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

First, find the differences between consecutive terms:
- First differences: \(9 - 2 = 7\), \(20 - 9 = 11\), \(35 - 20 = 15\), \(54 - 35 = 19\).
- Second differences: \(11 - 7 = 4\), \(15 - 11 = 4\), \(19 - 15 = 4\).

Since the second differences are constant, the sequence is quadratic and has the general form \(a n^2 + b n + c\).
The coefficient \(a\) is half of the second difference:
\(2a = 4 \implies a = 2\).

Subtract \(2n^2\) from each term of the original sequence:
- For \(n = 1\): \(2 - 2(1)^2 = 0\)
- For \(n = 2\): \(9 - 2(2)^2 = 1\)
- For \(n = 3\): \(20 - 2(3)^2 = 2\)
- For \(n = 4\): \(35 - 2(4)^2 = 3\)

The resulting linear sequence is \(0, 1, 2, 3, \dots\), which has the formula \(n - 1\).

Combining these, we get the expression for the \(n\)-th term:
\(2n^2 + n - 1\).

M1 for finding first and second differences.
M1 for setting up the quadratic coefficient \(a = 2\).
A1 for the correct final expression \(2n^2 + n - 1\).

To clear the fractions, multiply the entire equation by the common denominator \(x(x+1)\):
\(4(x+1) - 3x = 1 \cdot x(x+1)\)

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

Subtract \(x\) from both sides:
\(x^2 = 4\)

Take the square root of both sides to obtain the solutions:
\(x = 2\) or \(x = -2\).

M1 for multiplying by \(x(x+1)\) to clear denominators.
M1 for simplifying to \(x^2 = 4\).
A1 for both correct solutions \(x = 2\) and \(x = -2\).

Substitute \(t = 30\) into the given formula:
\(M = 400 \times 2^{-0.1(30)}\)
\(M = 400 \times 2^{-3}\)

Since \(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\), we calculate:
\(M = 400 \times \frac{1}{8}\)
\(M = 50\) grams.

M1 for substituting \(t = 30\) to get \(400 \times 2^{-3}\).
M1 for evaluating \(2^{-3} = \frac{1}{8}\).
A1 for the correct answer of \(50\).

(a) The line of best fit must pass through the mean point \((\bar{x}, \bar{y}) = (5, 8)\). Substituting these coordinates into the equation:
\(8 = m(5) + 18\)
\(5m = 8 - 18\)
\(5m = -10 \implies m = -2\).

(b) Now substitute \(m = -2\) and \(x = 7.5\) into the line of best fit equation:
\(y = -2(7.5) + 18\)
\(y = -15 + 18 = 3\).

M1 for substituting \((5, 8)\) into the line equation.
A1 for obtaining \(m = -2\).
A1 for substituting \(x = 7.5\) to find \(y = 3\).

The vertex form of a quadratic curve is given by \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex.
Given the vertex is \((3, -6)\), we can write the equation as:
\(y = a(x - 3)^2 - 6\)

Expanding the equation:
\(y = a(x^2 - 6x + 9) - 6\)
\(y = ax^2 - 6ax + (9a - 6)\)

Now, compare this with the given equation \(y = ax^2 + bx + 12\):
By equating the constant terms:
\(9a - 6 = 12\)
\(9a = 18 \implies a = 2\).

By equating the \(x\)-coefficients:
\(b = -6a\)
\(b = -6(2) = -12\).

M1 for using the vertex form to set up \(y = a(x-3)^2 - 6\) (or using \(-\frac{b}{2a} = 3\)).
M1 for comparing constant terms or substituting the point to get \(9a - 6 = 12\).
A1 for both correct values: \(a = 2\) and \(b = -12\).

(a) Substitute the coordinates of the point \((2, 3)\) into the equation \(y = \frac{k}{x^2}\):
\(3 = \frac{k}{2^2}\)
\(3 = \frac{k}{4} \implies k = 12\).

(b) Substitute \(k = 12\) and \(x = \frac{1}{2}\) into the equation:
\(y = \frac{12}{(1/2)^2}\)
\(y = \frac{12}{1/4}\)
\(y = 12 \times 4 = 48\).

M1 for substituting \((2, 3)\) to find \(k = 12\).
M1 for substituting \(x = \frac{1}{2}\) into \(y = \frac{12}{x^2}\).
A1 for both correct answers: \(k = 12\) and \(y = 48\).

(a) First, calculate \(g(4)\):
\(g(4) = \frac{6}{4-1} = \frac{6}{3} = 2\).
Now, find \(f(g(4)) = f(2)\):
\(f(2) = 2(2) - 3 = 4 - 3 = 1\).

(b) To find the inverse function \(f^{-1}(x)\), set \(y = 2x - 3\) and solve for \(x\):
\(y + 3 = 2x\)
\(x = \frac{y+3}{2}\)

Therefore, \(f^{-1}(x) = \frac{x+3}{2}\).

M1 for finding \(g(4) = 2\) and evaluating \(f(2)\).
M1 for rearranging \(y = 2x - 3\) to make \(x\) the subject.
A1 for both correct answers: \(1\) and \(\frac{x+3}{2}\).

First, find the gradient of \(L_1\):
\(m_1 = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2\).

The gradient of the line perpendicular to \(L_1\), denoted by \(m_2\), is the negative reciprocal:
\(m_2 = -\frac{1}{-2} = \frac{1}{2}\).

Next, find the coordinates of the midpoint of \(AB\):
\(M = \left(\frac{2+6}{2}, \frac{5 + (-3)}{2}\right) = \left(\frac{8}{2}, \frac{2}{2}\right) = (4, 1)\).

Now, use the point-slope form with the midpoint \((4, 1)\) and gradient \(m = \frac{1}{2}\):
\(y - 1 = \frac{1}{2}(x - 4)\)
\(y - 1 = \frac{1}{2}x - 2\)
\(y = \frac{1}{2}x - 1\).

M1 for finding the gradient of \(L_1\) is \(-2\) and therefore the perpendicular gradient is \(\frac{1}{2}\).
M1 for finding the midpoint of \(AB\) is \((4, 1)\).
A1 for the correct equation \(y = \frac{1}{2}x - 1\) (or equivalent form in decimals: \(y = 0.5x - 1\)).

The differences between consecutive terms are: 12 - 5 = 7, 25 - 12 = 13, 44 - 25 = 19, 69 - 44 = 25. The second differences are: 13 - 7 = 6, 19 - 13 = 6, 25 - 19 = 6. Since the second difference is constant, the sequence is quadratic of the form \(an^2 + bn + c\) where \(2a = 6\), so \(a = 3\). Subtracting \(3n^2\) from each term of the sequence gives: \(5 - 3 = 2\), \(12 - 12 = 0\), \(25 - 27 = -2\), \(44 - 48 = -4\). The linear sequence \(2, 0, -2, -4, \dots\) has first term 2 and common difference \(-2\), which is given by \(-2n + 4\). Combining these, the \(n\)-th term is \(3n^2 - 2n + 4\).

M1 for finding second difference of 6. A1 for finding the coefficient of \(n^2\) is 3. A1 for the correct full expression \(3n^2 - 2n + 4\).

Multiply the first equation by 3: \(9x + 6y = 12\). Multiply the second equation by 2: \(10x - 6y = 26\). Add the two equations: \(19x = 38\), which gives \(x = 2\). Substitute \(x = 2\) into the first equation: \(3(2) + 2y = 4\), which simplifies to \(6 + 2y = 4\), so \(2y = -2\) and \(y = -1\).

M1 for a valid method to eliminate one variable. A1 for \(x = 2\). A1 for \(y = -1\).

Substitute \(M = 8\) into the equation: \(8 = 64 \times 2^{-0.25t}\). Divide both sides by 64: \(8/64 = 2^{-0.25t}\), which simplifies to \(1/8 = 2^{-0.25t}\). Since \(1/8 = 2^{-3}\), we can write the equation as \(2^{-3} = 2^{-0.25t}\). Equating the indices: \(-3 = -0.25t\), which gives \(t = 12\).

M1 for substituting \(M=8\) and isolating the exponential term to get \(1/8 = 2^{-0.25t}\). M1 for writing both sides with base 2 to get \(2^{-3} = 2^{-0.25t}\). A1 for \(t = 12\).

The line of best fit always passes through the mean point \((\bar{h}, \bar{s}) = (8, 65)\). Substituting these values into the equation of the line gives: \(65 = m(8) + 25\). Subtract 25 from both sides to get \(40 = 8m\). Dividing by 8 gives \(m = 5\).

M1 for identifying that the line passes through the mean point \((8, 65)\). M1 for substituting the mean values into the equation \(s = mh + 25\). A1 for \(m = 5\).

Using the vertex form of a quadratic function, \(y = a(x - h)^2 + k\) with vertex \((2, 5)\), we get \(y = a(x - 2)^2 + 5\). Substituting the point \((4, 13)\) into this equation: \(13 = a(4 - 2)^2 + 5\), which simplifies to \(13 = 4a + 5\). Subtracting 5 from both sides gives \(8 = 4a\), so \(a = 2\). Expanding the equation: \(y = 2(x - 2)^2 + 5 = 2(x^2 - 4x + 4) + 5 = 2x^2 - 8x + 8 + 5 = 2x^2 - 8x + 13\). Thus, \(a = 2\), \(b = -8\), and \(c = 13\).

M1 for writing the equation in vertex form \(y = a(x - 2)^2 + 5\). A1 for finding \(a = 2\). A1 for expanding to get \(b = -8\) and \(c = 13\).

A vertical asymptote occurs where the denominator is equal to zero. For the line \(x = 3\) to be the vertical asymptote, the denominator \(x - p\) must be zero when \(x = 3\). This gives \(3 - p = 0\), so \(p = 3\). The equation becomes \(y = \frac{k}{x - 3}\). To find \(k\), we substitute the coordinates of the given point \((5, 4)\) into the equation: \(4 = \frac{k}{5 - 3}\), which simplifies to \(4 = \frac{k}{2}\). Thus, \(k = 8\).

M1 for identifying \(p = 3\) from the vertical asymptote. M1 for substituting \(p = 3\) and \((5, 4)\) into the function. A1 for \(k = 8\).

First, find the composite function \(f(g(x)) = f\left(\frac{5}{x+1}\right) = 2\left(\frac{5}{x+1}\right) - 3 = \frac{10}{x+1} - 3\). Now, set this expression equal to 1: \(\frac{10}{x+1} - 3 = 1\). Add 3 to both sides: \(\frac{10}{x+1} = 4\). Multiply both sides by \(x+1\): \(10 = 4(x+1)\), which gives \(10 = 4x + 4\). Subtract 4 from both sides to get \(6 = 4x\), so \(x = 1.5\).

M1 for composing the functions to find \(f(g(x)) = \frac{10}{x+1} - 3\). M1 for solving the equation \(\frac{10}{x+1} = 4\). A1 for \(x = 1.5\) or \(3/2\).

First, calculate the gradient of \(L_1\): \(m_1 = \frac{11 - 3}{5 - 1} = \frac{8}{4} = 2\). Since \(L_2\) is perpendicular to \(L_1\), its gradient is the negative reciprocal of \(m_1\): \(m_2 = -\frac{1}{2} = -0.5\). Use the point-slope formula with the point \((4, -1)\) to find the equation of \(L_2\): \(y - (-1) = -0.5(x - 4)\). This simplifies to \(y + 1 = -0.5x + 2\). Subtracting 1 from both sides gives \(y = -0.5x + 1\).

M1 for finding the gradient of \(L_1\) to be 2. M1 for finding the perpendicular gradient of \(L_2\) to be \(-0.5\) (or \(-1/2\)). A1 for the correct equation \(y = -0.5x + 1\) (or equivalent form with fractions).

Find the differences between consecutive terms:
12 - 5 = 7
23 - 12 = 11
38 - 23 = 15

Find the second differences (the differences between the first differences):
11 - 7 = 4
15 - 11 = 4

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

Subtract \(2n^2\) from each term of the original sequence:
- For \(n = 1\): \(5 - 2(1)^2 = 3\)
- For \(n = 2\): \(12 - 2(2)^2 = 4\)
- For \(n = 3\): \(23 - 2(3)^2 = 5\)
- For \(n = 4\): \(38 - 2(4)^2 = 6\)

The sequence of differences is 3, 4, 5, 6, ..., which is a linear sequence with the \(n\)-th term formula \(n + 2\).

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

M1 for finding the constant second difference is 4 (or identifying the coefficient of \(n^2\) is 2).
M1 for subtracting \(2n^2\) from the sequence to find the linear part \(n+2\) (or for setting up correct simultaneous equations).
A1 for \(2n^2 + n + 2\).

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

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

Rearrange all terms to one side to form a quadratic equation:
\(x^2 - 3x - 4 = 0\)

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

This gives the solutions:
\(x = 4\) or \(x = -1\).

M1 for multiplying by the common denominator to obtain \(2(x+2) + 2(x-1) = (x-1)(x+2)\) or equivalent.
M1 for expanding and simplifying to a standard quadratic form \(x^2 - 3x - 4 = 0\).
A1 for both solutions \(x = 4\) and \(x = -1\).

Let the multiplier for one year be \(1 + \frac{r}{100}\).
Since the growth is exponential, the value after 2 years is given by:
\(400 \times \left(1 + \frac{r}{100}\right)^2 = 484\)

Divide both sides by 400:
\(\left(1 + \frac{r}{100}\right)^2 = \frac{484}{400}\)

Simplify the fraction:
\(\frac{484}{400} = \frac{121}{100}\)

Take the square root of both sides (since \(r > 0\), we take the positive root):
\(1 + \frac{r}{100} = \sqrt{\frac{121}{100}} = \frac{11}{10} = 1.1\)

Subtract 1 from both sides:
\[\frac{r}{100} = 0.1\]

Multiply by 100:
\(r = 10\)

M1 for setting up the equation \(400(1 + \frac{r}{100})^2 = 484\) or equivalent.
M1 for taking the square root to get \(1 + \frac{r}{100} = 1.1\) (or \(\frac{11}{10}\)).
A1 for \(r = 10\).

The vertex form of a quadratic function is given by:
\(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex.

Substituting the vertex \((2, 5)\):
\(y = a(x - 2)^2 + 5\)

Since the graph passes through \((4, -3)\), substitute \(x = 4\) and \(y = -3\) to find \(a\):
\(-3 = a(4 - 2)^2 + 5\)
\(-3 = a(2)^2 + 5\)
\(-3 = 4a + 5\)
\(-8 = 4a \implies a = -2\)

Now substitute \(a = -2\) back into the equation and expand to find the standard form:
\(y = -2(x - 2)^2 + 5\)
\(y = -2(x^2 - 4x + 4) + 5\)
\(y = -2x^2 + 8x - 8 + 5\)
\(y = -2x^2 + 8x - 3\)

M1 for writing the equation in vertex form as \(y = a(x - 2)^2 + 5\) (or equivalent system of equations).
M1 for substituting \((4, -3)\) into their equation to find \(a = -2\).
A1 for \(y = -2x^2 + 8x - 3\).

First, find the gradient of the given line \(3x - 4y = 8\) by writing it in gradient-intercept form:
\(4y = 3x - 8 \implies y = \frac{3}{4}x - 2\)

The gradient of this line is \(m_1 = \frac{3}{4}\).

Since the two lines are perpendicular, the product of their gradients is \(-1\):
\(m_2 = -\frac{1}{m_1} = -\frac{4}{3}\)

Now, use the point-slope formula with the point \((6, -1)\):
\(y - y_1 = m_2(x - x_1)\)
\(y - (-1) = -\frac{4}{3}(x - 6)\)
\(y + 1 = -\frac{4}{3}x + 8\)
\(y = -\frac{4}{3}x + 7\)

M1 for finding the gradient of the given line is \(\frac{3}{4}\) (or rearranging to find the slope).
M1 for using the perpendicular gradient rule to obtain \(m = -\frac{4}{3}\).
A1 for \(y = -\frac{4}{3}x + 7\) (or equivalent in the form \(y = mx + c\)).

The total number of balls in the bag is \(5 + 3 = 8\).
We want to find the probability that both balls are of the same colour, which can happen in two mutually exclusive ways:

1. Both balls are red:
\(P(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\)

2. Both balls are blue:
\(P(\text{Blue, Blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}\)

Add the probabilities to find the combined probability:
\(P(\text{Same colour}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56}\)

Simplify the fraction:
\(\frac{26}{56} = \frac{13}{28}\)

M1 for calculating either \(P(\text{Red, Red}) = \frac{20}{56}\) or \(P(\text{Blue, Blue}) = \frac{6}{56}\).
M1 for adding the two probabilities: \(\frac{20}{56} + \frac{6}{56}\).
A1 for \(\frac{13}{28}\) (or equivalent fraction).

Multiply the numerator and denominator by the conjugate of the denominator, which is \(3 + \sqrt{3}\):
\(\frac{6}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{6(3 + \sqrt{3})}{(3)^2 - (\sqrt{3})^2}\)

Simplify the denominator using the difference of squares:
\((3)^2 - (\sqrt{3})^2 = 9 - 3 = 6\)

Substitute the simplified denominator back into the expression:
\(\frac{6(3 + \sqrt{3})}{6} = 3 + \sqrt{3}\)

M1 for multiplying both numerator and denominator by \(3 + \sqrt{3}\).
M1 for correctly simplifying the denominator to \(6\).
A1 for \(3 + \sqrt{3}\).

When \(t = 0\), \(P = A = 350\).
When \(t = 5\), \(P = 350 \cdot b^5 = 580\).
This gives \(b^5 = \frac{580}{350} = \frac{58}{35}\), so \(b = \left(\frac{58}{35}\right)^{0.2} \approx 1.106325\).
When \(t = 12\), the population is \(P = 350 \cdot (1.106325)^{12} \approx 350 \cdot 3.37734 \approx 1182.07\).
Rounding to the nearest whole number gives 1182.

M1 for setting up the equation \(350 \cdot b^5 = 580\) or equivalent
A1 for finding \(b \approx 1.11\) or \(b^5 \approx 1.66\)
M1 for calculating \(P(12) = 350 \cdot b^{12}\)
A1 for 1182 (to nearest whole number)

Using \(u_2 = 15\):
\(a(2)^2 + b(2) + 5 = 15 \implies 4a + 2b = 10 \implies 2a + b = 5\)

Using \(u_4 = 49\):
\(a(4)^2 + b(4) + 5 = 49 \implies 16a + 4b = 44 \implies 4a + b = 11\)

Subtracting the first equation from the second:
\((4a + b) - (2a + b) = 11 - 5 \implies 2a = 6 \implies a = 3\)

Substituting \(a = 3\) back into \(2a + b = 5\):
\(2(3) + b = 5 \implies b = -1\)

So, the \(n\)-th term is \(u_n = 3n^2 - n + 5\).

For \(n = 10\):
\(u_{10} = 3(10)^2 - 10 + 5 = 300 - 10 + 5 = 295\).

M1 for setting up two simultaneous equations in terms of \(a\) and \(b\)
A1 for finding \(a = 3\) and \(b = -1\)
M1 for substituting \(n = 10\) into their expression for \(u_n\)
A1 for 295

Substitute \(y = 2x - 3\) into \(2x^2 - y^2 = 7\):
\(2x^2 - (2x - 3)^2 = 7\)
\(2x^2 - (4x^2 - 12x + 9) = 7\)
\(-2x^2 + 12x - 9 = 7\)
\(-2x^2 + 12x - 16 = 0\)
Divide by \(-2\):
\(x^2 - 6x + 8 = 0\)
\((x - 2)(x - 4) = 0\)
So \(x = 2\) or \(x = 4\).

When \(x = 2\), \(y = 2(2) - 3 = 1 \implies P(2, 1)\).
When \(x = 4\), \(y = 2(4) - 3 = 5 \implies Q(4, 5)\).

Distance \(PQ = \sqrt{(4 - 2)^2 + (5 - 1)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47\).

M1 for substituting \(y = 2x - 3\) into the curve equation
A1 for finding the x-coordinates \(x = 2\) and \(x = 4\)
M1 for finding the corresponding y-coordinates and using the distance formula
A1 for 4.47

The equation of the regression line of \(y\) on \(x\) is of the form \(y - \bar{y} = m(x - \bar{x})\).
Substituting the given values:
\(y - 120 = -4.5(x - 14)\)
\(y = -4.5x + 63 + 120\)
\(y = -4.5x + 183\)

Substitute \(x = 10\) to estimate \(y\):
\(y = -4.5(10) + 183 = -45 + 183 = 138\).

M1 for using the property that the regression line passes through \((\bar{x}, \bar{y})\)
A1 for finding the equation of the regression line: \(y = -4.5x + 183\) (or equivalent)
M1 for substituting \(x = 10\) into their line equation
A1 for 138

Using the vertex form of a quadratic function, we can write:
\(f(x) = a(x - h)^2 + k\)
where \((h, k) = (3, -4)\) is the vertex.
So, \(f(x) = a(x - 3)^2 - 4\).

Since the graph passes through \((1, 8)\):
\(8 = a(1 - 3)^2 - 4\)
\(12 = 4a \implies a = 3\).

Therefore, the function is \(f(x) = 3(x - 3)^2 - 4\).

To find \(f(6)\):
\(f(6) = 3(6 - 3)^2 - 4 = 3(3)^2 - 4 = 27 - 4 = 23\).

M1 for writing the function in vertex form \(a(x - 3)^2 - 4\) (or setting up equations with vertex properties)
A1 for finding \(a = 3\)
M1 for substituting \(x = 6\) into their quadratic function
A1 for 23

To find the local maximum, we can plot the graph of \(y = x^4 - 4x^3 + 2x^2 + 5\) on a GDC.
Using the maximum analyzer tool on the calculator, we look for the peak between the two minimums.
The local maximum is located at \(x \approx 0.381966\).
Rounding to 3 significant figures, we get \(0.382\).

Alternatively, solving \(f'(x) = 4x^3 - 12x^2 + 4x = 0\) gives \(x = 0\), \(x = \frac{3-\sqrt{5}}{2} \approx 0.382\), and \(x = \frac{3+\sqrt{5}}{2} \approx 2.62\).
The local maximum occurs at \(x = 0.382\).

M1 for sketching/visualizing the graph with GDC and identifying the local maximum point
A1 for finding \(x \approx 0.382\)

First, find the midpoint of \(AB\):
\(M = \left(\frac{-2 + 4}{2}, \frac{5 + 8}{2}\right) = (1, 6.5)\).

Next, find the gradient of \(L_1\):
\(m_1 = \frac{8 - 5}{4 - (-2)} = \frac{3}{6} = 0.5\).

Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) satisfies:
\(m_2 \cdot m_1 = -1 \implies m_2 = -2\).

Now, find the equation of \(L_2\) using the point \(M(1, 6.5)\):
\(y - 6.5 = -2(x - 1)\)
\(y = -2x + 2 + 6.5 \implies y = -2x + 8.5\).

The \(y\)-intercept of \(L_2\) is \(8.5\).

M1 for finding the midpoint \((1, 6.5)\)
M1 for finding the perpendicular gradient \(-2\)
M1 for substituting their midpoint and gradient into a linear equation
A1 for 8.5 (or equivalent fraction)

The probability of not rolling a 6 on any single roll is \(1 - p\).
Therefore, the probability of rolling no 6s in two rolls is \((1 - p)^2\).

The probability of rolling at least one 6 is given by:
\(P(\text{at least one 6}) = 1 - P(\text{no 6s}) = 1 - (1 - p)^2\).

We are given that this probability is \(0.36\):
\(1 - (1 - p)^2 = 0.36\)
\((1 - p)^2 = 0.64\)
\(1 - p = 0.8\) (since \(0 \le p \le 1\), we take the positive root)
\(p = 0.2\).

M1 for expressing \(P(\text{no 6s}) = (1 - p)^2\)
M1 for setting up the equation \(1 - (1 - p)^2 = 0.36\)
A1 for finding \(1 - p = 0.8\) or solving the quadratic equation
A1 for \(p = 0.2\)

Let the population after \(t\) hours be \(P(t) = P_0 \cdot k^t\), where \(P_0 = 1200\).

When \(t = 3\), \(P(3) = 1200 \cdot k^3 = 2100\).
This gives:
\(k^3 = \frac{2100}{1200} = 1.75 \implies k = (1.75)^{\frac{1}{3}} \approx 1.20507\)

The population after \(8\) hours is:
\(P(8) = 1200 \cdot k^8 = 1200 \cdot (1.75)^{\frac{8}{3}}\)

Using a graphics calculator:
\(P(8) \approx 1200 \cdot 4.417387... \approx 5300.86\)

Rounding to the nearest hundred gives \(5300\).

M1: For writing a correct exponential growth equation, e.g., \(1200 \cdot k^3 = 2100\) or finding \(k \approx 1.205\).
M1: For evaluating the growth factor over 8 hours, e.g., \(k^8 \approx 4.42\) or evaluating \(1200 \times 1.205^8\).
A1.57: For \(5300\) (correct to the nearest hundred).

We use the formula \(u_n = a n^2 + b n + 5\) with the given terms:

For \(n = 2\):
\(a(2)^2 + b(2) + 5 = 21 \implies 4a + 2b = 16 \implies 2a + b = 8\)

For \(n = 5\):
\(a(5)^2 + b(5) + 5 = 90 \implies 25a + 5b = 85 \implies 5a + b = 17\)

We solve this system of simultaneous linear equations by subtracting the first simplified equation from the second:
\((5a + b) - (2a + b) = 17 - 8\)
\(3a = 9 \implies a = 3\)

Substitute \(a = 3\) back into \(2a + b = 8\):
\(2(3) + b = 8 \implies b = 2\)

Now, substitute \(a = 3\), \(b = 2\), and \(n = 10\) into the general formula to find \(u_{10}\):
\(u_{10} = 3(10)^2 + 2(10) + 5 = 300 + 20 + 5 = 325\).

M1: Formulates two simultaneous equations in terms of \(a\) and \(b\) (e.g., \(4a+2b = 16\) and \(25a+5b = 85\)).
M1: Correctly solves the system to obtain \(a = 3\) and \(b = 2\).
A1.57: Correctly computes \(u_{10} = 325\).

Plot the function \(f(x) = x^3 - 2x^2 - 5x + 1\) on your graphics calculator.

By analyzing the graph or using the equation solver function, we find three x-intercepts (roots):
\(x \approx -1.58\)
\(x \approx 0.188\)
\(x \approx 3.39\)

The largest solution is the rightmost intersection point, which has the value of \(x \approx 3.3888...\).

To 3 significant figures, this is \(3.39\).

M1: Sketching or indicating the use of a GDC to find the roots of the cubic equation, or using systematic trial and error / interval bisection.
M1: Identifying that there are multiple roots and selecting the largest positive root.
A1.57: Final answer of \(3.39\) (accept \(3.39\) or \(3.389\); do not accept \(3.4\) without first showing a more accurate value).

The equation of a quadratic curve with vertex \((h, k)\) can be written in vertex form:
\(y = a(x - h)^2 + k\)

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

To determine the coefficient \(a\), substitute the point \((1, 8)\):
\(8 = a(1 - 3)^2 - 4\)
\(8 = 4a - 4\)
\(12 = 4a \implies a = 3\)

Thus, the equation of the quadratic curve is:
\(y = 3(x - 3)^2 - 4\)

To find where the curve crosses the \(y\)-axis, we set \(x = 0\):
\(y = 3(0 - 3)^2 - 4 = 3(9) - 4 = 27 - 4 = 23\).

M1: For using the vertex form of the quadratic equation, \(y = a(x - 3)^2 - 4\).
M1: For substituting \((1, 8)\) to solve for \(a\) and obtaining \(a = 3\).
A1.57: For setting \(x = 0\) to find the \(y\)-intercept of \(23\).

The total number of balls is \(5 + 3 = 8\).

We can find the probability of selecting at least one blue ball by subtracting the probability of selecting no blue balls (i.e., both balls are red) from 1.

First ball is red:
\(P(R_1) = \frac{5}{8}\)

Since the selection is without replacement, there are now \(7\) balls remaining, \(4\) of which are red.
Second ball is red:
\(P(R_2 | R_1) = \frac{4}{7}\)

The probability of selecting two red balls is:
\(P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\)

Therefore, the probability of selecting at least one blue ball is:
\(P(\text{at least one blue}) = 1 - P(\text{both red}) = 1 - \frac{5}{14} = \frac{9}{14}\).

M1: For calculating the probability of selecting two red balls: \(\frac{5}{8} \times \frac{4}{7}\) (or alternative correct tree diagram branches).
M1: For calculating the complement from 1: \(1 - \frac{5}{14}\) (or summing the probabilities of \((R, B)\), \((B, R)\), and \((B, B)\)).
A1.57: For the simplified fraction \(\frac{9}{14}\) (or equivalent decimal \(0.643\) rounded to 3 s.f.).

To subtract the fractions, we write them over a common denominator, which is \((x-2)(x+3)\):

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

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

Expand the numerator:
\(3(x+3) - 2(x-2) = 3x + 9 - 2x + 4 = x + 13\)

Expand the denominator:
\((x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6\)

Thus, the simplified single fraction is:
\(\frac{x+13}{x^2+x-6}\)

M1: For writing the expression over a common denominator: \(\frac{3(x+3) - 2(x-2)}{(x-2)(x+3)}\).
M1: For expanding the numerator correctly to obtain \(x+13\).
A1.57: For expanding the denominator and writing the final fraction as \(\frac{x+13}{x^2+x-6}\) (accept \((x+13)/(x^2+x-6)\)).

First, find the midpoint, \(M\), of the line segment \(AB\):
\(M = \left(\frac{2+6}{2}, \frac{5+(-3)}{2}\right) = (4, 1)\)

Next, find the gradient, \(m_{AB}\), of the line \(AB\):
\(m_{AB} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2\)

The gradient, \(m\), of the perpendicular bisector is the negative reciprocal of \(m_{AB}\):
\(m = -\frac{1}{-2} = \frac{1}{2} = 0.5\)

Now, use the point-slope formula with the midpoint \((4, 1)\) and the perpendicular gradient \(m = 0.5\):
\(y - 1 = 0.5(x - 4)\)
\(y - 1 = 0.5x - 2\)
\(y = 0.5x - 1\)

M1: For calculating the midpoint \((4, 1)\).
M1: For finding the gradient of the line segment, which is \(-2\), and determining the perpendicular gradient is \(0.5\) (or \(\frac{1}{2}\)).
A1.57: For the correct equation of the perpendicular bisector \(y = 0.5x - 1\) (or \(y = \frac{1}{2}x - 1\)).

The formula for the total surface area of a cone is:
\(A = \pi r^2 + \pi r l\)
where \(r\) is the radius of the base and \(l\) is the slant height.

First, find the slant height \(l\) using Pythagoras' theorem:
\(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}\)

Substitute the values \(r = 6\text{ cm}\) and \(l = 10\text{ cm}\) into the surface area equation:
\(A = \pi (6)^2 + \pi (6)(10)\)
\(A = 36\pi + 60\pi = 96\pi\)

Using a calculator to compute the decimal value:
\(A \approx 96 \times 3.14159265... \approx 301.59289...\text{ cm}^2\)

Rounding to 3 significant figures gives \(302\text{ cm}^2\).

M1: For calculating the slant height \(l = 10\text{ cm}\).
M1: For substituting the correct radius and slant height into the total surface area formula, i.e., \(\pi \times 6^2 + \pi \times 6 \times 10\).
A1.57: For \(302\) (accept answers in the range \(301\) to \(302\), or \(96\pi\)).

Let the equation of the quadratic graph be \(y = ax^2 + bx + c\). Using the coordinates of the given points, we obtain three equations: (1) \(a + b + c = 5\), (2) \(4a + 2b + c = 12\), and (3) \(a - b + c = 3\). Subtracting equation (3) from equation (1) gives \(2b = 2\), which simplifies to \(b = 1\). Substituting \(b = 1\) into equation (1) yields \(a + c = 4\). Substituting \(b = 1\) into equation (2) yields \(4a + c = 10\). Subtracting these two resulting equations, we get \(3a = 6\), which gives \(a = 2\). Substituting \(a = 2\) back into \(a + c = 4\) gives \(c = 2\). Thus, the quadratic function is \(y = 2x^2 + x + 2\). For \(x = 4\), we calculate \(y = 2(4)^2 + 4 + 2 = 32 + 4 + 2 = 38\).

M1: For setting up simultaneous equations using the coordinates. M1: For solving to find at least one coefficient correctly. A1: For finding the correct quadratic function equation or all coefficients correctly. A1: For substituting \(x = 4\) to find the final answer of 38.

Let \(t\) be the number of years. The value of the machine after \(t\) years is given by \(V(t) = 8000 \times (1 - 0.075)^t = 8000 \times 0.925^t\). We need to solve the inequality \(8000 \times 0.925^t < 4500\), which simplifies to \(0.925^t < 0.5625\). Taking logarithms on both sides, we get \(t \ln(0.925) <
\ln(0.5625)\). Since \(\ln(0.925)\) is negative, reversing the inequality yields \(t > \frac{\ln(0.5625)}{\ln(0.925)} \approx 7.38\). Since we are looking for the number of complete years, we round up to the next integer, which gives 8 years.

M1: For writing the correct exponential decay expression \(8000 \times 0.925^t\). M1: For setting up the inequality or equation and solving for \(t\) using GDC or logarithms. A1: For finding the critical value of \(t \approx 7.38\) and correctly rounding up to get 8.

Using the formula for the \(n\)-th term: For \(n = 2\), \(u_2 = a(2^2) + b(2) - 3 = 9 \implies 4a + 2b = 12 \implies 2a + b = 6\). For \(n = 4\), \(u_4 = a(4^2) + b(4) - 3 = 29 \implies 16a + 4b = 32 \implies 4a + b = 8\). Subtracting the first simplified equation from the second gives \((4a + b) - (2a + b) = 8 - 6 \implies 2a = 2 \implies a = 1\). Substituting \(a = 1\) back into \(2a + b = 6\) gives \(2(1) + b = 6 \implies b = 4\). Therefore, the general term is \(u_n = n^2 + 4n - 3\). For the 10th term (\(n = 10\)), we compute \(u_{10} = 10^2 + 4(10) - 3 = 100 + 40 - 3 = 137\).

M1: For substituting \(n = 2\) and \(n = 4\) to establish two linear equations in \(a\) and \(b\). M1: For solving the simultaneous equations to find the values of both \(a\) and \(b\). A1: For finding \(a = 1\) and \(b = 4\). A1: For substituting \(n = 10\) into the resulting formula to find 137.

The total number of marbles in the box is \(x + 4\). The probability of drawing a red marble first is \(\frac{x}{x+4}\), and the probability of drawing a second red marble is \(\frac{x-1}{x+3}\). The probability that both are red is given by \(\frac{x}{x+4} \times \frac{x-1}{x+3} = \frac{1}{3}\). Clearing the fractions by cross-multiplying yields \(3x(x-1) = (x+4)(x+3)\). Expanding both sides gives \(3x^2 - 3x = x^2 + 7x + 12\). Rearranging into a standard quadratic form gives \(2x^2 - 10x - 12 = 0\), which simplifies to \(x^2 - 5x - 6 = 0\). Factoring the quadratic, we get \((x - 6)(x + 1) = 0\). Since the number of marbles \(x\) must be positive, we reject \(x = -1\) and find that \(x = 6\).

M1: For writing the probability equation \(\frac{x}{x+4} \times \frac{x-1}{x+3} = \frac{1}{3}\). M1: For expanding and simplifying to a quadratic equation of the form \(2x^2 - 10x - 12 = 0\) or equivalent. A1: For solving the quadratic equation to get \(x = 6\) (and rejecting the negative solution).

First, we find the coordinates of the midpoint \(M\) of the line segment \(AB\): \(M = \left(\frac{2+6}{2}, \frac{5-3}{2}\right) = (4, 1)\). Next, we find the gradient of \(AB\): \(m_{AB} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2\). The gradient of the perpendicular bisector is the negative reciprocal of \(-2\), which is \(m = -\frac{1}{-2} = 0.5\). Using the point-gradient form with the midpoint \((4, 1)\) and gradient \(0.5\): \(y - 1 = 0.5(x - 4) \implies y - 1 = 0.5x - 2 \implies y = 0.5x - 1\).

M1: For calculating the midpoint \((4, 1)\). M1: For calculating the gradient of \(AB\) as \(-2\) and finding the perpendicular gradient as \(0.5\). A1: For substituting the midpoint and perpendicular gradient into the line equation. A1: For the final correct equation \(y = 0.5x - 1\) (or equivalent form with fractions, e.g., \(y = \frac{1}{2}x - 1\)).

For \(n=1\), there is only 1 unit triangle, and it points upwards, so \(u_1(1) = 1\). For \(n=2\), there are 3 unit triangles pointing upwards, so \(u_1(2) = 3\). For \(n=3\), there are 6 unit triangles pointing upwards, so \(u_1(3) = 6\).

Award 1.5 marks for any two correct values. Award 2.5 marks for all three correct values.

The sequence of \(u_1(n)\) is 1, 3, 6, 10, ... which represents the triangular numbers. The standard formula for the \(n\)-th triangular number is \(\frac{n(n+1)}{2}\).

Award 1.5 marks for identifying the quadratic nature or setting up a system of equations. Award 1 mark for the correct final simplified expression.

The sequence of values for \(d_1(n)\) for \(n = 1, 2, 3, \dots\) is 0, 1, 3, 6, ... This is the sequence of triangular numbers shifted by 1 position (i.e. \(T_{n-1}\)). Thus, the expression is \(\frac{(n-1)((n-1)+1)}{2} = \frac{n(n-1)}{2}\).

Award 1.5 marks for realizing that the sequence is shifted triangular numbers or setting up differences. Award 1 mark for the correct final formula.

We substitute the formulas found in the previous parts: \(t_1(n) = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n^2+n + n^2-n}{2} = \frac{2n^2}{2} = n^2\).

Award 1 mark for correct substitution of the two formulas. Award 1.5 marks for complete correct expansion and algebraic simplification showing the final result \(n^2\).

The sequence for \(n = 2, 3, 4, \dots\) is 1, 3, 6, ... which is the triangular numbers shifted by 1. Substituting \(n-1\) into the triangular number formula yields \(\frac{(n-1)n}{2}\).

Award 1.5 marks for recognizing the sequence of triangular numbers starting at \(n=2\). Award 1 mark for the correct algebraic expression.

For \(n \ge 4\), the non-zero values of \(d_2(n)\) are 1, 3, 6, ... which is the sequence of triangular numbers. For \(n=4\), we want the 1st triangular number, so we substitute \(n-3\) into the formula: \(\frac{(n-3)((n-3)+1)}{2} = \frac{(n-3)(n-2)}{2}\).

Award 1.5 marks for identifying the sequence of triangular numbers with the shift. Award 1 mark for the correct final formula in terms of \(n\).

For \(n=3\), the possible sizes of triangles are 1, 2, and 3.
- Size 1: \(u_1(3) = 6\) and \(d_1(3) = 3\).
- Size 2: \(u_2(3) = 3\) and \(d_2(3) = 0\).
- Size 3: \(u_3(3) = 1\) and \(d_3(3) = 0\).
Total = \(6 + 3 + 3 + 0 + 1 + 0 = 13\).

Award 1.5 marks for listing or showing the sum of all components: 6, 3, 3, and 1. Award 1 mark for the correct total of 13.

We first calculate the sums for small values of \(n\):
\(U(1) = u_1(1) = 1\)
\(U(2) = u_1(2) + u_2(2) = 3 + 1 = 4\)
\(U(3) = u_1(3) + u_2(3) + u_3(3) = 6 + 3 + 1 = 10\)
We substitute these into the formula:
For \(n=1\): \(\frac{1(1+a)(1+b)}{c} = 1 \implies (1+a)(1+b) = c\)
For \(n=2\): \(\frac{2(2+a)(2+b)}{c} = 4 \implies (2+a)(2+b) = 2c\)
Using \(a=1, b=2, c=6\):
For \(n=1\): \(\frac{1 \times 2 \times 3}{6} = 1\) (correct).
For \(n=2\): \(\frac{2 \times 3 \times 4}{6} = 4\) (correct).
For \(n=3\): \(\frac{3 \times 4 \times 5}{6} = 10\) (correct).

Award 1 mark for finding \(U(1)=1\), \(U(2)=4\), and \(U(3)=10\). Award 1.5 marks for finding the correct parameters \(a=1, b=2, c=6\) (accept \(a=2, b=1, c=6\)).

Let \(S_n\) be the number of tiles in Pattern \(n\). We are given: \(S_1 = 1\), \(S_2 = 5\) (a difference of \(+4\)), and \(S_3 = 13\) (a difference of \(+8\)). Following this pattern of first differences (\(4, 8, 12, 16, \dots\)): \(S_4 = S_3 + 12 = 13 + 12 = 25\), and \(S_5 = S_4 + 16 = 25 + 16 = 41\). Therefore, the number of tiles in Pattern 5 is 41.

M1 for identifying the pattern of first differences: \(+4, +8, +12, \dots\) or finding \(S_4 = 25\). A1.5 for the correct final answer of 41.

Using the formula \(S_n = an^2 + bn + c\): For \(n = 1\): \(a + b + c = 1\) (Equation 1). For \(n = 2\): \(4a + 2b + c = 5\) (Equation 2). For \(n = 3\): \(9a + 3b + c = 13\) (Equation 3). Subtracting Equation 1 from Equation 2 gives: \(3a + b = 4\) (Equation 4). Subtracting Equation 2 from Equation 3 gives: \(5a + b = 8\) (Equation 5). Subtracting Equation 4 from Equation 5 gives: \(2a = 4 \implies a = 2\). Substituting \(a = 2\) into Equation 4 gives: \(3(2) + b = 4 \implies b = -2\). Substituting \(a = 2\) and \(b = -2\) into Equation 1 gives: \(2 - 2 + c = 1 \implies c = 1\). Thus, the expression for \(S_n\) is \(2n^2 - 2n + 1\).

M1 for calculating the second difference as \(4\) to find \(a = 2\) (or attempting to set up a system of three linear equations). M1 for correctly finding \(b = -2\) and \(c = 1\). A0.5 for writing the final expression as \(2n^2 - 2n + 1\) (or equivalent).

When \(t = 0\), \(P = 120 \implies c = 120\). When \(t = 4\), \(P = 200 \implies 4m + 120 = 200 \implies 4m = 80 \implies m = 20\). Thus, the linear model is \(P = 20t + 120\).

1 mark for finding \(c = 120\). 1 mark for setting up the gradient equation and finding \(m = 20\). 0.5 marks for writing the final equation in the required form.

At \(t = 0\), \(P = 120 \implies 120 = A \cdot b^0 \implies A = 120\). At \(t = 4\), \(P = 200 \implies 120 \cdot b^4 = 200 \implies b^4 = \frac{5}{3} \implies b = \left(\frac{5}{3}\right)^{0.25} \approx 1.1362\). To 3 significant figures, \(b = 1.14\).

1 mark for finding \(A = 120\). 1 mark for setting up the equation \(b^4 = 5/3\). 0.5 marks for rounding \(b\) correctly to 3 significant figures.

At \(t = 0\), \(P = 120 \implies c = 120\). At \(t = 4\), \(16a + 4b + 120 = 200 \implies 4a + b = 20\). At \(t = 8\), \(64a + 8b + 120 = 240 \implies 8a + b = 15\). Subtracting the first linear equation from the second gives \(4a = -5 \implies a = -1.25\). Substituting \(a\) back gives \(4(-1.25) + b = 20 \implies -5 + b = 20 \implies b = 25\).

1 mark for substituting \(c = 120\) and setting up simultaneous equations. 1 mark for correctly solving for \(a = -1.25\). 0.5 marks for finding \(b = 25\).

The maximum of a quadratic \(P(t) = at^2 + bt + c\) occurs at \(t = -\frac{b}{2a}\). Here, \(t = -\frac{25}{2(-1.25)} = 10\) months. The maximum population is \(P(10) = -1.25(10)^2 + 25(10) + 120 = -125 + 250 + 120 = 245\).

1 mark for finding \(t = 10\). 1 mark for substituting \(t = 10\) into the quadratic expression. 0.5 marks for obtaining the population maximum of 245.

Linear: \(P = 20(12) + 120 = 360\). Exponential: \(P = 120 \cdot \left(\frac{5}{3}\right)^{12/4} = 120 \cdot \left(\frac{5}{3}\right)^3 = 120 \cdot \frac{125}{27} \approx 555.56 \approx 556\). Quadratic: \(P = -1.25(12)^2 + 25(12) + 120 = -180 + 300 + 120 = 240\).

1 mark for the linear model value (360). 1 mark for the exponential model value rounded to the nearest integer (556). 0.5 marks for the quadratic model value (240).

As \(t \to \infty\), \(e^{-0.2t} \to 0\). Therefore, \(P \to \frac{600}{1 + 0} = 600\). The horizontal asymptote is \(P = 600\). This represents the carrying capacity of the lake (the maximum population that the lake environment can support indefinitely).

1 mark for determining the horizontal asymptote equation \(P = 600\). 1 mark for explaining the carrying capacity or maximum population concept. 0.5 marks for coherent reasoning.

A reduction of \(15\%\) means the remaining population is \(85\%\) of the original. Thus, \(P_{\text{new}}(t) = 0.85 P(t)\). The original carrying capacity was \(600\), so the new carrying capacity is \(0.85 \times 600 = 510\).

1 mark for the relationship \(P_{\text{new}}(t) = 0.85 P(t)\) or equivalent. 1 mark for calculating the new capacity as \(510\). 0.5 marks for logical connection.

Set \(P(t) = 375 \implies 400 + 50 \cos\left(\frac{\pi t}{6}\right) = 375 \implies 50 \cos\left(\frac{\pi t}{6}\right) = -25 \implies \cos\left(\frac{\pi t}{6}\right) = -0.5\). The smallest positive angle for which \(\cos(\theta) = -0.5\) is \(\theta = \frac{2\pi}{3}\). Thus, \(\frac{\pi t}{6} = \frac{2\pi}{3} \implies t = 4\).

1 mark for simplifying to \(\cos\left(\frac{\pi t}{6}\right) = -0.5\). 1 mark for recognizing the angle \(\theta = \frac{2\pi}{3}\). 0.5 marks for the correct value \(t = 4\).

We set \(T(t) = 35\): \(18 + 72(0.88)^t = 35\). Subtracting 18 from both sides gives \(72(0.88)^t = 17\). Dividing by 72 gives \(0.88^t = \frac{17}{72}\). Taking logarithms of both sides: \(t \ln(0.88) = \ln\left(\frac{17}{72}\right)\). Solving for \(t\): \(t = \frac{\ln(17/72)}{\ln(0.88)} \approx 11.291\). To 1 decimal place, \(t = 11.3\) minutes.

M1 for isolating the exponential term: \(0.88^t = \frac{17}{72}\) (or decimal equivalent \(0.236...\)). M1 for a correct logarithmic method to solve for \(t\), e.g., \(t = \log_{0.88}(17/72)\). A0.5 for the final answer \(11.3\).

The maximum height is at the center where \(x = 0\), so \(y(0) = 5.4\). Substituting \(x = 0\) and \(y = 5.4\) into \(y = ax^2 + c\) gives \(c = 5.4\). The total width of the tunnel at ground level (\(y = 0\)) is 6 meters, which means it crosses the ground at \(x = 3\) and \(x = -3\). Substituting \(x = 3\) and \(y = 0\) into \(y = ax^2 + 5.4\) gives: \(0 = a(3)^2 + 5.4 \implies 9a = -5.4 \implies a = -0.6\). Expressing \(-0.6\) as a fraction in simplest form gives \(-\frac{3}{5}\).

M1 for identifying \(c = 5.4\) or substituting \(x=0, y=5.4\). M1 for using a correct horizontal boundary point \(x=3\) (or \(x=-3\)) with \(y=0\) to set up the equation \(9a + 5.4 = 0\). A0.5 for the correct simplest fraction \(-\frac{3}{5}\) (or \(-3/5\)).