2024 VCE Specialist Maths Paper 2 Solutions

In this article, we reveal 2024 VCE Specialist Maths Exam Paper 2 Solutions, complete with full explanations written by the Matrix Maths Team.

2024 VCE Specialist Maths Paper 2 Solutions

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2024 VCE Specialist Maths Paper 2 Solutions – Section A

2024 VCE Specialist Maths Paper 2 Solutions – Section B

Question 1

(a).

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

Thus stationary points occur at \(x=0,\pm\sqrt{2}\). Note that \(f(0)=1\) and \(f(\pm\sqrt{2})=-3\) and so the stationary points are \((0,1),(\sqrt{2},-3),(-\sqrt{2},-3)\). Also note that the vertical asymptotes have equations at \(x=\pm 1\). Thus the graph of \(f(x)\) is as follows:

(b).

(i).

The volume is given by
\[
V=\pi\int_{1}^{6}x^{2}\,dy.
\]

Thus we must rearrange the equation for \(x^{2}\). Letting \(u=x^{2}\) we find that
\(
y = \frac{u^{2} – u + 1}{1 – u} \Rightarrow u^{2} + (y – 1)u + (1 – y) = 0.
\)
and so by using the quadratic formula we have that
\[
x^{2}=u=\frac{1-y\pm\sqrt{y^{2}+2y-3}}{2}.
\]

Since \(1\leq y\leq 6\) we have that \((0,1)\) lies on the branch of the curve that we want and hence we will take the positive case.

Hence
\[
V=\pi\int_{1}^{6}\frac{1-y+\sqrt{y^{2}+2y-3}}{2}dy.
\]

(ii).

Using the calculator one finds that
\[
V=\pi\int_{1}^{6}\frac{1-y+\sqrt{y^{2}+2y-3}}{2}dy=11.2\text{ units}^{3}.
\]

(c).

If \(b=-1\) then

\begin{align*}
g(x)
&= \frac{x^{4}-1}{1-x^{2}} \\
&= \frac{(x^{2}+1)(x^{2}-1)}{1-x^{2}} \\
&= -(x^{2}+1), x \neq \pm 1
\end{align*}
which has no asymptotes.

(d).

(i).

We want to restrict the number of solutions to \(g'(x) = 0\) to only allow for one solution. Assuming \(x \not\in \{-1, 1\}\),

\begin{align*}
g'(x) &= 0 \\
\iff \frac{-2x\Big((x^2 – 1)^2 – (b + 1)\Big)}{(1 – x^2)^2} &= 0 \\
\iff -2x\Big((x^2 – 1)^2 – (b + 1)\Big) &= 0 \tag{as \(x \not\in \{-1, 1\}\)}.
\end{align*}

This will always have one solution for \(x = 0\), so we need
\begin{align*}
(x^2 – 1)^2 – (b + 1) &= 0 \\
\iff (x^2 – 1)^2 &= b + 1
\end{align*}
to have no real solutions for \(x \not\in \{-1, 0, 1\}\). This will happen if and only if
\(
\begin{aligned}
b + 1 &\leq 0 & \text{(the ‘equal to’ accounts for the fact that we don’t consider } x \in \{-1, 1\} \text{ valid solutions)} \\
\iff b &\leq -1.
\end{aligned}
\)
which is our final answer for this part.

(ii).

Slightly different from before, we need

\((x^2 – 1)^2 = b + 1\)

to have two real solutions for \(x \not\in \{-1, 0, 1\}\). This will happen only if
\begin{align*}
b + 1 &> 0 \\
b &> -1.
\end{align*}

Solving for \(x^2\) however, we find that
\begin{align*}
(x^2 – 1)^2 &= b + 1 \\
\iff x^2 – 1 &= \pm\sqrt{b + 1} \\
\iff x^2 &= 1 \pm \sqrt{b + 1}.
\end{align*}

For \(b > -1\), this gives two possible solutions to
\begin{align*}
x^2 &= 1 + \sqrt{b + 1} \\
\iff x &= \pm\sqrt{1 + \sqrt{b + 1}},
\end{align*}
and our three solutions have been achieved.
However, for there to be no extra solutions for \(x \not\in \{-1, 0, 1\}\) from

\(x^2 = 1 – \sqrt{b + 1}\),

it must be the case that

The ‘equal to’ accounts for the fact that we don’t consider \(x = 0\) a valid solution:

\begin{align*}
1 – \sqrt{b + 1} &\leq 0 \\
\iff \sqrt{b + 1} &\geq 1 \\
\iff b + 1 &\geq 1 \\
\iff b &\geq 0.
\end{align*}
which is our final answer for this part.

(iii).

One final adjustment from the previous part is to keep \(b > -1\) so that there are the three solutions

\(0, \sqrt{1 + \sqrt{b + 1}}, -\sqrt{1 + \sqrt{b + 1}}\)

for \(x \not\in \{-1, 1\}\), and to restrict \(b\) so that

\(x^2 = 1 – \sqrt{b + 1}\),

gives two more solutions for \(x \not\in \{-1, 0, 1\}\). This will happen if and only if

\begin{align*}
1 – \sqrt{b + 1} &> 0 \\
\iff \sqrt{b + 1} &< 1 \\
\iff b + 1 &< 1 \\
\iff b &< 0.
\end{align*}
Since we already had to assume \(b > -1\), this gives \(-1 < b < 0\).

Question 2

(a).

Substituting in \(z=x+iy,z_{1}=1+2i,z_{2}=4\) we have that

\begin{align*}
|z-z_{1}|^{2}=|z-z_{2}|^{2}&\iff|(x-1)+i(y-2)|^{2}=|(x-4)+iy|^{2}
\\&\iff(x-1)^{2}+(y-2)^{2}=(x-4)^{2}+y^{2}
\\&\iff y=\frac{3}{2}x-\frac{11}{4}.
\end{align*}

(b).

The centre of the circle will be given by the midpoint of the line joining \(z_{1}\) and \(z_{2}\). That is,

\[
z_{c}=\frac{1+2i+4}{2}=\frac{5}{2}+i.
\]

The diameter of the circle is given by the distance between \(z_{1}\) and \(z_{2}\) which is
\[
|(1+2i)-4| = |-3+2i| = \sqrt{13}.
\]

Hence the radius is given by \(r=\frac{\sqrt{13}}{2}\). The equation of the circle is then given by

\(\left|z – \left(\frac{5}{2}+i\right)\right|= \frac{\sqrt{13}}{2}\)

 

(c).

In Cartesian form the equation of the circle is given by
\[
(x-1)^{2}+(y-2)^{2}=4.
\]
Algebraically this allows one to set \(y=0\) to find the \(x\)-intercept, which is \((1,0)\). Similarly, by letting \(x=0\), one finds the \(y\)-intercepts by \((0,2\pm\sqrt{3})\). Thus a sketch of the circle is as follows:

(d).

(i).

Let us add in the line segment.

(ii).

The starting point is \(z_{0}=2-i\). The angle is the principal angle of
\[
(-2+3i)-(2-i)=-4+4i
\]
which is \(\theta=\frac{3\pi}{4}\). Thus the equation of the ray is given by

\(
\mathrm{Arg}(z – (2 – i)) = \frac{3\pi}{4}.
\)

 

(e).

Since the angle that the ray makes with the positive \(x\)-axis is \(\theta=\frac{3\pi}{4}\) then the gradient of the ray is given by \(\tan(\frac{3\pi}{4})=-1\). Also, since the ray passes through the point \(2-i\) the Cartesian equation of the ray can be found to be \(y=-x+1\) (for \(x\leq 2\)). Solving simultaneously with the Cartesian equation of the circle as found in previous parts we find that the points of intersection of the ray and the circle are \(-1+2i\) and \(1\).

Thus by the diagram, the angle of the sector is \(\theta=\frac{\pi}{2}\) and the radius of the circle is \(r=2\). Hence the area of the minor segment is given by

\[
A=\frac{1}{2}r^{2}(\theta-\sin(\theta))=\frac{1}{2}\times2^{2}\bigg(\frac{\pi}{2}-\sin\bigg(\frac{\pi}{2}\bigg)\bigg)=(\pi-2)\text{ units}^{2}.
\]

 

Question 3

(a).

We find that
\[
\frac{d^{2}V}{dt^{2}}=\frac{8(250+5t^{4})-8t(20t^{3})}{(240+5t^{4})^{2}}=\frac{8(-15t^{4}+240)}{(240+5t^{4})^{2}}.
\]

Hence
\[
\frac{d^{2}V}{dt^{2}}=0\iff8(-15t^{4}+240)=0\iff t=2
\]
since \(t\geq 0\). Thus the maximum occurs for \(t=2\).

(b).

The volume of the disc is given by

\(
V = 0.001\pi r^{2} \Rightarrow \frac{dV}{dr} = 0.002\pi r.
\)

 

Hence by the Chain Rule we have that
\[
\frac{dr}{dt}=\frac{dr}{dV}\times\frac{dV}{dt}=\frac{1}{0.002\pi r}\times\frac{8t}{240+5t^{4}}.
\]

Taking \(t=4\) and \(r=6.54\) this gives that
\[
\frac{dr}{dt}=0.51\text{ m day}^{-1}
\]
to two decimal places.

(c).

(i).

Letting \(u=\sqrt{5}t^{2}\) one has that \(du=2\sqrt{5}t\,dt\).

Hence
\begin{align*}
\int\frac{8t}{240+5t^{4}}\,dt&=\frac{8}{2\sqrt{5}}\int\frac{2\sqrt{5}t\,dt}{240+(\sqrt{5}t^{2})^{2}}
\\&=\frac{8}{2\sqrt{5}}\int\frac{du}{240+u^{2}}.
\end{align*}

(ii).

By the previous part we have that

\begin{align*}
V&=\frac{8}{2\sqrt{5}}\int\frac{du}{240+u^{2}}
\\&=\frac{8}{2\sqrt{5}}\times\frac{1}{\sqrt{240}}\tan^{-1}\bigg(\frac{u}{\sqrt{240}}\bigg)+C
\\&=\frac{1}{5\sqrt{3}}\tan^{-1}\bigg(\frac{t^{2}}{4\sqrt{3}}\bigg)+C.
\end{align*}

Since \(t=0\) when \(V=0\) this implies that \(C=0\) and so
\[
V=\frac{1}{5\sqrt{3}}\tan^{-1}\bigg(\frac{t^{2}}{4\sqrt{3}}\bigg).
\]

(d).

Note that

\begin{align*}
\lim_{t\rightarrow\infty}V&=\lim_{t\rightarrow\infty}\frac{1}{5\sqrt{3}}\tan^{-1}\bigg(\frac{t^{2}}{4\sqrt{3}}\bigg)
\\&=\frac{1}{5\sqrt{3}}\times\frac{\pi}{2}
\\&=\frac{\pi}{10\sqrt{3}}.
\end{align*}

For this volume we have that
\(
\frac{\pi}{10\sqrt{3}} = 0.001\pi r^{2} \Rightarrow \text{Area} = \pi r^{2} = \frac{100\pi}{\sqrt{3}} = 181.38 \, \text{m}^{2}.
\)

(e).

Since the cleanup begins after \(5\) days we have that the rate of change of the volume of the pollutant is given by

\[
\frac{dV}{dt}=\begin{cases} \frac{8t}{240+5t^{4}},&\text{ if }0\leq t<5, \\ \frac{8t}{240+5t^{4}}-0.05,&\text{ if }t\geq5.\end{cases}
\]

Hence at a time \(T\geq 5\) we have that the volume of the pollutant is given by

\begin{align*}
V&=\int_{0}^{T}\frac{dV}{dt}\,dt
\\&=\int_{0}^{5}\frac{8t}{240+5t^{4}}\,dt+\int_{5}^{T}\frac{8t}{240+5t^{4}}-0.05\,dt
\\&=\int_{0}^{T}\frac{8t}{240+5t^{4}}\,dt-\int_{5}^{T}0.05\,dt
\\&=\frac{1}{5\sqrt{3}}\tan^{-1}\bigg(\frac{T^{2}}{4\sqrt{3}}\bigg)-0.05(T-5).
\end{align*}

We wish to find \(T\geq 5\) such that \(V=0\). That is, we solve the equation

\[
\frac{1}{5\sqrt{3}}\tan^{-1}\bigg(\frac{T^{2}}{4\sqrt{3}}\bigg)-0.05(T-5)=0
\]
for \(T\geq 5\). Using the calculator we find the solution to be \(8.4\). Thus the number of days after the start of the cleanup at which the pond will be free of pollutant is \(8.4-5=3.4\) days.

Question 4

(a).

Here we simply calculate that
\[
\frac{x^{2}}{9}-\frac{y^{2}}{4}=\sec^{2}(t)-\tan^{2}(t)=1.
\]

(b).

(c).

(i).

By differentiating, the velocity of the yacht is given by

\[
\vec{v}(t)=\begin{pmatrix} 3\sec(t)\tan(t) \\ 2\sec^{2}(t) \end{pmatrix}.
\]

Hence the square of the speed is given by
\begin{align*}
|\vec{v}(t)|^{2}&=9\sec^{2}(t)\tan^{2}(t)+4\sec^{4}(t)
\\&=9\sec^{2}(t)(\sec^{2}(t)-1)+4\sec^{4}(t)
\\&=13\sec^{4}(t)-9\sec^{2}(t).
\end{align*}

(ii).

By minimising the above expression of \(|\vec{v}(t)|^{2}\) on the calculator we find that the minimum occurs for \(t=3.14159\dots\) and so by guessing we have that the minimum occurs for \(t=\pi\).

(iii).

For \(t=\pi\) we find that
\[
|\vec{v}(\pi)|^{2}=13\sec^{4}(\pi)-9\sec^{2}(\pi)=13-9=4
\]
and so the minimum speed of the yacht is given by \(2\) m min\(^{-1}\).

(iv).

When \(t=\pi\) the coordinates of the yacht are given by
\[
\vec{r}(\pi)=\begin{pmatrix} 3\sec(\pi) \\ 2\tan(\pi) \end{pmatrix}=\begin{pmatrix} -3 \\ 0 \end{pmatrix}.
\]

(d).

(i).

The distance travelled is given by
\[
\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\sqrt{\bigg(\frac{dx}{dt}\bigg)^{2}+\bigg(\frac{dy}{dt}\bigg)^{2}}\,dt=\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\sqrt{13\sec^{4}(t)-9\sec^{2}(t)}\,dt.
\]

(ii).

The calculator gives that
\[
\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\sqrt{13\sec^{4}(t)-9\sec^{2}(t)}\,dt=9.4\text{ m}.
\]

(e).

We need to minimise
\[
|\vec{r}_{D}(t)-\vec{r}_{Y}(t)|=\sqrt{(2-3t-3\sec(t))^{2}+(4t-1-2\tan(t))^{2}+(6-t)^{2}}
\]
over the intersection of \(0\leq t\leq 5\) and \(\frac{2\pi}{3}\leq t\leq\frac{4\pi}{3}\). Since \(\frac{4\pi}{3}\leq 5\) this is just the minimum over \(\frac{2\pi}{3}\leq t\leq\frac{4\pi}{3}\). Using the calculator, we find this minimum to be \(14.0\) m.

Question 5

(a).

A vector equation for the line passing through \(A\) and \(B\) may be given by

\(
\begin{align*}
\overrightarrow{\mathrm{OA}} + \lambda \overrightarrow{\mathrm{AB}}
&= (\mathbf{i} – 2\mathbf{j} + 3\mathbf{k}) + \lambda (\mathbf{i} – 3\mathbf{j} – 4\mathbf{k}) \\
&= (1 + \lambda)\mathbf{i} + (-2 – 3\lambda)\mathbf{j} + (3 – 4\lambda)\mathbf{k}.
\end{align*}
\)
where \(\lambda \in R\).

(b).

Assuming \(A\) doesn’t lie on the line \(L_1\), the shortest distance from \(A\) to \(L_1\) is achieved when the displacement vector from \(A\) to a point on \(L_1\) is perpendicular with the direction vector \( \vec{v} := -\mathbf{i} + 2\mathbf{j} + \mathbf{k}.\) of \(L_1\).

This happens when
\(
\begin{align*}
\left(\vec{\mathrm{r}}_1(t) – \overrightarrow{\mathrm{OA}}\right) \cdot \vec{v} &= 0, \\
\vec{\mathrm{r}}_1(t) \cdot \vec{v} &= \overrightarrow{\mathrm{OA}} \cdot \vec{v}, \\
\begin{pmatrix}2\\1\\-3\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\1\end{pmatrix} + t\begin{pmatrix}-1\\2\\1\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\1\end{pmatrix} &= \begin{pmatrix}1\\-2\\3\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\1\end{pmatrix}, \\
-3 + 6t &= -2, \\
t &= \frac{1}{6}.
\end{align*}
\)

The length of the displacement vector between \(A\) and the point \(r(\frac{1}{6})\) is the answer, and it is

\(
\begin{align*}
\left|\vec{\mathrm{r}}_1\left(\frac{1}{6}\right) – \overrightarrow{\mathrm{OA}}\right|
&= \left|\begin{pmatrix}2\\1\\-3\end{pmatrix} + \frac{1}{6}\begin{pmatrix}-1\\2\\1\end{pmatrix} – \begin{pmatrix}1\\-2\\3\end{pmatrix}\right| \\
&= \frac{5\sqrt{66}}{6}.
\end{align*}
\)

(c).

Two direction vectors which span the plane containing points \(A\), \(B\) and \(C\) are \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\).

We can express the equation of the plane as

\(
\begin{align*}
\overrightarrow{\mathrm{OA}} + \lambda \overrightarrow{\mathrm{AB}} + \mu \overrightarrow{\mathrm{AC}}
&= (\mathbf{i} – 2\mathbf{j} + 3\mathbf{k}) + \lambda (\mathbf{i} – 3\mathbf{j} – 4\mathbf{k}) + \mu (-\mathbf{i} + 4\mathbf{j} – 8\mathbf{k}) \\
&= (1 + \lambda – \mu)\mathbf{i} + (-2 – 3\lambda + 4\mu)\mathbf{j} + (3 – 4\lambda – 8\mu)\mathbf{k}.
\end{align*}
\)
where \(\lambda, \mu \in R\).

(d).

(i).

To find the \(x\)-axis intercept, let \(y = z = 0\) and see that
\begin{align*}
2x &= 12 \\
x &= 6.
\end{align*}

To find the \(y\)-axis intercept, let \(x = z = 0\) and see that
\begin{align*}
-3y &= 12 \\
y &= -4.
\end{align*}

To find the \(z\)-axis intercept, let \(x = y = 0\) and see that
\begin{align*}
4z &= 12 \\
z &= 3.
\end{align*}

The three points are thus \((6, 0, 0)\), \((0, -4, 0)\) and \((0, 0, 3)\).

(ii).

Call the points \((6, 0, 0)\), \((0, -4, 0)\) and \((0, 0, 3)\) by \(X\), \(Y\) and \(Z\), respectively. The area of the triangle \(XYZ\) will be half the area of the parallelogram spanned by the vectors \(\overrightarrow{\mathrm{XY}}\) and \(\overrightarrow{\mathrm{XZ}}\). The area of this parallelogram is the magnitude of the cross product of these two vectors.

Hence the area of the triangle is
\begin{align*}
\frac{1}{2}\left|\overrightarrow{\mathrm{XY}} \times \overrightarrow{\mathrm{XZ}}\right|
&= \frac{1}{2}\left|\begin{pmatrix}-6\\-4\\0\end{pmatrix} \times \begin{pmatrix}-6\\0\\3\end{pmatrix}\right| \\
&= 6\sqrt{29}.
\end{align*}

Question 6

(a).

The null hypothesis is \(H_{0}:\mu=1000\) mL and the alternative hypothesis is \(H_{1}:\mu<1000\) mL.

(b).

(i).

The \(p\) value is given by

\[
p=\operatorname{Pr}(\bar{X}<997.5|\mu=1000).
\]

Since in this setting we have that
\[
\mu_{\bar{X}}=1000
\]
and
\[
\sigma_{\bar{X}}=\frac{4.2}{\sqrt{9}}=1.4
\]

using the calculator one finds that
\[
p=\operatorname{Pr}(\bar{X}<997.5|\mu=1000)=0.037.
\]

(ii).

Since the \(p\)-value is \(0.037<0.05\) the machine should be paused.

(c).

First we find the cricical value \(\alpha\), which is the largest number such that

\[
\operatorname{Pr}(\bar{X}<\alpha|\mu=1000)=0.05.
\]

The calculator gives this value to be approximately \(\alpha=997.697\). Thus the probability of a type II error using \(9\) bottles is given by
\[
\operatorname{Pr}(\bar{X}>997.697|\mu=997)=0.31.
\]

(d).

Using the calculator we can find the values of \(a\) and \(b\) to be \(a=996.7\) mL and \(b=1003.3\) mL.

(e).

For a random sample of \(50\) bottles a \(95\%\) confidence interval is given by

\[
\bigg(1005-1.96\times\frac{4}{\sqrt{5}},1005+1.96\times\frac{4}{\sqrt{5}}\bigg)=(1003.9,1006.1).
\]

(f).

We would expect \(95\%\) of these confidence intervals to contain the population mean volume. That is, \(0.95\times 40=38\) of them.

(g).

Using \(z = 1.96\) for the confidence interval of \(95\%\), we want to find sample size \(n\) for which

\begin{align*}
z\frac{\sigma}{\sqrt n} &\leq 1 \\
1.96 \times \frac{4}{\sqrt n} &\leq 1 \\
n &\geq 61.4656,
\end{align*}
and \(n = 62\) is the minimum sample size required.

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