2025 Cambridge IGCSE International Mathematics (0607) 模擬試題連答案詳解

Observe the pattern of the numerators: \(3, 7, 11, 15, \dots\)
This is an arithmetic progression with a first term \(a = 3\) and a common difference \(d = 4\).
The \(n\)-th term of the numerator is:
\(3 + (n - 1)4 = 4n - 1\).

Observe the pattern of the denominators: \(2, 5, 10, 17, \dots\)
These numbers are each 1 more than a perfect square:
\(1^2 + 1 = 2\)
\(2^2 + 1 = 5\)
\(3^2 + 1 = 10\)
\(4^2 + 1 = 17\)
So the \(n\)-th term of the denominator is \(n^2 + 1\).

Combining these, the \(n\)-th term of the sequence is:
\(\frac{4n - 1}{n^2 + 1}\).

M1 for finding the numerator's term \(4n - 1\) (or showing a common difference of 4)
M1 for finding the denominator's term \(n^2 + 1\)
A1.41 for the correct final fraction

Using the given information, set up two simultaneous equations:
For \(T_2 = 5\):
\(a(2^2) + b(2) - 5 = 5 \implies 4a + 2b = 10 \implies 2a + b = 5\)

For \(T_5 = 50\):
\(a(5^2) + b(5) - 5 = 50 \implies 25a + 5b = 55 \implies 5a + b = 11\)

Subtract the first simplified equation from the second:
\((5a + b) - (2a + b) = 11 - 5\)
\(3a = 6 \implies a = 2\)

Substitute \(a = 2\) back into the first equation:
\(2(2) + b = 5 \implies 4 + b = 5 \implies b = 1\).

M1 for writing down at least one correct equation in terms of \(a\) and \(b\) (e.g. \(4a+2b = 10\) or \(25a+5b = 55\))
M1 for a complete and correct algebraic method to solve their simultaneous equations
A1.41 for both correct values: \(a = 2\) and \(b = 1\)

Since the curve has a \(y\)-intercept at \((0, 5)\), we substitute \(x = 0\) and \(y = 5\) into the equation:
\(5 = a(0)^2 + b(0) + c \implies c = 5\).

Now, the equation is \(y = ax^2 + bx + 5\).
Substitute the point \((1, -1)\):
\(-1 = a(1)^2 + b(1) + 5 \implies a + b = -6\).

Substitute the point \((3, -1)\):
\(-1 = a(3)^2 + b(3) + 5 \implies 9a + 3b = -6 \implies 3a + b = -2\).

Subtract the equation \(a + b = -6\) from \(3a + b = -2\):
\((3a + b) - (a + b) = -2 - (-6)\)
\(2a = 4 \implies a = 2\).

Substitute \(a = 2\) back into \(a + b = -6\):
\(2 + b = -6 \implies b = -8\).

Thus, the values are \(a = 2\), \(b = -8\), and \(c = 5\).

M1 for identifying \(c = 5\)
M1 for substituting points to get simultaneous equations in \(a\) and \(b\) (e.g., \(a + b = -6\) and \(9a + 3b = -6\)) and attempting to solve them
A1.41 for all three values correct

Using the vertex form of a quadratic equation: \(y = p(x - h)^2 + k\), where the vertex is \((h, k) = (3, 4)\):
\(y = p(x - 3)^2 + 4\).

To find \(p\), substitute the point \((1, -8)\) into this equation:
\(-8 = p(1 - 3)^2 + 4\)
\(-8 = p(-2)^2 + 4\)
\(-12 = 4p \implies p = -3\).

Now, expand the equation to rewrite it in standard form:
\(y = -3(x - 3)^2 + 4\)
\(y = -3(x^2 - 6x + 9) + 4\)
\(y = -3x^2 + 18x - 27 + 4\)
\(y = -3x^2 + 18x - 23\).

Comparing this to \(y = px^2 + qx + r\), we get:
\(p = -3\), \(q = 18\), and \(r = -23\).

M1 for utilizing the vertex form \(y = p(x-3)^2 + 4\) and substituting \((1, -8)\) to find \(p = -3\)
M1 for expanding their vertex form equation correctly to standard form
A1.41 for all three correct coefficients: \(p = -3\), \(q = 18\), \(r = -23\)

Group the terms into quadratic terms and linear terms:
\(12x^2 - 3y^2 + 8x - 4y = (12x^2 - 3y^2) + (8x - 4y)\)

Factor out common factors from each group:
\(3(4x^2 - y^2) + 4(2x - y)\)

Recognise that \(4x^2 - y^2\) is a difference of two squares:
\(4x^2 - y^2 = (2x - y)(2x + y)\)

Substitute this back into the expression:
\(3(2x - y)(2x + y) + 4(2x - y)\)

Factor out the common binomial term \((2x - y)\):
\((2x - y)[3(2x + y) + 4]\)
\(= (2x - y)(6x + 3y + 4)\)

M1 for grouping and identifying the difference of squares: \(3(2x-y)(2x+y)\)
M1 for factoring the linear part to get \(4(2x-y)\) and realizing the common factor \((2x-y)\)
A1.41 for the correct fully factorised expression: \((2x - y)(6x + 3y + 4)\)

First, multiply both sides by the denominator \(2w + 7\):
\(v(2w + 7) = 3w - 5\)

Expand the left-hand side:
\(2vw + 7v = 3w - 5\)

Rearrange the equation to group all terms containing \(w\) on one side, and the other terms on the opposite side:
\(7v + 5 = 3w - 2vw\)

Factorise \(w\) out of the right-hand side:
\(7v + 5 = w(3 - 2v)\)

Divide both sides by \((3 - 2v)\) to isolate \(w\):
\(w = \frac{7v + 5}{3 - 2v}\)

(Alternatively, \(w = \frac{-7v - 5}{2v - 3}\) is also correct.)

M1 for multiplying by the denominator to clear the fraction: \(v(2w + 7) = 3w - 5\)
M1 for expanding and grouping all terms with \(w\) on one side and factoring out \(w\)
A1.41 for \(w = \frac{7v + 5}{3 - 2v}\) or its exact equivalent

We want to solve \(f(g(x)) = 3\).

Let \(u = g(x)\). Then:
\(f(u) = 3 \implies \frac{2u + 1}{u - 3} = 3\)
\(2u + 1 = 3(u - 3)\)
\(2u + 1 = 3u - 9\)
\(u = 10\)

Since \(u = g(x)\), substitute \(g(x) = 10\):
\(5x - 2 = 10\)
\(5x = 12 \implies x = \frac{12}{5}\) (or \(2.4\)).

M1 for substituting \(g(x)\) into \(f(x)\) to form the composite equation \(\frac{2(5x-2)+1}{(5x-2)-3} = 3\) (or for solving \(f(u) = 3\) to find \(u=10\))
M1 for simplifying to a linear equation in \(x\) (e.g. \(10x - 3 = 15x - 15\) or \(5x - 2 = 10\))
A1.41 for the correct final answer \(x = \frac{12}{5}\) (or \(2.4\))

Let \(y = h(x)\):
\(y = \frac{4}{2x - 1}\)

To find the inverse, rearrange the equation to make \(x\) the subject:
Multiply both sides by \((2x - 1)\):
\(y(2x - 1) = 4\)
\(2xy - y = 4\)

Add \(y\) to both sides:
\(2xy = 4 + y\)

Divide by \(2y\):
\(x = \frac{4 + y}{2y}\)

Replace \(y\) with \(x\) to get the inverse function:
\(h^{-1}(x) = \frac{4 + x}{2x}\) (which can also be written as \(\frac{2}{x} + \frac{1}{2}\)).

M1 for setting up the equation \(y = \frac{4}{2x-1}\) and multiplying by \((2x - 1)\)
M1 for isolating terms with \(x\) on one side and factoring if necessary to make \(x\) the subject
A1.41 for the correct expression \(h^{-1}(x) = \frac{4 + x}{2x}\) or equivalent

The terms of the sequence are \( 3, 10, 21, 36 \). Let us find the differences between consecutive terms: First differences: \( 10 - 3 = 7 \), \( 21 - 10 = 11 \), \( 36 - 21 = 15 \). Second differences: \( 11 - 7 = 4 \), \( 15 - 11 = 4 \). Since the second difference is constant, the sequence is quadratic, of the form \( an^2 + bn + c \), where \( 2a = 4 \) so \( a = 2 \). Subtracting \( 2n^2 \) from each term: For \( n=1 \): \( 3 - 2(1)^2 = 1 \). For \( n=2 \): \( 10 - 2(2)^2 = 2 \). For \( n=3 \): \( 21 - 2(3)^2 = 3 \). For \( n=4 \): \( 36 - 2(4)^2 = 4 \). The remaining sequence is \( 1, 2, 3, 4 \), which has the \(n\)th term of \( n \). Therefore, the overall \(n\)th term is \( 2n^2 + n \).

M1 for finding the second difference is 4. M1 for finding the coefficient of \( n^2 \) is 2. A1 for the final expression \( 2n^2 + n \).

The vertex form of a quadratic equation is \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. Here, the vertex is \( (3, -4) \), so we have: \( y = a(x - 3)^2 - 4 \). To find \( a \), substitute the point \( (1, 8) \) into this equation: \( 8 = a(1 - 3)^2 - 4 \) which simplifies to \( 8 = a(-2)^2 - 4 \), so \( 8 = 4a - 4 \). This gives \( 12 = 4a \), which means \( a = 3 \). Expanding the equation: \( y = 3(x - 3)^2 - 4 = 3(x^2 - 6x + 9) - 4 = 3x^2 - 18x + 27 - 4 = 3x^2 - 18x + 23 \).

M1 for expressing the equation in vertex form as \( y = a(x - 3)^2 - 4 \). M1 for substituting the point \( (1, 8) \) to determine \( a = 3 \). A1 for correctly expanding to \( y = 3x^2 - 18x + 23 \).

Start by multiplying both sides by \( 2 - x \) to clear the fraction: \( y(2 - x) = 3x - 5 \). Expand the left side: \( 2y - xy = 3x - 5 \). Rearrange the terms to group all terms with \( x \) on one side of the equation: \( 2y + 5 = 3x + xy \). Factor out \( x \) on the right side: \( 2y + 5 = x(3 + y) \). Divide both sides by \( y + 3 \) to isolate \( x \): \( x = \frac{2y + 5}{y + 3} \).

M1 for multiplying to clear the denominator, obtaining \( y(2 - x) = 3x - 5 \). M1 for isolating terms containing \( x \) on one side and factoring to get \( x(3 + y) \). A1 for the correct final formula \( x = \frac{2y + 5}{y + 3} \) or equivalent.

To find \( \mathrm{g}(\mathrm{f}(4)) \), first evaluate the inner function \( \mathrm{f}(4) \): \( \mathrm{f}(4) = 2(4) - 3 = 8 - 3 = 5 \). Now substitute this result into the outer function \( \mathrm{g}(x) \): \( \mathrm{g}(5) = 5^2 + 5 = 25 + 5 = 30 \).

M1 for finding \( \mathrm{f}(4) = 5 \). M1 for substituting their value into \( \mathrm{g}(x) \), showing \( 5^2 + 5 \). A1 for the correct answer 30.

The formula for the \(n\)th term of a geometric sequence is \( u_n = u_1 r^{n-1} \). For \( u_4 \), we have: \( u_4 = 54 r^3 = -2 \). Solving for \( r^3 \): \( r^3 = -\frac{2}{54} = -\frac{1}{27} \). Taking the cube root, we find the common ratio: \( r = -\frac{1}{3} \). To find the fifth term \( u_5 \): \( u_5 = u_4 \times r = -2 \times \left(-\frac{1}{3}\right) = \frac{2}{3} \).

M1 for establishing the equation \( 54 r^3 = -2 \). A1 for finding the common ratio \( r = -\frac{1}{3} \). A1 for obtaining the fifth term \( u_5 = \frac{2}{3} \).

A quadratic function with roots at \( x = -2 \) and \( x = 5 \) can be written in factored form as: \( y = a(x + 2)(x - 5) \). We are given that the \(y\)-intercept is \( (0, -30) \). Substituting these coordinates gives: \( -30 = a(0 + 2)(0 - 5) \) which simplifies to \( -30 = -10a \), so \( a = 3 \). Now expand the function: \( y = 3(x + 2)(x - 5) = 3(x^2 - 3x - 10) = 3x^2 - 9x - 30 \).

M1 for writing the general factored form \( y = a(x + 2)(x - 5) \). M1 for substituting \( (0, -30) \) to find \( a = 3 \). A1 for expanding to get the final equation \( y = 3x^2 - 9x - 30 \).

Let \( y = \frac{4}{x - 1} \). To find the inverse function, swap the variables \( x \) and \( y \): \( x = \frac{4}{y - 1} \). Now, solve for \( y \). Multiply both sides by \( y - 1 \): \( x(y - 1) = 4 \). Expand the brackets: \( xy - x = 4 \). Add \( x \) to both sides: \( xy = 4 + x \). Finally, divide by \( x \): \( y = \frac{4 + x}{x} \). Therefore, \( \mathrm{f}^{-1}(x) = \frac{4 + x}{x} \).

M1 for setting up the relation with swapped variables, e.g., \( x = \frac{4}{y - 1} \). M1 for isolating the \( y \) term to obtain \( xy = 4 + x \). A1 for the correct inverse function \( \mathrm{f}^{-1}(x) = \frac{4 + x}{x} \) or equivalent, e.g., \( \frac{4}{x} + 1 \).

Analyze the numerator and denominator separately. The numerators are: \( 2, 5, 8, 11, 14, \dots \) This is an arithmetic sequence with a first term of 2 and a common difference of 3. The \(n\)th term for the numerator is: \( 2 + 3(n - 1) = 3n - 1 \). The denominators are: \( 3, 6, 11, 18, 27, \dots \) The first differences are \( 3, 5, 7, 9 \) and the second differences are constant at \( 2 \). This indicates a quadratic sequence of the form \( an^2 + bn + c \), where \( 2a = 2 \) so \( a = 1 \). Subtracting \( n^2 \) from each denominator term: For \( n=1 \): \( 3 - 1^2 = 2 \). For \( n=2 \): \( 6 - 2^2 = 2 \). For \( n=3 \): \( 11 - 3^2 = 2 \). The remaining constant sequence is 2, so the denominator is \( n^2 + 2 \). Combining these, the \(n\)th term of the sequence is \( \frac{3n - 1}{n^2 + 2} \).

M1 for finding the \(n\)th term of the numerator as \( 3n - 1 \). M1 for finding the \(n\)th term of the denominator as \( n^2 + 2 \). A1 for the correct combined fraction \( \frac{3n - 1}{n^2 + 2} \).

To find the \(n\)-th term, we can analyze the numerators and denominators separately.

1. **Numerators:**
The sequence of numerators is \(3, 7, 11, 15, 19, \dots\)
This is an arithmetic progression with first term \(a = 3\) and common difference \(d = 4\).
The \(n\)-th term of the numerator is:
\(3 + 4(n-1) = 4n - 1\)

2. **Denominators:**
The sequence of denominators is \(2, 5, 10, 17, 26, \dots\)
These numbers are all 1 more than a perfect square:
\(1^2 + 1 = 2\)
\(2^2 + 1 = 5\)
\(3^2 + 1 = 10\)
\(4^2 + 1 = 17\)
\(5^2 + 1 = 26\)
So the \(n\)-th term of the denominator is:
\(n^2 + 1\)

Combining these results, the \(n\)-th term of the sequence is:
\(\frac{4n - 1}{n^2 + 1}\)

M1 for identifying the sequence of numerators is arithmetic and finding \(4n - 1\) (or equivalent)
M1 for identifying the sequence of denominators is quadratic and finding \(n^2 + 1\) (or equivalent)
A1 for the correct combined fraction \(\frac{4n - 1}{n^2 + 1}\)

Let the \(n\)-th term of the sequence be represented by a quadratic expression of the form \(an^2 + bn + c\).

First, find the differences between successive terms:
- First differences: \(15-4=11\), \(32-15=17\), \(55-32=23\), \(84-55=29\)
- Second differences: \(17-11=6\), \(23-17=6\), \(29-23=6\)

Since the second differences are constant and equal to \(6\), we have:
\(2a = 6 \implies a = 3\)

Now, subtract the term \(3n^2\) from each term of the original sequence to find the linear part:
- For \(n=1\): \(4 - 3(1)^2 = 1\)
- For \(n=2\): \(15 - 3(2)^2 = 3\)
- For \(n=3\): \(32 - 3(3)^2 = 5\)
- For \(n=4\): \(55 - 3(4)^2 = 7\)
- For \(n=5\): \(84 - 3(5)^2 = 9\)

The resulting linear sequence is \(1, 3, 5, 7, 9, \dots\)
This is an arithmetic progression with first term \(1\) and common difference \(2\), which has the general term:
\(1 + 2(n-1) = 2n - 1\)

Combining the quadratic and linear parts, the \(n\)-th term of the sequence is:
\(3n^2 + 2n - 1\)

M1 for finding the constant second difference is 6 and establishing \(a = 3\)
M1 for subtracting \(3n^2\) from the terms and attempting to find the \(n\)-th term of the resulting linear sequence (or setting up and solving a system of simultaneous equations)
A1 for the final correct expression \(3n^2 + 2n - 1\)

The vertex form of a quadratic equation is given by:
\(y = a(x - h)^2 + k\)

Substituting the coordinates of the vertex \((h, k) = (3, -4)\):
\(y = a(x - 3)^2 - 4\)

Now, use the coordinates of the given point \((1, 8)\) to find the value of \(a\):
\(8 = a(1 - 3)^2 - 4\)
\(8 = a(-2)^2 - 4\)
\(8 = 4a - 4\)
\(12 = 4a \implies a = 3\)

Substitute \(a = 3\) back into the vertex form and expand the expression:
\(y = 3(x - 3)^2 - 4\)
\(y = 3(x^2 - 6x + 9) - 4\)
\(y = 3x^2 - 18x + 27 - 4\)
\(y = 3x^2 - 18x + 23\)

M1 for substituting the vertex into the form \(y = a(x - 3)^2 - 4\)
M1 for substituting the point \((1, 8)\) to find \(a = 3\)
A1 for expanding correctly to get the final equation \(y = 3x^2 - 18x + 23\)

Since the curve has \(x\)-intercepts at \(x = -2\) and \(x = 5\), its equation can be written in factored form as:
\(y = a(x + 2)(x - 5)\)

Since the curve passes through the point \((3, -30)\), substitute \(x = 3\) and \(y = -30\) into the equation:
\(-30 = a(3 + 2)(3 - 5)\)
\(-30 = a(5)(-2)\)
\(-30 = -10a \implies a = 3\)

Substitute \(a = 3\) back and expand to obtain the standard quadratic form:
\(y = 3(x + 2)(x - 5)\)
\(y = 3(x^2 - 3x - 10)\)
\(y = 3x^2 - 9x - 30\)

M1 for using the intercepts to write the factored form \(y = a(x + 2)(x - 5)\)
M1 for substituting the point \((3, -30)\) to find \(a = 3\)
A1 for expanding correctly to obtain the standard equation \(y = 3x^2 - 9x - 30\)

To make \(x\) the subject of the formula:

Multiply both sides of the equation by \(3x - 4\):
\(y(3x - 4) = 5x + 2\)

Expand the left side:
\(3xy - 4y = 5x + 2\)

Collect all terms containing \(x\) on one side and the rest of the terms on the other side:
\(3xy - 5x = 4y + 2\)

Factorize out \(x\):
\(x(3y - 5) = 4y + 2\)

Divide both sides by \(3y - 5\):
\(x = \t\frac{4y + 2}{3y - 5}\)

M1 for removing the fraction to get \(y(3x - 4) = 5x + 2\)
M1 for correctly isolating all \(x\) terms to one side, e.g., \(3xy - 5x = 4y + 2\)
A1 for the correct final formula \(x = \frac{4y + 2}{3y - 5}\) (or equivalent such as \(x = \frac{-4y - 2}{5 - 3y}\))

To find the value of \(x\) for which \(g(f(x)) = 3\), we can work from the outside in.

Let \(y = f(x)\). Then:
\(g(y) = 3\)
\(\frac{6}{y + 1} = 3\)

Solve for \(y\):
\(6 = 3(y + 1)\)
\(2 = y + 1\)
\(y = 1\)

Since \(y = f(x)\), substitute \(f(x) = 1\):
\(2x - 3 = 1\)
\(2x = 4\)
\(x = 2\)

M1 for substituting \(f(x)\) into \(g(x)\) to get \(\frac{6}{(2x - 3) + 1} = 3\) or for solving \(g(y) = 3\) to find the intermediate value \(1\)
M1 for attempting to solve the resulting equation \(2x - 3 = 1\) or \(\frac{6}{2x - 2} = 3\)
A1 for \(x = 2\)

(a) The given sequence is \(5, 11, 19, 29, \dots\).
Calculating the first differences:
\(11 - 5 = 6\)
\(19 - 11 = 8\)
\(29 - 19 = 10\)

The first differences are increasing by 2 each time. Thus, the next difference is \(10 + 2 = 12\).
The 5th term (Diagram 5) is \(29 + 12 = 41\).

(b) Since the second differences are constant and equal to 2, the sequence has a quadratic formula of the form \(an^2 + bn + c\), where \(2a = 2 \implies a = 1\).
Subtracting \(n^2\) from the terms of the sequence:
For \(n = 1\): \(5 - 1^2 = 4\)
For \(n = 2\): \(11 - 2^2 = 7\)
For \(n = 3\): \(19 - 3^2 = 10\)
For \(n = 4\): \(29 - 4^2 = 13\)

The sequence of remainders is \(4, 7, 10, 13, \dots\), which is a linear sequence with a common difference of 3.
The general term for this linear part is \(3n + 1\).
Combining these parts, the general expression for the \(n\)-th term is \(n^2 + 3n + 1\).

M1 for finding the differences 6, 8, 10 or continuing the sequence to find the next term.
A1 for 41.
M1 for recognising a quadratic sequence and finding the coefficient of \(n^2\) is 1.
M1 for subtracting \(n^2\) and finding the linear part \(3n + 1\).
A1 for \(n^2 + 3n + 1\) (or equivalent).

(a) We are given:
\(T_2 = a \cdot b = 12\)
\(T_4 = a \cdot b^3 = 27\)

Dividing \(T_4\) by \(T_2\):
\(\frac{a \cdot b^3}{a \cdot b} = \frac{27}{12}\)
\(b^2 = \frac{9}{4}\)

Since \(b > 0\), we take the positive square root:
\(b = \sqrt{\frac{9}{4}} = \frac{3}{2} = 1.5\).

(b) Using the equation \(a \cdot b = 12\) with \(b = 1.5\):
\(a \cdot 1.5 = 12\)
\(a = \frac{12}{1.5} = 8\).

(c) The 6th term is given by:
\(T_6 = a \cdot b^5\)
\(T_6 = 8 \cdot \left(\frac{3}{2}\right)^5\)
\(T_6 = 8 \cdot \frac{243}{32} = \frac{243}{4} = 60.75\).

M1 for writing down the equations \(ab = 12\) and \(ab^3 = 27\).
M1 for dividing to obtain \(b^2 = 2.25\) (or \(b^2 = \frac{27}{12}\)).
A1 for \(b = 1.5\).
A1 for \(a = 8\).
M1 for calculating \(T_6 = 8 \cdot (1.5)^5\).
A1 for \(\frac{243}{4}\) or \(60 \frac{3}{4}\) (accept 60.75).

(a) Substitute \(x = 0\) and \(y = 5\) into the quadratic equation:
\(5 = a(0)^2 + b(0) + c \implies c = 5\).

(b) Now the equation is \(y = ax^2 + bx + 5\).
Substitute \((2, 11)\):
\(11 = a(2)^2 + b(2) + 5 \implies 4a + 2b = 6 \implies 2a + b = 3\) (Equation 1).
Substitute \((-1, 8)\):
\(8 = a(-1)^2 + b(-1) + 5 \implies a - b + 5 = 8 \implies a - b = 3\) (Equation 2).

(c) Adding Equation 1 and Equation 2:
\((2a + b) + (a - b) = 3 + 3\)
\(3a = 6 \implies a = 2\).

Substitute \(a = 2\) into Equation 2:
\(2 - b = 3 \implies b = -1\).

Therefore, the equation of the curve is \(y = 2x^2 - x + 5\).

B1 for \(c = 5\).
M1 for substituting \((2, 11)\) to get \(4a + 2b = 6\) (or equivalent).
M1 for substituting \((-1, 8)\) to get \(a - b = 3\) (or equivalent).
M1 for a valid method to solve the simultaneous equations.
A1 for \(a = 2\) and \(b = -1\).
A1 for the final equation \(y = 2x^2 - x + 5\).

(a) A quadratic curve is symmetrical about the vertical line passing through its vertex. Here, the line of symmetry is \(x = 3\).
One \(x\)-intercept is at \((1, 0)\), which is 2 units to the left of the line of symmetry.
The other \(x\)-intercept must be 2 units to the right of \(x = 3\), which is at \(x = 3 + 2 = 5\).
Thus, the coordinates of the other intercept are \((5, 0)\).

(b) The intercepts are at \(x = 1\) and \(x = 5\), so we can set \(p = 1\) and \(q = 5\) (or vice versa).
The function is \(f(x) = k(x - 1)(x - 5)\).
Using the vertex \((3, -8)\) as a point on the curve:
\(-8 = k(3 - 1)(3 - 5)\)
\(-8 = k(2)(-2)\)
\(-8 = -4k \implies k = 2\).

(c) Expand the expression for \(f(x)\):
\(f(x) = 2(x - 1)(x - 5)\)
\(f(x) = 2(x^2 - 6x + 5)\)
\(f(x) = 2x^2 - 12x + 10\).

B1 for identifying \(x = 5\) or the point \((5, 0)\).
B1 for \(p = 1\) and \(q = 5\) (accept transposed values).
M1 for substituting \((3, -8)\) into \(y = k(x - p)(x - q)\) to solve for \(k\).
A1 for \(k = 2\).
M1 for expanding the brackets \(2(x-1)(x-5)\).
A1 for \(f(x) = 2x^2 - 12x + 10\).

(a) First, factorise out the common factor of 3:
\(12x^2 - 3y^2 = 3(4x^2 - y^2)\)

Now, use the difference of two squares formula \(a^2 - b^2 = (a - b)(a + b)\) for \(4x^2 - y^2\):
\(4x^2 - y^2 = (2x)^2 - (y)^2 = (2x - y)(2x + y)\)

Thus, the fully factorised form is:
\(3(2x - y)(2x + y)\).

(b) Factorise the numerator: \(2x^2 - 5x - 3\).
We look for numbers that multiply to \(-6\) and add to \(-5\), which are \(-6\) and \(1\):
\(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).

Factorise the denominator: \(4x^2 - 1\).
Using the difference of squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\).

Now rewrite the fraction and simplify by cancelling common factors:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

M1 for factorising 3 to get \(3(4x^2 - y^2)\).
A1 for \(3(2x - y)(2x + y)\).
M1 for factorising the quadratic numerator into \((2x + 1)(x - 3)\).
M1 for factorising the denominator into \((2x - 1)(2x + 1)\).
A1 for \(\frac{x - 3}{2x - 1}\).

(a) Start by multiplying both sides by \((5 - 2x)\) to clear the fraction:
\(y(5 - 2x) = 3x + 2\)
\(5y - 2xy = 3x + 2\)

Rearrange the equation to group all terms with \(x\) on one side:
\(5y - 2 = 3x + 2xy\)

Factorise \(x\) from the right-hand side:
\(5y - 2 = x(3 + 2y)\)

Divide by \((3 + 2y)\) to make \(x\) the subject:
\(x = \frac{5y - 2}{2y + 3}\).

(b) Multiply the entire equation by the common denominator \((x - 2)(x + 1)\):
\(4(x + 1) - 3(x - 2) = 1(x - 2)(x + 1)\)

Expand both sides:
\(4x + 4 - 3x + 6 = x^2 - x - 2\)
\(x + 10 = x^2 - x - 2\)

Rearrange to form a quadratic equation equal to 0:
\(x^2 - 2x - 12 = 0\)

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-12)}}{2(1)}\)
\(x = \frac{2 \pm \sqrt{4 + 48}}{2} = \frac{2 \pm \sqrt{52}}{2}\)

Calculate the numerical values:
\(x = \frac{2 + 7.2111}{2} \approx 4.61\)
\(x = \frac{2 - 7.2111}{2} \approx -2.61\).

M1 for clearing the fraction to get \(y(5 - 2x) = 3x + 2\).
M1 for isolating the \(x\) terms on one side of the equation.
M1 for factoring out \(x\) to get \(x(2y + 3) = 5y - 2\).
A1 for \(x = \frac{5y - 2}{2y + 3}\) (or equivalent).
M1 for expanding and eliminating the fractions to get \(4(x + 1) - 3(x - 2) = (x - 2)(x + 1)\).
M1 for simplifying to a standard quadratic equation: \(x^2 - 2x - 12 = 0\).
M1 for using a correct quadratic formula method or equivalent calculator solve.
A1 for \(x = 4.61\) and A1 for \(x = -2.61\) (both must be to 2 d.p.).

(a) Substitute \(x = 3\) into the function \(f(x)\):
\(f(3) = \frac{2(3) + 3}{3 - 1} = \frac{6 + 3}{2} = \frac{9}{2} = 4.5\).

(b) Set \(y = \frac{2x + 3}{x - 1}\) and swap/rearrange to make \(x\) the subject:
\(y(x - 1) = 2x + 3\)
\(xy - y = 2x + 3\)
\(xy - 2x = y + 3\)
\(x(y - 2) = y + 3\)
\(x = \frac{y + 3}{y - 2}\)

So, the inverse function is:
\(f^{-1}(x) = \frac{x + 3}{x - 2}\).

(c) Solve \(g(f(x)) = 7\):
\((f(x))^2 - 2 = 7\)
\((f(x))^2 = 9\)
\(f(x) = 3\) or \(f(x) = -3\)

Case 1: \(f(x) = 3\)
\(\frac{2x + 3}{x - 1} = 3\)
\(2x + 3 = 3(x - 1)\)
\(2x + 3 = 3x - 3 \implies x = 6\).

Case 2: \(f(x) = -3\)
\(\frac{2x + 3}{x - 1} = -3\)
\(2x + 3 = -3(x - 1)\)
\(2x + 3 = -3x + 3\)
\(5x = 0 \implies x = 0\).

Thus, the solutions are \(x = 0\) and \(x = 6\).

B1 for \(4.5\) (or \(\frac{9}{2}\)).
M1 for setting up the equation \(y(x - 1) = 2x + 3\).
M1 for factoring out \(x\) to get \(x(y - 2) = y + 3\).
A1 for \(f^{-1}(x) = \frac{x + 3}{x - 2}\).
M1 for setting \((f(x))^2 - 2 = 7\) and concluding \(f(x) = \pm 3\).
M1 for solving \(f(x) = 3\) or \(f(x) = -3\).
A1 for \(x = 6\) and A1 for \(x = 0\).

(a) First, find \(j(-2)\):
\(j(-2) = 2(-2) + 5 = -4 + 5 = 1\).
Now find \(h(1)\):
\(h(1) = 3^1 = 3\).

(b) We want to solve \(j(h(x)) = 59\):
\(2(3^x) + 5 = 59\)
\(2(3^x) = 54\)
\(3^x = 27\)
Since \(27 = 3^3\), we have:
\(x = 3\).

(c) We need to solve the inequality \(3^x > 2x + 5\) for \(x > 0\).
Let us find the point of intersection where \(3^x = 2x + 5\) for \(x > 0\).
By inspection or graphical calculator exploration:
If \(x = 2\):
\(3^2 = 9\)
\(2(2) + 5 = 9\)
Thus, \(x = 2\) is the intersection point.
Since the exponential function \(3^x\) grows much faster than the linear function \(2x + 5\) for positive values of \(x\), the curve \(y = 3^x\) lies above the line \(y = 2x + 5\) for all values of \(x\) greater than the intersection point.
Therefore, the solution to the inequality is \(x > 2\).

M1 for finding \(j(-2) = 1\).
A1 for \(3\).
M1 for writing \(2(3^x) + 5 = 59\).
M1 for obtaining \(3^x = 27\).
A1 for \(x = 3\).
M1 for finding the intersection point \(x = 2\) by setting up \(3^x = 2x + 5\).
A1 for the correct inequality \(x > 2\).

Let the \(n\)-th term be \(u_n = an^2 + bn + c\). (a) The terms are: \(u_1 = 4\), \(u_2 = 11\), \(u_3 = 22\), \(u_4 = 37\), \(u_5 = 56\). First differences: 7, 11, 15, 19. Second differences: 4, 4, 4. Since the second difference is constant, it is a quadratic sequence with \(2a = 4\) which gives \(a = 2\). Subtracting \(2n^2\) from each term gives the sequence: \(4 - 2(1)^2 = 2\), \(11 - 2(2)^2 = 3\), \(22 - 2(3)^2 = 4\). This sequence 2, 3, 4, ... has the linear formula \(n+1\). Thus, the \(n\)-th term is \(u_n = 2n^2 + n + 1\). (b) For the 15th term, substitute \(n = 15\) into the expression: \(u_{15} = 2(15)^2 + 15 + 1 = 2(225) + 16 = 450 + 16 = 466\). (c) Set the \(n\)-th term equal to 407: \(2n^2 + n + 1 = 407\), which simplifies to \(2n^2 + n - 406 = 0\). Using the quadratic formula, \(n = \frac{-1 \pm \sqrt{1^2 - 4(2)(-406)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 3248}}{4} = \frac{-1 \pm \sqrt{3249}}{4\). Since \(\sqrt{3249} = 57\), we get \(n = \frac{-1 + 57}{4} = 14\) (since \(n\) must be positive).

(a) M1 for finding second difference of 4 or stating form \(2n^2 + bn + c\). A1 for \(2n^2 + n + 1\). (b) B1 for substituting \(n = 15\) into their expression. B1 for 466. (c) M1 for setting their quadratic expression equal to 407 and trying to solve. A1 for 14.

(a) Sequence A is geometric with first term 3 and common ratio 2. The 6th term is \(3 \times 2^5 = 96\). Sequence B is arithmetic with first term 5 and common difference 4. The 6th term is \(5 + 5 \times 4 = 25\). Therefore, the 6th term of Sequence C is \(w_6 = \frac{96}{25}\). (b) The first term of Sequence A is \(a = 3\) and the common ratio is \(r = 2\). The \(n\)-th term is \(3 \times 2^{n-1}\). (c) The first term of Sequence B is \(b_1 = 5\) and common difference is \(d = 4\). The \(n\)-th term of Sequence B is \(5 + (n-1)4 = 4n + 1\). Therefore, the \(n\)-th term of Sequence C is \(w_n = \frac{3 \times 2^{n-1}}{4n + 1}\).

(a) B1 for 96/25 or equivalent decimal 3.84. (b) B1 for recognizing common ratio is 2. B1 for \(3 \times 2^{n-1}\). (c) B1 for identifying the \(n\)-th term of Sequence B as \(4n + 1\). B1 for final fraction.

(a) Substituting the points into \(f(x) = ax^2 + bx + c\): For \((1, -2)\), \(a + b + c = -2\) (Equation 1). For \((2, 3)\), \(4a + 2b + c = 3\) (Equation 2). For \((4, 25)\), \(16a + 4b + c = 25\) (Equation 3). Subtracting Equation 1 from Equation 2 gives \(3a + b = 5\), which means \(b = 5 - 3a\). Subtracting Equation 2 from Equation 3 gives \(12a + 2b = 22\), which simplifies to \(6a + b = 11\). Substituting \(b = 5 - 3a\) into \(6a + b = 11\) gives \(6a + 5 - 3a = 11\), so \(3a = 6\) and \(a = 2\). Then, \(b = 5 - 3(2) = -1\). Substituting \(a = 2\) and \(b = -1\) into Equation 1 gives \(2 - 1 + c = -2\), so \(c = -3\). Thus, \(a = 2, b = -1, c = -3\). (b) The \(x\)-coordinate of the vertex is \(x = -\frac{b}{2a} = -\frac{-1}{2(2)} = 0.25\). The \(y\)-coordinate is \(f(0.25) = 2(0.25)^2 - 0.25 - 3 = 0.125 - 0.25 - 3 = -3.125\). Therefore, the coordinates of the minimum point are \((0.25, -3.125)\). (c) Solve \(2x^2 - x - 3 = 0\): factoring gives \((2x - 3)(x + 1) = 0\), so \(x = 1.5\) or \(x = -1\).

(a) M1 for writing down at least two correct simultaneous equations. M1 for solving to find one coefficient. A1 for all three correct: \(a=2, b=-1, c=-3\). (b) M1 for finding \(x = -b/2a = 0.25\) or utilizing vertex form/differentiation. A1 for \((0.25, -3.125)\). (c) B1 for \(x = 1.5\) and B1 for \(x = -1\) (or factorizing correctly).

(a) Since the \(x\)-intercepts are at \(-2\) and \(5\), we have \(p = -2\) and \(q = 5\) (since \(p < q\)). So, \(g(x) = k(x + 2)(x - 5)\). Substitute \((3, -30)\) into the function: \(-30 = k(3 + 2)(3 - 5)\), which gives \(-30 = k(5)(-2)\), so \(-30 = -10k\) and \(k = 3\). (b) Expand the function: \(g(x) = 3(x + 2)(x - 5) = 3(x^2 - 3x - 10) = 3x^2 - 9x - 30\). So \(a = 3, b = -9, c = -30\). (c) The vertex is on the axis of symmetry, which lies midway between the two \(x\)-intercepts: \(x = \frac{-2 + 5}{2} = 1.5\). The \(y\)-coordinate of the vertex is: \(g(1.5) = 3(1.5 + 2)(1.5 - 5) = 3(3.5)(-3.5) = -36.75\). The vertex coordinates are \((1.5, -36.75)\).

(a) B1 for identifying \(p = -2, q = 5\). M1 for substituting \((3, -30)\) to find \(k\). A1 for \(k=3\). (b) M1 for expanding \(3(x+2)(x-5)\). A1 for \(a=3, b=-9, c=-30\). (c) M1 for \(x = 1.5\). A1 for \((1.5, -36.75)\).

(a) Multiply both sides by \((5 - 3t)\) to get \(S(5 - 3t) = 2t + 3\), so \(5S - 3St = 2t + 3\). Rearrange to group the terms with \(t\) on one side: \(5S - 3 = 2t + 3St\). Factor out \(t\): \(5S - 3 = t(2 + 3S)\). Divide by \(2 + 3S\) to get \(t = \frac{5S - 3}{3S + 2}\). (b) Factor the numerator and the denominator: Numerator is \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\) and Denominator is \(x^2 - 9 = (x - 3)(x + 3)\). Thus, \(\frac{2x^2 - 5x - 3}{x^2 - 9} = \frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}\). (c) Expand each part: \((2x - 3)^2 = 4x^2 - 12x + 9\) and \((x - 4)(x + 4) = x^2 - 16\). Subtract the second from the first: \((4x^2 - 12x + 9) - (x^2 - 16) = 4x^2 - 12x + 9 - x^2 + 16 = 3x^2 - 12x + 25\).

(a) M1 for multiplying by denominator to get \(5S - 3St = 2t + 3\). M1 for collecting terms in \(t\) on one side. A1 for \(t = \frac{5S - 3}{3S + 2}\) or equivalent. (b) M1 for factoring numerator to \((2x+1)(x-3)\). M1 for factoring denominator to \((x-3)(x+3)\). A1 for \(\frac{2x+1}{x+3}\). (c) B1 for \(4x^2 - 12x + 9\). B1 for \(x^2 - 16\). A1 for \(3x^2 - 12x + 25\).

(a) First find \(g(-3) = (-3)^2 + 2 = 9 + 2 = 11\). Then find \(f(11) = 3(11) - 5 = 33 - 5 = 28\). (b) Let \(y = \frac{4}{x - 1}\). Rearrange to make \(x\) the subject: \(y(x - 1) = 4 \implies xy - y = 4 \implies xy = y + 4 \implies x = \frac{y + 4}{y}\). So, \(h^{-1}(x) = \frac{x + 4}{x}\) (which can also be written as \(1 + \frac{4}{x}\)). (c) Substitute \(f(x)\) into \(g(x)\) and subtract 3: \(g(f(x)) - 3 = (3x - 5)^2 + 2 - 3 = (9x^2 - 30x + 25) - 1 = 9x^2 - 30x + 24\).

(a) M1 for finding \(g(-3) = 11\). A1 for 28. (b) M1 for setting up \(y = 4/(x-1)\) and multiplying by \(x-1\). M1 for making \(x\) the subject. A1 for \(\frac{x+4}{x}\) or equivalent. (c) M1 for expanding \((3x-5)^2\) to get \(9x^2 - 30x + 25\). A1 for \(9x^2 - 30x + 24\).

(a) The \(n\)-th term of a geometric sequence is \(u_n = ar^{n-1}\). We are given: \(u_3 = ar^2 = 18\) (Equation 1) and \(u_6 = ar^5 = 486\) (Equation 2). Dividing Equation 2 by Equation 1 gives \(\frac{ar^5}{ar^2} = \frac{486}{18}\), which simplifies to \(r^3 = 27\), so \(r = 3\). Substitute \(r = 3\) into Equation 1: \(a(3)^2 = 18 \implies 9a = 18 \implies a = 2\). So the common ratio is 3 and the first term is 2. (b) The sum of the first \(n\) terms is given by \(S_n = \frac{a(r^n - 1)}{r - 1}\). For \(n = 8\): \(S_8 = \frac{2(3^8 - 1)}{3 - 1} = \frac{2(6561 - 1)}{2} = 6560\). (c) Set the \(n\)-th term equal to 39366: \(u_n = 2 \times 3^{n-1} = 39366\), so \(3^{n-1} = 19683\). Since \(3^9 = 19683\), we have \(n - 1 = 9\), which gives \(n = 10\).

(a) M1 for setting up \(ar^2 = 18\) and \(ar^5 = 486\). M1 for dividing the equations to get \(r^3 = 27\). A1 for \(r=3, a=2\). (b) M1 for substituting \(a=2\), \(r=3\), and \(n=8\) into the sum formula. A1 for 6560. (c) M1 for setting up \(2 \times 3^{n-1} = 39366\) and dividing by 2. A1 for 10.

(a) The vertex form is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Since the vertex is at \((3, -4)\), we have \(h = 3\) and \(k = -4\). (b) The equation is \(y = a(x - 3)^2 - 4\). Substitute the point \((1, 8)\) into the equation: \(8 = a(1 - 3)^2 - 4 \implies 8 = a(-2)^2 - 4 \implies 8 = 4a - 4 \implies 12 = 4a \implies a = 3\). (c) Expand the vertex form: \(y = 3(x - 3)^2 - 4 = 3(x^2 - 6x + 9) - 4 = 3x^2 - 18x + 27 - 4 = 3x^2 - 18x + 23\). So the values are \(b = -18\) and \(c = 23\).

(a) B1 for \(h = 3, k = -4\). (b) M1 for substituting \((1, 8)\) into their vertex form. A1 for \(a = 3\). (c) M1 for expanding \(3(x-3)^2 - 4\). A1 for \(b = -18, c = 23\).

(a) Substituting \(x = 0\) and \(y = 6\) into \(y = ax^2 + bx + c\) gives:
\(6 = a(0)^2 + b(0) + c\)
\(c = 6\)

(b) Substituting \(c = 6\) and the points \((2, 10)\) and \((-1, 13)\) into \(y = ax^2 + bx + c\):
For \((2, 10)\):
\(10 = a(2)^2 + b(2) + 6 \implies 4a + 2b = 4 \implies 2a + b = 2\)

For \((-1, 13)\):
\(13 = a(-1)^2 + b(-1) + 6 \implies a - b = 7\)

We now solve these simultaneous equations:
Adding the equations \(2a + b = 2\) and \(a - b = 7\):
\((2a + b) + (a - b) = 2 + 7\)
\(3a = 9 \implies a = 3\)

Substituting \(a = 3\) into \(a - b = 7\):
\(3 - b = 7 \implies b = -4\)

(a) B1 for \(c = 6\)
(b) M1 for substituting at least one of the points \((2, 10)\) or \((-1, 13)\) to form an equation in \(a\) and \(b\) (e.g., \(4a + 2b = 4\) or \(a - b = 7\))
M1 for a correct method to solve their two simultaneous linear equations
A1 for both \(a = 3\) and \(b = -4\)

(a)(i) Since each ring increases by 6 dots, for Ring 4 we have \(r_4 = 18 + 6 = 24\).
(a)(ii) The cumulative total for 4 rings is \(T_4 = T_3 + r_4 = 36 + 24 = 60\).
(b) Since \(r_n\) is an arithmetic sequence with first term 6 and common difference 6, its general term is \(r_n = 6n\).
(c) The sum of the first \(n\) terms of the sequence is given by \(T_n = \sum_{k=1}^{n} 6k = 6 \times \frac{n(n+1)}{2} = 3n(n+1) = 3n^2 + 3n\).
(d) We set up and solve the equation \(3n^2 + 3n = 540 \implies n^2 + n - 180 = 0\). Factoring the quadratic, we get \((n - 12)(n + 15) = 0\). Since \(n\) must be positive, we reject \(-15\) to find \(n = 12\).

(a)(i) B1 for 24.
(a)(ii) B1 for 60.
(b) B2 for 6n (or M1 for showing common difference of 6).
(c) M1 for attempting to sum the arithmetic sequence or using quadratic differences. M1 for finding the coefficient of the quadratic term as 3. A1 for 3n^2 + 3n.
(d) M1 for setting 3n^2 + 3n = 540. M1 for factoring or using the quadratic formula. A1 for n = 12.

(a)(i) Following the pattern of increasing differences, \(P_4 = 7 + 4 = 11\).
(a)(ii) Likewise, \(P_5 = 11 + 5 = 16\).
(b) The terms are 1, 2, 4, 7, 11, 16. The first differences are 1, 2, 3, 4, 5. The second differences are all 1. Because the second differences are constant and non-zero, the relationship is quadratic.
(c) Let \(P_n = an^2 + bn + c\). The constant second difference is 1, so \(2a = 1 \implies a = 0.5\). For \(n = 0\), \(P_0 = c \implies c = 1\). For \(n = 1\), \(P_1 = 0.5(1)^2 + b(1) + 1 = 2 \implies b = 0.5\). Thus, \(P_n = 0.5n^2 + 0.5n + 1\).
(d) We solve \(0.5n^2 + 0.5n + 1 \ge 100 \implies n^2 + n - 198 \ge 0\). Solving \(n^2 + n - 198 = 0\) with the quadratic formula yields \(n = \frac{-1 + \sqrt{793}}{2} \approx 13.58\). Since \(n\) must be an integer, we round up to 14. Testing \(n = 13\) gives 92 pieces, and \(n = 14\) gives 106 pieces. Thus, 14 cuts are required.

(a)(i) B1 for 11.
(a)(ii) B1 for 16.
(b) M1 for finding the first differences. A1 for showing the second differences are all 1.
(c) M1 for calculating a = 0.5. M1 for setting up equations to find b and c. A1 for P_n = 0.5n^2 + 0.5n + 1.
(d) M1 for the inequality 0.5n^2 + 0.5n + 1 >= 100. A1 for 14.

(a)(i) \(S_4 = 4^2 = 16\).
(a)(ii) \(P_4 = 4 \times 4 = 16\).
(a)(iii) \(S_5 = 5^2 = 25\).
(a)(iv) \(P_5 = 4 \times 5 = 20\).
(b) Since \(S_n\) represents the perfect squares, we have \(S_n = n^2\).
(c) Since \(P_n\) represents multiples of 4, we have \(P_n = 4n\).
(d) From part (b), since \(n > 0\), we have \(n = \sqrt{S_n}\). Substituting this into the expression for \(P_n\) from part (c), we get \(P_n = 4\sqrt{S_n}\).
(e) If \(P_n = 320\), then \(4n = 320 \implies n = 80\). The number of blocks is \(S_{80} = 80^2 = 6400\).

(a) B1 for both S_n values (16 and 25). B1 for both P_n values (16 and 20).
(b) B1 for S_n = n^2.
(c) B1 for P_n = 4n.
(d) M1 for rearranging to get n in terms of S_n. A1 for P_n = 4*sqrt(S_n).
(e) M1 for finding n = 80. A1 for 6400.

(a)(i) Given the left endpoint is \((0,0)\) and the total width is 24 m, the right end is \((24, 0)\).
(a)(ii) By symmetry, the highest point occurs at \(x = 12\). The maximum height is 8 m, so the vertex is \((12, 8)\).
(b) Since the roots are at \(x = 0\) and \(x = 24\), we can write the function as \(y = ax(x - 24)\). Substituting the vertex \((12, 8)\) yields \(8 = a(12)(-12) \implies 8 = -144a \implies a = -\frac{1}{18}\). Thus, the equation is \(y = -\frac{1}{18}x(x - 24) = -\frac{1}{18}x^2 + \frac{4}{3}x\).
(c) A symmetrical 10 m wide barge centered at \(x = 12\) has its outer edges at \(x = 12 - 5 = 7\) and \(x = 12 + 5 = 17\). The height of the arch at \(x = 7\) is: \(y = -\frac{1}{18}(7)^2 + \frac{4}{3}(7) = -\frac{49}{18} + \frac{28}{3} = \frac{119}{18} \approx 6.611\) m. Subtracting the required safety clearance of 1.2 m gives a maximum height of \(\frac{119}{18} - 1.2 = \frac{487}{90} \approx 5.41\) metres.

(a)(i) B1 for (24, 0).
(a)(ii) B1 for (12, 8).
(b) M1 for using factored form y = ax(x - 24) or vertex form. M1 for substituting (12, 8) to find a. A2 for fully establishing the equation.
(c) M1 for identifying x = 7 or 17 as the critical point. M1 for calculating the arch height at this point as 119/18 or approx 6.61. A1 for subtracting 1.2. A1 for the final answer 5.41.

(a) The sum of the dimensions is \(x + 3x + h = 150 \implies 4x + h = 150 \implies h = 150 - 4x\).
(b) The volume of the rectangular box is \(V = \text{Length} \times \text{Width} \times \text{Height} = (3x)(x)(150 - 4x) = 3x^2(150 - 4x) = 450x^2 - 12x^3\).
(c) For a physical box, we must have both \(x > 0\) and \(h > 0\). Since \(h = 150 - 4x > 0 \implies 4x < 150 \implies x < 37.5\). Thus, the domain is \(0 < x < 37.5\).
(d)(i) Finding the derivative of the volume function: \(V'(x) = 900x - 36x^2\). Setting \(V'(x) = 0 \implies 36x(25 - x) = 0\). Since \(x > 0\), the maximum occurs at \(x = 25\) cm.
(d)(ii) Evaluating the volume at this point yields \(V(25) = 450(25)^2 - 12(25)^3 = 93750\) \(\text{cm}^3\).

(a) M1 for x + 3x + h = 150. A1 for h = 150 - 4x.
(b) M1 for using the formula V = l * w * h. A1 for correct expansion to get 450x^2 - 12x^3.
(c) M1 for setting h > 0. A1 for 0 < x < 37.5.
(d)(i) M1 for using differentiation or GDC maximum solver. A1 for x = 25.
(d)(ii) M1 for substituting x = 25 into volume expression. A1 for 93750.