2023 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

Using log laws, we combine the terms: \(\log_3(x(x - 2)) = 1\). Converting to exponential form gives \(x(x - 2) = 3^1\), which simplifies to the quadratic equation \(x^2 - 2x - 3 = 0\). Factoring this gives \((x - 3)(x + 1) = 0\), yielding solutions \(x = 3\) and \(x = -1\). Since the argument of a logarithm must be positive, we reject \(x = -1\), leaving \(x = 3\) as the only valid solution.

M1 for combining the logarithms to obtain \(x^2 - 2x = 3\). A1 for the final answer \(x = 3\) (rejecting \(x = -1\)).

First, find a common denominator, which is \((2x - 1)(x + 3)\). Express both fractions with this denominator: \(\frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\). Expand the numerator: \(\frac{3x + 9 - 4x + 2}{(2x - 1)(x + 3)}\). Simplify the numerator to get \(\frac{11 - x}{(2x - 1)(x + 3)}\).

M1 for putting both fractions over a common denominator with correct numerator expansion: \(3(x + 3) - 2(2x - 1)\). A1 for the fully simplified final fraction.

First, factor out the common term of 2 from the expression: \(2(6x^2 - 7xy - 3y^2)\). Next, factorise the quadratic trinomial inside the brackets: \(6x^2 - 7xy - 3y^2 = (3x + y)(2x - 3y)\). Thus, the fully factorised form is \(2(3x + y)(2x - 3y)\).

M1 for taking out the common factor of 2 or for a correct partial factorisation such as \((6x + 2y)(2x - 3y)\). A1 for the fully factorised expression.

Multiply the first equation by 5 and the second equation by 2 to eliminate \(y\): \(15x - 10y = 55\) and \(4x + 10y = 2\). Adding these equations gives \(19x = 57\), which simplifies to \(x = 3\). Substitute \(x = 3\) back into the second equation: \(2(3) + 5y = 1 \implies 6 + 5y = 1 \implies 5y = -5 \implies y = -1\).

M1 for a valid method to eliminate one variable (e.g., getting equations to \(15x - 10y = 55\) and \(4x + 10y = 2\)). A1 for both correct values of \(x\) and \(y\).

Multiply the coefficients and add the exponents: \((4 \times 8) \times 10^{5 + (-9)} = 32 \times 10^{-4}\). Convert this to standard form by writing 32 as \(3.2 \times 10^1\), so we get \(3.2 \times 10^1 \times 10^{-4} = 3.2 \times 10^{-3}\).

M1 for obtaining \(32 \times 10^{-4}\) or \(0.0032\) or showing a correct application of exponent laws. A1 for the final answer in correct standard form.

The number of students studying at least one of the two subjects is \(30 - 5 = 25\). Let \(x\) be the number of students who study both. Using the principle of inclusion-exclusion: \(18 + 15 - x = 25 \implies 33 - x = 25 \implies x = 8\).

M1 for finding that 25 students study at least one subject (\(30 - 5 = 25\)) or writing down the equation \(18 + 15 - x = 25\). A1 for 8.

After 1 year, the value is \(4000 \times 0.9 = 3600\). After 2 years, the value is \(3600 \times 0.9 = 3240\). Alternatively, \(4000 \times (0.9)^2 = 4000 \times 0.81 = 3240\).

M1 for working out the value after 1 year (\(\$3600\)) or for setting up the calculation \(4000 \times 0.9^2\). A1 for 3240.

The gradient of the given line is \(m_1 = 2\). The gradient of a line perpendicular to it is the negative reciprocal, \(m_2 = -\frac{1}{2}\). Using the equation of a straight line with the point \((4, 7)\), we get \(y - 7 = -\frac{1}{2}(x - 4)\), which simplifies to \(y - 7 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 9\).

M1 for identifying the perpendicular gradient as \(-\frac{1}{2}\). A1 for the correct final equation \(y = -\frac{1}{2}x + 9\) (or equivalent).

Apply the subtraction law of logarithms:
\(\log_2\left(\frac{x+14}{x-1}\right) = 4\)

Convert the logarithmic equation into exponential form:
\(\frac{x+14}{x-1} = 2^4\)
\(\frac{x+14}{x-1} = 16\)

Multiply both sides by \((x - 1)\):
\(x + 14 = 16(x - 1)\)
\(x + 14 = 16x - 16\)

Rearrange and solve for \(x\):
\(15x = 30\)
\(x = 2\)

M1 for writing \(\frac{x+14}{x-1} = 2^4\) or \(\frac{x+14}{x-1} = 16\)
A1 for \(x = 2\)

Express both fractions over a common denominator, \((x-2)(x+3)\):
\(\frac{3(x+3) - 2(x-2)}{(x-2)(x+3)}\)

Expand the numerator:
\(\frac{3x + 9 - (2x - 4)}{(x-2)(x+3)}\)
\(\frac{3x + 9 - 2x + 4}{(x-2)(x+3)}\)

Simplify the numerator:
\(\frac{x + 13}{(x-2)(x+3)}\)

M1 for writing the expression over a common denominator with at least one correct numerator expansion, e.g., \(\frac{3(x+3) - 2(x-2)}{(x-2)(x+3)}\)
A1 for \(\frac{x+13}{(x-2)(x+3)}\) or \(\frac{x+13}{x^2+x-6}\)

Start by applying the power law of logarithms to the first term:
\( \log_3(x-2)^2 - \log_3(x+4) = 1 \)

Next, use the quotient rule for logarithms:
\( \log_3 \left( \frac{(x-2)^2}{x+4} \right) = 1 \)

Convert the equation from logarithmic form to exponential form:
\( \frac{(x-2)^2}{x+4} = 3^1 \)

Multiply both sides by \( x+4 \):
\( (x-2)^2 = 3(x+4) \)

Expand both sides and simplify:
\( x^2 - 4x + 4 = 3x + 12 \)
\( x^2 - 7x - 8 = 0 \)

Factor the quadratic equation:
\( (x-8)(x+1) = 0 \)

This gives \( x = 8 \) or \( x = -1 \).

Finally, we must check the domain of the logarithmic terms:
For \( \log_3(x-2) \) to be defined, we require \( x-2 > 0 \implies x > 2 \).
For \( \log_3(x+4) \) to be defined, we require \( x+4 > 0 \implies x > -4 \).

Since \( x = -1 \) does not satisfy \( x > 2 \), we reject \( x = -1 \).
Thus, the only valid solution is \( x = 8 \).

M1 for applying the power law of logarithms to obtain \( \log_3(x-2)^2 \)
M1 for applying the subtraction law to obtain \( \log_3 \left(\frac{(x-2)^2}{x+4}\right) = 1 \) or equivalent
M1 for solving the resulting quadratic equation to find \( x = 8 \) and \( x = -1 \)
A1 for rejecting \( x = -1 \) and stating \( x = 8 \) as the only solution

First, factorise the denominators of both fractions:
\( x^2 - 9 = (x-3)(x+3) \)
\( x^2 + 3x = x(x+3) \)

Rewrite the expression with the factorised denominators:
\( \frac{2}{(x-3)(x+3)} - \frac{1}{x(x+3)} \)

The lowest common denominator (LCD) is \( x(x-3)(x+3) \). Express each fraction with the LCD:
\( \frac{2x}{x(x-3)(x+3)} - \frac{x-3}{x(x-3)(x+3)} \)

Subtract the numerators:
\( \frac{2x - (x-3)}{x(x-3)(x+3)} = \frac{2x - x + 3}{x(x-3)(x+3)} = \frac{x+3}{x(x-3)(x+3)} \)

Simplify the fraction by dividing both the numerator and denominator by the common factor \( (x+3) \):
\( \frac{1}{x(x-3)} \)

M1 for factorising the denominators to \( (x-3)(x+3) \) and \( x(x+3) \)
M1 for writing both fractions with the common denominator \( x(x-3)(x+3) \)
M1 for simplifying the numerator to \( x+3 \)
A1 for the final simplified fraction \( \frac{1}{x(x-3)} \) or \( \frac{1}{x^2-3x} \)

The total number of beads initially is \( 5 + n \).

The probability of selecting a red bead on the first draw is:
\( P(R_1) = \frac{5}{5+n} \)

Since the selection is without replacement, there are now 4 red beads left and a total of \( 4 + n \) beads left. The probability of selecting a red bead on the second draw is:
\( P(R_2 | R_1) = \frac{4}{4+n} \)

The combined probability of selecting two red beads is:
\( P(R_1 \cap R_2) = \frac{5}{5+n} \times \frac{4}{4+n} = \frac{20}{(5+n)(4+n)} \)

We are given that this probability is \( \frac{5}{14} \):
\( \frac{20}{(5+n)(4+n)} = \frac{5}{14} \)

Divide both sides of the equation by 5:
\( \frac{4}{(5+n)(4+n)} = \frac{1}{14} \)

Cross-multiply to get:
\( (5+n)(4+n) = 56 \)
\( n^2 + 9n + 20 = 56 \)
\( n^2 + 9n - 36 = 0 \)

Factorise the quadratic equation:
\( (n+12)(n-3) = 0 \)

This gives \( n = -12 \) or \( n = 3 \). Since the number of beads must be a positive integer, we reject \( n = -12 \).

Therefore, \( n = 3 \).

M1 for setting up the multiplication of probabilities: \( \frac{5}{5+n} \times \frac{4}{4+n} \)
M1 for equating to \( \frac{5}{14} \) and forming a quadratic equation, e.g., \( n^2 + 9n - 36 = 0 \)
M1 for solving the quadratic equation to find potential solutions (e.g., factorising to \( (n+12)(n-3) = 0 \))
A1 for the final answer \( n = 3 \) (rejecting \( n = -12 \))

Let \( \theta = \angle ABC \).
The area of triangle \( ABC \) is given by the formula:
\( \text{Area} = \frac{1}{2} a c \sin \theta \)
\( 10 = \frac{1}{2} \times 5 \times 6 \times \sin \theta \)
\( 10 = 15 \sin \theta \)
\( \sin \theta = \frac{10}{15} = \frac{2}{3} \)

We need to find \( \cos \theta \). We use the trigonometric identity:
\( \sin^2 \theta + \cos^2 \theta = 1 \)
\( \left(\frac{2}{3}\right)^2 + \cos^2 \theta = 1 \)
\( \frac{4}{9} + \cos^2 \theta = 1 \)
\( \cos^2 \theta = 1 - \frac{4}{9} = \frac{5}{9} \)

Since angle \( ABC \) is obtuse (\( 90^\circ < \theta < 180^\circ \)), the cosine of the angle must be negative.
\( \cos \theta = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3} \)

M1 for substituting given values into the area formula: \( 10 = \frac{1}{2} \times 6 \times 5 \times \sin\theta \)
A1 for finding \( \sin\theta = \frac{2}{3} \) (or equivalent)
M1 for using the identity \( \sin^2\theta + \cos^2\theta = 1 \) to find \( \cos^2\theta = \frac{5}{9} \)
A1 for the final answer \( -\frac{\sqrt{5}}{3} \) (negative sign must be included)

First, find the volume of the cone in terms of \( r \):
\( V_{\text{cone}} = \frac{1}{3} \pi r^2 (12r) = 4\pi r^3 \)

Let this volume equal the volume of the sphere of radius \( R \):
\( \frac{4}{3}\pi R^3 = 4\pi r^3 \)

Divide both sides of the equation by \( 4\pi \):
\( \frac{1}{3} R^3 = r^3 \)

Multiply both sides by 3:
\( R^3 = 3r^3 \)

Take the cube root of both sides to express \( R \) in terms of \( r \):
\( R = \sqrt[3]{3}r \)

We need to find the ratio \( r : R \):
\( r : R = r : \sqrt[3]{3}r \)

Divide both sides of the ratio by \( r \) to express it in the form \( 1 : k \):
\( r : R = 1 : \sqrt[3]{3} \)

Therefore, \( k = \sqrt[3]{3} \).

M1 for expressing the volume of the cone as \( 4\pi r^3 \)
M1 for equating the volume of the cone and the sphere: \( \frac{4}{3}\pi R^3 = 4\pi r^3 \)
M1 for simplifying to find \( R^3 = 3r^3 \) or \( R = \sqrt[3]{3}r \)
A1 for the final ratio \( 1 : \sqrt[3]{3} \) (or equivalent)

(a) To get three spheres of the same color, they must be either all red, all blue, or all green.
Total number of spheres initially is \(5 + 4 + 3 = 12\).
- Probability of 3 Red: \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320}\)
- Probability of 3 Blue: \(\frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320}\)
- Probability of 3 Green: \(\frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320}\)
Sum of probabilities: \(\frac{60 + 24 + 6}{1320} = \frac{90}{1320} = \frac{3}{44}\) (or approximately \(0.0682\)).

(b) "At least two blue" means either exactly 2 blue spheres or exactly 3 blue spheres are drawn.
- Probability of 3 Blue: \(\frac{24}{1320}\)
- Probability of exactly 2 Blue:
There are three ordered outcomes for 2 Blue and 1 Non-Blue (N): B-B-N, B-N-B, N-B-B.
The number of non-blue spheres is \(12 - 4 = 8\).
\(P(\text{B-B-N}) = \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} = \frac{96}{1320}\)
\(P(\text{B-N-B}) = \frac{4}{12} \times \frac{8}{11} \times \frac{3}{10} = \frac{96}{1320}\)
\(P(\text{N-B-B}) = \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{96}{1320}\)
Total for exactly 2 Blue = \(3 \times \frac{96}{1320} = \frac{288}{1320}\).
Total probability for at least 2 Blue = \frac{288 + 24}{1320} = \frac{312}{1320} = \frac{13}{55}\) (or approximately \(0.236\)).

(c) Given that the first sphere is green, we have 11 spheres left: 5 red, 4 blue, and 2 green.
We want to find the probability that the third sphere is green. This can happen in two scenarios:
1. The second sphere is also green, and then the third is green:
\(P(\text{G}_2 \text{ and G}_3 | \text{G}_1) = \frac{2}{11} \times \frac{1}{10} = \frac{2}{110}\)
2. The second sphere is not green, and then the third is green (the number of non-green spheres left is 9):
\(P(\text{N}_2 \text{ and G}_3 | \text{G}_1) = \frac{9}{11} \times \frac{2}{10} = \frac{18}{110}\)
Total conditional probability = \(\frac{2}{110} + \frac{18}{110} = \frac{20}{110} = \frac{2}{11}\) (or approximately \(0.182\)).

(a) [3 marks]:
- M1 for writing down product expressions for at least one color (e.g., \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\)).
- M1 for adding three correct product fractions.
- A1 for the correct answer of \(\frac{3}{44}\) (accept \(0.0682\) or \(0.06818...\)).

(b) [4 marks]:
- M1 for finding the probability of exactly 3 blue spheres (\(\frac{24}{1320}\)).
- M1 for finding the probability of one combination of 2 Blue and 1 Non-Blue (e.g., \(\frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} = \frac{96}{1320}\)).
- M1 for multiplying by 3 and adding the 3-blue case.
- A1 for correct final simplified fraction \(\frac{13}{55}\) (accept \(0.236\) or \(0.2363...\)).

(c) [4 marks]:
- M1 for identifying the reduced pool of 11 spheres (5 red, 4 blue, 2 green).
- M1 for calculating \(P(\text{G, G})\) or \(P(\text{N, G})\) from the remaining pool.
- M1 for sum of both cases (\(\frac{2}{110} + \frac{18}{110}\)).
- A1 for the correct probability of \(\frac{2}{11}\) (accept \(0.182\) or \(0.1818...\)).

(a) Let us determine the interior angle \(\angle ABC\).
The bearing of \(B\) from \(A\) is \(065^\circ\), which means the back bearing of \(A\) from \(B\) is \(65 + 180 = 245^\circ\).
The bearing of \(C\) from \(B\) is \(140^\circ\).
The interior angle \(\angle ABC = 245^\circ - 140^\circ = 105^\circ\).
Applying the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 18^2 + 25^2 - 2(18)(25)\cos(105^\circ)\)
\(AC^2 = 324 + 625 - 900\cos(105^\circ)\)
\(AC^2 = 949 - 900(-0.258819) = 1181.937\)
\(AC = \sqrt{1181.937} \approx 34.379 \approx 34.4\text{ km}\).

(b) To find the bearing of \(C\) from \(A\), we first find the angle \(\angle BAC\) using the Sine Rule:
\(\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC}\)
\(\sin(\angle BAC) = \frac{25 \sin(105^\circ)}{34.379} \approx \frac{25 \times 0.965926}{34.379} \approx 0.7024\)
\(\angle BAC = \sin^{-1}(0.7024) \approx 44.62^\circ\).
Since the bearing of \(B\) from \(A\) is \(065^\circ\), the bearing of \(C\) from \(A\) is:
\(\text{Bearing} = 65^\circ + 44.62^\circ = 109.62^\circ \approx 110^\circ\) (or \(109.6^\circ\)).

(c) The area of the triangle is:
\(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)\)
\(\text{Area} = \frac{1}{2} \times 18 \times 25 \times \sin(105^\circ) = 225 \times 0.965926 \approx 217.33 \approx 217\text{ km}^2\).

(a) [4 marks]:
- M1 for finding the correct interior angle \(\angle ABC = 105^\circ\) (allow working on a sketch showing alternate angles).
- M1 for substitution into the Cosine Rule: \(18^2 + 25^2 - 2(18)(25)\cos(105^\circ)\).
- A1 for \(AC^2 \approx 1182\).
- A1 for \(34.4\) or \(34.38\) (accept range \(34.3\) to \(34.5\)).

(b) [4 marks]:
- M1 for correct setup of the Sine Rule: \(\frac{\sin(\angle BAC)}{25} = \frac{\sin(105^\circ)}{\text{their } AC}\).
- A1 for \(\angle BAC \approx 44.6^\circ\).
- M1 for adding \(65^\circ\) to their angle \(\angle BAC\).
- A1 for \(110^\circ\) or \(109.6^\circ\) (accept range \(109.5^\circ\) to \(110^\circ\)).

(c) [3 marks]:
- M1 for using \(\frac{1}{2} a b \sin C\).
- M1 for substituting values: \(\frac{1}{2} \times 18 \times 25 \times \sin(105^\circ)\).
- A1 for \(217\) (accept range \(217\) to \(218\)).

(a) The volume of the cone is:
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h\)
Since \(h = 3r\), we substitute:
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 (3r) = \pi r^3\)
The volume of the hemisphere is:
\(V_{\text{hemisphere}} = \frac{2}{3} \pi r^3\)
Therefore, the total volume is:
\(V_{\text{total}} = \pi r^3 + \frac{2}{3} \pi r^3 = \frac{5}{3} \pi r^3\)
We are given that the total volume is \(360\pi\text{ cm}^3\):
\(\frac{5}{3} \pi r^3 = 360\pi\)
\(\frac{5}{3} r^3 = 360\)
\(r^3 = 360 \times \frac{3}{5} = 216\)
\(r = \sqrt[3]{216} = 6\text{ cm}\). (Shown)

(b) The total surface area consists of the curved surface area of the cone and the curved surface area of the hemisphere.
- Curved surface area of the hemisphere:
\(A_{\text{hemisphere}} = 2 \pi r^2 = 2 \pi (6^2) = 72\pi\)
- Curved surface area of the cone:
\(A_{\text{cone}} = \pi r l\), where \(l\) is the slant height of the cone.
Using Pythagoras' theorem:
\(l = \sqrt{r^2 + h^2}\)
Since \(r = 6\) and \(h = 3(6) = 18\):
\(l = \sqrt{6^2 + 18^2} = \sqrt{36 + 324} = \sqrt{360} = 6\sqrt{10}\)
Thus,
\(A_{\text{cone}} = \pi \times 6 \times 6\sqrt{10} = 36\sqrt{10}\pi\)
Total surface area:
\(A_{\text{total}} = 72\pi + 36\sqrt{10}\pi = (72 + 36\sqrt{10})\pi\text{ cm}^2\).

(c) Mass is calculated as:
\(\text{Mass} = \text{Density} \times \text{Volume}\)
\(\text{Mass} = 0.8 \times 360\pi\)
\(\text{Mass} = 288\pi \approx 904.78\text{ g}\).
To 3 significant figures, the mass is \(905\text{ g}\).

(a) [3 marks]:
- M1 for writing a correct formula for the total volume: \(\frac{1}{3} \pi r^2 (3r) + \frac{2}{3} \pi r^3\).
- M1 for setting \(\frac{5}{3} \pi r^3 = 360\pi\) and solving for \(r^3\).
- A1 for obtaining \(r^3 = 216\) leading to \(r = 6\).

(b) [5 marks]:
- M1 for formula of hemisphere curved surface: \(2 \pi r^2\) and getting \(72\pi\).
- M1 for finding slant height \(l = \sqrt{6^2 + 18^2}\).
- A1 for simplifying slant height to \(\sqrt{360}\) or \(6\sqrt{10}\).
- M1 for cone curved surface area: \(\pi \times 6 \times 6\sqrt{10}\).
- A1 for the exact answer \((72 + 36\sqrt{10})\pi\).

(c) [3 marks]:
- M1 for using \(\text{Mass} = \text{Density} \times \text{Volume}\).
- M1 for calculating \(0.8 \times 360\pi\).
- A1 for \(905\) (accept range \(904\) to \(905\)).

(a) The depreciation equation after 3 years is:
\(45000 \times \left(1 - \frac{r}{100}\right)^3 = 32805\)
\(\left(1 - \frac{r}{100}\right)^3 = \frac{32805}{45000} = 0.729\)
Taking the cube root of both sides:
\(1 - \frac{r}{100} = \sqrt[3]{0.729} = 0.9\)
\(\frac{r}{100} = 0.1 \implies r = 10\).

(b) (i) After the first 3 years, the value is \(\$32,805\).
With an appreciation of \(4.5\%\) per year for 5 years:
\(V = 32805 \times (1 + 0.045)^5\)
\(V = 32805 \times 1.045^5 \approx 32805 \times 1.246182 \approx 40881.04\)
To the nearest dollar, the value is \(\$40,881\).

(ii) Let \(n\) be the number of appreciation years after the first 3 years.
We want to find when the value first exceeds \(\$45,000\):
\(32805 \times (1.045)^n > 45000\)
\(1.045^n > \frac{45000}{32805} \approx 1.37174\)
Taking natural logarithms on both sides:
\(n \ln(1.045) > \ln(1.37174)\)
\(n \times 0.044017 > 0.316082\)
\(n > \frac{0.316082}{0.044017} \approx 7.18\)
Since \(n\) must be an integer (as compounding occurs annually), \(n = 8\) years.
Total years from purchase = \(3\text{ (depreciation years)} + 8\text{ (appreciation years)} = 11\text{ years}\).

(a) [4 marks]:
- M1 for setting up the equation: \(45000(1 - \frac{r}{100})^3 = 32805\).
- M1 for isolating the cubed term: \((1 - \frac{r}{100})^3 = 0.729\).
- M1 for taking the cube root: \(1 - \frac{r}{100} = 0.9\).
- A1 for \(r = 10\).

(b) (i) [3 marks]:
- M1 for using the compound formula: \(32805 \times (1.045)^5\).
- A1 for \(40881.04\).
- A1 for rounding to the nearest dollar: \(40881\).

(b) (ii) [4 marks]:
- M1 for setting up the inequality \(32805(1.045)^n > 45000\).
- M1 for solving using logarithms or trial and improvement (showing work for \(n=7\) and \(n=8\)).
- A1 for finding the number of appreciation years, \(n = 8\).
- A1 for the total years of \(11\).

(a) To find the \(y\)-intercept, let \(x = 0\):
\(f(0) = \frac{0^2 - 2(0) + 4}{0 - 2} = \frac{4}{-2} = -2\).
So the graph crosses the \(y\)-axis at \((0, -2)\).
To find the \(x\)-intercepts, let \(f(x) = 0\):
\(x^2 - 2x + 4 = 0\)
Discriminant: \(D = (-2)^2 - 4(1)(4) = 4 - 16 = -12 < 0\).
Since the discriminant is negative, there are no real roots, so there are no \(x\)-intercepts.

(b) The vertical asymptote occurs where the denominator is zero and the numerator is non-zero.
Denominator \(x - 2 = 0 \implies x = 2\).

(c) We perform algebraic division or rewrite the numerator:
\(f(x) = \frac{x(x - 2) + 4}{x - 2} = x + \frac{4}{x - 2}\).
As \(x \to \pm\infty\), the term \(\frac{4}{x - 2} \to 0\).
Therefore, the equation of the oblique asymptote is \(y = x\).

(d) To find stationary points, we find \(f'(x)\):
\(f(x) = x + 4(x-2)^{-1}\)
\(f'(x) = 1 - 4(x-2)^{-2} = 1 - \frac{4}{(x-2)^2}\)
Setting \(f'(x) = 0\):
\(1 - \frac{4}{(x-2)^2} = 0 \implies (x-2)^2 = 4\)
\(x - 2 = 2 \implies x = 4\)
\(x - 2 = -2 \implies x = 0\)
Evaluating \(y\)-values:
- For \(x = 4\): \(f(4) = \frac{4^2 - 2(4) + 4}{4 - 2} = \frac{12}{2} = 6\). This gives \((4, 6)\).
- For \(x = 0\): \(f(0) = -2\). This gives \((0, -2)\).
By checking the sign of the derivative or from the shape of the rational function:
\((4, 6)\) is a local minimum, and \((0, -2)\) is a local maximum.

(a) [3 marks]:
- M1 for setting \(x = 0\) to find \(y = -2\).
- M1 for setting numerator \(x^2 - 2x + 4 = 0\) to search for \(x\)-intercepts.
- A1 for correctly stating \((0, -2)\) as the only intercept and showing there are no \(x\)-intercepts.

(b) [1 mark]:
- A1 for \(x = 2\).

(c) [3 marks]:
- M1 for attempt to divide numerator by denominator.
- A1 for obtaining \(f(x) = x + \frac{4}{x-2}\).
- A1 for \(y = x\).

(d) [4 marks]:
- M1 for differentiating \(f(x)\) correctly (using quotient rule or index notation).
- M1 for setting \(f'(x) = 0\) and solving to get \(x = 0\) and \(x = 4\).
- A1 for both coordinate points: \((4, 6)\) and \((0, -2)\).
- A1 for correctly identifying \((4, 6)\) as the local minimum and \((0, -2)\) as the local maximum.

(a) Find a common denominator, which is \((2x - 1)(x + 3)\):
\(\frac{3}{2x - 1} - \frac{2}{x + 3} = \frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\)
Simplify the numerator:
\(3(x + 3) - 2(2x - 1) = 3x + 9 - 4x + 2 = 11 - x\)
Expand the denominator:
\((2x - 1)(x + 3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3\)
So, the simplified fraction is:
\(\frac{11 - x}{2x^2 + 5x - 3}\).

(b) Using the result from part (a):
\(\frac{11 - x}{2x^2 + 5x - 3} = \frac{1}{2}\)
Cross-multiply to eliminate the fractions:
\(2(11 - x) = 2x^2 + 5x - 3\)
\(22 - 2x = 2x^2 + 5x - 3\)
Rearrange into standard quadratic form:
\(2x^2 + 7x - 25 = 0\)
Solve using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-25)}}{2(2)}\)
\(x = \frac{-7 \pm \sqrt{49 + 200}}{4}\)
\(x = \frac{-7 \pm \sqrt{249}}{4}\)
Calculating the values:
- \(x_1 = \frac{-7 + 15.7797}{4} \approx 2.19\) (to 3 s.f.)
- \(x_2 = \frac{-7 - 15.7797}{4} \approx -5.69\) (to 3 s.f.)

(a) [4 marks]:
- M1 for establishing a common denominator \((2x - 1)(x + 3)\).
- M1 for correct algebraic expansion of numerator: \(3(x + 3) - 2(2x - 1)\).
- A1 for numerator simplified to \(11 - x\).
- A1 for correct denominator \(2x^2 + 5x - 3\) (or written in factored form).

(b) [7 marks]:
- M1 for setting their fraction from (a) equal to \(\frac{1}{2}\).
- M1 for cross-multiplying correctly: \(2(11 - x) = 2x^2 + 5x - 3\).
- A1 for obtaining the correct quadratic equation: \(2x^2 + 7x - 25 = 0\).
- M1 for substitution into the quadratic formula (or GDC equivalent): \(x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-25)}}{4}\).
- A1 for discriminant evaluation: \(\sqrt{249}\).
- A1 for \(x \approx 2.19\) (accept range \(2.19\) to \(2.20\)).
- A1 for \(x \approx -5.69\) (accept range \(-5.70\) to \(-5.69\)).

(a) Use laws of logarithms to combine the terms on the left side:
\(\log_3(x) + \log_3(x-2)^2 = 1 + \log_3(15)\)
\(\log_3(x(x-2)^2) = \log_3(3) + \log_3(15)\) (since \(1 = \log_3(3)\))
\(\log_3(x(x-2)^2) = \log_3(3 \times 15)\)
\(\log_3(x(x-2)^2) = \log_3(45)\)
By equating the arguments of the logs:
\(x(x-2)^2 = 45\)
\(x(x^2 - 4x + 4) = 45\)
\(x^3 - 4x^2 + 4x - 45 = 0\)
Testing integer factors of 45, we find \(x = 5\) is a root:
\(5^3 - 4(5^2) + 4(5) - 45 = 125 - 100 + 20 - 45 = 0\).
Factor out \((x-5)\):
\((x-5)(x^2 + x + 9) = 0\)
The quadratic factor \(x^2 + x + 9 = 0\) has no real roots since its discriminant is \(1^2 - 4(1)(9) = -35 < 0\).
We must also check the domain of the logarithm:
- \(x > 0\)
- \(x - 2 > 0 \implies x > 2\)
Since \(x = 5\) satisfies both conditions, the unique solution is \(x = 5\).

(b) From the first equation:
\(\log_2\left(\frac{y}{x}\right) = 3 \implies \frac{y}{x} = 2^3 = 8 \implies y = 8x\)
From the second equation:
\(3^y = 9^{x+1} \implies 3^y = (3^2)^{x+1} \implies 3^y = 3^{2x+2}\)
This gives:
\(y = 2x + 2\)
Substitute \(y = 8x\) into this equation:
\(8x = 2x + 2 \implies 6x = 2 \implies x = \frac{1}{3}\)
Now find \(y\):
\(y = 8 \left(\frac{1}{3}\right) = \frac{8}{3}\).

(a) [6 marks]:
- M1 for writing \(2\log_3(x-2)\) as \(\log_3(x-2)^2\).
- M1 for combining LHS: \(\log_3(x(x-2)^2)\).
- M1 for expressing \(1\) as \(\log_3(3)\) and combining RHS to \(\log_3(45)\).
- M1 for removing logarithms to get \(x(x-2)^2 = 45\).
- A1 for expanding and solving the cubic to obtain \(x = 5\).
- A1 for verifying the domain restrictions (explaining why other potential roots are invalid or showing \(x^2+x+9=0\) has no real solutions).

(b) [5 marks]:
- M1 for simplifying the log equation: \(\log_2(\frac{y}{x}) = 3\).
- A1 for \(y = 8x\).
- M1 for simplifying the exponential equation: \(y = 2x + 2\).
- M1 for substituting one equation into the other and solving for \(x\).
- A1 for both correct values: \(x = \frac{1}{3}\) and \(y = \frac{8}{3}\).

(a) First find the gradient of the line \(L_1\) passing through \(A(-2, 5)\) and \(B(4, -3)\):
\(m_1 = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3}\)
Now use the point-slope form with point \(A(-2, 5)\):
\(y - 5 = -\frac{4}{3}(x + 2)\)
\(y = -\frac{4}{3}x - \frac{8}{3} + 5\)
\(y = -\frac{4}{3}x + \frac{7}{3}\) (or \(4x + 3y = 7\)).

(b) To find the perpendicular bisector \(L_2\):
- Find the midpoint of \(AB\):
\(M = \left(\frac{-2 + 4}{2}, \frac{5 - 3}{2}\right) = (1, 1)\)
- Find the perpendicular gradient:
\(m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}\)
- Find the equation of \(L_2\) passing through \(M(1, 1)\):
\(y - 1 = \frac{3}{4}(x - 1)\)
\(y = \frac{3}{4}x - \frac{3}{4} + 1\)
\(y = \frac{3}{4}x + \frac{1}{4}\) (or \(3x - 4y = -1\)).

(c) To find the intersection of \(L_2\) and \(L_3\):
\(y = \frac{3}{4}x + \frac{1}{4}\)
\(y = 2x - 7\)
Equating the expressions for \(y\):
\(2x - 7 = \frac{3}{4}x + \frac{1}{4}\)
Multiply all terms by 4 to clear the fraction:
\(8x - 28 = 3x + 1\)
\(5x = 29\)
\(x = 5.8\) (or \(\frac{29}{5}\))
Now find the \(y\)-value:
\(y = 2(5.8) - 7 = 11.6 - 7 = 4.6\) (or \(\frac{23}{5}\))
So the intersection is \((5.8, 4.6)\).

(a) [3 marks]:
- M1 for finding the gradient of \(AB\): \(-\frac{4}{3}\).
- M1 for substituting their gradient and one point into a linear equation form.
- A1 for \(y = -\frac{4}{3}x + \frac{7}{3}\) (or equivalent form, e.g., \(4x + 3y - 7 = 0\)).

(b) [5 marks]:
- M1 for finding the midpoint \(M(1, 1)\).
- M1 for obtaining the perpendicular gradient \(m_2 = \frac{3}{4}\) (negative reciprocal of their \(m_1\)).
- M1 for substituting their midpoint and perpendicular gradient into the line equation.
- A1 for \(y = \frac{3}{4}x + \frac{1}{4}\) (or any correct simplified equivalent form, e.g., \(3x - 4y = -1\)).

(c) [3 marks]:
- M1 for setting their \(L_2\) equal to \(2x - 7\).
- A1 for finding \(x = 5.8\) (or \(\frac{29}{5}\)).
- A1 for finding \(y = 4.6\) (or \(\frac{23}{5}\)).

(a)(i) First red is \(\frac{5}{n+5}\), second red is \(\frac{4}{n+4}\). Probability both red is \(\frac{5}{n+5} \times \frac{4}{n+4} = \frac{20}{(n+5)(n+4)}\).

(ii) Probability of (Red, Blue) is \(\frac{5}{n+5} \times \frac{n}{n+4} = \frac{5n}{(n+5)(n+4)}\). Probability of (Blue, Red) is \(\frac{n}{n+5} \times \frac{5}{n+4} = \frac{5n}{(n+5)(n+4)}\). Total probability of one of each colour is \(\frac{5n}{(n+5)(n+4)} + \frac{5n}{(n+5)(n+4)} = \frac{10n}{(n+5)(n+4)}\).

(b)(i) We are given \(\frac{20}{(n+5)(n+4)} = \frac{5}{33}\).
Multiply by 33 and \((n+5)(n+4)\):
\(20 \times 33 = 5(n+5)(n+4)\)
\(660 = 5(n^2 + 9n + 20)\)
Divide by 5:
\(132 = n^2 + 9n + 20\)
\(n^2 + 9n - 112 = 0\) (as required).

(ii) Solve \(n^2 + 9n - 112 = 0\):
\((n + 16)(n - 7) = 0\)
\(n = -16\) or \(n = 7\).
Since the number of marbles must be positive, \(n = 7\).

(a)(i) [2 marks]
M1 for product of two probabilities with denominator decreasing by 1
A1 for correct simplified expression \(\frac{20}{(n+5)(n+4)}\) or \(\frac{20}{n^2+9n+20}\)

(a)(ii) [3 marks]
M1 for \(\frac{5}{n+5} \times \frac{n}{n+4}\)
M1 for multiplying by 2 (for two orders)
A1 for correct simplified expression \(\frac{10n}{(n+5)(n+4)}\) or \(\frac{10n}{n^2+9n+20}\)

(b)(i) [3 marks]
M1 for setting their expression (a)(i) equal to \(\frac{5}{33}\)
M1 for expanding \((n+5)(n+4)\) to \(n^2+9n+20\) and multiplying by 33
A1 for fully correct working showing clearly the steps to reach \(n^2 + 9n - 112 = 0\)

(b)(ii) [3 marks]
M1 for factorising \((n+16)(n-7)\) or correct use of quadratic formula
A1 for \(n = 7\) and \(n = -16\)
A1 for rejecting \(-16\) and giving final answer \(n = 7\)

(a) From 2010 to 2018 is 8 years.
Value = \(15000 \times (1.08)^8 = 27763.953...\)
To the nearest dollar, the value is $27,764.

(b) We want \(15000 \times (1.08)^t > 50000\).
\(1.08^t > \frac{50000}{15000}\)
\(1.08^t > 3.3333...\)
\(t \log(1.08) > \log(3.3333...)\)
\(t > \frac{\log(3.3333...)}{\log(1.08)} \approx 15.64\)
Since we need the number of complete years, \(t = 16\).
(Check: for \(t=15\), value is \(15000 \times 1.08^{15} = 47582.54\). For \(t=16\), value is \(15000 \times 1.08^{16} = 51389.14\).)

(c) Value after 10 years: \(20000 \times (1 - \frac{r}{100})^{10} = 12000\)
\((1 - \frac{r}{100})^{10} = 0.6\)
\(1 - \frac{r}{100} = 0.6^{0.1} \approx 0.95021\)
\(\frac{r}{100} = 1 - 0.95021 = 0.04979\)
\(r \approx 4.98\) (to 2 d.p.)

(a) [3 marks]
M1 for \(15000 \times (1.08)^8\)
A1 for 27763.95...
A1 for 27764 (nearest dollar)

(b) [4 marks]
M1 for setting up inequality or equation: \(15000 \times (1.08)^t = 50000\)
M1 for use of logs or trial and improvement showing values around \(t = 15\) and \(t = 16\)
A1 for \(15.6...\) or \(15.64\)
A1 for 16 (complete years)

(c) [4 marks]
M1 for setting up equation: \(20000 \times (1 - r/100)^{10} = 12000\) or \(20000 \times x^{10} = 12000\)
M1 for finding \(1 - r/100 = 0.6^{0.1}\) or \(x = 0.9502...\)
M1 for \(r/100 = 0.04979...\)
A1 for \(4.98\)

(a) Draw a North line at \(B\).
The line \(AB\) goes in bearing \(060^\circ\). Thus, the angle of the line \(BA\) to the South at \(B\) is \(60^\circ\) (alternate angles).
Therefore, the bearing of \(A\) from \(B\) is \(060^\circ + 180^\circ = 240^\circ\).
The bearing of \(C\) from \(B\) is \(130^\circ\).
The angle \(ABC\) is the difference between these two directions:
\(ABC = 240^\circ - 130^\circ = 110^\circ\).

(b) Using the Cosine Rule on triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\)
\(AC^2 = 12^2 + 15^2 - 2(12)(15)\cos(110^\circ)\)
\(AC^2 = 144 + 225 - 360(-0.34202)\)
\(AC^2 = 369 + 123.127 = 492.127\)
\(AC = \sqrt{492.127} \approx 22.184\text{ km}\)
To 3 significant figures, \(AC = 22.2\text{ km}\).

(c) Using the Sine Rule to find angle \(BAC\):
\(\frac{\sin(BAC)}{15} = \frac{\sin(110^\circ)}{22.184}\)
\(\sin(BAC) = \frac{15 \cdot \sin(110^\circ)}{22.184} \approx 0.63539\)
\(BAC = \arcsin(0.63539) \approx 39.45^\circ\).
The bearing of \(C\) from \(A\) is the bearing of \(B\) from \(A\) plus angle \(BAC\):
\(\text{Bearing} = 060^\circ + 39.45^\circ = 099.45^\circ \approx 099.5^\circ\).

(d) The distance \(AC\) in metres is \(22.184 \times 1000 = 22184\text{ m}\).
Using right-angled trigonometry on the vertical triangle above \(AC\):
\(\tan(15^\circ) = \frac{\text{height}}{AC}\)
\(\text{height} = 22184 \times \tan(15^\circ) \approx 5944.2\text{ m}\).
To 3 significant figures, the height of the helicopter is \(5940\text{ m}\).

(a) [2 marks]
M1 for finding the direction of \(BA\) as \(240^\circ\) or demonstrating alternate interior angle is \(60^\circ\)
A1 for complete correct reasoning leading to \(240^\circ - 130^\circ = 110^\circ\)

(b) [3 marks]
M1 for \(12^2 + 15^2 - 2(12)(15)\cos(110^\circ)\)
A1 for \(492.1...\)
A1 for \(22.2\) or \(22.18...\) (accept 22 km if correct working shown)

(c) [3 marks]
M1 for \(\frac{\sin(BAC)}{15} = \frac{\sin(110)}{22.2}\) (or using cosine rule to find angle)
A1 for angle \(BAC = 39.4^\circ\) to \(39.5^\circ\)
A1 for bearing \(099.4^\circ\) to \(099.5^\circ\) (accept \(99.4^\circ\) or \(99.5^\circ\))

(d) [3 marks]
M1 for converting their \(AC\) to metres (multiply by 1000)
M1 for \(AC \times \tan(15^\circ)\)
A1 for \(5940\) (allow answers in range \(5930\) to \(5950\) depending on rounding of \(AC\))

(a) Let the sequence be 1, 13, 37. The first differences are 12 and 24. Continuing this arithmetic sequence of differences, the next difference is 36. Thus, Pattern 4 has \(37 + 36 = 73\) dots. (b) Since the second difference is constant and equals 12, \(2a = 12 \implies a = 6\). Using \(n=1\), \(a+b+c=1 \implies 6+b+c=1 \implies b+c=-5\). Using \(n=2\), \(4(6)+2b+c=13 \implies 2b+c=-11\). Solving these gives \(b = -6\) and \(c = 1\). (c) Solve \(6n^2 - 6n + 1 = 1261 \implies 6n^2 - 6n - 1260 = 0 \implies n^2 - n - 210 = 0\). Factoring gives \((n - 15)(n + 14) = 0\). Since \(n > 0\), we have \(n = 15\).

Part (a) [2 marks]: 1 mark for finding the next difference of 36, 1 mark for obtaining 73. Part (b) [3.5 marks]: 1 mark for establishing a system of equations or finding second differences, 1 mark for a = 6, 1 mark for b = -6, 0.5 marks for c = 1. Part (c) [2 marks]: 1 mark for setting up the quadratic equation, 1 mark for solving to find the positive integer value n = 15.

(a) Since 3 and 5 share no common factors, \(N = 3 + 5 - 1 = 7\). (b) Since 11 and 17 share no common factors, \(N = 11 + 17 - 1 = 27\). (c) Finding the factors of 36 and 48 shows their greatest common divisor is \(g = 12\). Applying the given formula: \(N = 36 + 48 - 12 = 72\).

Part (a) [2 marks]: 1 mark for attempting a drawing or applying the formula, 1 mark for finding N = 7. Part (b) [2 marks]: 1 mark for identifying coprime state and substituting, 1 mark for N = 27. Part (c) [3.5 marks]: 1.5 marks for identifying the greatest common divisor g = 12, 1 mark for substituting into the formula, 1 mark for finding N = 72.

(a) For \(t=0\), \(P(0) = \frac{5000}{1 + 9e^0} = \frac{5000}{10} = 500\). (b) For \(t=10\), \(P(10) = \frac{5000}{1 + 9e^{-2}} \approx \frac{5000}{1 + 1.218018} \approx 2254.27\), which rounds to 2254. (c) Set \(P(t) = 4000 \implies 1 + 9e^{-0.2t} = 1.25 \implies 9e^{-0.2t} = 0.25 \implies e^{-0.2t} = \frac{1}{36}\). Taking natural logarithms on both sides: \(-0.2t = -\ln(36) \implies t = 5\ln(36) \approx 17.9\) months. (d) As \(t \to \infty\), \(e^{-0.2t} \to 0\), and therefore \(P(t) \to \frac{5000}{1 + 0} = 5000\).

Part (a) [1.5 marks]: 1 mark for setting t = 0, 0.5 marks for finding 500. Part (b) [2 marks]: 1 mark for substituting t = 10 and evaluating the exponential expression, 1 mark for rounding to 2254. Part (c) [2.5 marks]: 1 mark for algebraic manipulation to isolate the exponential term, 1 mark for applying logarithms, 0.5 marks for the answer 17.9. Part (d) [1.5 marks]: 1 mark for evaluating the limit as t approaches infinity, 0.5 marks for the limiting population of 5000.

(a) Let the point directly opposite \(A\) be \(A'\). The underwater path \(AP\) is the hypotenuse of the right-angled triangle \(AA'P\) with sides 100 and \(x\). By Pythagoras' theorem, \(AP = \sqrt{x^2 + 100^2} = \sqrt{x^2 + 10000}\). The land path \(PB = 400 - x\). Combining the costs: \(C(x) = 150\sqrt{x^2 + 10000} + 90(400 - x) = 150\sqrt{x^2 + 10000} + 36000 - 90x\). (b) If \(x = 400\), then \(C(400) = 150\sqrt{400^2 + 10000} + 36000 - 90(400) = 150\sqrt{170000} \approx 61846.58\). The cost is $61,847. (c) If \(x = 0\), then \(C(0) = 150\sqrt{10000} + 36000 - 0 = 150(100) + 36000 = 15000 + 36000 = 51000\). The cost is $51,000. (d) To find the minimum, set the derivative \(C'(x) = 0\): \(150 \times \frac{x}{\sqrt{x^2+10000}} - 90 = 0 \implies 5x = 3\sqrt{x^2+10000} \implies 25x^2 = 9x^2 + 90000 \implies 16x^2 = 90000 \implies x^2 = 5625 \implies x = 75\). The minimum cost is \(C(75) = 150\sqrt{75^2 + 10000} + 36000 - 90(75) = 150(125) + 36000 - 6750 = 48000\). the optimal distance is 75 meters with a minimum cost of $48,000.

Part (a) [2 marks]: 1 mark for expressing the hypotenuse using Pythagoras' theorem, 1 mark for summing the two costs and establishing the final expression. Part (b) [1.5 marks]: 1 mark for substituting x = 400 into the cost function, 0.5 marks for finding $61,847. Part (c) [1.5 marks]: 1 mark for substituting x = 0 or using direct logic, 0.5 marks for finding $51,000. Part (d) [2.5 marks]: 1.5 marks for finding x = 75 (either by differentiation or using a graphic display calculator), 1 mark for calculating the minimum cost of $48,000.

(a) At \(t = 0\), the number of users is \(N(0) = 500\). Substitute \(t = 0\) and \(N = 500\) into the formula: \( 500 = \frac{20000}{1 + B e^{0}} \implies 500 = \frac{20000}{1 + B} \implies 1 + B = \frac{20000}{500} = 40 \implies B = 39\). (b) At \(t = 3\), \(N(3) = 2500\). Substitute \(B=39\): \( 2500 = \frac{20000}{1 + 39 e^{-3k}} \implies 1 + 39 e^{-3k} = \frac{20000}{2500} = 8 \implies 39 e^{-3k} = 7 \implies e^{-3k} = \frac{7}{39} \). Taking the natural logarithm on both sides: \( -3k = \ln\left(\frac{7}{39}\right) \implies k = -\frac{1}{3}\ln\left(\frac{7}{39}\right) = \frac{1}{3}\ln\left(\frac{39}{7}\right)\). Evaluating this expression: \( k \approx \frac{1}{3}(1.71765) \approx 0.57255 \approx 0.573\). (c) We want to find \(t\) when \(N(t) = 15000\): \( 15000 = \frac{20000}{1 + 39 e^{-kt}} \implies 1 + 39 e^{-kt} = \frac{20000}{15000} = \frac{4}{3} \implies 39 e^{-kt} = \frac{4}{3} - 1 = \frac{1}{3} \implies e^{-kt} = \frac{1}{117} \implies -kt = \ln\left(\frac{1}{117}\right) = -\ln(117) \implies t = \frac{\ln(117)}{k}\). Using \(k \approx 0.57255\): \( t = \frac{4.76217}{0.57255} \approx 8.317\) months. To the nearest integer, \(t \approx 8\) months.

(a) [2 Marks] M1 for substituting \(t = 0\) and \(N = 500\) into the formula. A1 for showing clearly that \(B = 39\). (b) [4 Marks] M1 for substituting \(N = 2500\), \(t = 3\), and \(B = 39\). M1 for simplifying to \(e^{-3k} = \frac{7}{39}\). A1 for the exact value of \(k = \frac{1}{3}\ln\left(\frac{39}{7}\right)\). A1 for correct calculation to 3 decimal places (0.573). (c) [4 Marks] M1 for setting up the equation \(15000 = \frac{20000}{1 + 39 e^{-kt}}\). M1 for rearranging to get \(e^{-kt} = \frac{1}{117}\). M1 for calculating \(t = \frac{\ln(117)}{k}\) with their value of \(k\). A1 for 8 months.

(a) The volume of a cylinder is given by: \( V = \pi r^2 h = 355 \implies h = \frac{355}{\pi r^2} \). The total surface area of a closed cylinder is: \( A = 2\pi r^2 + 2\pi r h \). Substitute \(h\): \( A(r) = 2\pi r^2 + 2\pi r \left(\frac{355}{\pi r^2}\right) = 2\pi r^2 + \frac{710}{r} \). (b) Differentiating \(A(r)\) with respect to \(r\): \( \frac{\text{d}A}{\text{d}r} = 4\pi r - 710 r^{-2} = 4\pi r - \frac{710}{r^2} \). (c) Set \(\frac{\text{d}A}{\text{d}r} = 0\): \( 4\pi r - \frac{710}{r^2} = 0 \implies 4\pi r^3 = 710 \implies r^3 = \frac{355}{2\pi} \implies r = \left(\frac{355}{2\pi}\right)^{1/3} \approx 3.837 \text{ cm}\). To 2 decimal places, \(r = 3.84\text{ cm}\). (d) The height is \(h = \frac{355}{\pi r^2}\). Using \(r \approx 3.837\) (or \(h = 2r\)): \( h \approx 7.674 \text{ cm}\). To 2 decimal places, \(h = 7.67\text{ cm}\).

(a) [3 Marks] M1 for using the volume formula \(V = \pi r^2 h = 355\) to write \(h\) in terms of \(r\). M1 for writing the surface area formula. A1 for substituting and simplifying to obtain the given expression. (b) [2 Marks] M1 for differentiating at least one term correctly. A1 for the correct derivative. (c) [3 Marks] M1 for setting \(\frac{\text{d}A}{\text{d}r} = 0\). M1 for isolating \(r^3\). A1 for \(r \approx 3.84\text{ cm}\). (d) [2 Marks] M1 for substituting their value of \(r\) back to find \(h\) (or using \(h = 2r\)). A1 for \(h \approx 7.67\text{ cm}\).

(a) The differences between successive terms are: \(6-1=5\), \(16-6=10\), \(31-16=15\). The differences increase by 5 each time. Thus, the next differences are 20 and 25: \(P_5 = 31 + 20 = 51\), \(P_6 = 51 + 25 = 76\). (b) First differences: 5, 10, 15, 20, 25. Second differences: 5, 5, 5, 5. Since the second differences are constant, the sequence can be modelled by a quadratic expression of degree 2. (c) Let \(P_n = An^2 + Bn + C\). The constant second difference is \(2A = 5 \implies A = 2.5\). Using \(P_1 = 1 \implies 2.5 + B + C = 1 \implies B + C = -1.5\). Using \(P_2 = 6 \implies 10 + 2B + C = 6 \implies 2B + C = -4\). Solving these simultaneous equations gives \(B = -2.5\) and \(C = 1\). Therefore, \(P_n = 2.5n^2 - 2.5n + 1\). (d) For \(n = 50\): \(P_{50} = 2.5(50)^2 - 2.5(50) + 1 = 6250 - 125 + 1 = 6126\).

(a) [2 Marks] A1 for \(P_5 = 51\). A1 for \(P_6 = 76\). (b) [2 Marks] M1 for finding the second differences (all 5). A1 for explaining that a constant second difference indicates a quadratic formula of degree 2. (c) [4 Marks] M1 for establishing \(2A = 5 \implies A = 2.5\). M1 for setting up equations for \(B\) and \(C\). M1 for solving to get \(B = -2.5\) and \(C = 1\). A1 for \(P_n = 2.5n^2 - 2.5n + 1\) (or equivalent). (d) [2 Marks] M1 for substituting \(n = 50\) into their formula. A1 for 6126.