2024 Cambridge IGCSE International Mathematics (0607) 模拟试题及答案详解

First, expand the product of the two binomials:

\((3x - 4)(2x + 5) = 6x^2 + 15x - 8x - 20 = 6x^2 + 7x - 20\)

Next, subtract the remaining expression:

\((6x^2 + 7x - 20) - (x^2 - 3x) = 6x^2 + 7x - 20 - x^2 + 3x\)

Combine like terms:

\(= 5x^2 + 10x - 20\)

M1 for expansion of the brackets: \(6x^2 + 15x - 8x - 20\) (allow one error) or for correctly subtracting \(x^2 - 3x\) from a quadratic expression.
A1 for the final answer: \(5x^2 + 10x - 20\) (or equivalent).

Factorise the numerator using the difference of two squares:

\(4x^2 - 9 = (2x - 3)(2x + 3)\)

Factorise the quadratic denominator:

\(2x^2 - x - 3 = (2x - 3)(x + 1)\)

Simplify by cancelling the common factor \((2x - 3)\):

\(\frac{(2x - 3)(2x + 3)}{(2x - 3)(x + 1)} = \frac{2x + 3}{x + 1}\)

M1 for factorising the numerator to \((2x - 3)(2x + 3)\) OR the denominator to \((2x - 3)(x + 1)\).
A1 for the final answer: \(\frac{2x + 3}{x + 1}\).

Using the properties of logarithms:

1. Move the coefficient to the exponent:
\(2\log_{10} 5 = \log_{10} 5^2 = \log_{10} 25\)

2. Combine using multiplication and division laws:
\(\log_{10} 25 + \log_{10} 8 - \log_{10} 2 = \log_{10} \left(\frac{25 \times 8}{2}\right)\)
\(= \log_{10} \left(\frac{200}{2}\right) = \log_{10} 100\)

3. Since \(100 = 10^2\):
\(\log_{10} 100 = 2\)

M1 for a correct application of any logarithm law, e.g., obtaining \(\log_{10} 25\), or combining terms to get \(\log_{10} 100\).
A1 for the final answer: \(2\).

Multiply the numerical parts:
\(4 \times 3 = 12\)

Multiply the powers of 10:
\(10^7 \times 10^{-4} = 10^{7 + (-4)} = 10^3\)

This gives:
\(12 \times 10^3\)

Convert this into standard form:
\(1.2 \times 10^4\)

M1 for intermediate answers such as \(12 \times 10^3\), \(12000\), or \(1.2 \times 10^k\) (where \(k \neq 4\)).
A1 for the correct standard form answer: \(1.2 \times 10^4\).

Since \(y\) is inversely proportional to \(x^2\), we can write:

\(y = \frac{k}{x^2}\)

Using the values \(x = 3\) and \(y = 4\), find the constant \(k\):

\(4 = \frac{k}{3^2} \implies 4 = \frac{k}{9} \implies k = 36\)

So the equation connecting \(y\) and \(x\) is:

\(y = \frac{36}{x^2}\)

Substitute \(x = 2\) to find \(y\):

\(y = \frac{36}{2^2} = \frac{36}{4} = 9\)

M1 for finding the constant of proportionality \(k = 36\), or for writing a correct relational equation such as \(y_1 \cdot x_1^2 = y_2 \cdot x_2^2\).
A1 for the correct answer: \(9\).

Subtract 5 from both sides:

\(-2x \leqslant 6\)

Divide by -2 and flip the direction of the inequality sign because you are dividing by a negative number:

\(x \geqslant -3\)

M1 for obtaining \(-2x \leqslant 6\), or for getting the correct numerical value \(-3\) but with an incorrect or missing inequality sign (e.g. \(x = -3\) or \(x \leqslant -3\)).
A1 for the correct inequality: \(x \geqslant -3\) (or \(-3 \leqslant x\)).

Apply the exponent to both parts of the term inside the bracket:

\((8x^6)^{\frac{2}{3}} = 8^{\frac{2}{3}} \times (x^6)^{\frac{2}{3}}\)

Evaluate each part:

1. \(8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4\)

2. \((x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\)

Combine the results:

\(4x^4\)

M1 for finding \(8^{\frac{2}{3}} = 4\) or for simplifying the variable part to \(x^4\), or for expressing the term as \(\sqrt[3]{64x^{12}}\).
A1 for the final simplified answer: \(4x^4\).

Multiply the entire equation by the common denominator \((x - 2)(x + 1)\) to eliminate the fractions:

\(3(x + 1) + 2(x - 2) = 0\)

Expand the brackets:

\(3x + 3 + 2x - 4 = 0\)

Combine like terms:

\(5x - 1 = 0\)

Solve for \(x\):

\(5x = 1 \implies x = \frac{1}{5} = 0.2\)

M1 for eliminating denominators correctly to obtain \(3(x + 1) + 2(x - 2) = 0\) (or equivalent).
A1 for the correct answer: \(0.2\) or \(\frac{1}{5}\).

A rhombus is a quadrilateral with all four sides equal. It has exactly two lines of symmetry, which are its diagonals. (Note: A rectangle also has exactly two lines of symmetry, but they are the perpendicular bisectors of its opposite sides, not its diagonals).

B1 for Rhombus

A regular polygon with $n$ sides has rotational symmetry of order $n$ and exactly $n$ lines of symmetry. Since the order of rotational symmetry is 10, the polygon is a regular decagon, which has 10 lines of symmetry.

B1 for 10

1. Multiply the entire equation by the common denominator \(x(x+2)\) to clear the fractions:
\(2(x+2) + 3x = x(x+2)\)

2. Expand both sides of the equation:
\(2x + 4 + 3x = x^2 + 2x\)
\(5x + 4 = x^2 + 2x\)

3. Rearrange the terms to form a standard quadratic equation:
\(x^2 - 3x - 4 = 0\)

4. Factorise the quadratic equation:
\((x-4)(x+1) = 0\)

5. Solve for \(x\):
\(x = 4\) or \(x = -1\)

M1 for multiplying through by \(x(x+2)\) or expressing the left-hand side with a correct common denominator.
M1 for expanding and simplifying to a correct three-term quadratic equation, e.g., \(x^2 - 3x - 4 = 0\).
M1 for factorising their quadratic equation of the form \(ax^2+bx+c=0\) into two linear brackets.
A1 for both \(x = 4\) and \(x = -1\).

1. Write down the proportionality relation: \(P = \frac{k}{(q+1)^2}\) for some constant \(k\).

2. Substitute the given values \(q = 2\) and \(P = 5\) to find \(k\):
\(5 = \frac{k}{(2+1)^2}\)
\(5 = \frac{k}{9} \implies k = 45\)

3. Write the formula connecting \(P\) and \(q\):
\(P = \frac{45}{(q+1)^2}\)

4. Substitute \(q = 4\) into the formula:
\(P = \frac{45}{(4+1)^2}\)
\(P = \frac{45}{25} = \frac{9}{5} = 1.8\)

M1 for writing down a correct starting equation linking the variables, e.g., \(P = \frac{k}{(q+1)^2}\).
M1 for substituting \(q = 2, P = 5\) and solving to find \(k = 45\).
M1 for substituting \(q = 4\) into their equation with their non-zero value of \(k\).
A1 for \(1.8\) or \(\frac{9}{5}\).

Method 1:
1. The area of a sector can be related to the arc length \(L\) and the radius \(r\) by the formula:
\(\text{Area} = \frac{1}{2} r L\)

2. Substitute \(\text{Area} = 15\pi\) and \(r = 6\):
\(15\pi = \frac{1}{2} \times 6 \times L\)
\(15\pi = 3L \implies L = 5\pi\text{ cm}\)

3. The perimeter of the sector includes the arc length plus two radii:
\(\text{Perimeter} = L + 2r\)
\(\text{Perimeter} = 5\pi + 2(6) = 5\pi + 12\text{ cm}\)

Method 2:
1. Find the sector angle \(\theta\):
\(15\pi = \frac{\theta}{360} \times \pi \times 6^2\)
\(15\pi = \frac{36\theta}{360} \pi \implies 15 = \frac{\theta}{10} \implies \theta = 150^\circ\)

2. Find the arc length:
\(\text{Arc length} = \frac{150}{360} \times 2 \pi \times 6 = 5\pi\text{ cm}\)

3. Find the perimeter:
\(\text{Perimeter} = 5\pi + 2(6) = 5\pi + 12\text{ cm}\)

M1 for setting up a correct equation for the area of the sector, e.g., \(15\pi = \frac{\theta}{360} \times \pi \times 6^2\) or \(15\pi = \frac{1}{2} \times 6 \times L\).
A1 for finding the arc length \(5\pi\) or the angle \(\theta = 150^\circ\).
M1 for adding twice the radius to their arc length, e.g., \(\text{arc length} + 12\).
A1 for \(5\pi + 12\).

1. Use the power rule of logarithms to rewrite the first term:
\(2\log_3(x) = \log_3(x^2)\)

2. Use the quotient rule to combine the logarithms on the left-hand side:
\(\log_3(x^2) - \log_3(x - 2) = \log_3\left(\frac{x^2}{x - 2}\right)\)

3. Convert the logarithmic equation into exponential form:
\(\frac{x^2}{x - 2} = 3^2\)
\(\frac{x^2}{x - 2} = 9\)

4. Solve the resulting quadratic equation:
\(x^2 = 9(x - 2)\)
\(x^2 = 9x - 18\)
\(x^2 - 9x + 18 = 0\)

5. Factorise the quadratic equation:
\((x - 6)(x - 3) = 0\)
So, \(x = 6\) or \(x = 3\).
Both values are valid since they are greater than 2 (which keeps the arguments of the original logarithms positive).

M1 for correctly applying the power rule to write \(2\log_3(x)\) as \(\log_3(x^2)\).
M1 for correctly applying the quotient rule to express the left-hand side as a single logarithm: \(\log_3\left(\frac{x^2}{x - 2}\right)\).
M1 for converting to index form and obtaining the quadratic equation \(x^2 - 9x + 18 = 0\).
A1 for both solutions \(x = 6\) and \(x = 3\).

Apply the power law of logarithms to the first term:
\(2\log 6 = \log(6^2) = \log 36\)

Substitute this back into the expression:
\(\log 36 - \log 9 + \log 25\)

Apply the division and multiplication laws of logarithms:
\(\log 36 - \log 9 = \log\left(\frac{36}{9}\right) = \log 4\)

\(\log 4 + \log 25 = \log(4 \times 25) = \log 100\)

Assuming base 10:
\(\log_{10} 100 = 2\)

M1 for correct use of the power law, e.g. \(2\log 6 = \log 36\)
M1 for combining the terms correctly, e.g. \(\log 4\) or \(\log 100\)
A1 for the final answer 2

Substitute the exact trigonometric values into the expression:
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
\(\tan 30^\circ = \frac{1}{\sqrt{3}}\)
\(\cos 45^\circ = \frac{1}{\sqrt{2}}\)

We obtain:
\(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{3}}} + \left(\frac{1}{\sqrt{2}}\right)^2\)

Simplify each term:
First term: \(\frac{\sqrt{3}}{2} \times \sqrt{3} = \frac{3}{2}\)
Second term: \(\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\)

Add the two simplified terms:
\(\frac{3}{2} + \frac{1}{2} = 2\)

M1 for substituting at least two correct exact values (e.g. \(\frac{\sqrt{3}}{2}\), \(\frac{1}{\sqrt{3}}\), or \(\frac{1}{\sqrt{2}}\))
M1 for simplifying the division to \(\frac{3}{2}\) or the square to \(\frac{1}{2}\)
A1 for the final answer 2

(a) Since the total number of students is 160:
$$20 + 48 + x + 32 + y = 160 \implies x + y = 60 \quad \text{--- (Equation 1)}$$

To find the mean, we use the midpoints of the intervals:
- For \(10 < t \le 15\), midpoint \(m_1 = 12.5\)
- For \(15 < t \le 20\), midpoint \(m_2 = 17.5\)
- For \(20 < t \le 25\), midpoint \(m_3 = 22.5\)
- For \(25 < t \le 30\), midpoint \(m_4 = 27.5\)
- For \(30 < t \le 35\), midpoint \(m_5 = 32.5\)

Since the mean is 22.125:
$$\frac{20(12.5) + 48(17.5) + x(22.5) + 32(27.5) + y(32.5)}{160} = 22.125$$
$$250 + 840 + 22.5x + 880 + 32.5y = 3540$$
$$22.5x + 32.5y = 1570 \implies 45x + 65y = 3140 \implies 9x + 13y = 628 \quad \text{--- (Equation 2)}$$

Multiplying Equation 1 by 9:
$$9x + 9y = 540$$

Subtracting this from Equation 2:
$$4y = 88 \implies y = 22$$

Substituting \(y = 22\) back into Equation 1:
$$x = 60 - 22 = 38$$

(b) To estimate the standard deviation:
$$\sum f m^2 = 20(12.5^2) + 48(17.5^2) + 38(22.5^2) + 32(27.5^2) + 22(32.5^2)$$
$$\sum f m^2 = 3125 + 14700 + 19237.5 + 24200 + 23237.5 = 84500$$

$$\text{Variance} = \frac{\sum f m^2}{\sum f} - (\text{mean})^2$$
$$\text{Variance} = \frac{84500}{160} - 22.125^2 = 528.125 - 489.515625 = 38.609375$$
$$\text{Standard deviation} = \sqrt{38.609375} \approx 6.21 \text{ minutes}$$

(c) The number of students who took more than 25 minutes is the sum of frequencies for the last two intervals:
$$\text{Number of students} = 32 + y = 32 + 22 = 54$$
$$\text{Probability} = \frac{54}{160} = 0.3375$$

(a) [4 marks]
- **B1** for \(x + y = 60\) oe.
- **M1** for correct expression for the sum of midpoints times frequencies, e.g., \(250 + 840 + 22.5x + 880 + 32.5y = 3540\).
- **A1** for obtaining a correct linear equation in \(x\) and \(y\), e.g., \(9x + 13y = 628\) oe.
- **A1** for both \(x = 38\) and \(y = 22\) correct.

(b) [3 marks]
- **M1** for attempting to calculate \(\sum f m^2\) (at least 3 terms correct).
- **M1** for substituting their value of \(\sum f m^2\) and mean into standard deviation formula.
- **A1** for \(6.21\) (accept 6.213 to 6.214).

(c) [1 mark]
- **B1** for \(\frac{54}{160}\) oe (e.g., \(\frac{27}{80}\) or 0.3375).

(a) The median is the height corresponding to a cumulative frequency of \(200 / 2 = 100\).
The cumulative frequency 100 lies in the interval \(20 < h \le 30\) where cumulative frequency increases from 50 to 110.
Using linear interpolation:
$$\text{Median} = 20 + \frac{100 - 50}{110 - 50} \times (30 - 20) = 20 + \frac{50}{60} \times 10 = 20 + 8.33 = 28.33 \approx 28.3 \text{ cm}$$

(b) The lower quartile (LQ) is the height corresponding to a cumulative frequency of \(200 / 4 = 50\).
Since the cumulative frequency at \(h \le 20\) is exactly 50, the lower quartile is:
$$\text{LQ} = 20 \text{ cm}$$

The upper quartile (UQ) is the height corresponding to a cumulative frequency of \(3 \times 200 / 4 = 150\).
The cumulative frequency 150 lies in the interval \(30 < h \le 40\) where cumulative frequency increases from 110 to 170.
Using linear interpolation:
$$\text{UQ} = 30 + \frac{150 - 110}{170 - 110} \times (40 - 30) = 30 + \frac{40}{60} \times 10 = 30 + 6.67 = 36.67 \text{ cm}$$

Therefore, the interquartile range (IQR) is:
$$\text{IQR} = \text{UQ} - \text{LQ} = 36.67 - 20 = 16.67 \approx 16.7 \text{ cm}$$

(c) A height of 35 cm lies exactly halfway in the interval \(30 < h \le 40\).
Using linear interpolation, the cumulative frequency at \(h = 35\) is:
$$\text{CF} = 110 + \frac{35 - 30}{40 - 30} \times (170 - 110) = 110 + \frac{5}{10} \times 60 = 110 + 30 = 140$$

Thus, 140 plants have a height of 35 cm or less.
The number of plants with a height greater than 35 cm is:
$$200 - 140 = 60$$

The percentage of plants is:
$$\text{Percentage} = \frac{60}{200} \times 100\% = 30\%$$

(a) [2 marks]
- **M1** for attempting to interpolate for CF = 100, e.g., \(20 + \frac{100 - 50}{110 - 50} \times 10\) oe.
- **A1** for \(28.3\) (accept \(28\frac{1}{3}\) or 28.33).

(b) [3 marks]
- **B1** for finding LQ = 20.
- **M1** for attempting to interpolate for CF = 150, e.g., \(30 + \frac{40}{60} \times 10\) to find UQ (yielding 36.7 or \(36\frac{2}{3}\)).
- **A1** for \(16.7\) (accept \(16\frac{2}{3}\) or 16.67).

(c) [3 marks]
- **M1** for attempting to interpolate the CF at \(h = 35\) (yielding 140).
- **M1** for finding the number of plants greater than 35 cm: \(200 - \text{their } 140\) (yielding 60).
- **A1** for \(30\%\).

First, calculate Karim's investment value after 12 years: \(K = 8000 \times (1.025)^{12} \approx 10759.11\). Sofia's investment after 12 years is worth $1200 less than Karim's: \(S = 10759.11 - 1200 = 9559.11\). Set up the compound interest equation for Sofia: \(7500 \times (1 + r/100)^{12} = 9559.11\). Divide by 7500: \((1 + r/100)^{12} = 1.274548\). Take the 12th root of both sides: \(1 + r/100 = (1.274548)^{1/12} \approx 1.0203775\). Solve for r: \(r/100 = 0.0203775 \implies r \approx 2.04\).

M1 for \(8000 \times (1.025)^{12}\) oe. A1 for 10759.11 (or 10760). M1 for subtracting 1200 from their Karim's value to get 9559.11. M1 for setting up \(7500 \times (1 + r/100)^{12} = 9559.11\) (or their S). M1 for solving to \(1 + r/100 = (1.274548)^{1/12}\) oe. A1 for 2.04 (accept 2.038).

Let n be the number of years. The population of city P after n years is \(P = 250000 \times (1.018)^n\). The population of town Q after n years is \(Q = 320000 \times (0.988)^n\). We need to find the smallest integer n such that \(250000 \times (1.018)^n > 320000 \times (0.988)^n\). This simplifies to \((1.018 / 0.988)^n > 320000 / 250000\), which is \((1.030364)^n > 1.28\). Taking logs on both sides: \(n \log(1.030364) > \log(1.28)\). This gives \(n > 0.10721 / 0.01299 \approx 8.25\). Since n must be an integer, the population of P first exceeds Q after 9 years.

M1 for \(250000 \times (1.018)^n\) oe seen. M1 for \(320000 \times (0.988)^n\) oe seen. M1 for setting up inequality or equation: \(250000 \times (1.018)^n > 320000 \times (0.988)^n\). M1 for simplifying to \((1.030...)^n > 1.28\) or taking logs. M1 for finding \(n \approx 8.25\) or testing values for n = 8 and n = 9. A1 for 9.

Let \(B\) be the event that Liam takes the bus, and \(W\) be the event that Liam walks.
Let \(L\) be the event that Liam is late, and \(T\) be the event that Liam is on time.

We are given:
\(P(B) = 0.75\)
\(P(W) = 1 - 0.75 = 0.25\)
\(P(L | B) = 0.08\)
\(P(L | W) = 0.32\)

(a) To find \(P(L)\):
\(P(L) = P(B \cap L) + P(W \cap L)\)
\(P(L) = (0.75 \times 0.08) + (0.25 \times 0.32)\)
\(P(L) = 0.06 + 0.08 = 0.14\)

(b) To find the conditional probability \(P(B | L)\):
\(P(B | L) = \frac{P(B \cap L)}{P(L)}\)
\(P(B | L) = \frac{0.75 \times 0.08}{0.14}\)
\(P(B | L) = \frac{0.06}{0.14} = \frac{6}{14} = \frac{3}{7} \approx 0.429\) (to 3 significant figures)

(c) Let the probability of Liam being late on any day be \(p = 0.14\).
The probability of him being late on exactly one of 2 consecutive days is:
\(P(\text{exactly one late}) = P(L \cap T) + P(T \cap L)\)
\(P(\text{exactly one late}) = (0.14 \times 0.86) + (0.86 \times 0.14)\)
\(P(\text{exactly one late}) = 2 \times 0.14 \times 0.86 = 0.2408 \approx 0.241\) (to 3 significant figures)

(a) [3 marks]
M1 for \(0.75 \times 0.08\) oe (soi by 0.06)
M1 for \(0.25 \times 0.32\) oe (soi by 0.08)
A1 for 0.14 or \(\frac{7}{50}\) oe

(b) [3 marks]
M1 for \(\frac{P(B \cap L)}{P(L)}\) structured fraction with their values
A1 for numerator 0.06 (or their \(P(B \cap L)\))
A1 for \(\frac{3}{7}\) or awrt 0.429

(c) [2 marks]
M1 for \(2 \times \text{their}(a) \times (1 - \text{their}(a))\) oe
A1 for 0.2408 or awrt 0.241

(a) Using the Cosine Rule in triangle $PQR$:
$$PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\angle PQR)$$
$$28^2 = 15^2 + 22^2 - 2(15)(22)\cos(\angle PQR)$$
$$784 = 225 + 484 - 660\cos(\angle PQR)$$
$$784 = 709 - 660\cos(\angle PQR)$$
$$660\cos(\angle PQR) = -75$$
$$\cos(\angle PQR) = -\frac{75}{660} = -\frac{5}{44}$$
$$\angle PQR = \arccos\left(-\frac{5}{44}\right) \approx 104.908^\circ \approx 104.9^\circ$$

(b) The bearing of $Q$ from $P$ is $055^\circ$.
Therefore, the back-bearing (direction from $Q$ to $P$) is:
$$055^\circ + 180^\circ = 235^\circ$$
Since the bearing of $R$ from $Q$ is less than $180^\circ$, the line $QR$ must be counter-clockwise from $QP$. Thus, we subtract $\angle PQR$ from the back-bearing:
$$\text{Bearing} = 235^\circ - 104.9^\circ = 130.1^\circ$$

(c) First, we find angle $QPR$ (let us call it $\angle P$) using the Cosine Rule:
$$QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos(\angle P)$$
$$22^2 = 15^2 + 28^2 - 2(15)(28)\cos(\angle P)$$
$$484 = 225 + 784 - 840\cos(\angle P)$$
$$484 = 1009 - 840\cos(\angle P)$$
$$840\cos(\angle P) = 525$$
$$\cos(\angle P) = \frac{525}{840} = 0.625$$

At the point $H$ where the boat is closest to $Q$, the line $QH$ is perpendicular to $PR$. Therefore, triangle $PQH$ is a right-angled triangle with the right angle at $H$.
$$PH = PQ \cdot \cos(\angle P)$$
$$PH = 15 \times 0.625 = 9.375\text{ km}$$

The constant speed of the boat is $20\text{ km/h}$. The time taken to travel from $P$ to $H$ is:
$$\text{Time} = \frac{9.375\text{ km}}{20\text{ km/h}} = 0.46875\text{ hours}$$
$$\text{Time in minutes} = 0.46875 \times 60 = 28.125\text{ minutes}$$

Correct to the nearest minute, the travel time is $28\text{ minutes}$.
Adding $28\text{ minutes}$ to the departure time of 14:30:
$$\text{Time of arrival} = 14:58$$

**Part (a)**
* **M1**: Correct substitute into Cosine Rule formula, e.g., $28^2 = 15^2 + 22^2 - 2 \times 15 \times 22 \times \cos(\angle PQR)$
* **M1**: Rearranging correctly to get $\cos(\angle PQR) = -\frac{75}{660}$ or equivalent decimal (at least $-0.114$)
* **A1**: Establishing $104.908...^\circ$ leading to $104.9^\circ$

**Part (b)**
* **B1**: Finding the back-bearing of $P$ from $Q$ to be $235^\circ$ (or clearly showing $180^\circ - 55^\circ = 125^\circ$ as interior angle)
* **M1**: For a correct strategy utilizing the back-bearing (or interior angle), e.g., $235^\circ - 104.9^\circ$ (or $180^\circ - (125^\circ + (180^\circ - 104.9^\circ))$)
* **A1**: $130.1^\circ$ or $130.1$
* **B1**: Final answer written in standard 3-figure decimal bearing notation (allow $130.1$ or $130$)

**Part (c)**
* **M1**: Correct use of Cosine Rule to find $\cos(\angle P) = 0.625$ or $\angle P \approx 51.3^\circ$
* **M1**: Finding the distance $PH = 15 \times \cos(\angle P)$ or $9.375\text{ km}$
* **M1**: Finding the travel time in minutes: $\frac{\text{their } PH}{20} \times 60$
* **A1**: 14:58 (accept 2:58 pm)

(a) The graph consists of two branches separated by the vertical asymptote at \(x = 0\). The left branch (\(x < 0\)) is strictly increasing and crosses the negative x-axis. The right branch (\(x > 0\)) is U-shaped with a local minimum in the first quadrant.

(b) The vertical asymptote occurs where the denominator of the fractional term is zero, which is at \(x = 0\).

(c) Using a graphic display calculator, find the local minimum of the right branch. This gives \(x \approx 1.10\) and \(y \approx 4.95\). The coordinates of the local minimum are \((1.10, 4.95)\).

(d) To find where the graph crosses the x-axis, set \(f(x) = 0 \implies 3x + \frac{2}{x^2} = 0 \implies 3x^3 + 2 = 0 \implies x = \sqrt[3]{-\frac{2}{3}} \approx -0.874\). The coordinates are \((-0.874, 0)\).

(e) Any horizontal line \(y = k\) intersects the left branch exactly once because the function strictly increases from \(-\infty\) to \(\infty\) for \(x < 0\). To have exactly one solution overall, the line must not intersect the right branch. The minimum value of the right branch is \(4.95\), so the line will not intersect it if \(k < 4.95\).

(a) **[2 marks]**
- 1 mark for correct shape of both branches.
- 1 mark for indicating the vertical asymptote at \(x = 0\) and crossing the negative x-axis.

(b) **[1 mark]**
- 1 mark for \(x = 0\).

(c) **[2 marks]**
- 1 mark for \(x \approx 1.10\) (or \(1.1\)).
- 1 mark for \(y \approx 4.95\).

(d) **[2 marks]**
- 1 mark for \(x \approx -0.874\).
- 1 mark for \(y = 0\).

(e) **[3 marks]**
- 1 mark for identifying the local minimum y-value \(4.95\).
- 1 mark for recognizing that for \(k < \text{minimum}\) there is only one solution.
- 1 mark for the correct final inequality \(k < 4.95\).

(a) The graph is a cubic curve starting at \((-2, -14)\), rising to a local maximum, falling to a local minimum, and then rising again to end at \((4, 16)\).

(b)(i) Using a graphic display calculator to find the local maximum yields \(x \approx -0.155\) and \(y \approx 4.08\). The coordinates are \((-0.155, 4.08)\).
(ii) Using the calculator to find the local minimum yields \(x \approx 2.15\) and \(y \approx -2.08\). The coordinates are \((2.15, -2.08)\).

(c) Equating \(g(x)\) and the line: \(x^3 - 3x^2 - x + 4 = 2x - 1 \implies x^3 - 3x^2 - 3x + 5 = 0\). Using a graphic display calculator to find the roots of this cubic equation yields the solutions: \(x \approx -1.45\), \(x = 1\), and \(x \approx 3.45\).

(d) For the equation \(g(x) = k\) to have three distinct real solutions, the horizontal line \(y = k\) must intersect the curve exactly three times. This occurs when \(k\) lies strictly between the y-coordinates of the local minimum and the local maximum: \(-2.08 < k < 4.08\).

(a) **[2 marks]**
- 1 mark for a correct cubic shape with correct orientation.
- 1 mark for correct y-intercept at \((0, 4)\) and showing turning points in correct quadrants.

(b)(i) **[2 marks]**
- 1 mark for \(x \approx -0.155\).
- 1 mark for \(y \approx 4.08\).
(ii) **[2 marks]**
- 1 mark for \(x \approx 2.15\).
- 1 mark for \(y \approx -2.08\).

(c) **[2 marks]**
- 1 mark for any two correct x-coordinates.
- 2 marks for all three correct: \(x \approx -1.45\), \(x = 1\), \(x \approx 3.45\).

(d) **[2 marks]**
- 1 mark for identifying correct boundaries from turning points in (b).
- 1 mark for correct inequality \(-2.08 < k < 4.08\) (allow follow-through from their answers in (b)).

(a) The perimeter \(P\) of a rectangle with sides of length \(x\) and \(y\) is given by:
\(P = 2x + 2y\)
Substituting \(y = 12 - x^2\):
\(P = 2x + 2(12 - x^2) = 24 + 2x - 2x^2\)

(b) The area \(A\) of the rectangle is given by:
\(A = x y\)
Substituting \(y = 12 - x^2\):
\(A = x(12 - x^2) = 12x - x^3\)

(c) (i) To find the value of \(x\) that maximizes the area, we can differentiate \(A\) with respect to \(x\) and set the derivative to zero:
\(\frac{dA}{dx} = 12 - 3x^2 = 0\)
\(3x^2 = 12\)
\(x^2 = 4\)
Since \(x \ge 0\), we have:
\(x = 2\)

(ii) The maximum area is found by substituting \(x = 2\) into the area expression:
\(A = 12(2) - (2)^3 = 24 - 8 = 16\)

(a) [2 marks]
- M1 for substituting \(y = 12 - x^2\) into \(2x + 2y\)
- A1 for \(24 + 2x - 2x^2\) or equivalent

(b) [2 marks]
- M1 for substituting \(y = 12 - x^2\) into \(x y\)
- A1 for \(12x - x^3\) or equivalent

(c) (i) [3 marks]
- M1 for differentiating their \(A\) with respect to \(x\) (or attempting to find maximum using a GDC)
- M1 for setting their derivative equal to 0, e.g., \(12 - 3x^2 = 0\)
- A1 for \(x = 2\)

(ii) [2 marks]
- M1 for substituting their value of \(x\) into their area expression
- A1 for \(16\) (or FT from their parts)

(a) The perimeter of a sector of radius \(r\) and angle \(x^\circ\) is:
\(P = 2r + \frac{x}{360} \times 2\pi r = 120\)
Dividing the entire equation by 2:
\(r + \frac{x}{360} \pi r = 60\)
Rearranging to express \(\frac{x}{360} \pi r\) in terms of \(r\):
\(\frac{x}{360} \pi r = 60 - r\)

The area of the sector is given by:
\(A = \frac{x}{360} \times \pi r^2 = \left( \frac{x}{360} \pi r \right) \times r\)
Substituting the expression from the perimeter equation:
\(A = (60 - r)r = 60r - r^2\)

(b) (i) To maximize the area \(A\), we differentiate with respect to \(r\) and set the derivative to 0:
\(\frac{dA}{dr} = 60 - 2r = 0\)
\(2r = 60 \implies r = 30\)

(ii) The maximum area is:
\(A = 60(30) - 30^2 = 1800 - 900 = 900\text{ cm}^2\)

(c) Substitute \(r = 30\) into the rearranged perimeter equation:
\(\frac{x}{360} \pi (30) = 60 - 30\)
\(\frac{30\pi}{360} x = 30\)
\(\frac{\pi}{12} x = 30\)
\(x = \frac{360}{\pi} \approx 114.5915...\)
To 1 decimal place, \(x = 114.6^\circ\).

(a) [4 marks]
- M1 for writing down a correct perimeter equation: \(2r + \frac{x}{360} \times 2\pi r = 120\)
- M1 for rearranging to make \(\frac{x}{360} \pi r\) or \(x\) the subject
- M1 for substituting this into the area formula \(A = \frac{x}{360} \pi r^2\)
- A1 for convincing completion leading to \(A = 60r - r^2\)

(b) (i) [2 marks]
- M1 for finding the derivative \(60 - 2r\) and setting to 0 (or using GDC to find the vertex of the quadratic)
- A1 for \(r = 30\)

(ii) [1 mark]
- B1 for \(900\)

(c) [2 marks]
- M1 for substituting their \(r = 30\) into the perimeter relation to solve for \(x\)
- A1 for \(114.6\) (accept 115 from 3 sig figs, but 1 decimal place is requested)

(a) The distance squared \(d^2\) between \(P(x, y)\) and \(A(2, 9)\) is given by:
\(d^2 = (x - 2)^2 + (y - 9)^2\)
Since \(P\) lies on the line \(y = 2x - 5\), we substitute \(y\) into the distance squared formula:
\(d^2 = (x - 2)^2 + (2x - 5 - 9)^2\)
\(d^2 = (x - 2)^2 + (2x - 14)^2\)
Expanding the brackets:
\((x - 2)^2 = x^2 - 4x + 4\)
\((2x - 14)^2 = 4x^2 - 56x + 196\)
Adding them together:
\(d^2 = x^2 - 4x + 4 + 4x^2 - 56x + 196\)
\(d^2 = 5x^2 - 60x + 200\)

(b) (i) To minimize the distance (and hence the distance squared), we differentiate \(d^2\) with respect to \(x\) and set it to 0:
\(\frac{d(d^2)}{dx} = 10x - 60 = 0\)
\(10x = 60 \implies x = 6\)

(ii) Substituting \(x = 6\) into the equation of the line \(y = 2x - 5\):
\(y = 2(6) - 5 = 7\)
So the coordinates of \(P\) are \((6, 7)\).

(iii) Substituting \(x = 6\) into the expression for \(d^2\):
\(d^2 = 5(6)^2 - 60(6) + 200 = 180 - 360 + 200 = 20\)
Thus, the minimum distance is:
\(d = \sqrt{20} \approx 4.47\) (to 3 significant figures).

(a) [4 marks]
- M1 for using the distance formula: \(d^2 = (x - 2)^2 + (y - 9)^2\)
- M1 for substituting \(y = 2x - 5\) into the formula
- M1 for expanding the squared brackets correctly
- A1 for obtaining the final expression \(5x^2 - 60x + 200\)

(b) (i) [2 marks]
- M1 for differentiating \(5x^2 - 60x + 200\) or using GDC to find the minimum of the quadratic function
- A1 for \(x = 6\)

(ii) [1 mark]
- B1 for \((6, 7)\)

(iii) [2 marks]
- M1 for calculating \(\sqrt{20}\) or substituting \(x = 6\) into \(\sqrt{5x^2 - 60x + 200}\)
- A1 for \(4.47\) or \(2\sqrt{5}\)

(a) When x = 6 and y = 5, d = 6^2 - 5^2 = 36 - 25 = 11.
(b) When x = 12 and y = 11, d = 12^2 - 11^2 = 144 - 121 = 23.
(c) If x = y + 1, then:
d = (y + 1)^2 - y^2 = y^2 + 2y + 1 - y^2 = 2y + 1.
(d) Since y is a non-negative integer, 2y is always an even number. Adding 1 to any even number always results in an odd number. Therefore, d = 2y + 1 must always be odd.

B1 for 11
B1 for 23
M1 for substituting x = y + 1 into x^2 - y^2 or observing the linear sequence 1, 3, 5, 7, 9...
A1 for 2y + 1 (or equivalent simplified expression)
B1 for a clear explanation that 2y is even so 2y + 1 is odd

(a)(i) When x = 2, y = 0, d = 2^2 - 0^2 = 4.
(ii) When x = 3, y = 1, d = 3^2 - 1^2 = 9 - 1 = 8.
(iii) When x = 5, y = 3, d = 5^2 - 3^2 = 25 - 9 = 16.

(b) Substitute x = y + 2 into the expression for d:
d = (y + 2)^2 - y^2
d = (y^2 + 4y + 4) - y^2
d = 4y + 4.

B1 for all three correct values in part (a) (4, 8, 16)
M2 for expanding (y + 2)^2 to get y^2 + 4y + 4
A2 for subtracting y^2 and correctly obtaining 4y + 4

(a) For consecutive squares, d = 2y + 1.
Setting 2y + 1 = 35 => 2y = 34 => y = 17.
Since they are consecutive, x = y + 1 = 18.
Check: 18^2 - 17^2 = 324 - 289 = 35.

(b) For squares with a difference of 2, d = 4y + 4.
Setting 4y + 4 = 40 => 4y = 36 => y = 9.
Since x - y = 2, x = y + 2 = 11.
Check: 11^2 - 9^2 = 121 - 81 = 40.

M1 for setting 2y + 1 = 35 to find y
A1 for x = 18 and y = 17
M1 for setting 4y + 4 = 40 to find y
A1 for y = 9
A1 for x = 11

(a) Substitute x = y + k into the expression x^2 - y^2:
d = (y + k)^2 - y^2
d = y^2 + 2ky + k^2 - y^2
d = 2ky + k^2
Factor out k:
d = k(2y + k).

(b) When k = 3, the formula becomes:
d = 3(2y + 3).
Since y is an integer, 2y + 3 is also an integer. Therefore, d is 3 times an integer, which means d must always be a multiple of 3.

M1 for substituting x = y + k into (x - y)(x + y) or x^2 - y^2
M1 for expanding to y^2 + 2ky + k^2 - y^2
A1 for correctly factoring out k to get k(2y + k)
M1 for substituting k = 3 to get 3(2y + 3)
A1 for explaining that since 2y + 3 is an integer, 3(2y + 3) is a multiple of 3

We set (x - y)(x + y) = 45.
Let u = x - y and v = x + y, where u and v are factor pairs of 45, and u <= v.
The factor pairs of 45 are:
1) u = 1, v = 45:
x - y = 1 and x + y = 45. Adding these gives 2x = 46 => x = 23. Then y = 22.
So, 45 = 23^2 - 22^2.

2) u = 3, v = 15:
x - y = 3 and x + y = 15. Adding these gives 2x = 18 => x = 9. Then y = 6.
So, 45 = 9^2 - 6^2.

3) u = 5, v = 9:
x - y = 5 and x + y = 9. Adding these gives 2x = 14 => x = 7. Then y = 2.
So, 45 = 7^2 - 2^2.

The three pairs of (x, y) are (23, 22), (9, 6), and (7, 2).

M1 for identifying the factor pairs of 45: (1, 45), (3, 15), (5, 9)
M1 for setting up simultaneous equations x - y = u and x + y = v to solve for x and y
A1 for (23, 22)
A1 for (9, 6)
A1 for (7, 2)

Let's analyze the parity of x^2 - y^2:
- Case 1: One of x and y is even and the other is odd.
Then one of x^2 and y^2 is even and the other is odd. Their difference, x^2 - y^2, must be odd.
Since 4m + 2 is always an even number, x^2 - y^2 can never equal 4m + 2 in this case.

- Case 2: Both x and y are even.
Let x = 2a and y = 2b.
Then x^2 - y^2 = (2a)^2 - (2b)^2 = 4a^2 - 4b^2 = 4(a^2 - b^2).
This is always a multiple of 4.

- Case 3: Both x and y are odd.
Let x = 2a + 1 and y = 2b + 1.
Then x^2 - y^2 = (2a + 1)^2 - (2b + 1)^2 = (4a^2 + 4a + 1) - (4b^2 + 4b + 1) = 4(a^2 + a - b^2 - b).
This is also always a multiple of 4.

Therefore, any difference of squares x^2 - y^2 must be either odd or a multiple of 4.
Since numbers of the form 4m + 2 are even but not multiples of 4 (they leave a remainder of 2 when divided by 4), they can never be written as the difference of two squares.

B1 for explaining that if one is odd and one is even, x^2 - y^2 is odd (and thus not of the form 4m + 2)
M1 for considering x and y both even and showing x^2 - y^2 is a multiple of 4
M1 for considering x and y both odd and showing x^2 - y^2 is a multiple of 4
A1 for concluding that x^2 - y^2 is either odd or a multiple of 4
A1 for noting that 4m + 2 is even but not a multiple of 4, hence impossible

(a) The boundary of the entire shape consists of two straight sections of length \(x\) and the circumferences of two semicircles of diameter \(y\), which combine to form one full circle of diameter \(y\) (and circumference \(\pi y\)).

The total perimeter of this boundary is:
\(2x + \pi y = 400\)

Rearranging to express \(y\) in terms of \(x\):
\(\pi y = 400 - 2x\)
\(y = \frac{2}{\pi}(200 - x)\)

The area of the rectangular pitch is:
\(A = x \times y = x \left( \frac{2}{\pi}(200 - x) \right) = \frac{2}{\pi} x (200 - x)\)

(b) Since both dimensions \(x\) and \(y\) must be positive:
\(x > 0\) and \(200 - x > 0 \implies x < 200\)
So, the range is \(0 < x < 200\).

(c) The quadratic model \(A(x) = \frac{2}{\pi} x (200 - x)\) is a downward-opening parabola with roots at \(x = 0\) and \(x = 200\).
By symmetry, the maximum area occurs at the vertex:
\(x = \frac{0 + 200}{2} = 100\) m

Substituting \(x = 100\) into the model:
\(A_{\max} = \frac{2}{\pi} (100)(200 - 100) = \frac{20000}{\pi} \approx 6366.1977...\)

To 3 significant figures:
- \(x = 100\) m
- \(\text{Maximum Area} = 6370\) m\(^2\)

(a)
- M1 for setting up the perimeter equation: \(2x + \pi y = 400\)
- M1 for rearranging to make \(y\) the subject: \(y = \frac{400-2x}{\pi}\) or equivalent
- A1 for substituting \(y\) into \(A = xy\) and simplifying to the given expression

(b)
- B1 for \(0 < x < 200\) (allow \(x < 200\) if \(x > 0\) is implied)

(c)
- M1 for finding the vertex of the parabola at \(x = 100\) (or using differentiation \(\frac{dA}{dx} = 0 \implies \frac{2}{\pi}(200 - 2x) = 0\))
- A1 for \(x = 100\)
- A1 for \(\text{Maximum Area} = 6370\) (or 6366)

(a) The playing field has length \(x\) and width \(y\).
Adding a 5-metre wide safety zone all around means the outer dimensions of the safety zone are:
- Outer length \(= x + 2(5) = x + 10\)
- Outer width \(= y + 2(5) = y + 10\)

The perimeter of this outer boundary is:
\(2(x + 10) + 2(y + 10) = 300\)
\(2x + 20 + 2y + 20 = 300\)
\(2x + 2y = 260\)
\(x + y = 130 \implies y = 110 - x\)

The area of the playing field is:
\(B = x \times y = x(110 - x)\)

(b) The graph of \(B = x(110 - x)\) is a parabola opening downwards with \(x\)-intercepts at \((0, 0)\) and \((110, 0)\), and a vertex (maximum point) at \((55, 3025)\).

(c) The maximum area occurs at the line of symmetry:
\(x = \frac{110}{2} = 55\) m
\(B_{\max} = 55(110 - 55) = 55 \times 55 = 3025\) m\(^2\)

(a)
- M1 for writing outer dimensions as \(x + 10\) and \(y + 10\)
- M1 for setting up the perimeter equation \(2(x + 10) + 2(y + 10) = 300\)
- A1 for solving for \(y\) to get \(y = 110 - x\) and showing \(B = x(110 - x)\)

(b)
- B1 for a smooth downward-opening parabola starting at \((0,0)\) and ending at \((110,0)\)
- B1 for labelling the intercepts at \(0\) and \(110\) on the horizontal axis

(c)
- M1 for substituting \(x = 55\) into the formula
- A1 for \(3025\)

(a) The total cost consists of turf/soil costs, fence costs, and the fixed fee.
1. Turf and soil preparation cost per square metre \(= 1.20 + 0.80 = \$2.00\).
Total cost for area \(B = 2.00 \times B = 2x(110 - x) = 220x - 2x^2\).

2. The perimeter of the playing field itself is \(2x + 2y\).
Since \(y = 110 - x\), the perimeter is:
\(2x + 2(110 - x) = 220\) metres
This is a constant perimeter regardless of \(x\).
Fence cost \(= 15 \times 220 = \$3300\).

3. Fixed fee \(= \$5000\).

Total Cost \(C\):
\(C = (220x - 2x^2) + 3300 + 5000 = 220x - 2x^2 + 8300\)

(b) The playing field with maximum area has \(x = 55\) m.
Substituting \(x = 55\) into the cost model:
\(C = 220(55) - 2(55^2) + 8300\)
\(C = 12100 - 6050 + 8300 = \$14350\)

(c) The donation is given by:
\(D = 10(B - 2500) = 10(x(110 - x) - 2500) = 1100x - 10x^2 - 25000\)

The net cost \(N\) is the total cost minus the donation:
\(N = C - D\)
\(N = (220x - 2x^2 + 8300) - (1100x - 10x^2 - 25000)\)
\(N = 220x - 2x^2 + 8300 - 1100x + 10x^2 + 25000\)
\(N = 8x^2 - 880x + 33300\)

(a)
- M1 for finding the total turf and soil rate of $2.00/m\(^2\)
- M1 for calculating the constant fence cost: \(15 \times 2(x + 110 - x) = 3300\)
- A1 for combining all elements: \(2x(110 - x) + 3300 + 5000 = 220x - 2x^2 + 8300\)

(b)
- M1 for substituting \(x = 55\) into the model
- A1 for 14350

(c)
- M1 for writing the donation formula: \(10(x(110 - x) - 2500)\)
- M1 for subtracting the donation from the cost expression
- A1 for \(8x^2 - 880x + 33300\)

(a) In a right-angled triangle formed by the sides and the diagonal of length 30:
- One side \(= 30 \cos \theta\)
- The other side \(= 30 \sin \theta\)

The perimeter of the rectangular pool is:
\(P = 2(30 \cos \theta + 30 \sin \theta) = 60(\sin \theta + \cos \theta)\)

(b) The graph starts at \(\theta = 0^\circ\) with \(P = 60\), rises to a maximum at \(\theta = 45^\circ\), and symmetrically curves down to \(P = 60\) at \(\theta = 90^\circ\).

(c) The maximum of \(\sin \theta + \cos \theta\) occurs when \(\theta = 45^\circ\).
At \(\theta = 45^\circ\):
\(P_{\max} = 60(\sin 45^\circ + \cos 45^\circ) = 60\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) = 60\sqrt{2} \approx 84.85...\) metres

To 3 significant figures:
- Maximum perimeter \(= 84.9\) m
- Angle \(\theta = 45^\circ\)

(d) When \(\theta = 45^\circ\), the sides of the pool are:
\(\text{Length} = 30 \cos 45^\circ = 15\sqrt{2} \approx 21.21\) m
\(\text{Width} = 30 \sin 45^\circ = 15\sqrt{2} \approx 21.21\) m

Area of the pool:
\(\text{Area} = 15\sqrt{2} \times 15\sqrt{2} = 225 \times 2 = 450\) m\(^2\)

(a)
- M1 for expressing sides as \(30\sin\theta\) and \(30\cos\theta\)
- A1 for \(P = 2(30\sin\theta + 30\cos\theta) = 60(\sin\theta + \cos\theta)\)

(b)
- B1 for a curve with endpoints at \((0, 60)\) and \((90, 60)\)
- B1 for a symmetrical peak at \(\theta = 45^\circ\)

(c)
- B1 for \(\theta = 45^\circ\)
- B1 for \(84.9\) m (or \(60\sqrt{2}\))

(d)
- M1 for multiplying the sides: \((30\sin45^\circ)(30\cos45^\circ)\) or equivalent
- A1 for \(450\)