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

To factorise fully, find the highest common factor of both terms.
- The highest common numerical factor of \(12\) and \(18\) is \(6\).
- The highest common variable factor of \(x^2y\) and \(xy^2\) is \(xy\).
Therefore, the highest common factor is \(6xy\).
Factorising this out gives:
\(12x^2y - 18xy^2 = 6xy(2x - 3y)\).

M1 for identifying a common factor, such as \(3xy(4x - 6y)\) or \(6x(2xy - 3y^2)\) or \(6y(2x^2 - 3x)\).
A1 (0.66 marks) for the fully factorised correct expression \(6xy(2x - 3y)\).

Find the common difference between consecutive terms:
\(7 - 3 = 4\)
\(11 - 7 = 4\)

Since the common difference is \(4\), the general formula contains \(4n\).
Now compare the terms of \(4n\) with our sequence:
- For \(n = 1\): \(4(1) = 4\), but the first term is \(3\) (which is \(4 - 1\)).
- For \(n = 2\): \(4(2) = 8\), but the second term is \(7\) (which is \(8 - 1\)).

Thus, the expression for the \(n\)-th term is \(4n - 1\).

M1 for recognizing a constant difference of 4 (e.g. writing \(4n + c\) or showing \(4n\)).
A1 (0.66 marks) for the fully correct simplified expression \(4n - 1\).

The formula for the volume of a cuboid is:
\(\text{Volume} = \text{length} \times \text{width} \times \text{height}\)

Substitute the given values into the formula:
\(90 = 5 \times 3 \times \text{height}\)
\(90 = 15 \times \text{height}\)

Divide both sides by \(15\) to find the height:
\(\text{height} = \frac{90}{15} = 6\text{ cm}\).

M1 for setting up a correct equation, e.g. \(5 \times 3 \times h = 90\) or showing \(90 / 15\).
A1 (0.66 marks) for the correct answer of 6.

Multiply both sides of the equation by 5:
\(2x - 3 = 15\)

Add 3 to both sides:
\(2x = 18\)

Divide by 2:
\(x = 9\)

M1 for multiplying both sides by 5 to obtain \(2x - 3 = 15\) (or equivalent first step)
A0.66 for the final correct answer \(x = 9\)

First, calculate the area of the triangular cross-section:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10\text{ cm}^2\)

Next, calculate the volume by multiplying the cross-sectional area by the length of the prism:
\(\text{Volume} = \text{Area} \times \text{length} = 10 \times 10 = 100\text{ cm}^3\)

M1 for calculating the correct cross-sectional area of the triangle: \(\frac{1}{2} \times 4 \times 5 = 10\)
A0.66 for multiplying by the length to obtain the correct volume: \(100\)

Find the common difference between consecutive terms:
\(11 - 5 = 6\)
\(17 - 11 = 6\)
Since the difference is constant, the sequence is arithmetic with a common difference of \(6\). The \(n\)-th term is in the form \(6n + c\).

Substitute \(n = 1\) to find \(c\):
\(6(1) + c = 5 \implies c = -1\)

Therefore, the \(n\)-th term is \(6n - 1\).

M1 for recognizing the common difference is \(6\) (e.g. writing an expression like \(6n + c\) or finding \(d = 6\))
A0.66 for the correct final formula: \(6n - 1\)

To factorise the expression fully, find the highest common factor of both terms:

- The highest common factor of the numerical coefficients \(8\) and \(12\) is \(4\).
- The common algebraic variable in both \(ax\) and \(ay\) is \(a\).

Therefore, the overall highest common factor is \(4a\).

Divide each term by \(4a\):
\(8ax \div 4a = 2x\)
\(-12ay \div 4a = -3y\)

Thus, the factorised expression is:
\(4a(2x - 3y)\)

M1 for finding a partial common factor, such as \(2a(4x - 6y)\) or \(4(2ax - 3ay)\).
A1 for \(4a(2x - 3y)\).

First, calculate the area of the triangular cross-section:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times 3\text{ cm} \times 4\text{ cm} = 6\text{ cm}^2\)

Next, calculate the volume of the prism by multiplying the area of the cross-section by the length of the prism:
\(\text{Volume} = \text{Area of cross-section} \times \text{length}\)
\(\text{Volume} = 6\text{ cm}^2 \times 10\text{ cm} = 60\text{ cm}^3\)

M1 for calculating the area of the triangular cross-section: \(\frac{1}{2} \times 3 \times 4\) or showing \(6\).
A1 for \(60\).

First, find the differences between consecutive terms in the sequence:
\(11 - 7 = 4\)
\(15 - 11 = 4\)
\(19 - 15 = 4\)

Since the difference is a constant \(4\), this is an arithmetic sequence and the \(n\)-th term will contain \(4n\).

Let the \(n\)-th term expression be \(4n + c\).
For the first term (\(n = 1\)):
\(4(1) + c = 7\)
\(4 + c = 7 \implies c = 3\)

Therefore, the \(n\)-th term is \(4n + 3\).

M1 for recognizing a common difference of \(4\) (e.g., showing \(4n\) or \(4n + c\) where \(c \neq 3\)).
A1 for \(4n + 3\) (or equivalent).

Find the highest common factor of \(6x^2y\) and \(-15xy^2\). The highest common factor of \(6\) and \(15\) is \(3\). The highest common factor of \(x^2\) and \(x\) is \(x\). The highest common factor of \(y\) and \(y^2\) is \(y\). So the highest common factor is \(3xy\). Dividing both terms by \(3xy\) gives: \(6x^2y \div 3xy = 2x\) and \(-15xy^2 \div 3xy = -5y\). Thus, the fully factorised expression is \(3xy(2x - 5y)\).

M1 for finding any partial common factor, e.g. \(3(2x^2y - 5xy^2)\) or \(xy(6x - 15y)\). A1 for the correct fully factorised expression \(3xy(2x - 5y)\).

The terms decrease by \(3\) each time, so the common difference is \(-3\). This means the \(n\)-th term formula includes \(-3n\). The sequence can be represented as \(-3n + c\). Substituting \(n = 1\) gives \(-3(1) + c = 11\), which simplifies to \(-3 + c = 11\), so \(c = 14\). Thus, the \(n\)-th term is \(-3n + 14\) or \(14 - 3n\).

M1 for finding the common difference is \(-3\), or for an expression of the form \(-3n + c\). A1 for \(-3n + 14\) or \(14 - 3n\).

The area of the triangular cross-section is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 7 = 14\text{ cm}^2\). The volume of the prism is the cross-sectional area multiplied by its length: \(\text{Volume} = 14 \times 10 = 140\text{ cm}^3\).

M1 for calculating the area of the triangular cross-section: \(\frac{1}{2} \times 4 \times 7 = 14\). A1 for \(140\).

We want to find two numbers that multiply to \(-18\) and add up to \(-7\).
These two numbers are \(-9\) and \(2\) because:
\((-9) \times 2 = -18\) and \((-9) + 2 = -7\).
Therefore, the factorised expression is:
\((x-9)(x+2)\).

M1 for writing \((x+a)(x+b)\) where \(ab=-18\) or \(a+b=-7\).
A1 for \((x-9)(x+2)\) or equivalent (such as \((x+2)(x-9)\)).

First, calculate the area of the triangular cross-section:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 4 = 10\text{ cm}^2\).

Then, find the volume by multiplying this area by the length of the prism:
\(\text{Volume} = \text{Area of cross-section} \times \text{length} = 10 \times 12 = 120\text{ cm}^3\).

M1 for a correct method to find the area of the triangular cross-section, e.g., \(\frac{1}{2} \times 5 \times 4\), or for a correct volume calculation step, e.g., \(\frac{1}{2} \times 5 \times 4 \times 12\).
A1 for 120.

First, find the difference between consecutive terms of the sequence:
\(10 - 3 = 7\)
\(17 - 10 = 7\)
This is an arithmetic sequence with a common difference of \(7\). Therefore, the formula for the \(n\)-th term has the form \(7n + c\).

When \(n = 1\), the first term is \(3\):
\(7(1) + c = 3\)
\(7 + c = 3 \implies c = -4\).

Thus, the expression for the \(n\)-th term is \(7n - 4\).

M1 for identifying the common difference is 7, or writing an expression of the form \(7n + c\) (for any constant \(c\)), or writing \(3 + 7(n-1)\).
A1 for \(7n - 4\) (or equivalent simplified expression).

The sequence is arithmetic because the difference between consecutive terms is constant. The first term is \(a = 7\) and the common difference is \(d = 4\). Using the formula for the \(n\)-th term: \(u_n = a + (n - 1)d\), we get: \(u_n = 7 + (n - 1)4 = 7 + 4n - 4 = 4n + 3\).

M1 for finding the common difference of 4 (e.g. \(4n\) as part of the answer). A1 for the correct expression \(4n + 3\).

The total surface area of a cuboid is the sum of the areas of its six faces. There are three pairs of identical faces: two faces of size \(5\text{ cm} \times 3\text{ cm} = 15\text{ cm}^2\), two faces of size \(3\text{ cm} \times 2\text{ cm} = 6\text{ cm}^2\), and two faces of size \(5\text{ cm} \times 2\text{ cm} = 10\text{ cm}^2\). Total surface area = \(2 \times (15 + 6 + 10) = 2 \times 31 = 62\text{ cm}^2\).

M1 for attempting to find the sum of the areas of three different faces: \((5 \times 3) + (3 \times 2) + (5 \times 2)\) (or showing at least two correct face areas of 15, 6, or 10). A1 for 62.

Expand each bracket: \(3(2x - 5) = 6x - 15\) and \(-2(x - 4) = -2x + 8\). Now, combine the terms: \(6x - 15 - 2x + 8 = (6x - 2x) + (-15 + 8) = 4x - 7\).

M1 for correct expansion of at least one bracket (e.g. \(6x - 15\) or \(-2x + 8\)). A1 for \(4x - 7\).

To factorise completely:

1. Find the highest common factor (HCF) of the numerical coefficients \(10\) and \(15\), which is \(5\).
2. Find the highest common factor of the variable parts \(x^2y\) and \(xy^2\), which is \(xy\).
3. Combine these to find the overall HCF: \(5xy\).
4. Divide each term by this HCF to find the terms inside the bracket:
\(\frac{10x^2y}{5xy} = 2x\)
\(\frac{-15xy^2}{5xy} = -3y\)
5. Write the final factorised expression: \(5xy(2x - 3y)\).

M1 for identifying a common factor of \(5\), \(x\), \(y\), \(xy\), \(5x\), or \(5y\) (e.g. \(5(2x^2y - 3xy^2)\) or \(xy(10x - 15y)\))
A1 for the fully correct factorised expression: \(5xy(2x - 3y)\)

The volume of a prism is given by the formula:
\(\text{Volume} = \text{Area of cross-section} \times \text{length}\)

1. Calculate the area of the triangular cross-section:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times 4 \times 3 = 6\text{ cm}^2\)

2. Calculate the volume:
\(\text{Volume} = 6 \times 8 = 48\text{ cm}^3\).

M1 for calculating the area of the triangular cross-section: \(\frac{1}{2} \times 4 \times 3\)
A1 for the correct answer \(48\)

1. Identify the difference between consecutive terms:
\(13 - 17 = -4\)
\(9 - 13 = -4\)
Since the difference is constant, this is an arithmetic sequence with common difference \(d = -4\).

2. The formula for the \(n\)-th term of an arithmetic sequence is:
\(u_n = a + (n - 1)d\)
where \(a\) is the first term (\(a = 17\)) and \(d\) is the common difference (\(d = -4\)).

3. Substitute the values into the formula:
\(u_n = 17 + (n - 1)(-4)\)
\(u_n = 17 - 4n + 4\)
\(u_n = 21 - 4n\).

M1 for finding the common difference is \(-4\) (or writing an expression of the form \(-4n + k\), where \(k\) is any constant)
A1 for the correct \(n\)-th term expression: \(21 - 4n\) (or equivalent, such as \(-4n + 21\))

First, find the highest common factor of the coefficients 12 and 18, which is 6. Next, find the highest common factor of the variable parts \(p^2 q\) and \(pq^2\), which is \(pq\). The highest common factor of the terms is therefore \(6pq\). Divide each term by this factor: \(12p^2 q \div 6pq = 2p\) and \(-18pq^2 \div 6pq = -3q\). This gives the factorised form: \(6pq(2p - 3q)\).

M1 for finding a common factor of at least \(3pq\) or \(6p\) or \(6q\), e.g., \(6p(2pq - 3q^2)\). A1 for fully correct factorisation: \(6pq(2p - 3q)\).

The formula for the total surface area of a cuboid is \(A = 2(lw + lh + wh)\). Let \(h\) be the height of the cuboid. Given \(l = 5\), \(w = 4\), and \(A = 94\), we substitute these values into the formula: \(94 = 2(5 \times 4 + 5h + 4h)\). Simplify the terms inside the parentheses: \(94 = 2(20 + 9h)\). Divide both sides by 2: \(47 = 20 + 9h\). Subtract 20 from both sides: \(27 = 9h\). Divide by 9: \(h = 3\).

M1 for substituting values into the surface area formula, e.g., \(2(20 + 9h) = 94\). A1 for the correct height of 3.

To clear the denominators, multiply every term in the equation by 10 (the lowest common multiple of 5 and 2): \(10 \times \frac{2x - 3}{5} + 10 \times 1 = 10 \times \frac{x + 2}{2}\). This simplifies to: \(2(2x - 3) + 10 = 5(x + 2)\). Expand the brackets: \(4x - 6 + 10 = 5x + 10\). Simplify the left side: \(4x + 4 = 5x + 10\). Subtract \(4x\) from both sides: \(4 = x + 10\). Subtract 10 from both sides: \(x = -6\).

M1 for multiplying both sides by a common multiple (e.g., 10) to clear fractions: \(2(2x - 3) + 10 = 5(x + 2)\). M1 for expanding brackets and collecting like terms correctly: \(4x + 4 = 5x + 10\). A1 for the correct answer \(-6\).

First, group the first two terms as a difference of squares and the last two terms: \(2(a^2 - 4b^2) - (a - 2b)\). Factoring the difference of squares gives \(2(a - 2b)(a + 2b) - (a - 2b)\). We can now factor out the common binomial term \(a - 2b\), which results in \((a - 2b)[2(a + 2b) - 1]\). Simplifying the brackets gives the final factorised form: \((a - 2b)(2a + 4b - 1)\).

M1 for factorising the first two terms as \(2(a-2b)(a+2b)\). M1 for factorising out the common bracket \((a-2b)\). A1 for the final correct expression \((a - 2b)(2a + 4b - 1)\).

To rationalise the denominator, multiply both the numerator and denominator by the conjugate of the denominator, which is \(3\sqrt{2} + 4\): \(\frac{22(3\sqrt{2} + 4)}{(3\sqrt{2} - 4)(3\sqrt{2} + 4)}\). Expanding the denominator yields \((3\sqrt{2})^2 - 4^2 = 18 - 16 = 2\). Now, simplify the fraction: \(\frac{22(3\sqrt{2} + 4)}{2} = 11(3\sqrt{2} + 4) = 33\sqrt{2} + 44\). This can be written as \(44 + 33\sqrt{2}\).

M1 for multiplying both numerator and denominator by \(3\sqrt{2} + 4\). M1 for correctly simplifying the denominator to \(2\). A1 for the final answer \(44 + 33\sqrt{2}\) or equivalent order.

Using the laws of indices, we can rewrite \(3^{2x+1}\) as \(3 \cdot (3^x)^2\). Let \(y = 3^x\). The equation becomes a quadratic equation in terms of \(y\): \(3y^2 - 10y + 3 = 0\). Factorising this quadratic equation gives \((3y - 1)(y - 3) = 0\), which has solutions \(y = \frac{1}{3}\) or \(y = 3\). Substituting back \(y = 3^x\), we solve \(3^x = 3^{-1}\) which gives \(x = -1\), and \(3^x = 3^1\) which gives \(x = 1\).

M1 for expressing the equation in quadratic form such as \(3y^2 - 10y + 3 = 0\). M1 for solving the quadratic equation to get \(3^x = 1/3\) or \(3^x = 3\). A1 for both correct solutions \(x = -1\) and \(x = 1\).

To rationalise the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{5} + \sqrt{2}\):

\(\frac{6}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}}\)

\(= \frac{6(\sqrt{5} + \sqrt{2})}{5 - 2}\)

\(= \frac{6(\sqrt{5} + \sqrt{2})}{3}\)

\(= 2(\sqrt{5} + \sqrt{2})\)

\(= 2\sqrt{5} + 2\sqrt{2}\)

M1: For multiplying the numerator and denominator by \(\sqrt{5} + \sqrt{2}\)
A1: For simplifying the denominator to 3
A0.66: For the final correct simplified expression \(2\sqrt{5} + 2\sqrt{2}\) or \(2(\sqrt{5} + \sqrt{2})\)

First, factorise the numerator and the denominator:

Numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
Denominator: \(x^2 - 9 = (x - 3)(x + 3)\)

Now, substitute these back into the fraction and divide out the common factor \((x - 3)\):

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

M1: For correct factorisation of the numerator as \((2x + 1)(x - 3)\)
M1: For correct factorisation of the denominator as \((x - 3)(x + 3)\)
A0.66: For the final simplified fraction \(\frac{2x + 1}{x + 3}\)

Express both sides as powers of the same base, 2:

\(8 = 2^3\) and \(4 = 2^2\)

Substitute these bases into the equation:

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

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

\(2^{6x - 3} = 2^{2x + 6}\)

Since the bases are equal, equate the exponents:

\(6x - 3 = 2x + 6\)

Subtract \(2x\) from both sides:

\(4x - 3 = 6\)

Add 3 to both sides:

\(4x = 9\)

\(x = \frac{9}{4}\) (or \(2.25\))

M1: For converting to base 2: \(2^{3(2x-1)} = 2^{2(x+3)}\) or equivalent
M1: For equating powers: \(6x - 3 = 2x + 6\)
A0.66: For the final correct answer \(\frac{9}{4}\) or \(2.25\)

First, rationalize the denominator of the fraction by multiplying both the numerator and the denominator by \(\sqrt{5} + 1\):
\(\frac{4}{\sqrt{5} - 1} = \frac{4(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}\)
\(= \frac{4(\sqrt{5} + 1)}{5 - 1}\)
\(= \frac{4(\sqrt{5} + 1)}{4}\)
\(= \sqrt{5} + 1\)

Next, substitute this simplified expression back into the original equation:
\((\sqrt{5} + 1) - \sqrt{5} = 1\)

M1 for rationalizing the denominator, multiplying by \(\frac{\sqrt{5} + 1}{\sqrt{5} + 1}\)
A1 for reducing the fraction to \(\sqrt{5} + 1\)
A1 (0.66 marks) for the final answer 1

To write the expression as a single fraction, find a common denominator, which is \((2x - 1)(x + 3)\):
\(\frac{3(x + 3)}{(2x - 1)(x + 3)} - \frac{2(2x - 1)}{(2x - 1)(x + 3)}\)

Combine the numerators over the common denominator:
\(\frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\)

Expand the terms in the numerator:
\(\frac{3x + 9 - 4x + 2}{(2x - 1)(x + 3)}\)

Simplify the numerator by combining like terms:
\(\frac{11 - x}{(2x - 1)(x + 3)}\)

M1 for a common denominator of \((2x - 1)(x + 3)\) and appropriate numerators
M1 for expanding numerator correctly to get \(3x + 9 - 4x + 2\)
A1 (0.66 marks) for the correct final simplified fraction

Multiply all terms of the equation by the common denominator \(x(x + 2)\) to eliminate the fractions:
\(1(x + 2) + 2x = 1 \cdot x(x + 2)\)

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

Rearrange the equation into standard quadratic form \(ax^2 + bx + c = 0\):
\(x^2 - x - 2 = 0\)

Factor the quadratic expression:
\((x - 2)(x + 1) = 0\)

Find the solutions for \(x\):
\(x = 2\) or \(x = -1\)

M1 for multiplying throughout by \(x(x + 2)\) to obtain \(x + 2 + 2x = x^2 + 2x\) or equivalent
M1 for writing as standard quadratic form \(x^2 - x - 2 = 0\)
A1 (0.66 marks) for both correct solutions: \(x = 2\) and \(x = -1\)

First, write each base as a power of 3: \(9 = 3^2\) and \(27 = 3^3\). This gives \(3^{2x - 1} \times (3^2)^{x+2} = (3^3)^{x-3}\). Simplify the exponents using the index laws: \(3^{2x - 1} \times 3^{2x + 4} = 3^{3x - 9}\). Combine the terms on the left-hand side by adding the exponents: \(3^{2x - 1 + 2x + 4} = 3^{3x - 9}\), which simplifies to \(3^{4x + 3} = 3^{3x - 9}\). Since the bases are equal, we equate the exponents: \(4x + 3 = 3x - 9\). Subtracting \(3x\) and \(3\) from both sides gives \(x = -12\).

M1 for expressing all terms in base 3 (e.g., \(9^{x+2} = 3^{2x+4}\) or \(27^{x-3} = 3^{3x-9}\)) M1 for equating exponents and forming a linear equation: \(2x - 1 + 2(x + 2) = 3(x - 3)\) A0.66 for correct final answer \(-12\)

Multiply the numerator and the denominator by the conjugate of the denominator, which is \(3 + \sqrt{5}\): \(\frac{4(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}\). Expand the denominator: \((3 - \sqrt{5})(3 + \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4\). Now simplify the fraction: \(\frac{4(3 + \sqrt{5})}{4} = 3 + \sqrt{5}\). This is in the form \(a + \sqrt{b}\) where \(a = 3\) and \(b = 5\).

M1 for multiplying numerator and denominator by \(3 + \sqrt{5}\) M1 for simplifying the denominator to \(4\) A0.66 for final answer \(3 + \sqrt{5}\)

Find a common denominator, which is \((2x - 1)(x + 3)\). Write each fraction over this common denominator: \(\frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\). Expand the numerator: \(3(x + 3) - 2(2x - 1) = 3x + 9 - 4x + 2\). Simplify the numerator to get \(11 - x\). This gives the final fraction: \(\frac{11 - x}{(2x - 1)(x + 3)}\).

M1 for a common denominator of \((2x - 1)(x + 3)\) M1 for correct expansion of the numerator: \(3(x + 3) - 2(2x - 1)\) or \(3x + 9 - 4x + 2\) A0.66 for \(\frac{11 - x}{(2x - 1)(x + 3)}\) or \(\frac{11 - x}{2x^2 + 5x - 3}\)

Factorise the numerator: \(2x^2 - 7x - 15 = (2x + 3)(x - 5)\). Factorise the denominator: \(4x^2 - 9 = (2x - 3)(2x + 3)\). Divide out the common factor \(2x + 3\) to get \(\frac{x - 5}{2x - 3}\).

M1 for factorising numerator to \((2x + 3)(x - 5)\). M1 for factorising denominator to \((2x - 3)(2x + 3)\). A0.66 for the final simplified answer \(\frac{x - 5}{2x - 3}\).

Multiply the first equation by 5 and the second by 2: \(15x - 10y = 65\) and \(8x + 10y = 4\). Adding these equations gives \(23x = 69\), so \(x = 3\). Substituting \(x = 3\) into the first equation: \(3(3) - 2y = 13\) which simplifies to \(9 - 2y = 13\), so \(-2y = 4\) and \(y = -2\).

M1 for correctly eliminating one variable. A1 for either \(x = 3\) or \(y = -2\). A0.66 for finding the second variable correctly.

Multiply the numerator and denominator by \(\sqrt{2} - 1\): \(\frac{(3\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)}\). The numerator expands to \(3(2) - 3\sqrt{2} - \sqrt{2} + 1 = 6 - 4\sqrt{2} + 1 = 7 - 4\sqrt{2}\). The denominator simplifies to \(2 - 1 = 1\). Thus, the simplified expression is \(7 - 4\sqrt{2}\).

M1 for multiplying both numerator and denominator by \(\sqrt{2} - 1\). M1 for expanding numerator to \(7 - 4\sqrt{2}\). A0.66 for final correct expression.

(a) The sequence for the number of sticks starts with 5, 9, 13, ...
This is an arithmetic sequence with a common difference of 4.
Pattern 4: \(13 + 4 = 17\)
Pattern 5: \(17 + 4 = 21\)

(b) Since the common difference is 4, the expression has the form \(4n + c\).
For \(n = 1\): \(4(1) + c = 5 \implies c = 1\).
So, the \(n\)-th term expression is \(4n + 1\).

(c) Set the expression equal to 121 and solve for \(n\):
\(4n + 1 = 121\)
\(4n = 120\)
\(n = 30\).

(a) 2 marks: 1 mark for 17 and 1 mark for 21.
(b) 3.7 marks: M1 for identifying the common difference of 4, M1 for setting up \(4n\), A1.7 for the fully correct expression \(4n + 1\).
(c) 3 marks: M1 for setting their expression from (b) equal to 121, M1 for solving algebraic steps, A1 for \(30\).

(a) First convert radius to meters: \(35\text{ cm} = 0.35\text{ m}\).
Using the formula for the volume of a cylinder, \(V = \pi r^2 h\):
\(V = \pi \times 0.35^2 \times 1.2 = \pi \times 0.1225 \times 1.2 = 0.147\pi \approx 0.461814\text{ m}^3\).
Rounding to 3 decimal places gives \(0.462\text{ m}^3\).

(b) Using the formula for the total surface area of a closed cylinder, \(A = 2\pi r^2 + 2\pi rh\):
\(A = 2\pi(0.35)^2 + 2\pi(0.35)(1.2)\)
\(A = 2\pi(0.1225) + 2\pi(0.42)\)
\(A = 0.245\pi + 0.84\pi = 1.085\pi \approx 3.4086\text{ m}^2\).
Correct to 3 significant figures, the surface area is \(3.41\text{ m}^2\).

(c) Mass is calculated as \(\text{Volume} \times \text{Density}\).
Using the exact volume of \(0.147\pi \approx 0.461814\text{ m}^3\):
\(\text{Mass} = 0.461814 \times 850 \approx 392.54\text{ kg}\).
Rounding to the nearest whole number gives \(393\text{ kg}\).
(Note: using the rounded value of \(0.462\text{ m}^3\) also yields \(0.462 \times 850 = 392.7 \approx 393\text{ kg}\)).

(a) 2.7 marks: M1 for converting unit to \(0.35\text{ m}\), M1 for volume formula substitution \(\pi \times 0.35^2 \times 1.2\), A0.7 for obtaining a value of \(0.4618...\) showing the rounded value of \(0.462\).
(b) 3 marks: M1 for area of the two circular bases \(2\pi(0.35)^2\), M1 for curved area \(2\pi(0.35)(1.2)\), A1 for \(3.41\) (accept \(3.409\)).
(c) 3 marks: M1 for multiplying volume by 850, A1 for \(392.5\) to \(393\), A1 for rounding to nearest integer: \(393\).

(a) By plotting the function \(y = x^3 - 3x^2 + 2\) on the GDC and finding the local maximum using the analyze graph feature, we find the maximum point is at \((0, 2)\).

(b) Set \(f(x) = -2\):
\(x^3 - 3x^2 + 2 = -2 \implies x^3 - 3x^2 + 4 = 0\).
Using the GDC polynomial solver or finding the x-intercepts of \(y = x^3 - 3x^2 + 4\), we obtain the solutions:
\(x = -1\) and \(x = 2\).

(c) The local maximum is at \((0, 2)\).
Using the GDC to find the local minimum, we find it at \((2, -2)\).
We find the equation of the line passing through \((0, 2)\) and \((2, -2)\):
Gradient \(m = \frac{-2 - 2}{2 - 0} = \frac{-4}{2} = -2\).
Since the line passes through \((0, 2)\), the y-intercept \(c = 2\).
Thus, the equation is \(y = -2x + 2\).

(a) 2.7 marks: M1 for plotting curve on GDC, A1.7 for coordinates \((0, 2)\).
(b) 3 marks: M1 for setting up \(x^3 - 3x^2 + 4 = 0\) or equivalent, M1 for utilizing GDC to locate roots, A1 for both \(x = -1\) and \(x = 2\).
(c) 3 marks: M1 for locating the local minimum at \((2, -2)\), M1 for finding the gradient \(m = -2\), A1 for writing the final equation \(y = -2x + 2\).

(a) The volume of a cylinder is \(V_{\text{cylinder}} = \pi r^2 h\). Since \(h = 2r\), we have:
\(V_{\text{cylinder}} = \pi r^2 (2r) = 2\pi r^3\).
The volume of a hemisphere of radius \(r\) is:
\(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\).
Adding these together to find the total volume:
\(V = 2\pi r^3 + \frac{2}{3}\pi r^3 = \frac{8}{3}\pi r^3\). (Shown)

(b) Setting \(V = 450\):
\(\frac{8}{3}\pi r^3 = 450\)
\(r^3 = \frac{450 \times 3}{8\pi} = \frac{1350}{8\pi} \approx 53.7148\)
Using a GDC to find the cube root:
\(r = \sqrt[3]{53.7148} \approx 3.77\text{ cm}\) (to 3 s.f.)

(c) The total surface area of the toy consists of:
1. The flat circular base: \(\pi r^2\)
2. The curved surface of the cylinder: \(2\pi r h = 2\pi r (2r) = 4\pi r^2\)
3. The curved surface of the hemisphere: \(2\pi r^2\)

Total Surface Area = \(\pi r^2 + 4\pi r^2 + 2\pi r^2 = 7\pi r^2\).
Using the unrounded value of \(r \approx 3.773\):
Total Surface Area = \(7\pi (3.773)^2 \approx 313.04 \approx 313.0\text{ cm}^2\) (to 1 d.p.).

M1 for cylinder volume formula \(2\pi r^3\) or hemisphere volume formula \(\frac{2}{3}\pi r^3\)
A1 for adding to show \(\frac{8}{3}\pi r^3\)
M1 for setting up equation \(\frac{8}{3}\pi r^3 = 450\)
M1 for solving for \(r\)
A1 for \(3.77\) (accept answers in range \([3.77, 3.78]\))
M1 for summing three area components: \(\pi r^2 + 4\pi r^2 + 2\pi r^2\) (or \(7\pi r^2\))
M1 for substituting their value of \(r\)
A1 for \(313.0\) (accept answers in range \([312.0, 314.0]\) depending on rounding of \(r\))

(a) A correct sketch should show a cubic curve with a local maximum in the second/first quadrant, a local minimum in the fourth quadrant, crossing the \(y\)-axis at \((0, 2)\).

(b)(i) Using the GDC maximum finder, we get the local maximum at \((0, 2)\).

(b)(ii) Using the GDC minimum finder, we find the local minimum at \(x \approx 2.67\) and \(y \approx -7.48\). The coordinates are \((2.67, -7.48)\).

(b)(iii) To find the points of intersection with \(y = -3\), we solve \(x^3 - 4x^2 + 2 = -3\), which is equivalent to finding the roots of \(x^3 - 4x^2 + 5 = 0\).
Using the GDC equation solver or intersection finder with \(y_1 = x^3 - 4x^2 + 2\) and \(y_2 = -3\), we find the \(x\)-coordinates of the intersection points:
\(x = -1\)
\(x \approx 1.38\)
\(x \approx 3.62\)

G2 for a correct sketch showing the cubic shape with correct maximum and minimum locations (G1 if shape is correct but coordinates are visibly inaccurate)
B1 for \((0, 2)\)
M1 for finding minimum \(x \approx 2.67\) or \(y \approx -7.48\)
A1 for \((2.67, -7.48)\) (accept \(2.66\) to \(2.67\) and \(-7.49\) to \(-7.48\))
M1 for setting up the intersection or equation \(x^3 - 4x^2 + 2 = -3\)
A1 for \(x = -1\)
A1 for \(x \approx 1.38\) and \(x \approx 3.62\)

(a) Area = \(\text{length} \times \text{width} = (2x + 5)(x + 3)\)
Expanding the brackets:
\(2x(x) + 2x(3) + 5(x) + 5(3) = 2x^2 + 6x + 5x + 15 = 2x^2 + 11x + 15\).

(b) Since the area is \(112\text{ m}^2\):
\(2x^2 + 11x + 15 = 112\)
Subtracting 112 from both sides:
\(2x^2 + 11x + 15 - 112 = 0\)
\(2x^2 + 11x - 97 = 0\). (Shown)

(c) Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) or GDC polynomial solver:
\(x = \frac{-11 \pm \sqrt{11^2 - 4(2)(-97)}}{2(2)}
= \frac{-11 \pm \sqrt{121 + 776}}{4}
= \frac{-11 \pm \sqrt{897}}{4}\)
Since \(\sqrt{897} \approx 29.94995\):
\(x_1 = \frac{-11 + 29.94995}{4} \approx 4.737 \approx 4.74\)
\(x_2 = \frac{-11 - 29.94995}{4} \approx -10.237 \approx -10.24\)
Thus, the solutions are \(x = 4.74\) and \(x = -10.24\).

(d) Since the dimensions of the garden must be positive, \(x\) must be positive. Therefore, we use \(x \approx 4.737\).
Width = \(x + 3 = 4.737 + 3 = 7.737 \approx 7.74\text{ m}\) (to 3 s.f.).

M1 for attempting to expand \((2x + 5)(x + 3)\) (at least 3 terms correct)
A1 for \(2x^2 + 11x + 15\)
M1 for equating their expression to 112
A1 for completing the algebra to show \(2x^2 + 11x - 97 = 0\) clearly
M1 for using GDC solver or formula with correct substitution
A1 for \(4.74\) (or \(4.737...\))
A1 for \(-10.24\) (or \(-10.237...\))
M1 for choosing the positive value of \(x\) and adding 3
A1 for \(7.74\text{ m}\) (accept \(7.73\) to \(7.75\))

(a) Plot the curve \(y = 0.5x^3 - 2x^2 - x + 5\) on the graphic display calculator.
Use the maximum finder function to locate the turning point.
The local maximum is at \(x \approx -0.2301\) and \(y \approx 5.1181\).
To 3 significant figures, the coordinates are \((-0.230, 5.12)\).

(b) Plot the line \(y = 2\) on the same axes.
Use the intersection finder tool to locate the points of intersection.
The three intersections are at \(x \approx -1.24\), \(x \approx 1.11\), and \(x \approx 4.133\).
The largest \(x\)-coordinate is \(x \approx 4.13\) (correct to 3 significant figures).

(a) [5 marks total]:
- 2 marks for finding the correct \(x\)-coordinate of \(-0.230\) (or \(-0.23\))
- 2 marks for finding the correct \(y\)-coordinate of \(5.12\)
- 1 mark for presenting the coordinates in the form \((x, y)\)

(b) [3.7 marks total]:
- 2 marks for showing the setting up of the equation \(0.5x^3 - 2x^2 - x + 5 = 2\) or sketching the intersection of the curve and the line
- 1.7 marks for finding the correct largest root \(x = 4.13\)

(a) The volume of a cone is given by \(V_{\text{cone}} = \frac{1}{3} \pi r^2 h\).
\(V_{\text{cone}} = \frac{1}{3} \pi (3.0)^2 (7.0) = 21\pi \approx 65.973\text{ cm}^3\).

The volume of a hemisphere is given by \(V_{\text{hemi}} = \frac{2}{3} \pi r^3\).
\(V_{\text{hemi}} = \frac{2}{3} \pi (3.0)^3 = 18\pi \approx 56.549\text{ cm}^3\).

Total volume = \(21\pi + 18\pi = 39\pi \approx 122.522\text{ cm}^3\).
To 3 significant figures, the total volume is \(123\text{ cm}^3\).

(b) Mass = \(\text{Volume} \times \text{Density}\).
Using the unrounded volume of \(122.522\text{ cm}^3\):
\(\text{Mass} = 122.522 \times 4.5 = 551.349\text{ g}\).
To 3 significant figures, the mass is \(551\text{ g}\).

(a) [5 marks total]:
- 2 marks for correct substitution into the cone volume formula: \(\frac{1}{3} \times \pi \times 3^2 \times 7\)
- 2 marks for correct substitution into the hemisphere volume formula: \(\frac{2}{3} \times \pi \times 3^3\)
- 1 mark for correct total volume of \(123\text{ cm}^3\) (accept \(122.5\) to \(123\))

(b) [3.7 marks total]:
- 2 marks for correct method of multiplying volume by density (e.g. \(122.5 \times 4.5\) or \(123 \times 4.5\))
- 1.7 marks for correct mass of \(551\text{ g}\) (accept \(551\) to \(554\) depending on rounding of volume used)

(a) Calculate the first differences of the sequence:
\(11 - 5 = 6\)
\(19 - 11 = 8\)
\(29 - 19 = 10\)
\(41 - 29 = 12\)
The differences increase by 2 each time. The next differences are \(14\) and \(16\).
Next term = \(41 + 14 = 55\).
Following term = \(55 + 16 = 71\).

(b) Since the second differences are constant (always 2), the sequence is quadratic of the form \(an^2 + bn + c\), where \(a = \frac{2}{2} = 1\).
Subtracting \(n^2\) from the original terms:
- 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 linear with a common difference of 3 and first term 4. Its formula is \(3n + 1\).
Thus, the general formula is \(n^2 + 3n + 1\).

(c) Substitute \(n = 30\) into the formula:
\(30^2 + 3(30) + 1 = 900 + 90 + 1 = 991\).

(a) [2 marks total]:
- 1 mark for the first term of \(55\)
- 1 mark for the second term of \(71\)

(b) [4.7 marks total]:
- 2 marks for identifying second difference of 2 and identifying \(n^2\) as part of the formula
- 1.7 marks for identifying the linear sequence \(3n + 1\)
- 1 mark for combining to write the full term: \(n^2 + 3n + 1\) (allow follow through for correct quadratic method if mistake made)

(c) [2 marks total]:
- 1 mark for substituting \(30\) into their formula
- 1 mark for the correct answer of \(991\)

(a) Using the formula for the volume of a cylinder:
\[V = \pi r^2 h\]
\[V = \pi \times 4.2^2 \times 11.5\]
\[V = \pi \times 17.64 \times 11.5 = 202.86\pi \approx 637.31\text{ cm}^3\]
Correct to the nearest cubic centimetre, this is \(637\text{ cm}^3\).

(b) Volume of the sphere is given by:
\[V_{\text{sphere}} = \frac{4}{3}\pi R^3\]
Setting this equal to the volume of the cylinder:
\[\frac{4}{3}\pi R^3 = 637.31\]
\[R^3 = \frac{3 \times 637.31}{4\pi} \approx 152.145\]
\[R = \sqrt[3]{152.145} \approx 5.34\text{ cm}\]
(If using the rounded volume \(637\text{ cm}^3\), \(R = \sqrt[3]{\frac{3 \times 637}{4\pi}} \approx 5.34\text{ cm}\).)

(c) The surface area of the sphere is given by:
\[A = 4\pi R^2\]
If using the exact radius \(R \approx 5.3385\text{ cm}\):
\[A = 4\pi (5.3385)^2 \approx 358.1\text{ cm}^2\]
If using \(R = 5.34\text{ cm}\):
\[A = 4\pi (5.34)^2 \approx 358.3\text{ cm}^2\]

(a)
M1 for substituting correct values into cylinder volume formula: \(\pi \times 4.2^2 \times 11.5\)
A1 for showing intermediate calculation \(637.31...\) and concluding it rounds to \(637\)

(b)
M1 for equating the volume of a sphere formula to \(637\) or \(637.31\)
M1.7 for isolating \(R\) correctly, i.e., \(R = \sqrt[3]{\frac{3 \times 637}{4\pi}}\) or equivalent
A1 for \(5.34\) (accept \(5.338\) to \(5.34\))

(c)
M1 for substituting their value of \(R\) into surface area of sphere formula \(4\pi R^2\)
A1 for correct evaluation using their \(R\)
A1 for \(358.1\) or \(358.3\) (accept answers in the range \(358.0\) to \(358.4\))

(a) Profit is given by Revenue minus Cost:
\[P(x) = R(x) - C(x)\]
\[P(x) = 35x - (0.05x^2 + 12x + 450)\]
\[P(x) = -0.05x^2 + 23x - 450\]

(b) The maximum profit occurs at the vertex of the quadratic equation \(P(x) = -0.05x^2 + 23x - 450\).
The x-coordinate of the vertex is:
\[x = -\frac{b}{2a} = -\frac{23}{2(-0.05)} = 230\]
So, the company must sell 230 headphones.

(c) Substitute \(x = 230\) into the profit function \(P(x)\):
\[P(230) = -0.05(230)^2 + 23(230) - 450\]
\[P(230) = -0.05(52900) + 5290 - 450 = -2645 + 5290 - 450 = 2195\]
The maximum daily profit is $2195.

(d) The break-even points occur where \(P(x) = 0\):
\[-0.05x^2 + 23x - 450 = 0\]
Using a GDC or the quadratic formula:
\[x = \frac{-23 \pm \sqrt{23^2 - 4(-0.05)(-450)}}{2(-0.05)}\]
\[x = \frac{-23 \pm \sqrt{529 - 90}}{-0.1} = \frac{-23 \pm \sqrt{439}}{-0.1}\]
\[x_1 \approx 20.48 \approx 20\]
\[x_2 \approx 439.52 \approx 440\]
Thus, the break-even points are 20 and 440 headphones.

(a)
M1 for substituting \(R(x) - C(x)\)
A1 for \(-0.05x^2 + 23x - 450\)

(b)
M1.7 for utilizing GDC peak-finder, setting the derivative \(P'(x) = -0.1x + 23 = 0\), or applying the vertex formula \(x = -\frac{b}{2a}\)
A1 for \(230\)

(c)
M1 for substituting their answer from part (b) into their formula from part (a), or using GDC to find the y-coordinate of the vertex
A1 for \(2195\)

(d)
M1 for setting \(P(x) = 0\) or finding x-intercepts on GDC
A1 for \(20\) (accept \(20.5\))
A1 for \(440\) (accept \(439.5\))

(a) The total surface area of a closed rectangular box with a square base of side \(x\) and height \(h\) consists of 2 square faces of area \(x^2\) and 4 rectangular faces of area \(xh\). Thus, the surface area \(A = 2x^2 + 4xh\). Since \(A = 216\), we have \(2x^2 + 4xh = 216 \implies 4xh = 216 - 2x^2 \implies h = \frac{216 - 2x^2}{4x} = \frac{108 - x^2}{2x}\). (b) The volume of the box is given by \(V = \text{base area} \times \text{height} = x^2 h\). Substituting our expression for \(h\) from part (a): \(V = x^2 \left(\frac{108 - x^2}{2x}\right) = \frac{x(108 - x^2)}{2} = 54x - 0.5x^3\). (c)(i) To find the maximum volume, we can plot the graph of \(V = 54x - 0.5x^3\) on a GDC and find the coordinates of the local maximum for \(x > 0\). Alternatively, differentiating with respect to \(x\) gives \(\frac{dV}{dx} = 54 - 1.5x^2\). Setting this to zero for maximum volume gives \(1.5x^2 = 54 \implies x^2 = 36 \implies x = 6\) (since \(x > 0\)). (c)(ii) Substituting \(x = 6\) back into the volume formula yields \(V = 54(6) - 0.5(6)^3 = 324 - 108 = 216\) cm\(^3\).

(a) M1 for writing down the surface area equation: \(2x^2 + 4xh = 216\). A1 for correct algebraic simplification leading to \(h = \frac{108 - x^2}{2x}\). (b) M1 for substituting \(h\) into the volume formula \(V = x^2 h\). A1 for showing the simplification to the given formula. (c)(i) M1 for expressing or implying the derivative \(\frac{dV}{dx} = 54 - 1.5x^2\) or indicating GDC maximum search. M1 for setting derivative to 0: \(1.5x^2 = 54\). A1 for \(x = 6\). (c)(ii) M1 for substituting their value of \(x\) into the volume equation. A1 for \(216\).

(a) Observing the pattern of the terms: The coefficient of \(a\) doubles with each consecutive term: \(1, 2, 4, 8, 16, \dots\). The coefficient of \(b\) follows the pattern: \(1, 3, 7, 15, 31, \dots\). Thus, \(u_4 = 8a + 15b\) and \(u_5 = 16a + 31b\). (b) The coefficient of \(b\) is given by the sequence \(1, 3, 7, 15, \dots\). This is represented by \(2^n - 1\). (c) Using the given values: \(u_2 = 2a + 3b = 13\) and \(u_4 = 8a + 15b = 43\). Multiplying the first equation by 5 gives \(10a + 15b = 65\). Subtracting \(u_4\) from this: \((10a + 15b) - (8a + 15b) = 65 - 43 \implies 2a = 22 \implies a = 11\). Substituting \(a = 11\) into the first equation: \(2(11) + 3b = 13 \implies 22 + 3b = 13 \implies 3b = -9 \implies b = -3\). (d) Substituting \(a = 11\) and \(b = -3\) into the general term: \(u_n = 11(2^{n-1}) - 3(2^n - 1) = 11(2^{n-1}) - 6(2^{n-1}) + 3 = 5(2^{n-1}) + 3\). The sum of the first 8 terms is \(S_8 = \sum_{n=1}^{8} (5(2^{n-1}) + 3) = 5 \sum_{n=1}^{8} 2^{n-1} + \sum_{n=1}^{8} 3\). Since \(\sum_{n=1}^{8} 2^{n-1}\) is a geometric series with first term 1 and common ratio 2, its sum is \(\frac{2^8 - 1}{2 - 1} = 255\). Thus, \(S_8 = 5(255) + 3(8) = 1275 + 24 = 1299\).

(a) B1 for \(u_4 = 8a + 15b\). B1 for \(u_5 = 16a + 31b\). (b) M1 for identifying the sequence of \(b\) coefficients as involving powers of 2. A1 for \(2^n - 1\). (c) M1 for writing down the correct equations \(2a + 3b = 13\) and \(8a + 15b = 43\). M1 for attempting to solve the simultaneous equations. A1 for \(a = 11\). A1 for \(b = -3\). (d) M1 for expressing \(u_n\) in terms of \(n\) or calculating the first 8 terms individually. A1 for \(1299\).

(a) Draw a North-South line through \(L\). The bearing of \(L\) from \(P\) is \(060^\circ\), which means the interior angle between the line \(PL\) and the South direction at \(L\) is \(060^\circ\) (alternate angles). Since the bearing of \(B\) from \(L\) is \(135^\circ\), the angle between the South direction at \(L\) and the line \(LB\) is \(180^\circ - 135^\circ = 45^\circ\). Thus, \(\angle PLB = 60^\circ + 45^\circ = 105^\circ\). (b) Using the Cosine Rule in \(\triangle PLB\): \(PB^2 = PL^2 + LB^2 - 2(PL)(LB)\cos(\angle PLB) = 12^2 + 18^2 - 2(12)(18)\cos(105^\circ) = 144 + 324 - 432\cos(105^\circ)\). Since \(\cos(105^\circ) \approx -0.258819\), \(PB^2 \approx 468 - 432(-0.258819) = 579.81\). Thus, \(PB = \sqrt{579.81} \approx 24.079\) km. To 3 s.f., \(PB = 24.1\) km. (c) Let \(\theta = \angle LPB\). Using the Sine Rule in \(\triangle PLB\): \(\frac{\sin\theta}{18} = \frac{\sin(105^\circ)}{24.079} \implies \sin\theta = \frac{18\sin(105^\circ)}{24.079} \approx 0.72207\). Thus, \(\theta \approx 46.22^\circ\). The bearing of \(B\) from \(P\) is the bearing of \(L\) from \(P\) plus \(\angle LPB\), which is \(060^\circ + 46.22^\circ = 106.22^\circ\). To 1 d.p., the bearing is \(106.2^\circ\). (d) Distance \(PB \approx 24.079\) km. Speed = \(120\) km/h. Time taken in hours = \(\frac{24.079}{120}\). Time in minutes = \(\frac{24.079}{120} \times 60 = \frac{24.079}{2} = 12.0395\) minutes. To 3 s.f., this is \(12.0\) minutes.

(a) M1 for finding the direction angle of \(PL\) with the South line at \(L\) is \(60^\circ\) or finding the bearing of \(P\) from \(L\) is \(240^\circ\). A1 for showing clearly that \(\angle PLB = 105^\circ\). (b) M1 for correct substitution into the Cosine Rule. A1 for \(PB^2 \approx 579.8\). A1 for \(24.1\) (accept \(24.08\)). (c) M1 for using the Sine Rule or Cosine Rule to find \(\angle LPB\). A1 for \(\angle LPB \approx 46.2^\circ\). A1 for bearing of \(106.2^\circ\) (accept \(106^\circ\)). (d) M1 for using \(\text{time} = \frac{\text{distance}}{\text{speed}} \times 60\). A1 for \(12.0\) (or \(12\)).

Let \(r\) be the radius of both the hemisphere and the cone, and \(h = 3r\) be the height of the cone. The total surface area consists of the curved surface area of the hemisphere and the curved surface area of the cone. The curved surface area of the hemisphere is \(2\pi r^2\). The curved surface area of the cone is \(\pi r l\), where \(l\) is the slant height of the cone: \(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (3r)^2} = \sqrt{10r^2} = r\sqrt{10}\). Thus, the total surface area \(A\) is given by: \(A = 2\pi r^2 + \pi r(r\sqrt{10}) = \pi r^2(2 + \sqrt{10})\). Given that \(A = 120\text{ cm}^2\): \(120 = \pi r^2(2 + \sqrt{10})\) which gives \(r^2 = \frac{120}{\pi(2 + \sqrt{10})} \approx 7.39927\), so \(r \approx 2.72016\text{ cm}\). The total volume \(V\) of the paperweight is the sum of the volume of the hemisphere and the volume of the cone: \(V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2(3r) = \frac{5}{3}\pi r^3\). Substituting \(r \approx 2.72016\): \(V \approx \frac{5}{3}\pi (2.72016)^3 \approx 105.385\text{ cm}^3\). Correct to 3 significant figures, the volume is \(105\text{ cm}^3\).

M1 for expressing slant height of the cone as \(r\sqrt{10}\). M1 for setting up the total surface area equation: \(2\pi r^2 + \pi r l = 120\). A1 for obtaining \(r^2(2+\sqrt{10}) = \frac{120}{\pi}\). A1 for calculating \(r \approx 2.72\text{ cm}\) (at least 3 s.f.). M1 for volume formula of hemisphere: \(\frac{2}{3}\pi r^3\). M1 for volume formula of cone: \(\frac{1}{3}\pi r^2 h = \pi r^3\). A1 for total volume formula in terms of \(r\): \(V = \frac{5}{3}\pi r^3\). M1 for substituting their value of \(r\) into their volume formula. A2 for final answer 105 (allow 105.3 to 105.4).

The time taken by Amy is \(\frac{36}{x}\) hours. The time taken by Ben is \(\frac{36}{x-3}\) hours. Since Ben takes \(24\text{ minutes}\) longer than Amy, and \(24\text{ minutes} = \frac{24}{60} = \frac{2}{5}\) hours, we can write: \(\frac{36}{x-3} - \frac{36}{x} = \frac{2}{5}\). Divide both sides by 2: \(\frac{18}{x-3} - \frac{18}{x} = \frac{1}{5}\) Multiplying through by the common denominator \(5x(x-3)\) gives: \(90x - 90(x-3) = x(x-3)\) which simplifies to: \(90x - 90x + 270 = x^2 - 3x\), leading to: \(x^2 - 3x - 270 = 0\). Solving the quadratic equation by factoring: \((x - 18)(x + 15) = 0\). Since speed must be positive, \(x = 18\text{ km/h}\). The time Amy takes is \(\frac{36}{18} = 2\) hours.

M1 for Amy's time \(\frac{36}{x}\) and Ben's time \(\frac{36}{x-3}\). M1 for converting 24 minutes to hours: \(\frac{24}{60}\) or \(0.4\) or \(\frac{2}{5}\). M2 for setting up the correct equation: \(\frac{36}{x-3} - \frac{36}{x} = \frac{2}{5}\). M1 for eliminating denominators correctly to get \(90x - 90(x-3) = x(x-3)\). A1 for showing the step leading to the given quadratic equation: \(x^2 - 3x - 270 = 0\). M1 for solving the quadratic equation: \((x-18)(x+15)=0\). A1 for choosing the positive root \(x = 18\). M1 for calculating Amy's time: \(\frac{36}{18}\) hours. A1 for final answer of 2 hours.

Using the given terms for \(u_n = an^2 + bn + c\): For \(n=1\): \(a + b + c = 5\). For \(n=2\): \(4a + 2b + c = 14\). For \(n=3\): \(9a + 3b + c = 29\). Subtracting the first equation from the second: \(3a + b = 9\). Subtracting the second equation from the third: \(5a + b = 15\). Subtracting these two results: \(2a = 6 \implies a = 3\). Substituting \(a = 3\) into \(3a + b = 9\) gives \(9 + b = 9 \implies b = 0\). Substituting \(a = 3\) and \(b = 0\) into \(a + b + c = 5\) gives \(3 + c = 5 \implies c = 2\). Thus, \(u_n = 3n^2 + 2\). For the second sequence, \(v_n = d \cdot 2^n + e\). Given \(v_1 = u_1 = 5\): \(2d + e = 5\). Given \(v_2 = u_2 = 14\): \(4d + e = 14\). Subtracting the two equations: \(2d = 9 \implies d = 4.5\). Substituting \(d = 4.5\) into \(2d + e = 5\) gives \(9 + e = 5 \implies e = -4\). Therefore, \(v_n = 4.5 \cdot 2^n - 4\). To find \(v_5\): \(v_5 = 4.5 \cdot 2^5 - 4 = 4.5 \cdot 32 - 4 = 144 - 4 = 140\).

M1 for setting up simultaneous equations for \(a, b, c\). M1 for finding first differences: 9 and 15, and second difference: 6. A1 for \(a = 3\). A1 for \(b = 0\) and \(c = 2\). M1 for setting up equations for \(d\) and \(e\): \(2d + e = 5\) and \(4d + e = 14\). M1 for solving the simultaneous equations to find \(d\) or \(e\). A1 for \(d = 4.5\) and \(e = -4\). M1 for substituting \(n = 5\) into their formula for \(v_n\). A2 for the final answer 140.

\textbf{Part (a)}
The total volume \(V\) of the solid is the sum of the volume of the hemisphere and the volume of the cone:
\(V = V_{\text{hemisphere}} + V_{\text{cone}} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\).

We are given that \(V = 360\pi\), so:
\(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = 360\pi\).

Divide the entire equation by \(\pi\):
\(\frac{2}{3}r^3 + \frac{1}{3}r^2 h = 360\).

Multiply by 3 to clear the fractions:
\(2r^3 + r^2 h = 1080\).

Rearrange to make \(h\) the subject:
\(r^2 h = 1080 - 2r^3 \implies h = \frac{1080 - 2r^3}{r^2}\).

\textbf{Part (b)}
The total surface area \(A\) of the solid is the sum of the curved surface area of the hemisphere and the curved surface area of the cone (since their circular faces are joined together inside the solid):
\(A = A_{\text{hemisphere}} + A_{\text{cone}} = 2\pi r^2 + \pi r l\),
where \(l\) is the slant height of the cone: \(l = \sqrt{r^2 + h^2}\).

Substitute \(h = \frac{1080 - 2r^3}{r^2}\) into the expression for \(l\):
\(l = \sqrt{r^2 + \left(\frac{1080 - 2r^3}{r^2}\right)^2}\).

Therefore, the total surface area \(A\) is:
\(A = 2\pi r^2 + \pi r \sqrt{r^2 + \left(\frac{1080 - 2r^3}{r^2}\right)^2}\).

\textbf{Part (c)}
(i) Plotting the function \(y = 2\pi x^2 + \pi x \sqrt{x^2 + \left(\frac{1080 - 2x^3}{x^2}\right)^2}\) on a graphic display calculator (GDC) for \(x > 0\) and finding the coordinates of the minimum point:
The minimum point is located at approximately \((7.017, 542.31)\).
To 3 significant figures, the value of \(r\) is \(7.02\).
(ii) The minimum total surface area to 1 decimal place is \(542.3\text{ cm}^2\).

\textbf{Part (a)}
M1: For equating the sum of the formulas of hemisphere and cone to \(360\pi\): \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = 360\pi\).
M1: For algebraic steps to eliminate \(\pi\) and isolate the \(h\) term: \(r^2 h = 1080 - 2r^3\).
A1: Correct final expression shown clearly: \(h = \frac{1080 - 2r^3}{r^2}\).

\textbf{Part (b)}
M1: For using the slant height formula \(l = \sqrt{r^2 + h^2}\) and the total area formula \(A = 2\pi r^2 + \pi r l\).
A1: Correct substitution shown to obtain the required surface area expression.

\textbf{Part (c)(i)}
M1: For attempting to sketch/analyze the function on GDC (indicated by mentioning a minimum search or derivative setting).
A2: For \(r = 7.02\) (or \(7.017...\)) (A1 for \(7.0\)).

\textbf{Part (c)(ii)}
M1: For substituting the value of \(r\) back into the function or reading the y-value from GDC.
A1: For \(A = 542.3\) (accept \(542\)).

\textbf{Part (a)}
In 1 hour, Pump A alone empties \(\frac{1}{x}\) of the tank.
In 1 hour, Pump B alone empties \(\frac{1}{x+4}\) of the tank.
Together, in 1 hour, they empty:
\(\frac{1}{x} + \frac{1}{x+4} = \frac{(x+4) + x}{x(x+4)} = \frac{2x+4}{x^2+4x}\).

\textbf{Part (b)(i)}
Since both pumps together empty the tank in 3 hours, the fraction of the tank they empty together in 1 hour is \(\frac{1}{3}\).
Therefore:
\(\frac{2x+4}{x^2+4x} = \frac{1}{3}\).

Cross-multiply to clear the fractions:
\(3(2x+4) = 1(x^2+4x)\)
\(6x + 12 = x^2 + 4x\).

Rearrange into standard quadratic form:
\(x^2 + 4x - 6x - 12 = 0 \implies x^2 - 2x - 12 = 0\).

\textbf{Part (b)(ii)}
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for \(a = 1, b = -2, c = -12\):
\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-12)}}{2(1)}\)
\(x = \frac{2 \pm \sqrt{4 + 48}}{2} = \frac{2 \pm \sqrt{52}}{2}\).

Evaluating these:
\(x = \frac{2 + 7.2111}{2} \approx 4.61\) (to 3 s.f.)
\(x = \frac{2 - 7.2111}{2} \approx -2.61\) (to 3 s.f.).

\textbf{Part (b)(iii)}
Since time \(x\) must be positive, we choose the positive root \(x = 1 + \sqrt{13} \approx 4.60555\) hours.
Pump B alone takes \(x + 4\) hours:
\(\text{Time for Pump B} \approx 4.60555 + 4 = 8.60555 \text{ hours}\).

Convert the decimal part to minutes:
\(0.60555 \times 60 \approx 36.33\) minutes.
To the nearest minute, the time taken is 8 hours and 36 minutes.

\textbf{Part (a)}
M1: For writing \(\frac{1}{x} + \frac{1}{x+4}\).
A1: Correct simplified single fraction: \(\frac{2x+4}{x^2+4x}\) (or equivalent).

\textbf{Part (b)(i)}
M1: For equating the 1-hour fraction to \(\frac{1}{3}\): \(\frac{2x+4}{x^2+4x} = \frac{1}{3}\).
M1: For cross-multiplying correctly: \(6x + 12 = x^2 + 4x\).
A1: For reaching the final quadratic equation \(x^2 - 2x - 12 = 0\) with no errors seen.

\textbf{Part (b)(ii)}
M1: For correct substitution into the quadratic formula: \(\frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-12)}}{2(1)}\) (allow 1 sign slip).
A1: For \(x = 4.61\).
A1: For \(x = -2.61\).

\textbf{Part (b)(iii)}
M1: For calculating \(x + 4 \approx 8.61\) hours (must use positive root).
A1: For 8 hours and 36 minutes (allow 8h 36m).

\textbf{Part (a)(i)}
Sequence A: \(5, 11, 17, 23, 29, \dots\)
This is an arithmetic sequence with first term \(a = 5\) and common difference \(d = 6\).
The \(n\)-th term is:
\(a + (n-1)d = 5 + 6(n-1) = 6n - 1\).

\textbf{Part (a)(ii)}
Sequence B: \(3, 9, 19, 33, 51, \dots\)
Let's find the differences:
First differences: \(6, 10, 14, 18, \dots\)
Second differences: \(4, 4, 4, \dots\)
Since the second differences are constant, it is a quadratic sequence of the form \(An^2 + Bn + C\), where \(2A = 4 \implies A = 2\).
Subtracting \(2n^2\) from each term of Sequence B:
Term 1: \(3 - 2(1^2) = 1\)
Term 2: \(9 - 2(2^2) = 1\)
Term 3: \(19 - 2(3^2) = 1\)
Term 4: \(33 - 2(4^2) = 1\)
This is a constant sequence of \(1\).
So, the \(n\)-th term is \(2n^2 + 1\).

\textbf{Part (a)(iii)}
Sequence C: \(128, 64, 32, 16, 8, \dots\)
This is a geometric sequence with first term \(a = 128\) and common ratio \(r = 0.5\).
The \(n\)-th term is:
\(128 \times (0.5)^{n-1}\) (or \(2^{8-n}\)).

\textbf{Part (b)}
The \(k\)-th term of Sequence A is \(6k - 1\).
The \(k\)-th term of Sequence B is \(2k^2 + 1\).
We are given:
\(6k - 1 = (2k^2 + 1) - 110\)
\(6k - 1 = 2k^2 - 109\).

Rearranging the terms to form a quadratic equation:
\(2k^2 - 6k - 108 = 0\).
Divide by 2:
\(k^2 - 3k - 54 = 0\).
Factorizing:
\((k - 9)(k + 6) = 0\).

Thus, \(k = 9\) or \(k = -6\).
Since \(k\) represents the position of a term in the sequence, it must be a positive integer.
Therefore, \(k = 9\).

\textbf{Part (a)(i)}
M1: For identifying common difference of 6 or writing \(6n + c\).
A1: Correct expression: \(6n - 1\).

\textbf{Part (a)(ii)}
M1: For finding second differences of 4 (implies \(2n^2\)).
M1: For subtracting \(2n^2\) from the sequence terms to find the linear/constant part.
A1: Correct expression: \(2n^2 + 1\).

\textbf{Part (a)(iii)}
M1: For identifying common ratio of 0.5 or \(2^{-1}\).
A1: Correct expression: \(128 \times (0.5)^{n-1}\) or \(2^{8-n}\) (or equivalent).

\textbf{Part (b)}
M1: For setting up the equation: \(6k - 1 = 2k^2 + 1 - 110\).
M1: For simplifying to a quadratic equation and solving (e.g. factoring to \((k-9)(k+6)=0\) or using the quadratic formula).
A1: For \(k = 9\) (must reject \(k = -6\)).

(a) Volume of a cone: \(V = \frac{1}{3}\pi r^2 h\). Total surface area of a cone: \(A = \pi r^2 + \pi r l\), where \(l = \sqrt{r^2 + h^2}\) is the slant height. Since the volume is numerically equal to the total surface area: \(\frac{1}{3}\pi r^2 h = \pi r^2 + \pi r\sqrt{r^2 + h^2}\). Dividing both sides by \(\pi r\) (since \(r > 0\)): \(\frac{1}{3}rh = r + \sqrt{r^2 + h^2}\). (b) Substitute \(h = 12\) into the equation: \(\frac{1}{3}r(12) = r + \sqrt{r^2 + 12^2}\) which simplifies to \(4r = r + \sqrt{r^2 + 144}\). Rearranging terms: \(3r = \sqrt{r^2 + 144}\). Squaring both sides: \(9r^2 = r^2 + 144\) which gives \(8r^2 = 144\). Solving for \(r^2\): \(r^2 = 18\). Since \(r > 0\), \(r = \sqrt{18} = 3\sqrt{2}\). Thus, \(a = 3\) and \(b = 2\), so \(r = 3\sqrt{2}\). (c) The curved surface area is given by \(A_{\text{curved}} = \pi r l = \pi r \sqrt{r^2 + h^2}\). From part (b), \(3r = \sqrt{r^2 + 144}\), so \(l = 3r = 3(3\sqrt{2}) = 9\sqrt{2}\). Alternatively, \(l = \sqrt{18 + 144} = \sqrt{162} = 9\sqrt{2}\). The curved surface area is \(\pi (3\sqrt{2})(9\sqrt{2}) = 54\pi\).

(a) M1: For equating volume and total surface area formulas, \(\frac{1}{3}\pi r^2 h = \pi r^2 + \pi r l\). M1: For substituting \(l = \sqrt{r^2 + h^2}\). A1: For dividing by \(\pi r\) to obtain the required show-that equation. (b) M1: For substituting \(h = 12\) and simplifying to \(3r = \sqrt{r^2 + 144}\). M1: For squaring both sides to get \(9r^2 = r^2 + 144\). A1: For obtaining \(8r^2 = 144\) or \(r^2 = 18\). A1: For final exact answer \(3\sqrt{2}\). (c) M1: For formula of curved surface area \(\pi r l\). M1: For finding slant height \(l = 9\sqrt{2}\) or \(\sqrt{162}\). A1: For final answer \(54\pi\).

First, factorise the first equation: \(x^2 - xy - 6y^2 = 0 \implies (x - 3y)(x + 2y) = 0\). This gives two cases: Case 1: \(x = 3y\) or Case 2: \(x = -2y\). Now, simplify the second equation: \(x^2 + 2xy + y^2 = 36 \implies (x + y)^2 = 36\), which means \(x + y = 6\) or \(x + y = -6\). Alternatively, we can substitute the cases directly into the second equation: Case 1: \(x = 3y\). Substitute into \(x + y = 6\): \(3y + y = 6 \implies 4y = 6 \implies y = 1.5\), which gives \(x = 3(1.5) = 4.5\). Substitute into \(x + y = -6\): \(3y + y = -6 \implies 4y = -6 \implies y = -1.5\), which gives \(x = 3(-1.5) = -4.5\). This gives two solution pairs: \((4.5, 1.5)\) and \((-4.5, -1.5)\). Case 2: \(x = -2y\). Substitute into \(x + y = 6\): \(-2y + y = 6 \implies -y = 6 \implies y = -6\), which gives \(x = -2(-6) = 12\). Substitute into \(x + y = -6\): \(-2y + y = -6 \implies -y = -6 \implies y = 6\), which gives \(x = -2(6) = -12\). This gives two more solution pairs: \((12, -6)\) and \((-12, 6)\). Thus, the four solution pairs are: \((4.5, 1.5)\), \((-4.5, -1.5)\), \((12, -6)\), and \((-12, 6)\).

M2: For factorising the first equation to get \((x-3y)(x+2y) = 0\) (M1 for partial factorisation or attempt). A1: For identifying the two cases \(x = 3y\) and \(x = -2y\). M1: For identifying that the second equation can be written as \((x+y)^2 = 36\) or \(x+y = \pm 6\). M2: For solving Case 1: substituting \(x = 3y\) and finding the values of \(y\) and \(x\) (M1 for one pair, A1 for both pairs: \((4.5, 1.5)\) and \((-4.5, -1.5)\)). M2: For solving Case 2: substituting \(x = -2y\) and finding the values of \(y\) and \(x\) (M1 for one pair, A1 for both pairs: \((12, -6)\) and \((-12, 6)\)). A2: For clearly presenting all four solution pairs (deduct 1 mark if only 2 or 3 correct pairs are given).

(a)(i) Sequence A is an arithmetic sequence with a common difference of 5. The next term is \(18 + 5 = 23\). (ii) Sequence B has terms 5, 11, 19, 29. The first differences are 6, 8, 10, so the next difference is 12. The next term is \(29 + 12 = 41\). (iii) Sequence C is a geometric sequence with a common ratio of 2. The next term is \(32 \times 2 = 64\). (b)(i) For Sequence A, the common difference is \(d = 5\) and the first term is \(u_1 = 3\). The \(n\)-th term is \(u_n = 3 + (n-1)5 = 5n - 2\). (b)(ii) For Sequence B, let the \(n\)-th term be \(v_n = an^2 + bn + c\). The second differences are constant and equal to 2, so \(2a = 2 \implies a = 1\). Subtracting \(n^2\) from each term gives the sequence \(4, 7, 10, 13\), which is linear with \(n\)-th term \(3n + 1\). Thus, the \(n\)-th term is \(v_n = n^2 + 3n + 1\). (b)(iii) For Sequence C, the terms are powers of 2: \(2^2, 2^3, 2^4, 2^5, \dots\). Thus, the \(n\)-th term is \(2^{n+1}\) (or \(2 \times 2^n\)).

(a)(i) B1: For 23. (ii) B1: For 41. (iii) B1: For 64. (b)(i) M1: For finding the common difference of 5 or writing \(5n + k\). A1: For \(5n - 2\). (b)(ii) M1: For identifying a quadratic sequence and finding the second difference is 2, hence \(a = 1\). M1: For a correct method to find \(b\) and \(c\) (e.g., set of equations). A1: For \(n^2 + 3n + 1\). (b)(iii) M1: For identifying a geometric sequence with ratio 2 or writing in the form \(k \cdot 2^n\). A1: For \(2^{n+1}\) or \(2 \times 2^n\).

(a) For \(n=4\), \(S = 4^2 = 16\), and \(M = 40\). For \(n=5\), \(S = 5^2 = 25\), and \(M = 60\). (b) Since the number of small squares is the square of \(n\), the expression is \(S = n^2\). (c) When \(n = 6\), \(M = 2(6)^2 + 2(6) = 72 + 12 = 84\). (d) Using \(n=1, M=4 \implies a + b = 4\). Using \(n=2, M=12 \implies 4a + 2b = 12 \implies 2a + b = 6\). Subtracting the first equation from the second gives \(a = 2\). Substituting back gives \(b = 2\). (e) If \(S = 144\), then \(n^2 = 144 \implies n = 12\). The number of matchsticks is \(M = 2(12)^2 + 2(12) = 288 + 24 = 312\). (f) Set \(2n^2 + 2n = 220 \implies 2n^2 + 2n - 220 = 0 \implies n^2 + n - 110 = 0 \implies (n + 11)(n - 10) = 0\). Since \(n > 0\), \(n = 10\).

(a) 2 marks: 1 mark for S row [16, 25], 1 mark for M row [40, 60]. (b) 1 mark: \(S = n^2\). (c) 1 mark: 84. (d) 3 marks: 1 mark for setting up simultaneous equations (e)g. \(a+b=4\) and \(4a+2b=12\)), 1 mark for \(a = 2\), 1 mark for \(b = 2\). (e) 2 marks: 1 method mark for finding \(n = 12\), 1 accuracy mark for \(M = 312\). (f) 3 marks: 1 method mark for setting up the quadratic equation \(2n^2 + 2n = 220\), 1 method mark for factorising or solving, 1 accuracy mark for \(n = 10\).

(a) For \(n=4\), \(S = 13\) and \(P = 28\). For \(n=5\), \(S = 16\) and \(P = 34\). (b) The difference in \(S\) is 3, so \(S = 3n + k\). When \(n=1\), \(3(1) + k = 4 \implies k = 1\). Thus \(S = 3n + 1\). (c) The difference in \(P\) is 6, so \(P = 6n + c\). When \(n=1\), \(6(1) + c = 10 \implies c = 4\). Thus \(P = 6n + 4\). (d) Substitute \(S = 3n + 1\) into the right-hand side: \(2S + 2 = 2(3n + 1) + 2 = 6n + 2 + 2 = 6n + 4 = P\). (e) Using \(P = 2S + 2\), set \(184 = 2S + 2 \implies 182 = 2S \implies S = 91\). (f) Since \(P = 2S + 2 = 2(S + 1)\), the perimeter must always be an even number. Since 101 is an odd number, it is not possible. Alternatively, using \(P = 6n + 4\), \(101 = 6n + 4 \implies 97 = 6n \implies n = 16.17\), which is not an integer.

(a) 2 marks: 1 mark for S row [13, 16], 1 mark for P row [28, 34]. (b) 2 marks: 1 method mark for recognizing common difference of 3 (e.g. \(3n + c\)), 1 accuracy mark for \(S = 3n + 1\). (c) 2 marks: 1 method mark for recognizing common difference of 6 (e.g. \(6n + c\)), 1 accuracy mark for \(P = 6n + 4\). (d) 2 marks: 1 mark for substitution of \(S\) and 1 mark for expanding and simplifying to show equality. (e) 2 marks: 1 method mark for \(184 = 2S + 2\) or setting up with \(n\), 1 accuracy mark for \(S = 91\). (f) 2 marks: 1 mark for stating that \(P\) must be even (or \(n\) must be an integer) and 1 mark for showing that 101 violates this condition.

(a) For \(n=4\), \(B = 10\), \(J = 12\). For \(n=5\), \(B = 15\), \(J = 20\). (b) For \(n=6\), \(B = 15 + 6 = 21\), and \(J = 20 + 10 = 30\). (c) Using \(n=1, B=1 \implies 0.5(1)^2 + k(1) = 1 \implies 0.5 + k = 1 \implies k = 0.5\). (d) The values of \(J\) are 0, 2, 6, 12, 20. The first differences are 2, 4, 6, 8, and the second differences are constant at 2. This is a quadratic sequence: \(J = an^2 + bn + c\). Since second difference is 2, \(a = 1\). Thus \(J = n^2 + bn + c\). For \(n=1, J=0 \implies 1 + b + c = 0\). For \(n=2, J=2 \implies 4 + 2b + c = 2\). Subtracting: \(3 + b = 2 \implies b = -1\). Then \(c = 0\). So \(J = n^2 - n\). (e) \(B + J = (0.5n^2 + 0.5n) + (n^2 - n) = 0.5n^2 + n^2 + 0.5n - n = 1.5n^2 - 0.5n\). (f) For \(n=10\), \(B + J = 1.5(10)^2 - 0.5(10) = 1.5(100) - 5 = 150 - 5 = 145\).

(a) 2 marks: 1 mark for B row [10, 15], 1 mark for J row [12, 20]. (b) 2 marks: 1 mark for \(B = 21\), 1 mark for \(J = 30\). (c) 2 marks: 1 method mark for substituting a coordinate (e.g. \(n=1, B=1\)), 1 accuracy mark for \(k = 0.5\). (d) 2 marks: 1 method mark for recognizing quadratic sequence or writing \(n(n-1)\), 1 accuracy mark for \(J = n^2 - n\) (or equivalent). (e) 2 marks: 1 mark for setting up \((0.5n^2 + 0.5n) + (n^2 - n)\), 1 mark for correct simplification to \(1.5n^2 - 0.5n\). (f) 2 marks: 1 method mark for substituting \(n=10\) into the expression, 1 accuracy mark for 145.

**Part (a)**
For \(n = 5\), the columns have heights 5, 4, 3, 2, 1.
The number of internal vertical edges between adjacent columns is:
- between col 1 and col 2: 4
- between col 2 and col 3: 3
- between col 3 and col 4: 2
- between col 4 and col 5: 1
Total vertical internal edges = \(4 + 3 + 2 + 1 = 10\).
By symmetry, the number of internal horizontal edges is also 10.
Total internal edges \(I(5) = 10 + 10 = 20\).
*(Alternatively, using the formula \(I(n) = n(n-1)\): \(I(5) = 5 \times 4 = 20\).)*

**Part (b)**
By analyzing the boundary of Pattern \(n\):
- The bottom edge has length \(n\).
- The left vertical edge has length \(n\).
- The horizontal step segments have total length \(n\).
- The vertical step segments have total length \(n\).
Thus, the perimeter is \(P(n) = n + n + n + n = 4n\).

**Part (c)**
The total number of unit segments is the sum of the perimeter and the internal edges:
\(E(n) = P(n) + I(n)\)
\(E(n) = 4n + (n^2 - n) = n^2 + 3n\).

**Part (d)**
Set \(E(n) = 180\):
\(n^2 + 3n = 180 \implies n^2 + 3n - 180 = 0\)
\((n + 15)(n - 12) = 0\)
Since \(n\) must be positive, \(n = 12\).

**Part (a)** [1 Mark]
- B1 for 20

**Part (b)** [2 Marks]
- M1 for recognizing perimeter grows by 4 each step or finding \(P(1)=4, P(2)=8, P(3)=12\)
- A1 for \(4n\)

**Part (c)** [2 Marks]
- M1 for writing \(E(n) = P(n) + I(n)\) using their expression for \(P(n)\)
- A1 for fully simplifying to \(n^2 + 3n\)

**Part (d)** [3.5 Marks]
- M1 for setting up the equation \(n^2 + 3n = 180\)
- M1 for solving the quadratic equation (e.g., factorizing to \((n-12)(n+15)=0\))
- A1 for \(n = 12\)
- A0.5 for rejecting \(n = -15\)

**Part (a)**
From the volume formula:
\(100 = \frac{1}{3}\pi r^2 h \implies h = \frac{300}{\pi r^2}\).

Substitute \(h\) into the curved surface area formula:
\(A = \pi r \sqrt{r^2 + \left(\frac{300}{\pi r^2}\right)^2}\)
\(A = \sqrt{(\pi r)^2 \left(r^2 + \frac{90000}{\pi^2 r^4}\right)}\)
\(A = \sqrt{\pi^2 r^2 \left(r^2 + \frac{90000}{\pi^2 r^4}\right)}\)
\(A = \sqrt{\pi^2 r^4 + \frac{90000}{r^2}}\).

**Part (b)**
Substitute \(r = 3\):
\(A(3) = \sqrt{\pi^2 (3)^4 + \frac{90000}{3^2}}\)
\(A(3) = \sqrt{81\pi^2 + 10000}\)
\(A(3) \approx \sqrt{799.438 + 10000} = \sqrt{10799.438} \approx 103.92 \approx 104\text{ cm}^2\).

**Part (c)**
Using a graphics display calculator to sketch \(y = \sqrt{\pi^2 x^4 + \frac{90000}{x^2}}\) for \(x > 0\):
(i) The minimum point occurs at \(r \approx 4.09435\text{ cm}\), which is \(4.09\text{ cm}\) to 2 decimal places.
(ii) The minimum curved surface area at this radius is \(A \approx 90.235\text{ cm}^2\), which is \(90.2\text{ cm}^2\) to 1 decimal place.

**Part (a)** [3 Marks]
- M1 for writing \(h\) in terms of \(r\): \(h = \frac{300}{\pi r^2}\)
- M1 for substituting \(h\) into \(A = \pi r \sqrt{r^2 + h^2}\)
- A1 for correct algebraic steps leading to the final given equation

**Part (b)** [2 Marks]
- M1 for substituting \(r = 3\) into the equation
- A1 for \(104\) (accept \(103.9\) to \(104.0\))

**Part (c)** [3.5 Marks]
- M1 for using GDC to find the minimum point of the function
- A1.5 for \(r = 4.09\) (award 1 mark for 4.1)
- A1 for \(A = 90.2\) (accept 90 or 90.2)

**Part (a)**
When \(t = 0\):
\(T(0) = 20 + 70 \times (0.92)^0 = 20 + 70 \times 1 = 90^\circ\text{C}\).

**Part (b)**
When \(t = 10\):
\(T(10) = 20 + 70 \times (0.92)^{10} \approx 20 + 70 \times 0.434388 = 50.407 \approx 50.4^\circ\text{C}\).

**Part (c)**
We set \(T(t) = 45\):
\(20 + 70 \times (0.92)^t = 45\)
\(70 \times (0.92)^t = 25\)
\((0.92)^t = \frac{25}{70} = \frac{5}{14}\)
Taking logarithms on both sides:
\(t \ln(0.92) = \ln\left(\frac{5}{14}\right)\)
\(t = \frac{\ln(5/14)}{\ln(0.92)} \approx \frac{-1.0296}{-0.08338} \approx 12.348 \approx 12.3\text{ minutes}\).

**Part (d)**
After \(t = 30\) minutes:
Original cup temperature:
\(T(30) = 20 + 70 \times (0.92)^{30} \approx 20 + 70 \times 0.08190 = 25.733^\circ\text{C}\).
Travel mug temperature:
\(T_{\text{flask}}(30) = 20 + 70 \times (0.97)^{30} \approx 20 + 70 \times 0.40101 = 48.071^\circ\text{C}\).
The difference in temperature is:
\(48.071 - 25.733 = 22.338 \approx 22^\circ\text{C}\) to the nearest degree.

**Part (a)** [1 Mark]
- B1 for \(90^\circ\text{C}\) (or 90)

**Part (b)** [1.5 Marks]
- M1 for substituting \(t = 10\)
- A0.5 for \(50.4\) (accept 50.4 or 50.41)

**Part (c)** [3 Marks]
- M1 for setting up the equation \(20 + 70 \times (0.92)^t = 45\) or \((0.92)^t = \frac{25}{70}\)
- M1 for taking logarithms to solve for \(t\) (or utilizing GDC)
- A1 for \(12.3\) (accept 12.3 to 12.4)

**Part (d)** [3 Marks]
- M1 for calculating \(T(30) \approx 25.7\)
- M1 for calculating \(T_{\text{flask}}(30) \approx 48.1\)
- A1 for \(22\) (accept 22 or 22.3)

Part (a):
For \(n=5\), following the pattern of differences:
- \(C_2 - C_1 = 3\)
- \(C_3 - C_2 = 6\)
- \(C_4 - C_3 = 9\)
The next difference is \(12\), so \(C_5 = C_4 + 12 = 18 + 12 = 30\).

Part (b):
The sequence of differences is an arithmetic progression: \(3, 6, 9, 12, \dots\)
This indicates a quadratic relationship of the form \(C_n = an^2 + bn + c\).
Using \(C_1 = 0, C_2 = 3, C_3 = 9\):
1) \(a + b + c = 0\)
2) \(4a + 2b + c = 3\)
3) \(9a + 3b + c = 9\)
Subtracting (1) from (2): \(3a + b = 3\).
Subtracting (2) from (3): \(5a + b = 6\).
Subtracting these two results: \(2a = 3 \implies a = 1.5\).
Thus, \(b = 3 - 3(1.5) = -1.5\).
And \(c = 0\).
Therefore, \(C_n = 1.5n^2 - 1.5n = \frac{3n(n-1)}{2}\).

Part (c):
Set \(C_k = 360\):
\(\frac{3k(k-1)}{2} = 360\)
\(3k(k-1) = 720\)
\(k(k-1) = 240\)
\(k^2 - k - 240 = 0\)
\((k - 16)(k + 15) = 0\)
Since \(k > 0\), \(k = 16\).

Part (a): [2 Marks]
- B1 for identifying the difference of 12.
- B1 for 30.

Part (b): [3.5 Marks]
- M1 for setting up simultaneous equations or using second differences.
- M1 for finding \(a = 1.5\) or equivalent.
- A1.5 for the fully correct formula \(\frac{3n(n-1)}{2}\) or \(1.5n^2 - 1.5n\).

Part (c): [3 Marks]
- M1 for setting their expression equal to 360.
- M1 for solving the quadratic equation.
- A1 for \(k = 16\).

Part (a):
The total surface area \(S\) of the capsule is the sum of the curved surface area of the cylinder and the surface areas of the two hemispheres:
\(S = 2\pi r h + 4\pi r^2\).
Given \(S = 12\pi\):
\(2\pi r h + 4\pi r^2 = 12\pi\)
Divide through by \(2\pi\):
\(rh + 2r^2 = 6\)
\(rh = 6 - 2r^2\)
\(h = \frac{6 - 2r^2}{r}\).

Part (b):
The total volume \(V\) is the sum of the volume of the cylinder and the volume of the two hemispheres:
\(V =
\pi r^2 h + \frac{4}{3}\pi r^3\).
Substitute \(h = \frac{6 - 2r^2}{r}\):
\(V = \pi r^2 \left(\frac{6 - 2r^2}{r}\right) + \frac{4}{3}\pi r^3\)
\(V = \pi r(6 - 2r^2) + \frac{4}{3}\pi r^3\)
\(V = 6\pi r - 2\pi r^3 + \frac{4}{3}\pi r^3\)
\(V = 6\pi r - \frac{2}{3}\pi r^3\).

Part (c):
To find the maximum volume, differentiate \(V\) with respect to \(r\):
\(\frac{dV}{dr} = 6\pi - 2\pi r^2\).
Set \(\frac{dV}{dr} = 0\):
\(6\pi - 2\pi r^2 = 0 \implies 2\pi r^2 = 6\pi \implies r^2 = 3 \implies r = \sqrt{3} \approx 1.73\) m.
Substitute \(r = \sqrt{3}\) back into the volume formula to find the maximum volume:
\(V = 6\pi(\sqrt{3}) - \frac{2}{3}\pi(\sqrt{3})^3 = 6\pi\sqrt{3} - 2\pi\sqrt{3} = 4\pi\sqrt{3} \approx 21.766\) m\(^3\).
To 3 significant figures, \(r = 1.73\) and \(V = 21.8\).

Part (a): [3 Marks]
- M1 for writing down the correct surface area formula: \(2\pi r h + 4\pi r^2 = 12\pi\).
- M1 for rearranging the equation to isolate the term with \(h\).
- A1 for showing the final expression clearly.

Part (b): [2.5 Marks]
- M1 for writing the volume formula \(V = \pi r^2 h + \frac{4}{3}\pi r^3\).
- M1 for substituting \(h\) from part (a).
- A0.5 for arriving at the correct shown expression.

Part (c): [3 Marks]
- M1 for differentiating \(V\) correctly to get \(6\pi - 2\pi r^2\).
- A1 for setting to 0 and finding \(r = \sqrt{3}\) (or 1.73).
- A1 for finding the maximum volume \(V \approx 21.8\).

Part (a):
Using Pythagoras' theorem, the length of the cable under water is \(\sqrt{x^2 + 400^2} = \sqrt{x^2 + 160000}\) meters.
The length of the cable over land is \(1000 - x\) meters.
The cost of the underwater section is \(50\sqrt{x^2 + 160000}\).
The cost of the overland section is \(30(1000 - x)\).
Therefore, the total cost \(C\) is:
\(C = 50\sqrt{x^2 + 160000} + 30(1000 - x)\).

Part (b):
When \(x = 0\):
\(C = 50\sqrt{0 + 160000} + 30(1000 - 0)\)
\(C = 50(400) + 30(1000) = 20000 + 30000 = 50000\) dollars.

Part (c):
To find the minimum cost, differentiate \(C\) with respect to \(x\):
\(\frac{dC}{dx} = 50 \cdot \frac{1}{2\sqrt{x^2 + 160000}} \cdot 2x - 30\)
\(\frac{dC}{dx} = \frac{50x}{\sqrt{x^2 + 160000}} - 30\)
Set \(\frac{dC}{dx} = 0\):
\(\frac{50x}{\sqrt{x^2 + 160000}} = 30\)
\(5x = 3\sqrt{x^2 + 160000}\)
Square both sides:
\(25x^2 = 9(x^2 + 160000)\)
\(25x^2 = 9x^2 + 1440000\)
\(16x^2 = 1440000\)
\(x^2 = 90000 \implies x = 300\) meters.
Substitute \(x = 300\) into the cost function:
\(C = 50\sqrt{300^2 + 160000} + 30(1000 - 300)\)
\(C = 50(500) + 30(700) = 25000 + 21000 = 46000\) dollars.

Part (a): [3 Marks]
- M1 for using Pythagoras to find the underwater length \(\sqrt{x^2 + 160000}\).
- M1 for identifying the overland length \(1000 - x\).
- A1 for the correct cost formula.

Part (b): [1.5 Marks]
- M1 for substituting \(x = 0\) into their cost equation.
- A0.5 for 50000.

Part (c): [4 Marks]
- M1 for differentiating the cost function correctly.
- M1 for setting derivative to 0 and rearranging.
- A1 for finding \(x = 300\).
- A1 for calculating minimum cost of 46000.

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6`.

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).

### Part A:
1. For \(n = 5\), the boundary dots value is \(12\).
For \(n = 6\), total dots \(T_6 = 21\), boundary dots \(B_6 = 15\), inner dots \(I_6 = 6\).
Thus, `[a] = 12`, `[b] = 21`, `[c] = 15`, `[d] = 6\).

2. The total number of dots is the sum of the first \(n\) positive integers:
\[T_n = \frac{n(n+1)}{2}\]

3. For \(n \ge 3\), the boundary is made of 3 sides of length \(n\). Each of the 3 vertices is counted twice, so:
\[B_n = 3n - 3\]

4. Since \(T_n = B_n + I_n\), we have:
\[I_n = T_n - B_n = \frac{n(n+1)}{2} - (3n - 3) = \frac{n^2 + n - 6n + 6}{2} = \frac{1}{2}(n^2 - 5n + 6)\]
This matches the required formula.

### Part B:
5. Setting \(I_n = 15\):
\[\frac{1}{2}(n^2 - 5n + 6) = 15 \implies n^2 - 5n + 6 = 30 \implies n^2 - 5n - 24 = 0\]
Factorising gives:
\[(n - 8)(n + 3) = 0\]
Since \(n\) must be positive and \(n \ge 3\), we have \(n = 8\).

6. Setting \(B_n = 4 I_n\):
\[3n - 3 = 4 \times \frac{1}{2}(n^2 - 5n + 6)\]
\[3(n - 1) = 2(n^2 - 5n + 6)\]
\[3n - 3 = 2n^2 - 10n + 12 \implies 2n^2 - 13n + 15 = 0\]
Factorising gives:
\[(2n - 3)(n - 5) = 0\]
Since \(n\) must be an integer, \(n = 5\).

### Part A:
- **Q1 [1.5 marks]**:
- Award 0.5 marks for the correct value \([a] = 12\).
- Award 1 mark for all three remaining correct values: \([b] = 21\), \([c] = 15\), \([d] = 6\). Award 0.5 marks if only one or two of these are correct.
- **Q2 [1.5 marks]**:
- **M1** for attempting to find a quadratic pattern or recognizing the sum of an arithmetic series.
- **A1** for the correct formula \(T_n = \frac{n(n+1)}{2}\) or equivalent.
- **Q3 [1.5 marks]**:
- **M1** for identifying a linear relationship with a common difference of 3 (e.g., \(3n + c\)).
- **A1** for the correct formula \(B_n = 3n - 3\) or equivalent.
- **Q4 [1.5 marks]**:
- **M1** for setting up \(I_n = T_n - B_n\).
- **A1** for showing complete algebraic steps leading to the given expression \(\frac{1}{2}(n^2 - 5n + 6)\).

### Part B:
- **Q5 [1.5 marks]**:
- **M1** for setting \(I_n = 15\) and forming a valid quadratic equation (e.g., \(n^2 - 5n - 24 = 0\)).
- **A1** for finding the correct solution \(n = 8\) (must reject \(n = -3\)).
- **Q6 [1 mark]**:
- **M1** for writing \(3n - 3 = 2(n^2 - 5n + 6)\) or equivalent and solving.
- **A1** for finding the correct integer solution \(n = 5\).