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Congratulations Year 12 on finishing the HSC exam season! Here are the 2024 Maths Advanced HSC Solutions put together by the Matrix Maths Team. Read on to see how you went!
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In this article, we reveal 2024 HSC Maths Advanced Exam Solutions, complete with full explanations written by the Matrix Maths Team.
Have you seen the 2022 HSC Mathematics Advanced exam paper yet? For reference, you can find it, here.
Doing past papers? Find all the solutions to the 2018 – 2021 HSC Maths Advance Exams, here.
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Question Number | Answer | Solutions |
1 | C | We can see that the line has a positive \(y-intercept\), so \(c=3\). The line also has a downward slope so \(m<0\), which means \(m=-2\). Therefore: \( |
2 | B | First we construct a Venn diagram to aid in our solution. There are \(60-5=55\) students in total that play either basketball or hockey. To find the number of students who play both we subtract \(55\) from \(38+35=73\) and arrive at \(18\). |
3 | A | First, we need to find the \(z\)-score for Pia’s mark in each subject: English: Mathematics: Science: History: Since the English \(z\)-score is the greatest, Pia performed best in English. |
4 | C | We can find the equation by sketching the transformations The final plot is \(C\). |
5 | A | We use reverse chain rule to integrate: \( \begin{aligned} \int (6x+1)^3 dx = \frac{(6x+1)^4}{4\times 6} +C \\ = \frac{1}{24}(6x+1)^4 +C \\ \end{aligned} \) \(A\) is the correct answer. |
6 | D | We notice two restrictions: 1. The denominator of a fraction cannot be 0. Thus we arrive at the condition: \(\begin{aligned} x^2-1&>0 \\ (x-1)(x+1)&>0. \end{aligned} \) Graphing to solve the inequality: The region corresponds with \((-\infty,-1)\cup(1,\infty)\). Our answer is \(A\). |
7 | C | We can consider the transformation for \(f(x) \rightarrow f(2x-1)\) in two ways: 1. Shift right by 1 Alternatively: 1. Factorise as \(2(x-\frac{1}{2})\) We take the second approach in this example, The blue graph illustrates the dilation and the red graph represents the shift, and is the final answer. Thus, we arrive at C. |
8 | D | From the box plot: \(Q_1 – \text{Min} = \text{Max} – Q_4\). Similarly, \(Q_2 – Q_1 = Q_4 – Q_3\). Additionally, \(Q_2 – Q_1\) is approximately double \(Q_1 – \text{ Min}\). In each histogram there are \(16\) scores. From the box plot we can find the position of each statistic in the five number summary: 1. Min: 1st score We label the bottom axis as follows to calculate the quartiles of each graph. A: Min = 1, \(Q_1=2, Q_2=4, Q_3=6\), Max = 7 For \(D\): that \(Q_3 – Q_2 = 1\) and Max \( – Q_3 = 2\). |
9 | B | We wish to find \(P(\text{two reds }| \text{ one red})\). \( \begin{aligned} P(\text{two reds }| \text{ one red}) &= \frac{P(\text{two red})}{P(\text{one red})} \\ &= \frac{P(\text{two red})}{1 – P(\text{two white})} \\ &= \frac{\frac{2}{5} \times \frac{1}{4}}{1 – \frac{3}{5} \times \frac{2}{4}} \\ &= \frac{1}{7} \end{aligned} \) Thus we arrive at \(B\). |
10 | B | We note that by the Fundamental Theorem of Calculus \(f(x)= A'(x)\). Additionally, the points of inflection of \(A(x)\) correspond to where the sign of the gradient of \(f(x)\) changes. Thus we examine the gradient of the graph of \(f(x)\). The graph is green for positive gradients and red for negative: As the graph changes between red and green 3 times \(A(x)\) has 3 points of inflection. Thus we arrive at \(B\). |
In the graph below, blue represents sections of the graph with a positive gradient, and red with negative. The pink points represent stationary points, where the gradient of the function is zero.
This allows us to fill the first column of the table:
In the graph below, blue represents sections of the graph which are concave up (positive second derivative), and red with concave down (negative second derivative). The pink points represent points of inflexion, where the concavity of the function is zero.
We can then finish off the table:
We’ll call the individual terms of this arithmetic series \(T_n\), where \(T_1 = 50\) and \(T_N = 2024\) for some \(N\). We can see that the common difference is \(d=7\), and so we may use the formula \(T_n = T_1 + (n – 1) d\) to find \(N\) as
\(
\begin{aligned}
T_N &= 2024 \\
50 + (N – 1) \times 7 &= 2024 \\
7 (N – 1) &= 2024 – 50 \\
N – 1 &= 1974 \div 7 \\
N &= 283
\end{aligned}
\)
We can then use the formula \(S_n = \frac{n}{2} (T_1 + T_n)\) for arithmetic series to see that
The vertical intercept of the function for Population \(K\) is \(280\). Thus we have:
\(Percentage change in the yearly population:
Population \(W\):
\(
\begin{aligned}
1.055-1 &= 0.055\\
&= 5.5\%\\
\end{aligned}
\)
Population \(K\):
\(
\begin{aligned}
1-0.97 &= 0.03\\
&= 3\%\\
\end{aligned}
\)
When \(x=50\):
\(
\begin{aligned}
y &= 35 \times 1.055^{50}\\
&\approx 494\\
\end{aligned}
\)
Summarising in a table:
Part (a)
Solving for \(x\) when these equations are equal:
\(Thus, the graphs intersect at \(x = -1\) and \(x = 2\).
Part (b)
We can use the formula \(\mathrm{Area} = \int_a^b (\operatorname{top}(x) – \operatorname{bottom}(x)) \, dx\) to see that
\(
\begin{aligned}
\mathrm{Area}
&= \int_{-1}^2 \left( (5 – x^2) – (x – 1)^2 \right) \, dx \\
&= \int_{-1}^2 \left( 4 + 2 x – 2 x^2 \right) \, dx \\
&= \left[ 4 x + x^2 – \frac{2}{3} x^3 \right]_{-1}^2 \\
&= 4 \times 2 + 2^2 – \frac{2}{3}\times 2^3 – \left(4 \times (-1) + (-1)^2 – \frac{2}{3}\times (-1)^3 \right) \\
&= \text{$9$ units$^2$}.
\end{aligned}
\)
We wish to find the first time at which \(\frac{dV}{dx} = 0\). Thus:
\(We then solve the differential equation by integrating:
\(
\begin{aligned}
\frac{dV}{dx} &= 300 – 7.5 t \\
V &= \int ( 300 – 7.5 t ) \, dt \\
&= 300 t – 3.75 t^2 + C,
\end{aligned}
\)
for some constant \(C\). We know that initially, when \(t = 0\), the volume in the tank is \(V = 350\). We can use this in the above equation to solve for \(C\) as
We can finally substitute our value of \(t = 40\) to find that
\(
Garden A is negatively skewed, has a median of approximately \(171\), and an interquartile range of \(176-154=22\).
Garden B is positively skewed, has a median of approximately \(152\), and an interquartile range of \(158-149=9\).
Therefore, Garden B has consistently lower heights since both the IQR and median are lower than Garden A. Also, for Garden B most of the scores are in the lower range since it is positively skewed.
Part (a)
Part (b)
Using \(V(1) = 2.6\) and our equation for \(V(t)\) gives
\(Part (c)
We want to find \(\frac{dV}{dt}\) when \(t = 2\), so we will first find \(\frac{dV}{dt}\) as
\(
Letting \(t = 2\) gives
\(
\begin{aligned}
\left. \frac{dV}{dt} \right\vert_{t = 2}
&= \left. 6.5 k e^{-k t} \right\vert_{t = 2} \\
&= -6.5 \ln(0.6) e^{\ln(0.6) (2)} \\
&= \text{1.195 volts/second}
\end{aligned}
\)
Part (a)
\(Part (b)
Let \(P(\text{at least 1 goal}) = 0.8\), and the number of trials be \(n\), then:
\(
\begin{aligned}
1-P(0\text{ goals}) &= 0.8 \\
1-0.85^n &= 0.8 \\
0.85^n &= 1- 0.8 \\
0.85^n &= 0.2 \\
n &= \log_{0.85}(0.2) \\
&= \frac{\log(0.2)}{\log(0.85)} \\
&\approx 9.903 …
\end{aligned}
\)
Therefore, to guarantee a probability of \(0.8\), he should make \(10\) attempts.
First, we can find the stationary points by solving \(y’=0\):
\(
\begin{aligned}
y’&=4x^3-6x^2 \\
0&=2x^2(2x-3) \\
2x^2=0 \,&\text{ or } \,2x-3=0 \\
x=0 \,&\text{ or } \,x=\frac{3}{2}
\end{aligned}
\)
When \(x=0: \, y=2\). Therefore, one stationary point is at \((0,2)\).
When \(x=\frac{3}{2}: \, y=\frac{5}{16}\). Therefore, another stationary point is at \(\left(\frac{3}{2},\frac{5}{16}\right)\).
We can check the nature of each stationary point by using the second derivative:
\(When \(x=0:\, y”=0\). Therefore, \((0,2)\) is a possible point of inflection. We need to check the concavity around that point.
When \(x=-0.1\):
\(
\begin{aligned}
y”&=12\times 0.01+12\times 0.1 \\
&=0.12+1.2 >0
\end{aligned}
\)
When \(x=0.1\):
\(
\begin{aligned}
y”&=12\times 0.01-12\times 0.1 \\
&=0.12-1.2 <0
\end{aligned}
\)
Therefore, the concavity changes on either side of \((0, 2)\), so it is a point of inflection.
When \(x=\frac{3}{2}:\, y”=9>0\). Therefore, \(\left(\frac{3}{2},\frac{5}{16}\right)\) is a local minimum.
To check if there are any other points of inflection we need to solve \(y”=0\):
\(
\begin{aligned}
12x^2-12x &=0 \\
12x(x-1)&=0 \\
x=0 \,&\text{ or } \,x=1.
\end{aligned}
\)
Therefore, there is a possible point of inflection when \(x=1\). We need to check the concavity around that point.
When \(x=0.9\):
\(
\begin{aligned}
y”&=12\times 0.81-12\times 0.9 \\
&=-1.08 <0
\end{aligned}
\)
When \(x=1.1\):
\(
\begin{aligned}
y”&=12\times 1.21-12\times 1.1 \\
&=1.32 >0
\end{aligned}
\)
When \(x=1, y=1\). Therefore, the concavity changes on either side \((1, 1)\), so it is a point of inflection.
The \(y\)-intercept is at \(y=2\).
Part (a)
In \(\triangle TCA\):
\(
\begin{aligned}
\tan 35^\circ &= \frac{40}{AC} \\
AC &= \frac{40}{\tan 35^\circ} \\
&\approx 57.13\text{ m } (2\text{ d.p.})
\end{aligned}
\)
Part (b)
In \(\triangle TCB\):
\(Let \(\theta=\angle BCA\),
\(
\begin{aligned}
\cos \theta &= \frac{AC^2+BC^2-100^2}{2\times AC\times BC}\\
\theta &= \cos^{-1}\left(\frac{57.13^2+69.28^2-100^2}{2\times 57.13\times 69.28}\right) \\
&=104.161…
\end{aligned}
\)
Part (a)
The second derivative of \(f(x)\) is
\(For \(f(x)\) to be concave up, we need \(f”(x)\) to be positive. The denominator \((1 + x^2)^2\) is always positive, so we need to show that \((1-x)(1+x)\) is positive. Since \(-1 < x < 1\),
\(
\begin{aligned}
x^2 &< 1 \\
0 &< 1 – x^2,
\end{aligned}
\) as required.
So, for \(-1 < x < 1\), \(f”(x)\) is positive and \(f(x)\) is concave up.
Part (b)
We can first see that the left area (red) and right area (blue) are both equal, since \(\ln(1 + x^2)\) is an even function. Denote both of these areas by \(A\) such that the total shaded area is \(2 A\).
Since \(\ln (1 + x^2)\) is non-negative for the values we’re concerned about, we can say that
\(So, the total shaded area is
\(2 A = \text{0.538275 units^2}\).
Part (c)
We’ve shown that \(\ln(1 + x^2)\). is a concave up function for \(-1 < x < 1\). This means that the tops of the trapezoids in part (b) will lie above the curve. So, the trapezoidal rule overestimates the actual area.
Part (a)
For \(x=58: z=0\).
For \(x=70:\)
Thus we have:
\(Part (b)
For \(x=46:\)
\(Now,
\(
\begin{aligned}
P(46<x<70) &= P(-0.8 < z<0.8) \\
&= P(-0.8 < z<0) + P(0 < z<0.8) \\
&= 2\times P(0 < z<0.8)
\end{aligned}
\)
The area under the normal curve between \(z =-0.8\) and \(z=0\) is equal to the area under the normal curve between \(z =0\) and \(z=0.8\). Therefore, the probability is equal to double the area under the normal curve between \(z =0\) and \(z=0.8\).
Part (c)
From the table, when \(P=0.9032: z=1.3\).
\(Therefore, the approximate minimum score is \(77.5\).
Part (a)
Let \(A_n\) be the amount of money in Jack’s account at the end of the \(nth\) month. We can see that \(A_0 = 0\), as in the very beginning, he has no money. At the beginning of each month, Jack will deposit \($80\), and the amount will grow by a factor of \(1 + \frac{0.06}{12} = 1.005\). Writing the first few terms and generalising from there,
\(
Part (b)
The time-frame of \(24\) periods with \(0.5%\) growth per period lines up with the scenario described in part (a). Since an annuity of \($80\) grows to a value of \($2,044.73\) (from part (b)), \($1\) will grow to
\(
Part (a)
For a probability density function, the area under the curve equals 1.
\[
\int_0^h \left(1 – \frac{x}{h}\right) \, dx = 1
\]
\[
\left[ x – \frac{x^2}{2h} \right]_0^h = 1
\]
\[
h – \frac{h}{2} = 1
\]
\[
\frac{h}{2} = 1
\]
\[
h = 2
\]
Part (b)
\[
F(x) = \int_{0}^{x} \left(1 – \frac{t}{2}\right) dt = x – \frac{x^2}{4} \quad (0 \leq x \leq 2)
\]
\[
F(x) =
\begin{cases}
0, & \text{for } x < 0, \\
x – \frac{x^2}{4}, & \text{for } 0 \leq x \leq 2, \\
1, & \text{for } x > 2
\end{cases}
\]
The graph of \( F(x) \) is shown below:
Part (c)
To find the median we let \(F(x) = \frac{1}{2}\).
\[
x – \frac{x^2}{4} = \frac{1}{2}
\]
\[
4x – x^2 = 2
\]
\[
x^2 – 4x + 2 = 0
\]
Solving for \( x \):
\[
x = \frac{4 \pm \sqrt{(-4)^2 – 4 \cdot 1 \cdot 2}}{2 \cdot 1}
\]
\[
x = 2 \pm \sqrt{2}
\]
Since \( 0 \leq x \leq 2 \), we take \( x = 2 – \sqrt{2} \approx 0.586 \)
For the first \(15\) years:
\(To find the present value for the next 10 years, we can subtract the present value of 15 years from the present value of 25 years using \(r=0.002\) and withdrawals of \($1200\).
For 25 years:
\(
\begin{aligned}
n &= 25\times 12 \\
&= 300
\end{aligned}
\)
For 15 years:
\(
\begin{aligned}
n &= 15\times 12 \\
&= 180
\end{aligned}
\)
\(
Therefore, the minimum sum to deposit is:
\(302072 + 89227.47 \approx \$391344.15\)
Part (a)
\(\frac{d}{dx}(x^2 \tan x) = 2x \tan x + x^2 \sec^2 x\)
Part (b)
\(Part (a)
\(c\) is the midpoint of the minimum and maximum height:
\(c = \frac{39 + 3}{2} = 21\)\(k\) is the amplitude which we can find by halving the difference between the minimum and maximum height:
\(k = \frac{39 – 3}{2} = 18\)
Part (b)
\(T = \frac{2\pi}{\frac{\pi}{24}} = 48 \text{ seconds}\).
Part (c)
\(A(t) = 21 – 18 \cos \left( \frac{\pi t}{24} \right)\)
Then \(B(t) = A(t)\) when Anna and Billie are at the same height:
They will be at the same height when \(t=3\)and \(t=3+\frac{48}{2}=27\).
\(
Consider \(y=ax^2+bx+c, a\neq 0\). The tangent will have the following equation:
\[ \frac{dy}{dx} = 2ax+b \]
When \(x = -4\), \(\frac{dy}{dx} = 0\):
\(Let \(t(x) = n(x)\):
\(
\begin{aligned}
2x+3 &= -\frac{1}{2}x-2\\
\frac{5}{2}x&=-5\\
x &= -2\\
\end{aligned}
\)
When \(x = -2, \frac{dy}{dx} = 2\):
\[ -4a + b = 2, \tag{2}\]
Now, (1) – (2):
\(
\begin{aligned}
-4a &= -2\\
a &= \frac{1}{2}
\end{aligned}
\)
Therefore, solving for b:
\(
\begin{aligned}
-8 \times \frac{1}{2} + b &= 0\\
b &= 4\\
\end{aligned}
\)
Substituting \(x=-2\) into the tangent equation:
\(
\begin{aligned}
t(-2) &= 2(-2)+3 \\
&= -1
\end{aligned}
\)
Substituting the point \((-2,-1)\) into \(y\) to solve for \(c\):
\(
Since \(|x| < 1:\)
\[
-1 < x < 1
\]
When \(x = -1\):
\(
As \(x \to 1\) from below 1:
\(\(
Part (a)
\(
Now,
\(
\begin{aligned}
P(x) &= 2x + \theta + (1 + x) \theta \\
&= 2x + \theta(2 + x) \\
&= 2x + \frac{2A}{x(x + 2)} (2 + x) \\
&= 2x + \frac{2A}{x} \quad \text{as required.}
\end{aligned}
\)
Part (b)
Taking the derivative and setting it to zero to find the critical points:
\[
P'(x) = 2 – \frac{2A}{x^2} = 0
\]
Solving for \( A \):
\[
2x^2 – 2A = 0
\]
\[
A = x^2
\]
Substituting this into our equation for \(\theta\):
\(
Since \( x > 0 \):
\[
\frac{x}{x + 2} < 1, \tag{as \( x < x + 2 \)}
\]
We have:
\(\(
Check:
Second derivative test:
\[
P”(x) = \frac{4A}{x^3}
\]
Substitute \( A = x^2 \):
\(
\begin{aligned}
P”(x) &= \frac{4x^2}{x^3} \\
&= \frac{4}{x}
\end{aligned}
\)
Since \(x > 0\):
\(Therefore, when \(\theta < 2\), the perimeter is minimised.
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Written by Oak Ukrit
Oak is the Head of Mathematics at Matrix Education and has been teaching for over 12 years and has been helping students at Matrix since 2016. He has 1st class honours in Aeronautical Engineering from UNSW where he taught for over 4 years while he was undertaking a PhD. When not plane spotting he enjoys landscape photography.© Matrix Education and www.matrix.edu.au, 2023. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.