2025 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

To find \( f^{-1}(3) \), we set \( f(x) = 3 \) and solve for \( x \):

\( \frac{4x - 3}{x + 2} = 3 \)

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

\( 4x - 3 = 3x + 6 \)

\( x = 9 \)

Therefore, \( f^{-1}(3) = 9 \).

M1 for setting \( \frac{4x - 3}{x + 2} = 3 \) (or attempting to find general \( f^{-1}(x) \))
M1 for clearing the fraction to obtain \( 4x - 3 = 3x + 6 \) (or \( f^{-1}(x) = \frac{2x + 3}{4 - x} \))
A1 for 9

Let the \(n\)-th term be \( u_n \).

The terms are: \( 3, \ 10, \ 21, \ 36, \ 55 \)
First differences: \( 7, \ 11, \ 15, \ 19 \)
Second differences: \( 4, \ 4, \ 4 \)

Since the second differences are constant and equal to \( 4 \), the sequence is quadratic with leading term \( an^2 \), where \( 2a = 4 \implies a = 2 \).

Subtracting \( 2n^2 \) from each term:
- For \( n = 1 \): \( 3 - 2(1)^2 = 1 \)
- For \( n = 2 \): \( 10 - 2(2)^2 = 2 \)
- For \( n = 3 \): \( 21 - 2(3)^2 = 3 \)

This gives the linear sequence \( 1, \ 2, \ 3, \ \dots \), which is simply \( n \).

Combining these parts, the \(n\)-th term is \( 2n^2 + n \).

M1 for finding first differences (7, 11, 15, 19) and second differences (4)
M1 for identifying the coefficient of \( n^2 \) as 2
A1 for \( 2n^2 + n \) (or equivalent standard form)

Group the terms in pairs:
\( 8x^3 - 20x^2 - 18x + 45 = 4x^2(2x - 5) - 9(2x - 5) \)

Factor out the common bracket \( (2x - 5) \):
\( = (4x^2 - 9)(2x - 5) \)

Factorise the difference of two squares \( 4x^2 - 9 \):
\( = (2x - 3)(2x + 3)(2x - 5) \).

M1 for grouping terms to obtain \( 4x^2(2x - 5) - 9(2x - 5) \) or equivalent
M1 for factorising into two brackets: \( (4x^2 - 9)(2x - 5) \)
A1 for complete factorisation: \( (2x - 3)(2x + 3)(2x - 5) \) (any order of brackets)

Using the Cosine Rule:
\( AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC) \)

Substitute the given values:
\( AC^2 = 3^2 + 5^2 - 2(3)(5)\cos(120^\circ) \)

Since \( \cos(120^\circ) = -\frac{1}{2} \):
\( AC^2 = 9 + 25 - 30\left(-\frac{1}{2}\right) \)
\( AC^2 = 34 + 15 = 49 \)

\( AC = \sqrt{49} = 7 \text{ cm} \).

M1 for correct substitution into the Cosine Rule: \( 3^2 + 5^2 - 2(3)(5)\cos(120^\circ) \)
M1 for substituting \( \cos(120^\circ) = -\frac{1}{2} \) and simplifying to \( AC^2 = 49 \)
A1 for 7

At the time of purchase, \( t = 0 \):
\( V = A \times r^0 \implies A = 200 \)

After 2 years, \( t = 2 \):
\( 242 = 200 \times r^2 \)
\( r^2 = \frac{242}{200} = 1.21 \)

Since \( r > 0 \), we take the positive square root:
\( r = \sqrt{1.21} = 1.1 \)

Now, find the value after 3 years (\( t = 3 \)):
\( V = 200 \times (1.1)^3 \)
\( V = 200 \times 1.331 = 266.2 \)

The value of the coin is \( \$266.20 \).

M1 for identifying \( A = 200 \) and setting up \( 200 \times r^2 = 242 \)
M1 for finding \( r = 1.1 \)
A1 for 266.2 or 266.20

We can complete the square for \( y = 2x^2 - 12x + 11 \):

\( y = 2(x^2 - 6x) + 11 \)

\( y = 2\left[(x - 3)^2 - 3^2\right] + 11 \)

\( y = 2(x - 3)^2 - 18 + 11 \)

\( y = 2(x - 3)^2 - 7 \)

This is in the vertex form \( y = a(x - h)^2 + k \), where the vertex (minimum point) is at \( (h, k) \).
Therefore, the coordinates of the minimum point are \( (3, -7) \).

M1 for attempting to complete the square, e.g., \( 2(x - 3)^2 + c \), or finding the x-coordinate using \( x = -\frac{b}{2a} = 3 \)
M1 for substituting \( x = 3 \) back into the equation to find the y-coordinate
A1 for \( (3, -7) \)

First, factorise all numerators and denominators completely:
1. \( 2x^2 - 5x - 3 = (2x + 1)(x - 3) \)
2. \( x^2 - 9 = (x - 3)(x + 3) \)
3. \( x^2 - 3x = x(x - 3) \)
4. \( 2x^2 + x = x(2x + 1) \)

Substitute these factored forms back into the expression:
\( \frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} \times \frac{x(x - 3)}{x(2x + 1)} \)

Cancel out common terms from the numerators and denominators:
- Cancel \( (2x + 1) \)
- Cancel \( x \)
- Cancel \( (x - 3) \)

This leaves:
\( \frac{x - 3}{x + 3} \).

M1 for factorising \( 2x^2 - 5x - 3 \) to \( (2x+1)(x-3) \) or \( x^2-9 \) to \( (x-3)(x+3) \)
M1 for factorising both \( x^2 - 3x \) to \( x(x-3) \) and \( 2x^2+x \) to \( x(2x+1) \)
A1 for \( \frac{x-3}{x+3} \) or equivalent fraction

We apply the transformations to the coordinates of the maximum point \( (4, -2) \):

1. The transformation \( f(x - 1) \) translates the graph 1 unit horizontally to the right.
This changes the \(x\)-coordinate: \( x_{\text{new}} = 4 + 1 = 5 \).

2. The transformation \( 3f(x - 1) \) stretches the graph vertically by a factor of 3.
This multiplies the \(y\)-coordinate by 3: \( -2 \times 3 = -6 \).

3. The transformation \( 3f(x - 1) + 5 \) translates the graph 5 units vertically upwards.
This adds 5 to the \(y\)-coordinate: \( -6 + 5 = -1 \).

Thus, the new coordinates are \( (5, -1) \).

M1 for finding the new x-coordinate: \( 4 + 1 = 5 \)
M1 for applying the vertical transformations to the y-coordinate: \( 3(-2) + 5 \)
A1 for \( (5, -1) \)

To find \(g^{-1}(4)\), let \(g(y) = 4\).
\(\frac{y+1}{2} = 4\)
\(y + 1 = 8\)
\(y = 7\)
So, \(g^{-1}(4) = 7\).
Now find \(f(7)\):
\(f(7) = 2(7) - 3 = 14 - 3 = 11\).

M1 for finding \(g^{-1}(4) = 7\) or finding expression \(g^{-1}(x) = 2x - 1\)
M1 for substituting their value into \(f(x)\)
A1 for 11

1. A stretch parallel to the \(y\)-axis with scale factor \(3\) transforms \(f(x)\) to \(3f(x)\).
2. A translation of \(\begin{pmatrix} 0 \\ -5 \end{pmatrix}\) shifts the graph vertically downwards by \(5\) units, yielding \(3f(x) - 5\).

M1 for \(3f(x)\) seen
M1 for subtracting \(5\) from their expression
A1 for \(3f(x) - 5\)

For the vertical asymptote, find where the denominator is zero:
\(2x + 4 = 0 \implies x = -2\).
For the horizontal asymptote, look at the ratio of the coefficients of \(x\) as \(x \to \infty\):
\(y = \frac{3}{2} = 1.5\).

M1 for setting denominator to 0 to find \(x = -2\)
M1 for finding horizontal limit \(y = 1.5\)
A1 for both correct equations

Group the terms in pairs:
\(4a(3x - 2y) - 3b(3x - 2y)\)
Factor out the common term \((3x - 2y)\):
\((4a - 3b)(3x - 2y)\).

M1 for grouping and factorising first pair: \(4a(3x - 2y)\)
M1 for grouping and factorising second pair: \(-3b(3x - 2y)\)
A1 for \((4a - 3b)(3x - 2y)\)

Find a common denominator:
\(\frac{2(2x+1) - 3(x-3)}{(x-3)(2x+1)}\)
Expand the numerator:
\(4x + 2 - 3x + 9 = x + 11\)
Combine to write the single fraction:
\(\frac{x + 11}{(x - 3)(2x + 1)}\).

M1 for putting over a common denominator \((x-3)(2x+1)\)
M1 for correct expansions of both numerator products: \(4x + 2\) and \(-3x + 9\)
A1 for final simplified fraction \(\frac{x + 11}{(x - 3)(2x + 1)}\)

Using \(T_2 = 14\):
\(a(2)^2 + b(2) = 14 \implies 4a + 2b = 14 \implies 2a + b = 7\)
Using \(T_3 = 33\):
\(a(3)^2 + b(3) = 33 \implies 9a + 3b = 33 \implies 3a + b = 11\)
Subtract the first simplified equation from the second:
\((3a + b) - (2a + b) = 11 - 7 \implies a = 4\)
Substitute \(a = 4\) back into the first equation:
\(2(4) + b = 7 \implies 8 + b = 7 \implies b = -1\).

M1 for writing down two equations in \(a\) and \(b\): \(4a+2b=14\) and \(9a+3b=33\)
M1 for solving the simultaneous equations correctly for one variable
A1 for both \(a = 4\) and \(b = -1\)

Use the cosine rule:
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)\)
\(AC^2 = 5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\)
Since \(\cos(60^\circ) = 0.5\):
\(AC^2 = 25 + 64 - 80(0.5)\)
\(AC^2 = 89 - 40 = 49\)
\(AC = \sqrt{49} = 7\).

M1 for applying the cosine rule correctly
M1 for substituting \(\cos(60^\circ) = 0.5\) to find \(AC^2 = 49\)
A1 for \(7\)

Given \(P_0 = 200\), when \(t = 3\), \(P = 1600\):
\(1600 = 200 \times k^3\)
Divide both sides by 200:
\(k^3 = 8\)
Find the cube root:
\(k = 2\).

M1 for substituting known values: \(1600 = 200 \times k^3\)
M1 for isolating \(k^3 = 8\)
A1 for \(k = 2\)

First differences: \(11-3=8\), \(25-11=14\), \(45-25=20\), \(71-45=26\). Second differences: \(14-8=6\), \(20-14=6\), \(26-20=6\). Since the second difference is constant, the sequence is quadratic and has the form \(an^2 + bn + c\) where \(2a = 6 \implies a = 3\). Substituting \(a=3\) gives \(3n^2 + bn + c\). For \(n=1\), \(3(1)^2 + b + c = 3 \implies b+c=0\). For \(n=2\), \(3(2)^2 + 2b + c = 11 \implies 12 + 2b + c = 11 \implies 2b+c=-1\). Subtracting these equations gives \(b = -1\), which then gives \(c = 1\). Thus, the \(n\)-th term is \(3n^2 - n + 1\).

M1: For finding second differences of 6 (or setting up \(2a = 6\))
M1: For substituting \(a = 3\) into \(3n^2 + bn + c\) and setting up equations to solve for \(b\) and \(c\)
A1: For the correct expression \(3n^2 - n + 1\)

By the Cosine Rule: \(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)\). Substituting the given values: \(PR^2 = 3^2 + 5^2 - 2(3)(5)\cos(120^\circ)\). Since \(\cos(120^\circ) = -\frac{1}{2}\), this becomes: \(PR^2 = 9 + 25 - 30\left(-\frac{1}{2}\right) = 34 + 15 = 49\). Taking the square root gives \(PR = \sqrt{49} = 7\text{ cm}\).

M1: For correct substitution into the Cosine Rule: \(3^2 + 5^2 - 2(3)(5)\cos(120^\circ)\)
M1: For using \(\cos(120^\circ) = -\frac{1}{2}\) to simplify the equation to \(PR^2 = 49\)
A1: For \(7\)

Let \(y = \frac{3x + 4}{2x - 1}\). Multiply both sides by \(2x - 1\): \(y(2x - 1) = 3x + 4 \implies 2xy - y = 3x + 4\). Rearrange to collect all \(x\) terms on one side: \(2xy - 3x = y + 4 \implies x(2y - 3) = y + 4\). Solve for \(x\): \(x = \frac{y + 4}{2y - 3}\). Thus, \(\mathrm{f}^{-1}(x) = \frac{x+4}{2x-3}\).

M1: For setting \(y = \frac{3x+4}{2x-1}\) and multiplying to get \(y(2x-1) = 3x+4\)
M1: For rearranging terms to group \(x\) and factoring: \(x(2y-3) = y+4\)
A1: For \(\frac{x+4}{2x-3}\) (or equivalent algebraic form)

To find the inverse function \(f^{-1}(x)\), we set \(y = f(x)\) and solve for \(x\):
\(y = \frac{ax + b}{x - 3}\)
\(y(x - 3) = ax + b\)
\(xy - 3y = ax + b\)
\(xy - ax = 3y + b\)
\(x(y - a) = 3y + b\)
\(x = \frac{3y + b}{y - a}\)

Thus, \(f^{-1}(x) = \frac{3x + b}{x - a}\).
Since \(f(x) = f^{-1}(x)\), we compare this to our original function and obtain \(a = 3\).

Now, use \(f(5) = 11\) with \(a = 3\):
\(11 = \frac{3(5) + b}{5 - 3}\)
\(11 = \frac{15 + b}{2}\)
\(22 = 15 + b\)
\(b = 7\).

So, \(a = 3\) and \(b = 7\).

M1: Set \(y = \frac{ax+b}{x-3}\) and make \(x\) the subject.
M1: Obtain expression for \(f^{-1}(x) = \frac{3x+b}{x-a}\).
A1: Deduce \(a = 3\) by comparing functions.
M1: Substitute \(x = 5, a = 3\) into \(f(5) = 11\).
A1: Solve to find \(b = 7\).

Let the width of the lawn be \(w\). Then its length is \(2w + 3\).
The path of width \(1.5\text{ m}\) is added on all sides, so:
Total width including the path \(= w + 1.5 \times 2 = w + 3\).
Total length including the path \(= (2w + 3) + 1.5 \times 2 = 2w + 6\).

The total area is the product of the total dimensions:
\((w + 3)(2w + 6) = 128\)
\(2w^2 + 6w + 6w + 18 = 128\)
\(2w^2 + 12w + 18 = 128\)
\(2w^2 + 12w - 110 = 0\)

Divide the entire equation by 2:
\(w^2 + 6w - 55 = 0\)

Factorise the quadratic equation:
\((w + 11)(w - 5) = 0\)

Since the width must be positive, \(w = 5\text{ m}\).

M1: Express total dimensions as \(w + 3\) and \(2w + 6\) (or equivalent).
M1: Set up the area equation \((w+3)(2w+6) = 128\).
A1: Expand and simplify to \(2w^2 + 12w - 110 = 0\) or \(w^2 + 6w - 55 = 0\).
M1: Factorise or solve using the quadratic formula.
A1: State the correct positive solution \(w = 5\) (ignore negative root).

We write down successive differences:
Terms: 5, 14, 35, 74, 137
1st differences: 9, 21, 39, 63
2nd differences: 12, 18, 24
3rd differences: 6, 6

Since the 3rd differences are constant, the sequence is cubic of the form \(an^3 + bn^2 + cn + d\).
Here, \(6a = 6 \Rightarrow a = 1\).

Now we subtract \(n^3\) from each term:
\(n=1: 5 - 1^3 = 4\)
\(n=2: 14 - 2^3 = 6\)
\(n=3: 35 - 3^3 = 8\)
\(n=4: 74 - 4^3 = 10\)
\(n=5: 137 - 5^3 = 12\)

The remaining terms form an arithmetic sequence 4, 6, 8, 10, 12 with a first term of 4 and common difference of 2.
The \(n\)-th term of this part is \(2n + 2\).

Therefore, the overall \(n\)-th term of the sequence is \(n^3 + 2n + 2\).

M1: Calculate first and second differences of the sequence.
M1: Find third differences and deduce the sequence is cubic with leading term \(n^3\).
M1: Subtract \(n^3\) from terms to find the remaining linear sequence.
M1: Determine the linear formula \(2n + 2\).
A1: State the correct final expression \(n^3 + 2n + 2\).

Step 1: Use the Cosine Rule to find the length of side \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 7.5^2 + 12^2 - 2(7.5)(12)\cos(62^\circ)\)
\(AC^2 = 56.25 + 144 - 180 \cos(62^\circ)\)
\(AC^2 \approx 200.25 - 84.505 = 115.745\)
\(AC \approx \sqrt{115.745} \approx 10.758\text{ cm}\).

Step 2: Find the area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2}(AB)(BC)\sin(\angle ABC)\)
\(\text{Area} = \frac{1}{2}(7.5)(12)\sin(62^\circ) \approx 39.73\text{ cm}^2\).

Step 3: The shortest distance from \(B\) to \(AC\) is the perpendicular height \(h\) relative to base \(AC\):
\(\text{Area} = \frac{1}{2}(AC)(h)\)
\(39.73 \approx \frac{1}{2}(10.758)(h)\)
\(h \approx \frac{2 \times 39.73}{10.758} \approx 7.39\text{ cm}\) (to 3 significant figures).

M1: Set up the Cosine Rule to find \(AC^2\).
A1: Obtain \(AC \approx 10.76\text{ cm}\).
M1: Calculate the area of triangle \(ABC\) using \(\frac{1}{2} a b \sin C\).
M1: Relate the area to the perpendicular height on \(AC\).
A1: Obtain the final answer \(7.39\) (accept \(7.38\) to \(7.40\)).

Let \(t\) be the number of years.
The value of the first investment is given by \(V_1 = 6000(1.045)^t\).
The value of the second investment is given by \(V_2 = 8000(1.021)^t\).

We set up the inequality where the first investment exceeds the second:
\(6000(1.045)^t > 8000(1.021)^t\)

Divide both sides by \(6000(1.021)^t\):
\(\left(\frac{1.045}{1.021}\right)^t > \frac{8000}{6000}\)
\((1.023506)^t > 1.333333\)

Take logarithms of both sides:
\(t \log(1.023506) > \log(1.333333)\)
\(t > \frac{\log(1.333333)}{\log(1.023506)}\)
\(t > 12.38\)

Since \(t\) must represent a complete number of years, we round up to the next integer. Thus, it will take 13 complete years.

M1: Set up the exponential expression for the first investment: \(6000(1.045)^t\).
M1: Set up the exponential expression for the second investment: \(8000(1.021)^t\).
M1: Formulate the inequality or equation: \(6000(1.045)^t > 8000(1.021)^t\).
M1: Solve for \(t\) using logarithms or a GDC to find the critical value \(t \approx 12.4\).
A1: Deduce the correct complete years as 13.

To find the points of intersection, equate the two functions:
\(2^x - 3 = 12 - x^2\)
\(2^x + x^2 - 15 = 0\)

Using a Graphic Display Calculator (GDC) to find the roots of \(f(x) = 2^x + x^2 - 15\):
- There is one negative root where \(x \approx -3.86\).
- There is one positive root where \(x \approx 2.82\).

Therefore, the two \(x\)-coordinates are \(-3.86\) and \(2.82\).

M1: Equate the expressions to get \(2^x - 3 = 12 - x^2\).
M1: Demonstrate a calculator-based numerical or graphical method to solve the equation.
A1: Identify one correct root to 3 s.f.
A1: Identify the other correct root to 3 s.f.
(Max 4 marks overall if solutions are correct but not rounded to 3 s.f.)

Multiply both sides of the equation by the common denominator \((x - 2)(x + 1)\):
\(5(x + 1) - 3(x - 2) = 2(x - 2)(x + 1)\)

Expand both sides:
\(5x + 5 - 3x + 6 = 2(x^2 - x - 2)\)
\(2x + 11 = 2x^2 - 2x - 4\)

Rearrange into standard quadratic form:
\(2x^2 - 4x - 15 = 0\)

Use the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{4 \pm \sqrt{(-4)^2 - 4(2)(-15)}}{2(2)}\)
\(x = \frac{4 \pm \sqrt{16 + 120}}{4}\)
\(x = \frac{4 \pm \sqrt{136}}{4}\)

Calculating the two values:
\(x_1 = \frac{4 + \sqrt{136}}{4} \approx 3.92\)
\(x_2 = \frac{4 - \sqrt{136}}{4} \approx -1.92\)

Thus, \(x = -1.92\) and \(x = 3.92\).

M1: Multiply by common denominator to obtain \(5(x+1) - 3(x-2) = 2(x-2)(x+1)\).
M1: Correctly expand both sides.
A1: Reduce the equation to \(2x^2 - 4x - 15 = 0\) (or equivalent).
M1: Apply the quadratic formula correctly to their quadratic equation.
A1: Obtain both final answers correct to 2 decimal places: \(x = -1.92\) and \(x = 3.92\).

Let the first term of the arithmetic progression (AP) be \(a = 8\) and its common difference be \(d\).
The 1st term is \(T_1 = 8\).
The 3rd term is \(T_3 = 8 + 2d\).
The 11th term is \(T_{11} = 8 + 10d\).

Since these three terms are consecutive terms of a geometric progression (GP), they share a common ratio \(r\):
\(\frac{8 + 2d}{8} = \frac{8 + 10d}{8 + 2d}\)

Cross-multiplying yields:
\((8 + 2d)^2 = 8(8 + 10d)\)
\(64 + 32d + 4d^2 = 64 + 80d\)
\(4d^2 - 48d = 0\)
\(4d(d - 12) = 0\)

Since the common difference \(d\) is non-zero, we have \(d = 12\).

The first three terms of our geometric progression are:
\(G_1 = T_1 = 8\)
\(G_2 = T_3 = 8 + 2(12) = 32\)
\(G_3 = T_{11} = 8 + 10(12) = 128\)

The common ratio is \(r = \frac{32}{8} = 4\).

To find the 10th term of this GP:
\(G_{10} = G_1 \cdot r^9 = 8 \cdot 4^9 = 8 \cdot 262144 = 2097152\).

M1: Write expressions for AP terms: \(8\), \(8+2d\), \(8+10d\).
M1: Set up geometric progression ratio equation: \((8+2d)^2 = 8(8+10d)\).
A1: Solve the equation to obtain the correct common difference \(d = 12\).
M1: Determine the common ratio of the GP is \(r = 4\).
A1: Find the 10th term as \(2097152\).

First, find \(f(3) = \frac{3(3) + 1}{3 - 2} = 10\). Then, \(g(f(3)) = g(10) = 2(10) + 3 = 23\). Set up the equation \(f(g(x)) = 23\). This gives \(f(2x + 3) = \frac{3(2x + 3) + 1}{(2x + 3) - 2} = 23\), which simplifies to \(\frac{6x + 10}{2x + 1} = 23\). Solving this gives \(6x + 10 = 23(2x + 1) \implies 6x + 10 = 46x + 23 \implies 40x = -13 \implies x = -0.325\).

M1 for finding \(f(3) = 10\). M1 for finding \(g(f(3)) = 23\). M1 for setting up the equation \(\frac{6x + 10}{2x + 1} = 23\). M1 for simplifying to \(40x = -13\). A1 for the correct answer \(x = -0.325\) (or equivalent fraction \(-\frac{13}{40}\)).

The sequence is 3, 11, 25, 45. The first differences are 8, 14, 20. The second differences are constant and equal to 6. Thus, the sequence is quadratic of the form \(T_n = an^2 + bn + c\). Since the second difference is \(2a = 6\), we have \(a = 3\). Subtracting \(3n^2\) from each term gives the sequence: \(3 - 3(1)^2 = 0\), \(11 - 3(2)^2 = -1\), \(25 - 3(3)^2 = -2\), \(45 - 3(4)^2 = -3\). This linear part has the first term 0 and a common difference of \(-1\), which is given by \(-n + 1\). Therefore, the \(n\)-th term of the sequence is \(3n^2 - n + 1\).

M1 for finding the second differences are 6. M1 for deducing the coefficient of \(n^2\) is \(a = 3\). M1 for subtracting \(3n^2\) from the terms to get \(0, -1, -2, -3\). M1 for finding the linear sequence term \(-n + 1\). A1 for the final correct expression \(3n^2 - n + 1\) (or equivalent).

First, use the Cosine Rule to find the length of \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(118^\circ) = 72^2 + 58^2 - 2(72)(58)\cos(118^\circ) = 5184 + 3364 - 8352(-0.46947) = 12469.01\), so \(AC \approx 111.66\text{ m}\). Next, calculate the area of the triangle using \(\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(ABC) = \frac{1}{2}(72)(58)\sin(118^\circ) \approx 1843.59\text{ m}^2\). The shortest distance \(h\) from \(B\) to \(AC\) is the perpendicular height: \(\text{Area} = \frac{1}{2} \cdot AC \cdot h \implies 1843.59 = \frac{1}{2}(111.66)h \implies h = \frac{2 \times 1843.59}{111.66} \approx 33.02\text{ m}\). To 3 significant figures, this is \(33.0\text{ m}\).

M1 for using the cosine rule to write an expression for \(AC^2\). A1 for \(AC \approx 111.7\) (or 112). M1 for using the area formula \(\frac{1}{2} a c \sin B\). M1 for equating the area to \(\frac{1}{2} \times AC \times h\). A1 for the correct distance 33.0 (accept 33).

Let \(t\) be the number of years. The value of the painting after \(t\) years is \(12000 \times (1.045)^t\). We set up the inequality: \(12000 \times (1.045)^t > 30000 \implies (1.045)^t > 2.5\). Taking natural logarithms on both sides: \(t \ln(1.045) > \ln(2.5) \implies t > \frac{\ln(2.5)}{\ln(1.045)} \approx 20.817\). Since we want the number of complete years, we round up to the next integer, which is 21 years.

M1 for writing the correct formula/expression \(12000(1.045)^t\). M1 for setting up the equation or inequality \(1.045^t > 2.5\). M1 for taking logarithms to solve: \(t = \log(2.5) / \log(1.045)\). A1 for finding the decimal value \(t \approx 20.8\) (or 20.817...). A1 for concluding 21 complete years (must be integer).

From the linear equation, we can express \(y\) as \(y = 3 - 2x\). Substitute this into the second equation: \(3x^2 - x(3 - 2x) - (3 - 2x)^2 = 1 \implies 3x^2 - 3x + 2x^2 - (9 - 12x + 4x^2) = 1\). Simplifying this gives: \(5x^2 - 3x - 9 + 12x - 4x^2 = 1 \implies x^2 + 9x - 10 = 0\). Factoring the quadratic equation gives \((x + 10)(x - 1) = 0\), which yields \(x = 1\) or \(x = -10\). If \(x = 1\), \(y = 3 - 2(1) = 1\). If \(x = -10\), \(y = 3 - 2(-10) = 23\). The solutions are \((1, 1)\) and \((-10, 23)\).

M1 for rearranging to \(y = 3 - 2x\) (or \(x = \frac{3-y}{2}\)). M1 for substituting into the quadratic equation and expanding correctly. M1 for simplifying to form the correct quadratic equation \(x^2 + 9x - 10 = 0\) (or equivalent in \(y\)). A1 for finding both correct \(x\) values 1 and -10 (or both \(y\) values). A1 for matching both pairs of solutions: \(x = 1, y = 1\) and \(x = -10, y = 23\).

We find the intersections of \(y = 2^x - 3\) and \(y = 5 - x^2\) using a GDC. Setting \(2^x - 3 = 5 - x^2\) gives \(2^x + x^2 - 8 = 0\). Plotting these two curves on a GDC, we see they intersect at two points. The positive intersection is exactly at \(x = 2.00\) (since \(2^2 - 3 = 1\) and \(5 - 2^2 = 1\)). The negative intersection is found at \(x \approx -2.80\).

M1 for indicating the use of GDC to plot both graphs. M1 for identifying two points of intersection. A1 for the positive solution \(x = 2\) (or 2.00). M2 for finding the negative solution \(x \approx -2.80\) (M1 for finding a value between -2.7 and -2.9).

Combine the fractions over a common denominator: \(\frac{4(x + 2) - 3(2x - 1)}{(2x - 1)(x + 2)} = 2 \implies \frac{4x + 8 - 6x + 3}{2x^2 + 3x - 2} = 2 \implies 11 - 2x = 2(2x^2 + 3x - 2)\). Expand and rearrange into a standard quadratic form: \(11 - 2x = 4x^2 + 6x - 4 \implies 4x^2 + 8x - 15 = 0\). Apply the quadratic formula: \(x = \frac{-8 \pm \sqrt{8^2 - 4(4)(-15)}}{2(4)} = \frac{-8 \pm \sqrt{304}}{8}\). This yields \(x = \frac{-8 + 17.4356}{8} \approx 1.18\) and \(x = \frac{-8 - 17.4356}{8} \approx -3.18\).

M1 for finding the common denominator and writing the equation as \(4(x+2) - 3(2x-1) = 2(2x-1)(x+2)\). M1 for expanding correctly to \(11 - 2x = 4x^2 + 6x - 4\). M1 for simplifying to \(4x^2 + 8x - 15 = 0\). M1 for applying the quadratic formula correctly with their coefficients. A1 for both correct solutions \(x = 1.18\) and \(x = -3.18\) (to 3 s.f.).

\(\textbf{Part (a)}\)
For \(n = 5\), the triangular grid has 5 dots along each of its three sides. Since the 3 corner dots are shared between adjacent sides, the total number of unique border dots is \(3 \times (5 - 1) = 12\). Thus, \(\text{i} = 12\).
For \(n = 6\), by the same method, the number of border dots is \(3 \times (6 - 1) = 15\). Thus, \(\text{ii} = 15\).

\(\textbf{Part (b)}\)
The sequence \(3, 6, 9, 12, 15, \dots\) is an arithmetic progression with a first term of 3 (for \(n=2\)) and a common difference of 3.
Thus, \(B_n = 3 + (n - 2) \times 3 = 3n - 3 = 3(n - 1)\).

\(\textbf{Part (c)}\)
Substituting \(n = 150\) into the expression: \(B_{150} = 3(150 - 1) = 3 \times 149 = 447\).

M1 for finding both table values: 12 and 15 (accept 1 mark for each if one is incorrect).
A1 for the correct expression \(3n - 3\) or \(3(n - 1)\).
M1 for substituting \(n = 150\) into their formula.
A1 for the correct final answer of 447.
C1.5 communication marks for showing clear workings of the differences or derivation.

\(\textbf{Part (a)}\)
For \(n = 6\), the total number of dots in the grid is: \(T_6 = \frac{6(6+1)}{2} = \frac{42}{2} = 21\).
From Question 1, the number of border dots is \(B_6 = 15\).
The number of inside dots is: \(I_6 = T_6 - B_6 = 21 - 15 = 6\). (Shown)

\(\textbf{Part (b)}\)
Using the relation \(I_n = T_n - B_n\):
\(I_n = \frac{n(n+1)}{2} - (3n - 3)\)
Expressing with a common denominator of 2:
\(I_n = \frac{n^2 + n - 2(3n - 3)}{2}\)
\(I_n = \frac{n^2 + n - 6n + 6}{2}\)
\(I_n = \frac{n^2 - 5n + 6}{2}\)
Factoring the numerator: \(n^2 - 5n + 6 = (n - 1)(n - 2)\).
Therefore, \(I_n = \frac{(n-1)(n-2)}{2}\). (Shown)

\(\textbf{Part (c)}\)
Set \(I_n = 190\):
\(\frac{(n-1)(n-2)}{2} = 190 \implies (n-1)(n-2) = 380\)
\(n^2 - 3n + 2 = 380 \implies n^2 - 3n - 378 = 0\)
Factoring the quadratic equation:
\((n - 21)(n + 18) = 0\)
Since \(n\) must be a positive integer, we discard \(n = -18\) to get \(n = 21\).

M1 for calculating \(T_6 = 21\) and stating \(B_6 = 15\) to find \(I_6 = 6\).
M2 for algebraic manipulation in part (b) (M1 for putting over a common denominator, M1 for factorising \(n^2 - 5n + 6\)).
M1 for setting up the quadratic equation \((n-1)(n-2) = 380\) or \(n^2 - 3n - 378 = 0\).
A1 for solving to get \(n = 21\).
C1.5 communication marks for showing clear intermediate algebraic steps.

\(\textbf{Part (a)}\)
A square grid of size \(k\) has 4 outer sides, each containing \(k\) dots. The 4 corner dots are shared between adjacent sides, so the total number of border dots is:
\(S\_border_k = 4(k - 1) = 4k - 4\).

\(\textbf{Part (b)}\)
Equating the two border expressions:
\(B_n = S\_border_k\)
\(3(n - 1) = 4(k - 1)\)
\(n - 1 = \frac{4}{3}(k - 1)\)
\(n = \frac{4}{3}(k - 1) + 1 = \frac{4k - 4 + 3}{3} = \frac{4k - 1}{3}\).

\(\textbf{Part (c)}\)
Substitute \(k = 100\) into the formula:
\(n = \frac{4(100) - 1}{3} = \frac{399}{3} = 133\).

M1 for finding the border formula \(4k-4\) or \(4(k-1)\).
M2 for equating \(3(n-1) = 4(k-1)\) and solving for \(n\) (M1 for isolating \(n-1\), M1 for expressing \(n\) fully in terms of \(k\)).
M1 for substituting \(k = 100\) into their formula.
A1 for the correct answer \(n = 133\).
C1.5 communication marks for clear step-by-step rearrangement of formulas.

\(\textbf{Part (a)}\)
Substitute the formulas \(B_n = 3(n-1)\) and \(I_n = \frac{(n-1)(n-2)}{2}\) into the ratio:
\(R_n = \frac{3(n-1)}{\frac{(n-1)(n-2)}{2}}\)
Multiply numerator and denominator by 2:
\(R_n = \frac{6(n-1)}{(n-1)(n-2)}\)
Since \(n \ge 3\), we have \(n - 1 \neq 0\). Cancelling the common factor of \(n-1\) gives:
\(R_n = \frac{6}{n-2}\). (Shown)

\(\textbf{Part (b)}\)
The number of border dots is equal to the number of inside dots when the ratio \(R_n = 1\):
\(\frac{6}{n-2} = 1 \implies n - 2 = 6 \implies n = 8\).

\(\textbf{Part (c)}\)
We want \(B_n > I_n\), which is equivalent to \(R_n > 1\):
\(\frac{6}{n-2} > 1\)
Since \(n \ge 3\), we have \(n - 2 > 0\), so we can multiply both sides by \(n - 2\) without changing the direction of the inequality:
\(6 > n - 2 \implies n < 8\).
Given the constraint \(n \ge 3\), the possible integer values of \(n\) are \(n = 3, 4, 5, 6, 7\).

M1 for setting up the fraction \(\frac{3(n-1)}{\frac{(n-1)(n-2)}{2}}\) and simplifying it to \(\frac{6}{n-2}\).
M1 for setting \(\frac{6}{n-2} = 1\) and solving to get \(n = 8\).
M2 for the inequality part (M1 for setting up \(\frac{6}{n-2} > 1\) or equivalent, M1 for solving to get \(n < 8\)).
A1 for listing the correct integer values: \(3, 4, 5, 6, 7\) (or writing \(3 \le n \le 7\)).
C1.5 communication marks for explicitly stating the constraint \(n \ge 3\) and solving the inequality step-by-step.

(a) Comparing the model parameters with the given values:
\(T_s = 20\)
\(T_0 = 95\)

(b) Substituting \(t = 5\) and \(T(5) = 65\):
\(65 = 20 + (95 - 20) a^{-5k}
45 = 75 a^{-5k}
a^{-5k} = \frac{45}{75} = 0.6\) (as required)

(c) After 15 minutes, \(t = 15\):
\(T(15) = 20 + 75 \times a^{-15k} = 20 + 75 \times (a^{-5k})^3\)
Substitute \(a^{-5k} = 0.6\):
\(T(15) = 20 + 75 \times (0.6)^3 = 20 + 75 \times 0.216 = 20 + 16.2 = 36.2^\circ\text{C}

(d) Set \)T(t) = 35\):
\(35 = 20 + 75 a^{-kt}
15 = 75 a^{-kt}
a^{-kt} = \frac{15}{75} = 0.2\)
Since \(a^{-kt} = (a^{-5k})^{t/5} = 0.6^{t/5}\):
\(0.6^{t/5} = 0.2\)
Taking logs on both sides:
\(\frac{t}{5} \log(0.6) = \log(0.2)\)
\(t = 5 \times \frac{\log(0.2)}{\log(0.6)} \approx 5 \times 3.1507 \approx 15.75 \approx 15.8\text{ minutes}\)

(a) B1 for both values correct.
(b) M1 for substituting \(t = 5, T = 65, T_s = 20, T_0 = 95\) into the equation. A1 for showing step-by-step reduction to \(a^{-5k} = 0.6\).
(c) M1 for expressing \(a^{-15k}\) as \((a^{-5k})^3\). M1 for calculating \(20 + 75(0.216)\). A1 for \(36.2^\circ\text{C}\).
(d) M1 for setting up the equation \(20 + 75 a^{-kt} = 35\). M1 for writing the index equation as \(0.6^{t/5} = 0.2\). M1 for taking logarithms to solve for \(t\). A1 for \(15.8\text{ minutes}\) (accept \(15.75\) to \(15.8\)).

(a) The volume of the box is given by:
\(V = x^2 h = 4000 \implies h = \frac{4000}{x^2}\)
The total surface area \(A\) of an open-topped box consists of one square base and four rectangular sides:
\(A = x^2 + 4xh\)
Substitute \(h = \frac{4000}{x^2}\) into the surface area formula:
\(A = x^2 + 4x \left(\frac{4000}{x^2}\right) = x^2 + \frac{16000}{x}\) (as required)

(b)(i) To find the minimum surface area, differentiate \(A\) with respect to \(x\) and set to \(0\):
\(\frac{\text{d}A}{\text{d}x} = 2x - \frac{16000}{x^2} = 0\)
\(2x = \frac{16000}{x^2} \implies 2x^3 = 16000 \implies x^3 = 8000\)
\(x = \sqrt[3]{8000} = 20\text{ cm}\)

(b)(ii) Substitute \(x = 20\) back into the expression for \(A\):
\(A = 20^2 + \frac{16000}{20} = 400 + 800 = 1200\text{ cm}^2\)

(c) The cost of the base is:
\(C_{\text{base}} = 0.05 \times x^2 = 0.05x^2\)
The cost of the four sides is:
\(C_{\text{sides}} = 0.03 \times (4xh) = 0.12xh\)
Substitute \(h = \frac{4000}{x^2}\):
\(C_{\text{sides}} = 0.12 \times \frac{4000}{x} = \frac{480}{x}\)
Therefore, the total cost \(C\) is:
\(C = 0.05x^2 + \frac{480}{x}\)

(a) M1 for expressing \(h\) in terms of \(x\) (e.g., \(h = \frac{4000}{x^2}\)). M1 for writing down the total surface area formula \(A = x^2 + 4xh\). A1 for substituting and showing clear algebraic steps to get the final given expression.
(b)(i) M1 for differentiating \(A\) (getting \(2x - \frac{16000}{x^2}\)). M1 for equating the derivative to 0 and solving \(x^3 = 8000\). A1 for \(x = 20\text{ cm}\). (Note: sketching/graphing calculator methods showing minimum at \(x=20\) can receive full marks if well-explained).
(b)(ii) M1 for substituting their \(x\) value into the surface area equation. A1 for \(1200\text{ cm}^2\).
(c) M1 for writing a cost equation containing both base and side components (e.g., \(C = 0.05x^2 + 0.12xh\)). A1 for substituting \(h\) and simplifying to \(C = 0.05x^2 + \frac{480}{x}\).

(a) Since the towers are \(120\text{ m}\) apart, the left tower is at \(x = 0\) and the right tower is at \(x = 120\).
The lowest point is midway, so \(x = \frac{120}{2} = 60\).
The height at this lowest point is \(6\text{ m}\), so \(y = 6\).
Therefore, the coordinates of the lowest point are \((60, 6)\).

(b) The quadratic curve has vertex at \((60, 6)\). We can use the vertex form of a quadratic equation:
\(y = a(x - 60)^2 + 6\)
Since the left tower is at \(x = 0\) and has height \(30\text{ m}\), the point \((0, 30)\) lies on the curve.
\(30 = a(0 - 60)^2 + 6
24 = 3600 a
a = \frac{24}{3600} = \frac{1}{150}\) (or \(\approx 0.00667\))
Now expand the vertex form to find \(b\) and \(c\):
\(y = \frac{1}{150}(x^2 - 120x + 3600) + 6
y = \frac{1}{150}x^2 - \frac{120}{150}x + \frac{3600}{150} + 6
y = \frac{1}{150}x^2 - 0.8x + 24 + 6
y = \frac{1}{150}x^2 - 0.8x + 30\)
Comparing with \(y = a x^2 + b x + c\):
\(a = \frac{1}{150}\) (or \(0.0067\)), \(b = -0.8\), \(c = 30\).

(c) For a distance of \(30\text{ m}\) from the left tower, substitute \(x = 30\):
\(y = \frac{1}{150}(30 - 60)^2 + 6
y = \frac{1}{150}(-30)^2 + 6
y = \frac{900}{150} + 6 = 6 + 6 = 12\text{ m}\).

(d) Set \(y = 15\) and solve for \(x\):
\(15 = \frac{1}{150}(x - 60)^2 + 6
9 = \frac{1}{150}(x - 60)^2
(x - 60)^2 = 1350
x - 60 = \pm\sqrt{1350} \approx \pm 36.74
x = 60 - 36.74 \approx 23.3\text{ m}
x = 60 + 36.74 \approx 96.7\text{ m}\)

(a) B1 for \(x\)-coordinate \(60\). B1 for \(y\)-coordinate \(6\).
(b) M1 for using the vertex form \(y = a(x-60)^2 + 6\) or substitution into \(y=ax^2+bx+c\) with known points. M1 for substituting \((0,30)\) to find \(a\). A1 for \(a = \frac{1}{150}\) (or \(0.0067\)). A1 for \(b = -0.8\) and \(c = 30\).
(c) M1 for substituting \(x = 30\) into their quadratic model. A1 for \(12\text{ m}\).
(d) M1 for setting up the equation \(\frac{1}{150}(x-60)^2+6=15\) or equivalent. A1 for \(23.3\text{ m}\) and \(96.7\text{ m}\) (accept either or both, 1 decimal place accuracy).