2024 HSC Maths Advanced Exam Solutions

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!

Written by:
Oak Ukrit
Notebook with pencils, erasers, ruler, and protractor on a pink background

In this article, we reveal 2024 HSC Maths Advanced Exam Solutions, complete with full explanations written by the Matrix Maths Team.

 

2024 HSC Maths Advanced Exam Solutions

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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Multiple Choice Questions

Question NumberAnswerSolutions
1CWe 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:

\(
y=-2x+3.
\)

2BFirst 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 \(18\).

3AFirst, we need to find the \(z\)-score for Pia’s mark in each subject:

English:
\(
\begin{aligned}
z &= \frac{78-66}{6} \\ & = 2
\end{aligned}
\)

Mathematics:
\(
\begin{aligned}
z &= \frac{80-71}{10} \\ & = 0.9
\end{aligned}
\)

Science:
\(
\begin{aligned}
z &= \frac{77-70}{15} \\ & = \frac{7}{15}
\end{aligned}
\)

History:
\(
\begin{aligned}
z &= \frac{85-72}{9} \\ & = \frac{13}{9}
\end{aligned}
\)

Since the English \(z\)-score is the greatest, Pia performed best in English.

4CWe can find the equation by sketching the transformations

The final plot is \(C\).

5AWe 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.

6DWe notice two restrictions:

1. The denominator of a fraction cannot be 0.
2. The number under a square root cannot be negative.

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 \(A\).

7CWe can consider the transformation for \(f(x) \rightarrow f(2x-1)\) in two ways:

1. Shift right by 1
2. Dilate by \(\frac{1}{2}\)

Alternatively:

1. Factorise as \(2(x-\frac{1}{2})\)
2. Dilate by \(\frac{1}{2}\)
3. Shift right by \(\frac{1}{2}\)

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.

8DFrom 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
2. \(Q_1\): 4th & 5th score
3. \(Q_2\): 8th & 9th score
4. \(Q_3\): 12th & 13th score
5. Max: 16th 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
B: Min = 1, \(Q_1=2, Q_2=4, Q_3=6\), Max = 7
C: Min = 1, \(Q_1=2, Q_2=4, Q_3=6\), Max = 7
D: Min = 1, \(Q_1=3, Q_2=4, Q_3=5\), Max = 7

For \(D\): that \(Q_3 – Q_2 = 1\) and Max \( – Q_3 = 2\).
This does not reflect the properties of the box plot. Thus we arrive at our solution \(D\).

9BWe 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 \(B\).

10BWe 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 \(B\).

2024 Maths Advanced Exam Solutions – Long Response Questions

Question 11

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:

Question 12

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

\(
\begin{aligned}
S_{283} &= \frac{283}{2} (50 + 2024) \\
&= 293\,471
\end{aligned}
\)

 

Question 13

The vertical intercept of the function for Population \(K\) is \(280\). Thus we have:

\(
\begin{aligned}
B &= 280
\end{aligned}
\)

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:

Question 14

Part (a)

Solving for \(x\) when these equations are equal:

\(
\begin{aligned}
(x – 1)^2 &= 5 – x^2 \\
x^2 – 2x + 1 &= 5 – x^2 \\
2x^2 – 2x – 4 &= 0 \\
x^2 – x – 2 &= 0 \\
(x – 2)(x + 1) &= 0.
\end{aligned}
\)

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
\(
\begin{aligned}
\mathrm{Area}
&= \int_{-1}^2 \left( (5 – x^2) – (x – 1)^2 \right) \, dx \\
&= \int_{-1}^2 \left( 4 + 2 x – 2 x^2 \right) \, dx \\
&= \left[ 4 x + x^2 – \frac{2}{3} x^3 \right]_{-1}^2 \\
&= 4 \times 2 + 2^2 – \frac{2}{3}\times 2^3 – \left(4 \times (-1) + (-1)^2 – \frac{2}{3}\times (-1)^3 \right) \\
&= \text{$9$ units$^2$}.
\end{aligned}
\)

Question 15

We wish to find the first time at which \(\frac{dV}{dx} = 0\). Thus:

\(
\begin{aligned}
\frac{dV}{dx} &= 0 \\
300 – 7.5 t &= 0 \\
7.5 t &= 300 \\
t &= \text{40 hours}
\end{aligned}
\)

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

\(
\begin{aligned}
V &= 300 t – 3.75 t^2 + C \\
(350) &= 300 (0) – 3.75 (0)^2 + C \\
C &= 350
\end{aligned}
\)

We can finally substitute our value of \(t = 40\) to find that

\(
\begin{aligned}
V(40)
&= 300 (40) – 3.75 (40)^2 + 350 \\
&= \text{6350 litres.}
\end{aligned}
\)

 

Question 16

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.

Question 17

Part (a)

Part (b)

Using \(V(1) = 2.6\) and our equation for \(V(t)\) gives

\(
\begin{aligned}
V(1) &= 2.6 \\
6.5 (1 – e^{-k (1)}) &= 2.6 \\
1 – e^{-k} &= 0.4 \\
e^{-k} &= 0.6 \\
-k &= \ln(0.6) \\
k &= -\ln(0.6) \\
&= \text{0.511 (3 decimal places)}
\end{aligned}
\)

Part (c)

We want to find \(\frac{dV}{dt}\) when \(t = 2\), so we will first find \(\frac{dV}{dt}\) as

\(
\begin{aligned}
\frac{dV}{dt}
&= \frac{d}{dt} \left[ 6.5 – 6.5 e^{-k t} \right] \\
&= 6.5 k e^{-k t}
\end{aligned}
\)

 

Letting \(t = 2\) gives
\(
\begin{aligned}
\left. \frac{dV}{dt} \right\vert_{t = 2}
&= \left. 6.5 k e^{-k t} \right\vert_{t = 2} \\
&= -6.5 \ln(0.6) e^{\ln(0.6) (2)} \\
&= \text{1.195 volts/second}
\end{aligned}
\)

Question 18

Part (a)

\(
\begin{aligned}
P(\overline{\text{goal}}) &= 1-0.15 \\
&=0.85 \\
P(\overline{\text{goal}},\overline{\text{goal}}) &= 0.85 \times 0.85 \\
&=0.7225\\
\end{aligned}
\)

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.

Question 19

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:

\(
\begin{aligned}
y”&=12x^2-12x
\end{aligned}
\)

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 \(y=2\).

Question 20

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 \(\triangle TCB\):

\(
\begin{aligned}
\tan 30^\circ &= \frac{40}{BC} \\
BC &= \frac{40}{\tan 30^\circ} \\
&\approx 69.282 …
\end{aligned}
\)

Let \(\theta=\angle BCA\),
\(
\begin{aligned}
\cos \theta &= \frac{AC^2+BC^2-100^2}{2\times AC\times BC}\\
\theta &= \cos^{-1}\left(\frac{57.13^2+69.28^2-100^2}{2\times 57.13\times 69.28}\right) \\
&=104.161…
\end{aligned}
\)

\(
\begin{aligned}
\text{Bearing of B from C }&= 90 + 104.161… \\ &= 194 ^\circ\text{T (nearest degree)}
\end{aligned}
\)

Question 21

  1. As both genders of anacondas age, female anacondas display a significantly faster growth rate compared to male anacondas, as indicated by their steeper upward trend in length on the scatterplot.
  2. Both male and female anacondas continue to grow in length after approximately 4 years, which is evident from the data points for older anacondas.
  3. Female anacondas are consistently longer than male anacondas across all ages, as demonstrated by their correspondingly higher data points.

Question 22

Part (a)

The second derivative of \(f(x)\) is

\(
\begin{aligned}
f”(x)
&= \frac{d^2}{dx^2} \left[ \ln(1 + x^2) \right] \\
&= \frac{d}{dx} \left[ \frac{2 x}{1 + x^2} \right] \\
&= \frac{(2) (1 + x^2) – (2 x) (2 x)}{(1 + x^2)^2} \quad \text{(by quotient rule)} \\
&= \frac{2(1 – x^2)}{(1 + x^2)^2} \\
\end{aligned}
\)

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)

Graph from 2024 Maths Advanced Exam

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

\(
\begin{aligned}
A
&= \int_0^1 \ln(1 + x^2) \, dx \\
&\approx \frac{1 – 0}{2 (4)} [ 0.6931 + 0 + 2 (0.0606 + 0.2231 + 0.4463) ] \quad \text{(using the trapezoidal rule)} \\
&= \text{0.2691375 units^2}
\end{aligned}
\)

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.

Question 23

Part (a)

For \(x=58: z=0\).
For \(x=70:\)

\(
\begin{aligned}
z &= \frac{70-58}{15} \\
&= 0.8
\end{aligned}
\)

Thus we have:

\(
\begin{aligned}
P(58<x<70) &= P(0 < z<0.8) \\
&= P(z=0.8)-P(z=0) \\
&= 0.7881 – 0.5 \\
&= 0.2881
\end{aligned}
\)

Part (b)

For \(x=46:\)

\(
\begin{aligned}
z &= \frac{46-58}{15} \\
&= -0.8
\end{aligned}
\)

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 \(P=0.9032: z=1.3\).

\(
\begin{aligned}
1.3 &\approx \frac{x-58}{15} \\
19.5 &\approx x-58 \\
x &\approx 77.5
\end{aligned}
\)

Therefore, the approximate minimum score is \(77.5\).

Question 24

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,

\(
\begin{aligned}
A_0 &= 0 \\
A_1 &= 80 (1.005) \\
A_2 &= 80 (1.005)^2 + 80 (1.005) \\
A_3 &= 80 (1.005)^3 + 80 (1.005)^2 + 80 (1.005) \\
&\vdots \\
A_{24} &= 80 (1.005)^{24} + 80 (1.005)^{23} + \cdots + 80 (1.005)^2 + 80 (1.005) \\
&= 80 (1.005) \left[ (1.005)^{23} + (1.005)^{22} + \cdots + (1.005) + 1 \right] \\
&= 80 (1.005) \times \frac{(1.005)^{24} – 1}{(1.005) – 1} \quad \text{(from geometric series)} \\
&= \$2,044.73 \text{ (nearest cent)}
\end{aligned}
\)

 

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

\(
\frac{2044.73}{4} = \text{25.559 (3 decimal places)}
\)

 

Question 25

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:

Graph from 2024 Maths Advanced Exam

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 \( x = 2 – \sqrt{2} \approx 0.586 \)

Question 26

For the first \(15\) years:

\(
\begin{aligned}
n &= 15\times 12 \\
&= 180 \\
r &= 0.024\div 12 \\
&= 0.002
\end{aligned}
\) \(
\begin{aligned}
PV &= 151.036 \times 2000 \\
&= \$302072
\end{aligned}
\)

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:
\(
\begin{aligned}
n &= 25\times 12 \\
&= 300
\end{aligned}
\)

\(
\begin{aligned}
PV &= 225.43 \times 1200 \\
&= \$270516
\end{aligned}
\)

 

For 15 years:
\(
\begin{aligned}
n &= 15\times 12 \\
&= 180
\end{aligned}
\)

\(
\begin{aligned}
PV &= 151.036 \times 1200 \\
&= \$181243.20
\end{aligned}
\)

 

\(
\begin{aligned}
\text{Total }PV &= 302072 + 270516 – 181243.20\\
&= \$391344.80
\end{aligned}
\)

Therefore, the minimum sum to deposit is:

\(302072 + 89227.47 \approx \$391344.15\)

 

Question 27

Part (a)

\(\frac{d}{dx}(x^2 \tan x) = 2x \tan x + x^2 \sec^2 x\)

 

Part (b)

\(
\begin{aligned}
\int (x^2 \tan^2 x + 2x \tan x + 1) \, dx &= \int \left( x^2 (\sec^2 x – 1) + 2x \tan x + 1 \right) \, dx \\
&= \int \left( x^2 \sec^2 x + 2x \tan x – x^2 + 1 \right) \, dx \\
&= x^2 \tan x – \frac{x^3}{3} + x + C \,\, \text{ (Using the result from (a)) }
\end{aligned}
\)

Question 28

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:

\(
\begin{aligned}
21 – 18 \cos \left[ \frac{\pi}{24} (t – 6) \right] &= 21 – 18 \cos \left( \frac{\pi t}{24} \right) \\
\cos \left( \frac{\pi t}{24} – \frac{\pi}{4} \right) &= \cos \left( \frac{\pi t}{24} \right) \\
\cos \left( \frac{\pi t}{24} – \frac{\pi}{4} \right)&= \cos \left( -\frac{\pi t}{24} \right) \\
\frac{\pi t}{24} – \frac{\pi}{4} &= -\frac{\pi t}{24} \\
\frac{\pi t}{12} &= \frac{\pi}{4} \\
t &= 3
\end{aligned}
\)

They will be at the same height when \(t=3\)and \(t=3+\frac{48}{2}=27\).

\(
\begin{aligned}
A(3) &=21 + 18 \cos \left( \frac{\pi}{8} \right) \\
&\approx 37.63 \, \text{m} \\
A(27) &=21 + 18 \cos \left( \frac{9\pi}{8} \right) \\
&\approx 4.37 \, \text{m} \\
\end{aligned}
\)

 

Question 29

Consider \(y=ax^2+bx+c, a\neq 0\). The tangent will have the following equation:

\[ \frac{dy}{dx} = 2ax+b \]

When \(x = -4\), \(\frac{dy}{dx} = 0\):

\(
\begin{align}
2a(-4) + b &= 0 \\
-8a + b &= 0 \tag{1}
\end{align}
\)

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 \(c\):

\(
\begin{aligned}
y &= \frac{1}{2} x^2 + 4x + c \\
-1&= \frac{1}{2} (-2)^2 + 4(-2) + c \\
-1&= 2-8+c \\
c&=5
\end{aligned}
\)

 

Question 30

\(
\begin{aligned}
S_{\infty} &= \frac{x}{1-x} \\
&= \frac{-(1-x)-1}{1-x} \\
&= -1 – \frac{1}{x-1}
\end{aligned}
\)

 

Since \(|x| < 1:\)
\[
-1 < x < 1
\]

When \(x = -1\):

\(
\begin{aligned}
y &= -1 – \frac{1}{-1-1} \\
&= -1 – \frac{1}{-2} \\
&= -\frac{1}{2}
\end{aligned}
\)

 

As \(x \to 1\) from below 1:

\(
\begin{aligned}
\frac{1}{x-1} &\to -\infty \\
-\frac{1}{x-1} &\to \infty \\
-1-\frac{1}{x-1} &\to \infty \\
\end{aligned}
\)

 

\(
∴\text{ Range of } S = \left(-\frac{1}{2}, \infty\right).
\)

 

Question 31

Part (a)

\(
\begin{aligned}
A &= \frac{1}{2} (1 + x)^2 \theta – \frac{1}{2} (1) \theta \\
&= \frac{1}{2} \theta [1 + 2x + x^2 – 1] \\
&= \frac{1}{2} \theta (x^2 + 2x) \\
2A &= \theta (x^2 + 2x) \\
\theta &= \frac{2A}{x^2 + 2x} \\
\theta &= \frac{2A}{x(x + 2)}
\end{aligned}
\)

 

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 \(\theta\):

\(
\begin{aligned}
\theta &= \frac{2A}{x(x + 2)} \\
&= \frac{2x^2}{x(x + 2)} \\
&= \frac{2x}{x + 2}
\end{aligned}
\)

 

Since \( x > 0 \):

\[
\frac{x}{x + 2} < 1, \tag{as \( x < x + 2 \)}
\]

We have:

\(
\begin{aligned}
2 \cdot \frac{x}{x + 2} &< 2 \\
\frac{2x}{x + 2} &< 2
\end{aligned}
\)

 

\(
∴\, \theta < 2 \quad \text{as required.}
\)

 

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 \(x > 0\):

\(
∴\, P”(x) > 0
\)

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.

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