\\
∴ \angle ABD &= \pi – E\\
∴ B &= \angle ABD + \angle CBD = \angle CBD + \pi – E\\
∴ \angle CBD &= B + E – \pi\\
C &= \angle DCA + \angle ACB\\
&= \pi – E + \angle ACB \textrm{ (opposite angles on cyclic quad)}\\
∴ \angle ACB &= C + E – \pi\\
\textrm{Consider } \Delta ABC &\textrm{ and } \Delta BCD\\
\frac{a}{\sin\angle CBD}&=\frac{e}{\sin\angle BDC}\\
∴ e &= \frac{a\sin\angle BDC}{\sin\angle CBD}\\
\frac{d}{\sin\angle ACB}&=\frac{e}{\sin\angle BAC}\\
∴ e &= \frac{d\sin\angle BAC}{\sin\angle ACB}\\
∴ \frac{a\sin\angle BDC}{\sin\angle CBD} &= \frac{d\sin\angle BAC}{\sin\angle ACB}\\
\textrm{Since } \angle BAC &= \angle BDC\\
\sin\angle BAC &= \sin\angle BDC\\
∴ \frac{a}{\sin\angle CBD} &= \frac{d}{\sin\angle ACB}\\
\frac{a}{\sin(B+E-\pi)} &= \frac{d}{\sin(C+E-\pi)}\\
\frac{a}{-\sin(\pi-(B+E))} &= \frac{d}{-\sin(\pi-(C+E))}\\
\frac{a}{\sin(B+E)} &= \frac{d}{\sin(C+E)} \textrm{ as required}
\end{align*}

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