2024 VCE Specialist Maths Exam Paper 1 Solutions

In this article, we reveal 2024 VCE Specialist Maths Exam Paper 1 Solutions, complete with full explanations written by the Matrix Maths Team.

2024 VCE Specialist Maths Exam Paper 1 Solutions

Have you seen the 2024 VCE Specialist maths exam paper yet? For reference, you can find it here.

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Question 1

(a).

We simply check that \(f(-\frac{2i}{3})=0\). Indeed,

\begin{align*}
f\bigg(-\frac{2i}{3}\bigg)&=3\bigg(-\frac{2i}{3}\bigg)^{3}+2i\bigg(-\frac{2i}{3}\bigg)^{2}+3\bigg(-\frac{2i}{3}\bigg)+2i
\\&=\frac{8}{9}i-\frac{8}{9}i-2i+2i
\\&=0.
\end{align*}

(b).

Polynomial long division gives that

\[
f(z)=(3z+2i)(z+i)(z-i)
\]

and so \(f(z)=0\) has solutions \(z=-\frac{2}{3}i,i,-i\).

(c).

Question 2

Suppose that \(x\) is an odd integer. That is, \(x=2k+1\) for some \(k\in\mathbb{Z}\). Then

\begin{align*}
2x^{2}-3x-7&=2(2k+1)^{2}-3(2k+1)-7
\\&=8k^{2}+2k-8
\\&=2(4k^{2}+k-4)
\end{align*}

which is even since \(4k^{2}+k-4\in\mathbb{Z}\).

Question 3

(a).

We simply check that

\[
1+\frac{-4}{x+1}+\frac{4}{(x+1)^{2}}=\frac{x^{2}-2x+1}{(x+1)^{2}}=\frac{(x-1)^{2}}{(x+1)^{2}}.
\]

(b).

The derivative of \(f(x)\) is given by

\[
f^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{8}{(x+1)^{3}}
\]

and so
\[
f^{\prime}(x)=0\iff \frac{4}{(x+1)^{2}}-\frac{8}{(x+1)^{3}}=0\iff x=1.
\]

Since \(f(1)=0\) this means that \(f\) has a stationary point at the point \((1,0)\). Since there is only one stationary point and the question says that the graph has one turning point, this must also be the turning point.

(c).

Question 4

(a).

We compute that \(|\vec{a}|=3\sqrt{2},|\vec{b}|=3\) and \(\vec{a}\cdot\vec{b}=-9\).

Hence
\[
\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}=-\frac{1}{\sqrt{2}}
\]
and so \(\theta=\frac{3\pi}{4}\).

(b).

Note that \(|\vec{c}|=(n^{2}+5)^{1/2}\) and that \(\vec{a}\cdot\vec{c}=9\).

Thus
\[
|\vec{a}\times\vec{c}|=|\vec{a}||\vec{c}||\sin\theta|=3(n^{2}+5)^{1/2}.
\]

Hence we want
\[
3(n^{2}+5)^{1/2}=9
\]
which rearranges to give \(n=\pm 4\).

Question 5

The volume of the solid of revolution is given by

\begin{align*}
V&=\pi\int_{1}^{\frac{k}{2}}k-\frac{1}{x^{2}}dx
\\&=\pi\bigg[kx+\frac{1}{x}\bigg]_{1}^{\frac{k}{2}}
\\&=\pi\bigg(\frac{k^{2}}{2}+\frac{2}{k}-k-1\bigg).
\end{align*}

Hence
\[
\pi\bigg(\frac{k^{2}}{2}+\frac{2}{k}-k-1\bigg)=\frac{7\pi}{2}
\]

which rearranges to give
\[
k^{3}-2k^{2}-9k+4=0.
\]

Question 6

(a).

The expected value is given by

\[
E(W_{1}+W_{2}+W_{3})=E(W_{1})+E(W_{2})+E(W_{3})=1+1.5+2=4.5
\]

and the variance is given by (recall that variance is additive over independent random variables)
\[
Var(W_{1}+W_{2}+W_{3})=Var(W_{1})+Var(W_{2})+Var(W_{3})=0.3^{2}+0.4^{2}+0.5^{2}=0.5.
\]

(b).

The variance of the cost to produce one weed trimmer is given by (again, recall that variance is additive over independent random variables)

\[
Var(10W_{1}+20W_{2}+15W_{3})=10^{2}Var(W_{1})+20^{2}Var(W_{2})+15^{2}Var(W_{3})=\frac{517}{4}\$/\text{h}.
\]

(c).

We want to find \(Pr(W_{2}<W_{1})=Pr(W_{2}-W_{1}<0)\). To this end, note that \(W_{2}-W_{1}\) is also normally distributed with \(E(W_{2}-W_{1})=E(W_{2})-E(W_{1})=0.5\) and \(Var(W_{2}-W_{1})=Var(W_{2})+Var(W_{1})=0.25\).

Hence
\[
Pr(W_{2}<W_{1})=Pr\bigg(Z<\frac{0-0.5}{\sqrt{0.25}}\bigg)=Pr(Z<-1)=0.16.
\]

Question 7

We rearrange the differential equation to
\(
\begin{align*}
2y \, dy = \frac{-x}{\sqrt{x^{2} + 1}} \, dx &\Rightarrow \int 2y \, dy = \int \frac{-x}{\sqrt{x^{2} + 1}} \, dx, \\
&\Rightarrow y^{2} = -\sqrt{x^{2} + 1} + C.
\end{align*}
\)

The initial conditions \(y(0)=-2\) imply that \(C=5\).

Hence the solution is given by
\(
y^{2} = 5 – \sqrt{x^{2} + 1} \Rightarrow y = -\sqrt{5 – \sqrt{x^{2} + 1}}.
\)

where we took the negative square root since \(y(0)\) is negative.

Question 8

(a).

Differentiating both sides with respect to \(x\) gives that

\[
2xy^{2}+2x^{2}y\frac{dy}{dx}+y+x\frac{dy}{dx}=0
\]

which rearranges to give
\[
\frac{dy}{dx}=\frac{-y(2xy+1)}{x(2xy+1)}=-\frac{y}{x}
\]
provided that \(2xy\neq -1\).

(b).

We want \(-\frac{y}{x}=-1\) which rearranges to \(y=x\).

Hence
\(
x^{4} + x^{2} = 2 \Rightarrow (x^{2} + 2)(x^{2} – 1) = 0 \Rightarrow x = \pm 1.
\)

Since \(y=x\) this means that \(y=\pm1\). That is, the points \((1,1)\) and \((-1,-1)\) have tangents with a gradient of \(-1\).

Question 9

(a).

When \(x=0\) we find that

\[
v^{2}=1600+\frac{672}{\pi}\cos^{-1}(0)=1936.
\]

Note that the speed detection device will be activated if the speed of the car is at least \(1.1\times 40=44 \text{ km h}^{-1}\). Since \(44^{2}=1936\) this means that the speed detection device will be activated.

(b).

Here we use the fact that the acceleration is given by \(a=\frac{d}{dx}(\frac{1}{2}v^{2})\).

Hence
\[
a=\frac{d}{dx}\bigg(\frac{1}{2}v^{2}\bigg)=\frac{d}{dx}\bigg(800+\frac{336}{\pi}\cos^{-1}(x)\bigg)=-\frac{336}{\pi\sqrt{400-x^{2}}}.
\]

When \(x=12\) this is
\[
a=-\frac{336}{\pi\sqrt{400-12^{2}}}=-\frac{21}{\pi}.
\]

Question 10

First we compute the normal vector that is perpendicular to both lines. This is given by

\[
\vec{n}=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\times\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}=\begin{pmatrix} 1 \\ -3 \\ 5 \end{pmatrix}.
\]

Next compute that \(|\vec{n}|=\sqrt{35}\) so that a unit normal vector is given by

\[
\hat{\vec{n}}=\frac{1}{\sqrt{35}}\begin{pmatrix} 1 \\ -3 \\ 5 \end{pmatrix}.
\]

We then put \(\vec{v}\) to be the difference in starting points of the two lines. That is,

\[
\vec{v}=\begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}-\begin{pmatrix} 1 \\ 0 \\ m \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ -1-m \end{pmatrix}.
\]

The shortest distance between the two lines is then given by the length of the projection of \(\vec{v}\) onto \(\hat{\vec{n}}\). That is, the shortest distance between the two lines is

\[
|\text{proj}_{\hat{\vec{n}}}(\vec{v})|=|\vec{v}\cdot\hat{\vec{n}}|=\frac{1}{\sqrt{35}}|1-5(1+m)|.
\]

Hence we want
\[
\frac{1}{\sqrt{35}}|1-5(1+m)|=\frac{14}{\sqrt{35}}
\]

which rearranges to give
\[
m=2,-\frac{18}{5}.
\]

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