Question 1 (7 Marks)
Below are seven puzzles, each with a missing section. From the draggable items, select the correct equivalent option to complete each puzzle.
Question 1(a) : 1 marks
▶️Answer/Explanation
Ans:
1. \( \sqrt{x^3} \) is equivalent to \( x^{\frac{3}{2}} \).
2. \( \sqrt[3]{x^2} \) is equivalent to \( x^{\frac{2}{3}} \).
3. \( \frac{x^2}{3} \) remains not equivalent to any of the provided expressions.
Question 1(b) : 1 marks
▶️Answer/Explanation
Ans:
Expression: \(\frac{9^x \times 3^{2x}}{3^x}\)
Here, we have a combination of exponential terms and division involving the same base, which is 3. To simplify this expression, we can use the properties of exponents:
1. \(9^x\) can be written as \(3^{2x}\), since \(9 = 3^2\).
2. Now we have \(3^{2x} \times 3^{2x}\) in the numerator and \(3^x\) in the denominator. We can combine the bases by adding the exponents.
\(3^{2x} \times 3^{2x} = 3^{2x + 2x} = 3^{4x}\).
3. So, the expression becomes \(\frac{3^{4x}}{3^x}\).
4. When you have the same base and you’re dividing, you can subtract the exponents: \(3^{4x – x} = 3^{3x}\).
Now, based on the simplified expression \(3^{3x}\), let’s choose the correct option from the draggable items:
- \(9^{2x}\)
- \(9^x\)
- \(27^x\)
The correct option is: \(27^x\)
Question 1(c) : 1 marks
▶️Answer/Explanation
Ans:
Expression: \((e^{2x})^3\)
Using the property of exponentiation \((a^b)^c = a^{bc}\), we can simplify the expression:
\((e^{2x})^3 = e^{2x \times 3} = e^{6x}\)
Now, based on the simplified expression \(e^{6x}\), let’s choose the correct option from the draggable items:
- \(e^{5x}\)
- \(e^{6x}\)
- \(e^{8x^3}\)
The correct option is: \(e^{6x}\)
Question 1(d) : 1 marks
▶️Answer/Explanation
Ans:
Expression: \(\cos \theta \tan \theta\)
1. Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
2. Substituting this value into the expression, we get: \(\cos \theta \cdot \frac{\sin \theta}{\cos \theta}\).
3. The \(\cos \theta\) terms cancel out, leaving us with just \(\sin \theta\).
So, \(\cos \theta \tan \theta = \sin \theta\).
Now, let’s choose the correct option from the draggable items:
- \(1\)
- \(\frac{\cos ^2 \theta}{\sin \theta}\)
- \(\sin \theta\)
The correct option is: \(\sin \theta\)
Question 1(e) : 1 marks
▶️Answer/Explanation
Ans:
Expression: \(\log \left(\frac{x^2}{y}\right)\)
Using logarithmic properties, we know that \(\log \left(\frac{a}{b}\right) = \log a – \log b\).
1. Apply this property to the expression: \(\log \left(\frac{x^2}{y}\right) = \log x^2 – \log y\).
2. Further, according to the power rule of logarithms, \(\log a^b = b \log a\), we can simplify \(\log x^2\) to \(2 \log x\).
So, \(\log \left(\frac{x^2}{y}\right) = 2 \log x – \log y\).
Now, let’s choose the correct option from the draggable items:
- \(2 \log x – \log y\)
- \(\frac{2 \log x}{\log y}\)
- \(\log 2x – \log y\)
The correct option is: \(2 \log x – \log y\)
Question 1(f) : 1 marks
▶️Answer/Explanation
Ans:
Expression: \(1 – \sin^2 \theta + \cos^2 \theta\)
1. Recall the Pythagorean identity for sine and cosine: \(\sin^2 \theta + \cos^2 \theta = 1\).
2. Substitute this identity into the expression: \(1 – \sin^2 \theta + \cos^2 \theta = 1 – 1\).
3. \(1 – 1\) simplifies to \(0\).
So, \(1 – \sin^2 \theta + \cos^2 \theta = 0\).
Now, let’s choose the correct option from the draggable items:
- \(0\)
- \(2 \sin^2 \theta\)
- \(2 \cos^2 \theta\)
The correct option is: \(0\)
Question 1(g) : 1 marks
▶️Answer/Explanation
Ans:
The given expressions \(e^x, e^{2x}, e^{3x}, e^{4x}\) form a geometric sequence..
In a geometric sequence, each term is obtained by multiplying the previous term by a fixed ratio. In this case, if we divide any term by the previous term, we get the same value:
\(\frac{e^{2x}}{e^x} = e^{x}\)
\(\frac{e^{3x}}{e^{2x}} = e^{x}\)
\(\frac{e^{4x}}{e^{3x}} = e^{x}\)
The common ratio here is \(e^{x}\), which is the same for all pairs of consecutive terms. This property characterizes a geometric sequence.
Question 2 (8 Marks)
Question 2(a) : 3 marks
Let $f(x)=\ln x$ and $g(x)=x-6$.
Given that $h(x)=2 f(g(x))$, write down $h(x)$.
▶️Answer/Explanation
Ans:
\(f(x) = \ln x\) and \(g(x) = x – 6\).
2. Now, we need to find \(h(x) = 2 f(g(x))\):
$\bullet$ Substitute the expression for \(g(x)\) into \(f(x)\):
\(f(g(x)) = \ln (x – 6)\).
$\bullet$ Multiply the result by 2:
\(2 \cdot \ln (x – 6)\).
So, the expression for \(h(x)\) is \(2 \ln (x – 6)\).
Question 2(b) : 1 marks
The $h(x)$ from part (a) is the transformation of $f(x)$ by:
$\bullet$ a vertical dilation with a scale factor $s$
$\bullet$ a horizontal translation of $t$ units to the right,
Write down the values of $s$ and $t$.
▶️Answer/Explanation
Ans:
In the given expression \(h(x) = 2 \ln (x – 6)\), we can identify the transformation of \(f(x) = \ln x\) as described. Let’s break down the transformation in terms of \(s\) and \(t\):
1. Vertical Dilation with Scale Factor \(s\):
The factor of 2 in front of \(\ln (x – 6)\) indicates a vertical dilation by a scale factor of \(s = 2\). This means that the function is stretched vertically by a factor of 2.
2. Horizontal Translation of \(t\) Units to the Right:
The term \(\ln (x – 6)\) involves \(x – 6\), which represents a horizontal translation of 6 units to the right. The value of \(t\) is \(6\).
So, the transformation is:
$\bullet$ Vertical dilation with \(s = 2\)
$\bullet$ Horizontal translation to the right by \(t = 6\) units.
Question 2(c) : 2 marks
Solve $h(x)=f(x)$.
▶️Answer/Explanation
Ans:
We are given that \(h(x) = 2 \ln (x – 6)\) and \(f(x) = \ln x\), and we need to find the values of \(x\) for which \(h(x) = f(x)\).
Setting up the equation and solving for \(x\):
\[2 \ln (x – 6) = \ln x\]
First, let’s simplify the equation by using properties of logarithms:
\[2 \ln (x – 6) = \ln x \implies \ln ((x – 6)^2) = \ln x\]
Now, we can remove the natural logarithm by taking the exponential of both sides:
\[(x – 6)^2 = x\]
Expand the left side:
\[x^2 – 12x + 36 = x\]
Rearrange the equation:
\[x^2 – 13x + 36 = 0\]
Now, we have a quadratic equation. To solve for \(x\), we can factor the quadratic or use the quadratic formula:
\[(x – 9)(x – 4) = 0\]
This gives us two possible solutions:
1. \(x – 9 = 0 \implies x = 9\)
2. \(x – 4 = 0 \implies x = 4\)
So, the solutions for \(x\) that satisfy \(h(x) = f(x)\) are \(x = 9\) and \(x = 4\). But ,the domain of the natural logarithm function is \(x-6 > 0\), ( argument of log must be positive ) so \(x\) cannot be 4.
Question 3 (8 Marks)
One of four main blood types can be found in a human body. They are known as $\mathbf{A}, \mathbf{B}, \mathbf{A B}$ and $\mathbf{O}$. Each blood type can be further classified as either a Rhesus positive (+) or Rhesus negative $(-)$.
For example, a possible combination is blood type $\mathbf{O}$ and Rhesus negative which is written as $\mathbf{O}$ –
The pie charts below shows the distribution of the blood types and Rhesus types for a blood donor centre recorded in 2019.
Question 3(a) : 2 marks
People with blood type $\mathbf{O}-$ are known as universal donors. They can donate their blood to patients with any blood type.
Show that the probability that a randomly selected person has blood type $\mathbf{O}-$ is 0.1 .
▶️Answer/Explanation
Ans:
To find the expected number of people with blood type O-, we can multiply the probability of having blood type O- by the total number of people (30). The expected value (\(E\)) is given by:
\[E = \text{Probability} \times \text{Total Number}\]
Given that the probability of having blood type O- is \(0.1\) and the total number of people is \(30\), we can calculate the expected number of people with blood type O-:
\[E = 0.1 \times 30 = 3\]
So, the expected number of people with blood type $\mathbf{O}-$ is 3.
Question 3(b) : 1 marks
Thirty people donate blood one morning. Determine the expected number of people that have blood type $\mathbf{O}-$.
▶️Answer/Explanation
Ans:
To determine the expected number of people that have blood type $\mathbf{O}-$ among the 30 people donating blood, we need to consider the probability of each person having that blood type.
From the previous answer, we found that the probability of a randomly selected person having blood type $\mathbf{O}-$ is 0.1 or $10\%$.
The expected number of people with blood type $\mathbf{O}-$ can be calculated by multiplying the probability of each person having that blood type by the total number of people donating blood.
Expected number of people with blood type $\mathbf{O}- = \text{Probability of blood type }\mathbf{O}- \times\text{Total number of people donating blood}$
Expected number of people with blood type $\mathbf{O}- = 0.1 \times 30 = 3$
Therefore, the expected number of people that have blood type $\mathbf{O}-$ among the 30 people donating blood is 3.
Question 3(c) : 3 marks
Two of the thirty donors are selected at random. Write down the missing values in the tree diagram below.
▶️Answer/Explanation
Ans:
Question 3(d) : 3 marks
Given that the second donor has blood type $\mathbf{O}-$, find the probability that the first donor does not have blood type $\mathbf{O}-$.
▶️Answer/Explanation
Ans:
$\bullet 1 \frac{\text { their } 3}{30} \times \frac{\text { their } 2}{\text { their } 29}$ or $\frac{1}{145} \mathrm{OE}$$\bullet 2 \frac{\text { their } 27}{30} \times \frac{\text { their } 3}{\text { their } 29}$ or $\frac{27}{290} \mathrm{OE}$
$\bullet 3 \frac{\text { their } \bullet 2}{\text { their } \bullet 1+\text { their } \bullet 2}$ OE
$\bullet 4 \frac{\text { their } 27}{\text { their } 29}$ OE
Question 4 (12 Marks)
An above-ground pool can be represented as a cylinder filled with water. If we look at the pool from above, we can see a circle.
The height of water in the pool is 1.17 metres (m). A pump is used to remove all water from the pool in 5 hours and 15 minutes. The pump removes water at a rate of 11200 litres (L) per hour.
$$
1000 \mathrm{~L}=1 \mathrm{~m}^3
$$
Question 4(a) : 1 marks
Show that the radius of the pool is $4 \mathrm{~m}$ to the nearest $\mathrm{m}$.
▶️Answer/Explanation
Ans:
To determine the radius of the pool, we can use the formula for the volume of a cylinder:
\[V = \pi r^2 h\]
Where:
- \(V\) is the volume of the cylinder (in cubic meters).
- \(\pi\) is the mathematical constant approximately equal to 3.14159.
- \(r\) is the radius of the cylinder’s base (in meters).
- \(h\) is the height of the cylinder (in meters).
We are given that the height of the water in the pool is \(h = 1.17\) meters and that the pump removes water at a rate of \(11200\) liters per hour.
First, let’s convert the given volume rate to cubic meters:
\[11200 \, \text{L/h} \times \frac{1 \, \text{m}^3}{1000 \, \text{L}} = 11.2 \, \text{m}^3/\text{h}\]
Next, we can use the fact that the entire volume of the pool (\(\pi r^2 h\)) is pumped out in \(5\) hours and \(15\) minutes, which is \(5.25\) hours.
So, we have:
\[11.2 \, \text{m}^3/\text{h} \times 5.25 \, \text{h} = \pi r^2 h\]
Now, we can solve for \(r\) by rearranging the formula:
\[r^2 = \frac{11.2 \, \text{m}^3/\text{h} \times 5.25 \, \text{h}}{\pi h}\]
\[r = \sqrt{\frac{11.2 \, \text{m}^3/\text{h} \times 5.25 \, \text{h}}{\pi \cdot 1.17 \, \text{m}}}\]
Calculating this gives us \(r \approx 4.00\) meters.
Therefore, the radius of the pool is approximately \(4\) meters to the nearest meter.
Question 4(b) : 3 marks
A cylindrical pool cover is placed over the pool. The pool cover has a radius of $4 \mathrm{~m}$ and a depth of $0.3 \mathrm{~m}$.
Find the outer surface area of the pool cover, to the nearest square metre.
▶️Answer/Explanation
Ans:
To find the outer surface area of the pool cover, we need to calculate the lateral surface area and the top surface area separately and then add them together.
The lateral surface area of a cylinder is given by the formula:
\[A_{\text{lateral}} = 2\pi rh\]
where:
$\bullet$ \(A_{\text{lateral}}\) is the lateral surface area,
$\bullet$ \(\pi\) is a mathematical constant approximately equal to 3.14159,
$\bullet$ \(r\) is the radius of the cylinder, and
$\bullet$ \(h\) is the height of the cylinder.
In this case, the radius of the pool cover is \(4\) m and the depth is \(0.3\) m.
Substituting these values into the formula, we get:
\[A_{\text{lateral}} = 2 \times 3.14159 \times 4 \mathrm{~m} \times 0.3 \mathrm{~m}\]
Simplifying the equation, we have:
\[A_{\text{lateral}} = 7.53982 \mathrm{~m}^2\]
The top surface area of the cylinder is given by the formula for the area of a circle:
\[A_{\text{top}} = \pi r^2\]
Substituting the radius of the pool cover, we get:
\[A_{\text{top}} = 3.14159 \times (4 \mathrm{~m})^2\]
Simplifying the equation, we have:
\[A_{\text{top}} = 50.26544 \mathrm{~m}^2\]
Now, we can add the lateral surface area and the top surface area to get the total outer surface area:
\[A_{\text{total}} = A_{\text{lateral}} + A_{\text{top}}\]
\[A_{\text{total}} = 7.53982 \mathrm{~m}^2 + 50.26544 \mathrm{~m}^2\]
\[A_{\text{total}} = 57.80526 \mathrm{~m}^2\]
Rounding this value to the nearest square meter, we get:
\[A_{\text{total}} \approx 58 \mathrm{~m}^2\]
Therefore, the outer surface area of the pool cover is approximately 58 square meters.
Question 4(c) : 1 marks
Material for the pool cover costs $\$ 3.40$ per square metre. Determine the cost of the pool cover.
▶️Answer/Explanation
Ans:
To determine the cost of the pool cover, we need to multiply the outer surface area of the pool cover by the cost per square meter.
Given that the outer surface area of the pool cover is approximately 58 square meters, and the cost per square meter is $\$ 3.40$, we can calculate the cost as follows:
Cost of the pool cover $=$ Outer surface area $\times$ Cost per square meter $=58$ square meters $\times \$ 3.40 /$ square meter $=\$ 197.20$
Therefore, the cost of the pool cover is approximately $\$ 197.20$.
Question 5 (15 marks)
The following video explains the use of sleeves for paper cups.Cardboard sleeves are used to reduce the discomfort of holding a hot paper cup.the sleeve forms part of a sector of a circle as shown.In this question, you will make calculations to find the area of the sleeve
The dimensions of the sleeve are provided on the diagram, in centimetres $(\mathrm{cm})$.
 |  |  |
Question 5(a) : 3 marks
Show that the value of angle $y$ is $70^{\circ}$, to the nearest degree.
▶️Answer/Explanation
Ans:
The expression \(12.55 – 10.6\) can be simplified as follows:
\[12.55 – 10.6 = 1.95\]
So, \(12.55 – 10.6 = 1.95\).
The equation \(\cos y = \frac{1.95}{5.7}\) implies that the cosine of angle \(y\) is equal to \(\frac{1.95}{5.7}\).
To find the value of \(y\), we can take the inverse cosine (arccos) of \(\frac{1.95}{5.7}\):
\[y = \cos^{-1} \left(\frac{1.95}{5.7}\right)\]
Therefore, \(y = \cos^{-1} \left(\frac{1.95}{5.7}\right)\Rightarrow 70^{\circ}\).
Question 5(b) : 2 marks
Hence, show that the value of angle $x$ is $\frac{2}{9} \pi$ radians.
▶️Answer/Explanation
Ans:
\( \begin{gathered}90^{\circ}-70^{\circ}=\frac{x }{ 2} \\ x=40^{\circ}\end{gathered} \)
Question 5(c) : 3 marks
Calculate the value of $v$.
▶️Answer/Explanation
Ans:
Starting with the given equation:
\(\sin 20 = \frac{10.6}{v}\)
We can rearrange the equation as:
\(v = \frac{10.6}{\sin 20}\)
Using a calculator, we can find the sine of 20 degrees:
\(\sin 20 \approx 0.3420\)
Substituting this value back into the equation, we have:
\(v = \frac{10.6}{0.3420}\)
Calculating this expression, we find:
\(v \approx 31.012 \)
Therefore, solving the equation \(\sin 20 = \frac{10.6}{v}\) gives us \(v \approx 31.012\).
Question 5(d) : 4 marks
Find the area of the sleeve.
▶️Answer/Explanation
Ans:
The area of a sector is given by the formula:
\[A = \frac{\theta}{360^\circ} \times \pi r^2\]
where \(\theta\) is the central angle in degrees and \(r\) is the radius of the sector.
Given that \(r_1 = 31 \, \text{cm}\), \(r_2 = 36.7 \, \text{cm}\), and \(\theta = 40^\circ\), we can calculate the areas as follows:
Area with radius \(r_1\):
\[A_1 = \frac{40^\circ}{360^\circ} \times \pi \times (31 \, \text{cm})^2\]
Area with radius \(r_2\):
\[A_2 = \frac{40^\circ}{360^\circ} \times \pi \times (36.7 \, \text{cm})^2\]
Calculating these values, we find:
\[A_1 \approx 335.45 \, \text{cm}^2\]
\[A_2 \approx 469.91 \, \text{cm}^2\]
To find the area of the sleeve (difference in areas), we subtract \(A_1\) from \(A_2\):
\[\Delta A = A_2 – A_1\]
\[\Delta A \approx 134.461 \, \text{cm}^2\]
Therefore, the correct area of the sleeve is approximately \(134.461 \, \text{cm}^2\).
Question 5(e) : 3 marks
The sleeve is cut out from a $25.1 \mathrm{~cm}$ by $7.6 \mathrm{~cm}$ rectangle. Calculate the percentage of material wasted.
▶️Answer/Explanation
Ans:
The area of the sleeve is given as approximately \(134.461 \, \text{cm}^2\).
The area of the original rectangle is calculated by multiplying its length and width:
\(\text{Area of rectangle} = 25.1 \, \text{cm} \times 7.6 \, \text{cm}\)
\(\text{Area of rectangle} = 190.76 \, \text{cm}^2\)
To find the percentage of material wasted, we subtract the area of the sleeve from the area of the rectangle, divide it by the area of the rectangle, and then multiply by 100:
\(\text{Percentage of material wasted} = \left(\frac{\text{Area of rectangle} – \text{Area of sleeve}}{\text{Area of rectangle}}\right) \times 100\)
\(\text{Percentage of material wasted} = \left(\frac{190.76 \, \text{cm}^2 – 134.461 \, \text{cm}^2}{190.76 \, \text{cm}^2}\right) \times 100\)
\(\text{Percentage of material wasted} \approx 29.50\%\)
Therefore, the correct percentage of material wasted is approximately 29.50%.
Question 6 (19 marks)
Many efforts have been made through the years to put into action solutions to resolve the problem of exposure to particulate matter in order to reach a globally agreed goal. In this question, you will make calculations and analyse exposure to particulate matter over a period of time.
Question 6(a) : 2 marks
Annual exposure to particulate matter across Europe in 1990 .
To visualize the level of exposure across Europe in 1990, use the table to construct a histogram.
Particulate matter for 1990
▶️Answer/Explanation
Ans:
To construct a histogram representing the level of exposure to particulate matter across Europe in 1990, we can use the given table. The x-axis will represent the different ranges of exposure, and the y-axis will represent the number of European countries falling into each range.
Here is the histogram:
In this histogram, the x’s represent the number of European countries falling into each range of exposure to particulate matter. The ranges are displayed on the x-axis, and the corresponding numbers of countries are shown on the y-axis.
Question 6(b) : 1 marks
Below is the histogram for the annual exposure to particulate matter across Europe in 2016.
Particulate matter for 2016
In comparison with your histogram in part (a), state one difference between annual exposure to particulate matter in 1990 and 2016.
▶️Answer/Explanation
Ans:
- in 1990, 3 countries had E between 0 and 10 while in 2016 there are 8
- Highest in 1990 is 17 while highest in 2016 is 24
Question 6(c) : 2 marks
Using the histogram for 2016 , show that an estimate for the mean of $E$ in 2016 is $15.3 \mu \mathrm{g} / \mathrm{m}^3$ to the nearest one decimal place.
▶️Answer/Explanation
Ans:
To estimate the mean of $E$ in 2016 using the given histogram, we can calculate the weighted average of the exposure ranges.
Let’s break down the calculation step by step:
1. Multiply each exposure range by the corresponding number of European countries:
$\bullet$ $5 \times 8 = 40$
$\bullet$ $15 \times 24 = 360$
$\bullet$ $25 \times 5 = 125$
$\bullet$ $35 \times 2 = 70$
2. Sum up the weighted exposures:
$40 + 360 + 125 + 70 = 595$
3. Divide the sum by the total number of European countries:
$\frac{595}{39} \approx 15.26$
Rounding this value to one decimal place, we get $15.3 \mu \mathrm{g} / \mathrm{m}^3$ as an estimate for the mean of $E$ in 2016.
Question 6(d) : 2 marks
The data from 1990 to 2016 can be modelled by a line of best fit as shown in Graph 1 .
The equation of the line of best fit can be written as $\mathrm{E}=a n+b$, where $\mathrm{E}$ is the annual exposure to particulate matter and $n$ is the number of years after 1990.
Determine the values of $a$ and $b$.
▶️Answer/Explanation
Ans:
$\begin{aligned}& (a=)-\frac{1}{5} \\
& (b=) 18
\end{aligned}$
Question 6(e) : 10 marks
You are a statistical adviser on a government panel that wants to reduce annual exposure to particulate matter in Europe to a level below $13 \mu \mathrm{g} / \mathrm{m}^3$ by 2030 .
Analyse the two models to make a prediction of annual exposure to particulate matter in 2030. In your answer, you should:
$\bullet$ identify a factor from the data to be considered when making your prediction
$\bullet$ predict the level of annual exposure to particulate matter in 2030 using models from both graphs
$\bullet$ comment on the accuracy of your predictions
$\bullet$ advise the government which is a better model to make predictions and justify your answer.
▶️Answer/Explanation
Ans:
To make a prediction of annual exposure to particulate matter in 2030, we will analyze the two models provided: the histogram data from 1990 to 2016 and the exponential model $E=6(0.91)^n+13.8$.
Factor to Consider:
One important factor to consider when making the prediction is the trend observed in the data. It is crucial to understand whether the trend is stable, increasing, or decreasing over time. This will help us assess the accuracy of the predictions and determine which model is more suitable.
Prediction using the Histogram Data:
Since we don’t have the specific data points for each year, we can’t directly use the histogram to predict the exposure in 2030. However, we can observe the general trend from the available data. If the exposure levels have been consistently decreasing over time, we can assume that the exposure in 2030 will be lower than in previous years. However, without the specific data points, it is difficult to make an accurate prediction using only the histogram.
Prediction using the Exponential Model:
Using the exponential model $E=6(0.91)^n+13.8$, we can make a prediction for 2030 by substituting $n=2030-1990=40$ into the equation:
$$E = 6(0.91)^{40} + 13.8$$
Calculating this, we find:
$$E \approx 13.14 \mu \mathrm{g} / \mathrm{m}^3$$
Comment on Accuracy:
The accuracy of the predictions depends on the underlying assumptions and the quality of the data used to create the models. The histogram data provides a general overview of exposure levels over time, but it lacks the specific data points necessary for precise predictions. On the other hand, the exponential model assumes a specific decay rate and provides a more precise estimate. However, its accuracy depends on the validity of the underlying assumptions and the fit of the model to the actual data.
Advice to the Government:
Based on the analysis, the exponential model is a better choice for making predictions for 2030. It provides a specific estimate of annual exposure to particulate matter, taking into account the decay rate over time. However, it is crucial to note that the accuracy of the exponential model depends on the validity of the assumptions and the fit to the actual data. Therefore, it is recommended that the government continues monitoring the actual data and adjusts the predictions accordingly as more information becomes available.
Question 7 (19 marks)
In this task you will investigate the area and perimeter of squares.
In the two squares below the vertex of the smaller square is the midpoint of the side of the larger square. The side length of the smaller square is 3 units.
Question 7(a) : 1 marks
Show that the side length of the larger square is $3 \sqrt{2}$ units.
▶️Answer/Explanation
Ans:
$\begin{aligned} & a^2+a^2=3^2 \\ & a^2=\frac{9}{2} \\ &\text{side length = 2a} \Rightarrow2 \times \frac{3}{\sqrt{2}}\end{aligned}$
Question 7(b) : 2 marks
Write down the missing values in the table up to row 6 .
▶️Answer/Explanation
Ans:
The pattern in the given table seems to involve doubling the area of the square at each stage. Let’s continue filling in the missing values:
The missing values have been filled in by doubling the area from the previous stage.
Question 7(c) : 2 marks
Describe in words two patterns in the table for area of square (A).
▶️Answer/Explanation
Ans:
There are two distinct patterns in the table for the area of the square (\(A\)):
1. Exponential Growth: The area of the square doubles with each successive stage. This exponential growth is evident in the fact that the area \(A\) is being multiplied by a factor of 2 for each increase in the stage number \(n\).
2. Multiplying by 9: The initial area of the square (\(A = 9\)) is being multiplied by 9 at each stage. This multiplication by 9 results in the new area for the next stage.
These patterns highlight the exponential nature of the growth and the repeated multiplication by 9, leading to the sequence of areas in the table.
Question 7(d) : 2 marks
Write down a general rule for $\mathrm{A}$ in terms of $n$.
▶️Answer/Explanation
Ans:
The correct general rule for \(A\) in terms of \(n\) is:
\[A = 9 \times 2^{n-1}\]
This formula accurately represents the relationship between the stage number \(n\) and the area \(A\) of the square.
Question 7(e) : 3 marks
Verify your general rule for $\mathrm{A}$.
▶️Answer/Explanation
Ans:
let’s verify the general rule \(A = 9 \times 2^{n-1}\) using the values from the table:
As we can see, the values calculated using the general rule \(A = 9 \times 2^{n-1}\) perfectly match the actual values of \(A\) in the table. This confirms that the general rule accurately represents the relationship between the stage number \(n\) and the area \(A\) of the square.
Question 7(f) : 20 marks
Investigate the values in the table to find a relationship for the perimeter of each square $(P)$ in terms of $n$. In your answer, you should:
$\bullet$ predict more values and record these in the table
$\bullet$ describe in words a pattern in the table for perimeter of square $(P)$
$\bullet$ write down a general rule for $P$ in terms of $n$
$\bullet$ test your general rule for $P$
$\bullet$ verify and justify your general rule for $P$ in relation to the squares
$\bullet$ ensure that you communicate all your working appropriately.
▶️Answer/Explanation
Ans:
let’s investigate the values in the table and find a relationship for the perimeter of each square (\(P\)) in terms of \(n\).
We have the following table:
\begin{tabular}{|c|c|}
\hline
\(n\) & \(P\) \\
\hline
1 & 12 \\
2 & \(12 \sqrt{2}\) \\
3 & 24 \\
4 & \(24 \sqrt{2}\) \\
5 & 48 \\
6 & \(48 \sqrt{2}\) \\
\hline
\end{tabular}
Predicting More Values and Recording in the Table:
\begin{tabular}{|c|c|}
\hline
\(n\) & \(P\) \\
\hline
1 & 12 \\
2 & \(12 \sqrt{2}\) \\
3 & 24 \\
4 & \(24 \sqrt{2}\) \\
5 & 48 \\
6 & \(48 \sqrt{2}\) \\
7 & 96 \\
8 & \(96 \sqrt{2}\) \\
\hline
\end{tabular}
Pattern in the Table for Perimeter of Square (\(P\)):
It appears that the perimeter \(P\) alternates between two types of values:
1. Values that are multiples of 12 (12, 24, 48, 96, …)
2. Values that are multiples of \(12 \sqrt{2}\) (\(12 \sqrt{2}\), \(24 \sqrt{2}\), \(48 \sqrt{2}\), \(96 \sqrt{2}\), …)
General Rule for \(P\) in Terms of \(n\):
From the pattern, we can observe that the perimeter \(P\) is doubling each time (\(P = 12 \times 2^{n-1}\)), and every other term is multiplied by \(\sqrt{2}\) (\(P = 12 \sqrt{2} \times 2^{n/2 – 1}\)).
So, the general rule for \(P\) in terms of \(n\) is:
\[P = \begin{cases} 12 \times 2^{n-1}, & \text{if } n \text{ is odd} \\ 12 \sqrt{2} \times 2^{n/2 – 1}, & \text{if } n \text{ is even} \end{cases}\]
Testing the General Rule for \(P\):
Let’s test the general rule with \(n = 5\) (odd) and \(n = 6\) (even):
For \(n = 5\):
\[P = 12 \times 2^{5-1} = 12 \times 16 = 192\]
For \(n = 6\):
\[P = 12 \sqrt{2} \times 2^{6/2 – 1} = 12 \sqrt{2} \times 2^2 = 48 \sqrt{2}\]
The values match the table.
Verifying and Justifying the General Rule for \(P\):
The general rule accurately captures the two alternating patterns observed in the table. When \(n\) is odd, the perimeter is a multiple of 12, and when \(n\) is even, the perimeter is a multiple of \(12 \sqrt{2}\). The multiplication by powers of 2 follows the doubling behavior in the table.
This rule is justified based on the patterns established in the table and the properties of squares and square roots in relation to their perimeters.