2025 Cambridge IGCSE International Mathematics (0607) 模拟试题及答案详解

First, calculate the profit: $156 - $120 = $36. Next, calculate the percentage profit by dividing the profit by the original cost and multiplying by 100: \(\frac{36}{120} \times 100 = \frac{3}{10} \times 100 = 30\)%.

M1 for finding the profit: 156 - 120 = 36. M1 for (their 36) / 120 * 100. A1 for 30.

Expand the first bracket: \(5(2x - 1) = 10x - 5\). Expand the second bracket: \(-2(3x - 4) = -6x + 8\). Combine like terms: \((10x - 6x) + (-5 + 8) = 4x + 3\).

M1 for 10x - 5. M1 for -6x + 8 (or +8 seen). A1 for 4x + 3 (or equivalent, e.g., 3 + 4x).

The sequence decreases by 3 each time, so the common difference is -3. This means the expression contains \(-3n\). The term before the first term (when \(n = 0\)) is \(17 + 3 = 20\). Therefore, the expression for the \(n\)-th term is \(20 - 3n\).

M1 for identifying a common difference of -3 (e.g. finding -3n). M1 for finding the constant term 20. A1 for 20 - 3n (or -3n + 20).

Cost price = $150. Total selling price = \(20 \times \$9.00 = \$180\). Profit = \(\$180 - \$150 = \$30\). Percentage profit = \(\frac{30}{150} \times 100\% = \frac{1}{5} \times 100\% = 20\%\).

M1 for finding the total selling price of \(20 \times 9 = 180\). M1 for finding the profit of \(180 - 150 = 30\) or setting up the percentage profit equation \(\frac{180 - 150}{150} \times 100\). A1 for 20% (or 20).

We look for two numbers that multiply to give \(-24\) and add to give \(5\). These numbers are \(8\) and \(-3\). Therefore, the completely factorised expression is \((x + 8)(x - 3)\).

M1 for identifying \(8\) and \(-3\) as the factors, or for writing an expression of the form \((x + a)(x + b)\) where \(ab = -24\) or \(a + b = 5\). A1 for \((x + 8)(x - 3)\) or \((x - 3)(x + 8)\).

The terms of the sequence decrease by \(3\) each time, so the common difference is \(-3\). This means the formula contains the term \(-3n\). To find the constant, we look at the first term \((n = 1)\): \(-3(1) + c = 17\), which gives \(c = 20\). Therefore, the \(n\)-th term is \(20 - 3n\).

M1 for finding the common difference of \(-3\) (or writing an expression containing \(-3n\)). A1 for \(20 - 3n\) or \(-3n + 20\) or equivalent.

1. Find the total selling price: \(20 \times \$11 = \$220\). 2. Find the total profit: \(\$220 - \$160 = \$60\). 3. Calculate the percentage profit: \(\frac{60}{160} \times 100 = \frac{3}{8} \times 100 = 37.5\%\).

M1 for finding total selling price of 220 or total profit of 60. M1 for dividing profit by cost price: 60/160. A1.53 for 37.5% (or 37.5).

1. Expand the first bracket: \(3(2x - 5) = 6x - 15\). 2. Expand the second bracket: \(-4(x - 2) = -4x + 8\). 3. Combine like terms: \(6x - 4x - 15 + 8 = 2x - 7\).

M1 for expansion 6x - 15. M1 for expansion -4x + 8 (or +8 seen). A1.53 for 2x - 7.

1. Find the common difference by subtracting consecutive terms: \(17 - 21 = -4\). This means the expression contains \(-4n\). 2. Find the zero-term or use the formula \(a + (n-1)d\): \(21 + (n-1)(-4) = 21 - 4n + 4 = 25 - 4n\).

M1 for identifying the common difference is -4 (e.g. finding -4n). M1 for finding the constant term of 25 (e.g. 21 - (-4)). A1.53 for 25 - 4n (or -4n + 25).

First, find the cost price of one pack: \(15 \div 20 = 0.75\) dollars. Next, find the profit on each pack: \(1.20 - 0.75 = 0.45\) dollars. Finally, calculate the percentage profit: \(\frac{0.45}{0.75} \times 100 = 60\%\).

M1 for finding cost price per pack as 0.75 or total selling price as 24. M1 for calculating profit as 0.45 or total profit as 9. A1 for 60 or 60%.

First, evaluate \(g(6) = \frac{6}{2} + 4 = 3 + 4 = 7\). Next, substitute this result into \(f(x)\) to get \(f(7) = 5(7) - 3 = 35 - 3 = 32\).

M1 for finding \(g(6) = 7\). M1 for substituting their 7 into \(f(x)\). A1 for 32.

The first term of the sequence is \(a = 11\) and the common difference is \(d = -4\). The formula for the \(n\)-th term is \(a + (n - 1)d\). Substituting the values gives \(11 + (n - 1)(-4) = 11 - 4n + 4 = 15 - 4n\).

M1 for finding the common difference of -4. M1 for using the formula \(a + (n-1)d\) with their values. A1 for 15 - 4n (or equivalent).

First, find the total revenue from selling all 24 watches:
\(24 \times \$22.50 = \$540\).

Next, calculate the profit:
\(\text{Profit} = \$540 - \$360 = \$180\).

Finally, calculate the percentage profit based on the cost price:
\(\text{Percentage Profit} = \frac{180}{360} \times 100 = 50\%\).

M1 for showing method to find total revenue, e.g. \(24 \times 22.50\) (or equivalent intermediate steps)
M1 for \(\frac{\text{their } 540 - 360}{360} \times 100\)
A1 for 50

We are given that \(f(x) = -5\). Set up the equation:
\(7 - 4x = -5\)

Subtract 7 from both sides:
\(-4x = -12\)

Divide by -4:
\(x = 3\)

M1 for setting up the equation \(7 - 4x = -5\)
M1 for isolating the term in \(x\), e.g., \(-4x = -12\) or \(4x = 12\)
A1 for 3

Find the highest common factor of the two terms, \(6x^2\) and \(15xy\):
- The highest common numerical factor of 6 and 15 is 3.
- The highest common variable factor of \(x^2\) and \(xy\) is \(x\).

Therefore, the highest common factor is \(3x\).

Divide each term by \(3x\) to find the terms inside the parentheses:
- \(6x^2 \div 3x = 2x\)
- \(-15xy \div 3x = -5y\)

This gives the fully factorised expression:
\(3x(2x - 5y)\).

M1 for finding a partial common factor of \(3\) or \(x\), e.g., \(3(2x^2 - 5xy)\) or \(x(6x - 15y)\)
A1 for \(3x(2x - 5y)\) or equivalent completely factorised form

First, convert the cost in the UK into dollars using the given exchange rate: \(\pounds 12 \times \$1.60 = \$19.20\). Next, find the difference in cost between the two countries: \(\$19.20 - \$18.00 = \$1.20\). Therefore, the difference in cost is \(\$1.20\).

M1 for converting \(\pounds 12\) to dollars: \(12 \times 1.60\) (or equivalent method) A1 for \(\$19.20\) A1 for final answer \(\$1.20\) (accept 1.2 or 1.20)

The sequence decreases by \(4\) each time, so the common difference is \(-4\). This means the expression for the \(n\)-th term will include \(-4n\). The 'zero term' (the term before the first term \(13\)) is \(13 + 4 = 17\). Thus, the expression for the \(n\)-th term is \(17 - 4n\). Alternatively, using the arithmetic sequence formula \(a + (n - 1)d\), we have \(13 + (n - 1)(-4) = 13 - 4n + 4 = 17 - 4n\).

M1 for identifying the common difference is \(-4\) (e.g. writing \(-4n\) as part of their expression) M1 for identifying the constant term is \(17\) A1 for the correct expression \(17 - 4n\) (accept \(-4n + 17\))

Simple interest is calculated using the formula \(I = \frac{P \times r \times t}{100}\).
The total amount in the account is \(A = P + I = P\left(1 + \frac{rt}{100}\right)\).
Substituting the given values:
\(1016 = x \left(1 + \frac{4.5 \times 6}{100}\right)\)
\(1016 = x \left(1 + \frac{27}{100}\right)\)
\(1016 = 1.27x\)
To solve for \(x\):
\(x = \frac{1016}{1.27} = \frac{101600}{127}\)
Since \(127 \times 8 = 1016\), we have:
\(x = 800\).

M1 for correct substitution into simple interest formula: \(1016 = x(1 + 0.045 \times 6)\) or equivalent
M1 for simplifying to \(1016 = 1.27x\)
M1 for attempt to divide \(1016\) by \(1.27\) (e.g., writing \(\frac{101600}{127}\))
A1 for \(800\)

Let us analyze the differences of the sequence:
Terms: \(3, \quad 8, \quad 17, \quad 30\)
First differences: \(5, \quad 9, \quad 13\)
Second differences: \(4, \quad 4\)
Since the second differences are constant, the sequence is quadratic and has the general form \(an^2 + bn + c\).
The coefficient \(a\) is given by half of the second difference:
\(2a = 4 \implies a = 2\).
We subtract \(2n^2\) from each term in the sequence to find the linear part:
- For \(n = 1\): \(3 - 2(1)^2 = 1\)
- For \(n = 2\): \(8 - 2(2)^2 = 0\)
- For \(n = 3\): \(17 - 2(3)^2 = -1\)
- For \(n = 4\): \(30 - 2(4)^2 = -2\)
The remaining sequence is \(1, 0, -1, -2, \dots\), which is an arithmetic progression with a first term of \(1\) and a common difference of \(-1\).
Its \(n\)-th term is \(1 + (n - 1)(-1) = 2 - n\).
Combining the two parts, the \(n\)-th term of the original sequence is \(2n^2 - n + 2\).

M1 for finding the constant second difference of 4
M1 for identifying the quadratic term as \(2n^2\)
M1 for finding the linear term \(2 - n\) (or setting up equations to solve for \(b\) and \(c\))
A1 for \(2n^2 - n + 2\) (or any equivalent expression)

Multiply both sides by the denominator \(2x + 7\):
\(y(2x + 7) = 3x - 5\)
Expand the left-hand side:
\(2xy + 7y = 3x - 5\)
Rearrange to group all terms containing \(x\) on one side and terms without \(x\) on the other side:
\(7y + 5 = 3x - 2xy\)
Factorise out \(x\) on the right-hand side:
\(7y + 5 = x(3 - 2y)\)
Divide by \(3 - 2y\) to isolate \(x\):
\(x = \frac{7y + 5}{3 - 2y}\)
Alternatively, this can be written as \(x = \frac{-7y - 5}{2y - 3}\).

M1 for multiplying by \(2x + 7\) to get \(y(2x + 7) = 3x - 5\)
M1 for expanding and grouping \(x\) terms: e.g., \(7y + 5 = 3x - 2xy\) or \(2xy - 3x = -7y - 5\)
M1 for factorising \(x\): \(x(3 - 2y) = 7y + 5\) or \(x(2y - 3) = -7y - 5\)
A1 for \(\frac{7y + 5}{3 - 2y}\) or equivalent

Find the amount of Euros to be converted after the 2% commission is deducted: Commission = 2% of €400 = 0.02 * 400 = €8. Amount to convert = 400 - 8 = €392. (Alternatively, 0.98 * 400 = 392 Euros). Convert the remaining Euros into Yen: 392 * 130 = 50960 Yen.

M1 for finding 2% of €400 (= €8) or 98% of €400 (= €392). M1 for multiplying their €392 by 130. A1 for 50960.

Let \( y = f(x) \), so \( y = \frac{3x - 1}{2x + 5} \). Multiply both sides by the denominator: \( y(2x + 5) = 3x - 1 \). Expand the bracket: \( 2xy + 5y = 3x - 1 \). Rearrange to collect the \( x \) terms on one side: \( 5y + 1 = 3x - 2xy \). Factorise \( x \) on the right side: \( 5y + 1 = x(3 - 2y) \). Divide by \( 3 - 2y \) to make \( x \) the subject: \( x = \frac{5y + 1}{3 - 2y} \). Replace \( y \) with \( x \) to write the inverse function: \( f^{-1}(x) = \frac{5x + 1}{3 - 2x} \).

M1 for setting \( y = \frac{3x - 1}{2x + 5} \) and multiplying to get \( y(2x + 5) = 3x - 1 \). M1 for collecting terms in \( x \) and factorising to obtain \( x(3 - 2y) = 5y + 1 \) or equivalent. A1 for \( f^{-1}(x) = \frac{5x + 1}{3 - 2x} \) or equivalent.

Find the first differences of the sequence: 12 - 7 = 5, 19 - 12 = 7, 28 - 19 = 9, 39 - 28 = 11. Find the second differences: 7 - 5 = 2, 9 - 7 = 2, 11 - 9 = 2. Since the second difference is constant at 2, the sequence is quadratic with leading term \( a n^2 \) where \( a = \frac{2}{2} = 1 \). Subtract \( n^2 \) from each term: for \( n = 1 \): 7 - 1 = 6; for \( n = 2 \): 12 - 4 = 8; for \( n = 3 \): 19 - 9 = 10; for \( n = 4 \): 28 - 16 = 12. The remaining linear sequence is 6, 8, 10, 12, ... which has the \( n \)-th term of \( 2n + 4 \). Combine the parts to get the final \( n \)-th term: \( n^2 + 2n + 4 \).

M1 for finding the constant second difference of 2. M1 for subtracting \( n^2 \) from the terms to get the sequence 6, 8, 10, 12... or setting up a system of equations. A1 for \( n^2 + 2n + 4 \).

We can calculate the amount Carla receives in two equivalent ways: Method 1: Deducting commission first. 1. Calculate the commission fee in Euros: \(\text{Commission} = 2\% \text{ of } 400\text{ EUR} = 0.02 \times 400 = 8\text{ EUR}\). 2. Subtract the commission to find the amount to be exchanged: \(400 - 8 = 392\text{ EUR}\). 3. Convert the remaining Euros to US Dollars: \(392 \times 1.15 = 450.80\text{ USD}\). Method 2: Converting first and then deducting commission. 1. Convert the total Euros to US Dollars: \(400 \times 1.15 = 460\text{ USD}\). 2. Calculate the commission fee in US Dollars: \(\text{Commission} = 2\% \text{ of } 460\text{ USD} = 0.02 \times 460 = 9.20\text{ USD}\). 3. Subtract the commission from the total US Dollars: \(460 - 9.20 = 450.80\text{ USD}\).

M1 for \(400 \times 1.15\) [460] or \(400 \times 0.98\) [392]. M1 for converting their reduced Euros to USD, e.g., \((400 - 8) \times 1.15\), or finding \(98\%\) of their converted USD, e.g., \(460 \times 0.98\). A1 for 450.8 or 450.80.

To find the inverse function \(f^{-1}(x)\): 1. Set \(y = f(x)\): \(y = \frac{3x + 1}{x - 2}\). 2. Rearrange the equation to make \(x\) the subject: \(y(x - 2) = 3x + 1\), which expands to \(xy - 2y = 3x + 1\). Grouping the \(x\) terms on one side: \(xy - 3x = 2y + 1\). Factoring out \(x\): \(x(y - 3) = 2y + 1\). Dividing by \((y - 3)\): \(x = \frac{2y + 1}{y - 3}\). 3. Replace \(x\) with \(f^{-1}(x)\) and \(y\) with \(x\): \(f^{-1}(x) = \frac{2x + 1}{x - 3}\).

M1 for setting \(y = \frac{3x + 1}{x - 2}\) and multiplying by \((x - 2)\). M1 for isolating terms in \(x\) to obtain \(x(y - 3) = 2y + 1\) or equivalent. A1 for \(\frac{2x + 1}{x - 3}\) or any algebraically equivalent expression.

Let the sequence of terms be \(T_n\): \(T_1 = 3\), \(T_2 = 10\), \(T_3 = 21\), \(T_4 = 36\). 1. Find the first differences between consecutive terms: \(10 - 3 = 7\), \(21 - 10 = 11\), \(36 - 21 = 15\). 2. Find the second differences: \(11 - 7 = 4\), \(15 - 11 = 4\). Since the second difference is a constant \(4\), the sequence is quadratic of the form \(an^2 + bn + c\), where \(2a = 4 \implies a = 2\). 3. Subtract \(2n^2\) from each term of the sequence to find the remaining linear part: For \(n = 1\): \(3 - 2(1)^2 = 1\). For \(n = 2\): \(10 - 2(2)^2 = 2\). For \(n = 3\): \(21 - 2(3)^2 = 3\). For \(n = 4\): \(36 - 2(4)^2 = 4\). The remaining sequence is \(1, 2, 3, 4, \dots\), which is simply \(n\). 4. Thus, the \(n\)-th term is \(2n^2 + n\).

M1 for finding second differences are constant and equal to 4, implying a lead term of \(2n^2\). M1 for subtracting \(2n^2\) from the terms to get the sequence \(1, 2, 3, 4, \dots\) or setting up a system of equations, e.g., \(a+b+c=3\), \(4a+2b+c=10\), \(9a+3b+c=21\). A1 for \(2n^2 + n\) or equivalent.

First, calculate the commission charged by the bank: \(1500 \times 0.02 = \$30\). Next, subtract the commission from the original amount to find the amount to be exchanged: \(1500 - 30 = \$1470\). Finally, convert the remaining US Dollars into Euros using the exchange rate: \(1470 \times 0.90 = 1323\text{ EUR}\).

M1 for calculating the commission as \(\$30\) or finding the remaining amount as \(0.98 \times 1500 = \$1470\). M1 for multiplying their remaining amount in USD by \(0.90\). A1 for the correct final answer of \(1323\).

Let us analyze the differences between consecutive terms of the sequence. First differences: \(7, 11, 15, 19\). Second differences: \(4, 4, 4\). Since the second differences are constant and equal to \(4\), the sequence is quadratic of the form \(an^2 + bn + c\), where \(2a = 4 \implies a = 2\). Now, subtract \(2n^2\) from each term of the original sequence to find the linear component: For \(n = 1\): \(5 - 2(1)^2 = 3\); For \(n = 2\): \(12 - 2(2)^2 = 4\); For \(n = 3\): \(23 - 2(3)^2 = 5\); For \(n = 4\): \(38 - 2(4)^2 = 6\). The remaining sequence is \(3, \ 4, \ 5, \ 6, \ \dots\), which has a first term of \(3\) and a common difference of \(1\). This linear part is represented by \(n + 2\). Combining these, the \(n\)-th term is \(2n^2 + n + 2\).

M1 for recognizing a quadratic sequence and finding the second differences are \(4\) (leading to a lead coefficient of \(2\)). M1 for subtracting \(2n^2\) from the sequence to obtain the linear sequence \(3, 4, 5, 6, \dots\) or setting up appropriate simultaneous equations. A1 for the correct expression \(2n^2 + n + 2\).

We are given \(g(f(x)) = 1\). First, apply the definition of \(g(x)\): \(5 - [f(x)]^2 = 1 \implies [f(x)]^2 = 4\). Taking the square root on both sides gives \(f(x) = 2\) or \(f(x) = -2\). Now, substitute the expression for \(f(x)\) to solve for \(x\): Case 1: \(\frac{3x - 1}{2} = 2 \implies 3x - 1 = 4 \implies 3x = 5 \implies x = \frac{5}{3}\). Case 2: \(\frac{3x - 1}{2} = -2 \implies 3x - 1 = -4 \implies 3x = -3 \implies x = -1\). Therefore, the values of \(x\) are \(-1\) and \(\frac{5}{3}\).

M1 for setting up the equation \(5 - \left(\frac{3x-1}{2}\right)^2 = 1\) or determining that \(f(x) = \pm 2\). M1 for solving at least one of the linear equations \(\frac{3x-1}{2} = 2\) or \(\frac{3x-1}{2} = -2\). A1 for both correct values: \(x = -1\) and \(x = \frac{5}{3}\) (or equivalent fractions/decimals).

First, calculate the commission fee on the initial amount: \(2\%\) of €\(400 = 0.02 \times 400 = 8\) Euros. Subtract the commission fee from the initial amount: €\(400 - 8 = 392\) Euros. Finally, convert the remaining Euros into Dollars: \(392 \times 1.25 = 392 \times \frac{5}{4} = 98 \times 5 = 490\) Dollars. Alternatively, convert the initial amount first: \(400 \times 1.25 = 500\) Dollars, and then apply the commission: \(500 \times (1 - 0.02) = 500 \times 0.98 = 490\) Dollars.

M1 for finding the commission of €8 (or $10) or using a multiplier of 0.98. M1 for multiplying the remaining Euros by 1.25 or multiplying $500 by 0.98. A1 for 490.

First, find the value of \(g(3)\): \(g(3) = 2(3) - 5 = 6 - 5 = 1\). We require \(f^{-1}(x) = 1\). By definition of the inverse function, this is equivalent to \(x = f(1)\). Substitute \(1\) into the expression for \(f(x)\): \(f(1) = \frac{3(1) + 1}{2} = \frac{4}{2} = 2\). Therefore, \(x = 2\). Alternatively, find the expression for the inverse function: let \(y = \frac{3x + 1}{2} \implies 2y = 3x + 1 \implies x = \frac{2y - 1}{3}\). So, \(f^{-1}(x) = \frac{2x - 1}{3}\). Set this equal to \(g(3)\): \(f^{-1}(x) = 1 \implies \frac{2x - 1}{3} = 1 \implies 2x - 1 = 3 \implies 2x = 4 \implies x = 2\).

M1 for finding \(g(3) = 1\). M1 for writing \(f^{-1}(x) = \frac{2x-1}{3}\) or setting up the equation \(x = f(1)\). A1 for 2.

First, find the differences between successive terms of the sequence: First differences: \(12 - 5 = 7\), \(21 - 12 = 9\), \(32 - 21 = 11\), \(45 - 32 = 13\). Second differences: \(9 - 7 = 2\), \(11 - 9 = 2\), \(13 - 11 = 2\). Since the second difference is a constant 2, the sequence is quadratic of the form \(T_n = an^2 + bn + c\), where \(a = \frac{2}{2} = 1\). Subtracting \(n^2\) from each term: For \(n=1\): \(5 - 1^2 = 4\); For \(n=2\): \(12 - 2^2 = 8\); For \(n=3\): \(21 - 3^2 = 12\); For \(n=4\): \(32 - 4^2 = 16\). The remaining terms form the sequence \(4, 8, 12, 16\), which is given by \(4n\). Thus, the \(n\)-th term of the original sequence is \(T_n = n^2 + 4n\). To find the 20th term, substitute \(n = 20\): \(T_{20} = 20^2 + 4(20) = 400 + 80 = 480\).

M1 for analyzing differences to find the second difference of 2. M1 for finding the \(n\)-th term expression \(n^2 + 4n\) or equivalent. A1 for 480.

Let the \(n\)-th term of the sequence be \(u_n\). Find the first differences between consecutive terms: 10 - 3 = 7, 21 - 10 = 11, 36 - 21 = 15, 55 - 36 = 19. Find the second differences: 11 - 7 = 4, 15 - 11 = 4, 19 - 15 = 4. Since the second differences are constant and equal to 4, the sequence is quadratic with a leading term of \(\frac{4}{2}n^2 = 2n^2\). Subtracting \(2n^2\) from each term in the sequence: For \(n = 1\): 3 - 2(1) = 1. For \(n = 2\): 10 - 8 = 2. For \(n = 3\): 21 - 18 = 3. For \(n = 4\): 36 - 32 = 4. For \(n = 5\): 55 - 50 = 5. The resulting sequence is 1, 2, 3, 4, 5, ... which has the general term \(n\). Thus, the \(n\)-th term is \(2n^2 + n\).

M1 for finding the second difference of 4 or identifying the leading term \(2n^2\). M1 for subtracting \(2n^2\) from terms to find the linear sequence \(1, 2, 3, \dots\). A1 for \(2n^2 + n\).

Calculate the value of Amara's investment after 2 years: After 1 year: \(\$500 + 10\% \text{ of } \$500 = \$550\). After 2 years: \(\$550 + 10\% \text{ of } \$550 = \$605\). Calculate the value of Ben's investment after 2 years: Simple interest per year: \(12\% \text{ of } \$500 = \$60\). Total simple interest for 2 years: \(\$60 \times 2 = \$120\). Total value after 2 years: \(\$500 + \$120 = \$620\). Calculate the difference in total values: \(\$620 - \$605 = \$15\).

M1 for calculating Amara's compound interest value after 2 years as \(\$605\). M1 for calculating Ben's simple interest value after 2 years as \(\$620\). A1 for correct difference of 15.

First, we can factor out the common numerical factor 2 from all terms: \(12x^2 - 18xy - 8x + 12y = 2(6x^2 - 9xy - 4x + 6y)\). Next, group the terms inside the parentheses: \(2 [ 3x(2x - 3y) - 2(2x - 3y) ]\). Factor out the common binomial factor \((2x - 3y)\): \(2(3x - 2)(2x - 3y)\). Alternatively, by grouping directly first: \(12x^2 - 18xy - 8x + 12y = 6x(2x - 3y) - 4(2x - 3y) = (6x - 4)(2x - 3y)\). Since \(6x - 4\) has a common factor of 2, we factorise it completely to get \(2(3x - 2)(2x - 3y)\).

M1 for grouping terms to obtain a partially factorised expression, e.g., \((6x - 4)(2x - 3y)\) or \(6x(2x - 3y) - 4(2x - 3y)\). M1 for recognizing that \(6x - 4\) can be further factorised as \(2(3x - 2)\). A1 for \(2(3x - 2)(2x - 3y)\) or equivalent completely factorised expression.

Interest earned in the first year: \(5\%\) of \(\$1200 = \frac{5}{100} \times 1200 = \$60\). Amount at the end of the first year: \(\$1200 + \$60 = \$1260\). Interest earned in the second year: \(5\%\) of \(\$1260 = \frac{5}{100} \times 1260 = \$63\). Total interest earned over 2 years: \(\$60 + \$63 = \$123\). Alternatively, using the compound interest formula: \(\text{Total Amount} = 1200 \times (1 + 0.05)^2 = 1200 \times 1.1025 = 1323\), so \(\text{Total Interest} = 1323 - 1200 = 123\).

M1 for finding the value of the investment after 1 year: \(\$1260\) (or interest for first year as \(\$60\)). M1 for a correct method to find the second year's interest or the total amount at the end of 2 years (e.g., \(1260 \times 1.05\) or \(1200 \times 1.05^2\)). A1 for \(123\).

First, calculate \(\text{g}(4) = 5 - 2(4) = -3\). We are given \(\text{f}^{-1}(x) = \text{g}(4)\), so \(\text{f}^{-1}(x) = -3\). By definition of the inverse function, this is equivalent to \(x = \text{f}(-3)\). Substitute \(-3\) into \(\text{f}(x)\): \(x = \frac{3(-3) + 1}{2} = \frac{-9 + 1}{2} = \frac{-8}{2} = -4\). Alternatively, find the inverse function: let \(y = \frac{3x+1}{2} \implies 2y = 3x + 1 \implies x = \frac{2y-1}{3}\), so \(\text{f}^{-1}(x) = \frac{2x-1}{3}\). Solving \(\frac{2x-1}{3} = -3 \implies 2x - 1 = -9 \implies 2x = -8 \implies x = -4\).

M1 for \(\text{g}(4) = -3\). M1 for setting up the equation \(\text{f}^{-1}(x) = -3\) or finding \(\text{f}^{-1}(x) = \frac{2x-1}{3}\). A1 for \(-4\).

Let the sequence be \(T_n\). The first differences are: \(8-3=5\), \(15-8=7\), \(24-15=9\), \(35-24=11\). The second differences are constant: \(7-5=2\), \(9-7=2\), \(11-9=2\). Since the second difference is constant at \(2\), the sequence is quadratic of the form \(an^2 + bn + c\) where \(2a = 2 \implies a = 1\). Subtracting \(n^2\) from each term of the sequence gives: \(3-1^2=2\), \(8-2^2=4\), \(15-3^2=6\), \(24-4^2=8\). The resulting sequence is \(2, 4, 6, 8, \dots\), which has \(n\)-th term \(2n\). Thus, the \(n\)-th term of the original sequence is \(n^2 + 2n\).

M1 for finding the second difference is \(2\) (or showing that the \(n^2\) term has coefficient \(1\)). M1 for finding the linear part by subtracting \(n^2\) to get \(2, 4, 6, 8\) or setting up equations. A1 for \(n^2 + 2n\) (or equivalent such as \(n(n+2)\)).

First, subtract the bank's flat fee from the initial amount of USD:
\(800 - 15 = 785\) USD.

Next, convert this remaining amount to SGD using the exchange rate:
\(785 \times 1.34 = 1051.90\) SGD.

M1 for subtracting the fee: \(800 - 15\)
M1 for multiplying by the exchange rate: \(\text{their } 785 \times 1.34\)
A1 for correct final answer 1051.90 (or 1051.9)

Find the common difference, \(d\), by subtracting consecutive terms:
\(13 - 7 = 6\).

The formula for the \(n\)th term of an arithmetic sequence is:
\(u_n = a + (n - 1)d\)
where \(a\) is the first term (7) and \(d\) is the common difference (6).

\(u_n = 7 + (n - 1) \times 6 = 6n + 1\)

Now, find the 80th term by substituting \(n = 80\):
\(u_{80} = 6(80) + 1 = 480 + 1 = 481\).

M1 for finding the common difference of 6
M1 for setting up the term expression \(7 + (80-1) \times 6\) or finding \(6n+1\)
A1 for 481

First, expand each part of the expression:
\(5(2x - 3) = 10x - 15\)
\(-3(x - 4) = -3x + 12\)

Now, group the like terms together:
\(10x - 15 - 3x + 12 = 10x - 3x - 15 + 12\)

Simplify:
\(7x - 3\).

M1 for expansion of first bracket: \(10x - 15\)
M1 for expansion of second bracket: \(-3x + 12\) (watch the sign!)
A1 for fully simplified expression \(7x - 3\)

The formula for compound interest is:
\(A = P \left(1 + \frac{r}{100}\right)^n\)
where:
- \(P = 1500\) (principal amount)
- \(r = 2.4\) (interest rate per year)
- \(n = 4\) (number of years)

Calculate the final amount \(A\):
\(A = 1500 \times \left(1 + \frac{2.4}{100}\right)^4\)
\(A = 1500 \times (1.024)^4\)
\(A \approx 1500 \times 1.099511628\)
\(A \approx 1649.2674\)

To find the total interest earned, subtract the principal from the final amount:
\(\text{Interest} = A - P\)
\(\text{Interest} = 1649.2674 - 1500 = 149.2674\)

Rounding to 2 decimal places gives $149.27.

M1 for writing down or using a correct compound interest calculation, e.g., \(1500 \times 1.024^4\) (or showing final value of 1649.27)
M1 for subtracting the original principal, 1500, from their calculated final amount
A1 for 149.27 (accept 149.26 to 149.27)

To expand \((3x - 4)(2x + 5)\), we multiply each term in the first bracket by each term in the second bracket:
\((3x - 4)(2x + 5) = 3x(2x + 5) - 4(2x + 5)\)
\(= 6x^2 + 15x - 8x - 20\)

Combine the like terms \(15x\) and \(-8x\):
\(= 6x^2 + 7x - 20\)

M1 for expanding to obtain at least 3 correct terms out of 4, e.g., \(6x^2 + 15x - 8x - 20\)
A1 for correct expansion of the middle terms, e.g., showing \(15x - 8x\)
A1 for the fully simplified expression \(6x^2 + 7x - 20\)

The first term \(a = 8\).
The common difference \(d\) is:
\(d = 14 - 8 = 6\)

The formula for the \(n\)-th term of an arithmetic sequence is:
\(a_n = a + (n-1)d\)

Substitute the values into the formula:
\(a_n = 8 + (n-1) \times 6\)
\(a_n = 8 + 6n - 6\)
\(a_n = 6n + 2\)

M1 for recognizing the common difference is 6 (e.g., writing \(6n + k\) where \(k\) is a constant)
M1 for attempting to find the constant term, e.g., \(8 - 6\) or setting up an equation \(6(1) + k = 8\)
A1 for \(6n + 2\) (or equivalent, e.g., \(2 + 6n\))

First, find the number of mangoes sold at each price.
75% of 40 = \( 0.75 \times 40 = 30 \) mangoes.
These are sold at $1.20 each: \( 30 \times 1.20 = 36 \) dollars.
The remaining mangoes are \( 40 - 30 = 10 \) mangoes.
These are sold at $0.80 each: \( 10 \times 0.80 = 8 \) dollars.
Total revenue = \( 36 + 8 = 44 \) dollars.
Profit = \( \text{Total Revenue} - \text{Cost Price} = 44 - 32 = 12 \) dollars.
Percentage profit = \( \frac{12}{32} \times 100 = 37.5 \)%.

M1 for finding 30 mangoes at $1.20 and 10 mangoes at $0.80
M1 for total revenue = $44 or profit = $12
A1 for 37.5%

First, evaluate \( f(4) \):
\( f(4) = 3(4) - 5 = 12 - 5 = 7 \).
Next, substitute this result into \( g(x) \):
\( g(f(4)) = g(7) = 7^2 + 2(7) = 49 + 14 = 63 \).

M1 for finding \( f(4) = 7 \)
M1 for substituting their 7 into \( g(x) \), i.e., \( 7^2 + 2(7) \)
A1 for 63

Identify the first term, \( a = 8 \), and the common difference, \( d = 13 - 8 = 5 \).
The formula for the \( n \)th term of an arithmetic sequence is \( u_n = a + (n - 1)d \).
For the 60th term, substitute \( n = 60 \):
\( u_{60} = 8 + (60 - 1) \times 5 \)
\( u_{60} = 8 + 59 \times 5 \)
\( u_{60} = 8 + 295 = 303 \).

M1 for recognizing common difference is 5
M1 for setting up the term expression: \( 8 + 59 \times 5 \) or finding the general term \( 5n + 3 \) and substituting \( n = 60 \)
A1 for 303

First, find the commission fee: \(450 \times 0.015 = 6.75\) EUR. Next, find the remaining amount to exchange: \(450 - 6.75 = 443.25\) EUR. Convert the remaining amount to JPY: \(443.25 \times 162.40 = 71983.8\) JPY. Rounding to the nearest whole JPY gives \(71984\).

M1 for \(450 \times 0.985\) or \(450 - 6.75\) (or equivalent method to subtract 1.5%). M1 for multiplying their remaining EUR by 162.40. A1 for 71984.

The first term is \(a = 7\) and the common difference is \(d = 13 - 7 = 6\). The formula for the \(n\)-th term of an arithmetic sequence is \(a + (n-1)d\). For the 40th term (\(n = 40\)): \(7 + (40-1) \times 6 = 7 + 39 \times 6 = 7 + 234 = 241\).

M1 for finding the common difference of 6. M1 for \(7 + 39 \times 6\) or equivalent. A1 for 241.

Multiply each term in the first bracket by each term in the second bracket: \((3x - 2)(2x + 5) = 3x(2x + 5) - 2(2x + 5) = 6x^2 + 15x - 4x - 10\). Combine the like terms to get: \(6x^2 + 11x - 10\).

M1 for any 3 correct terms out of 4 from \(6x^2\), \(15x\), \(-4x\), \(-10\) when expanding. A1 for \(6x^2 + 11x + c\) or \(ax^2 + 11x - 10\) (where \(a \neq 0\)). A1 for \(6x^2 + 11x - 10\).

We use the compound interest formula: \( A = P \left(1 + \frac{r}{100}\right)^t \) where \( P = 1500 \), \( r = 2.4 \), and \( t = 5 \). Substituting these values, we get: \( A = 1500 \times (1.024)^5 \). Calculating this gives: \( A = 1500 \times 1.1258999... \approx 1688.8498... \). Rounding to 2 decimal places gives \( 1688.85 \).

M1 for \( 1500 \times 1.024^5 \) or equivalent. A1 for 1688.85 or 1688.8498... A0.33 for rounding correct to 2 decimal places from their calculation.

The sequence decreases by 3 each time, which means the common difference is \( -3 \). Therefore, the formula for the \( n \)-th term has the form \( -3n + c \). Using the first term where \( n = 1 \): \( -3(1) + c = 11 \), which simplifies to \( c = 14 \). Thus, the \( n \)-th term is \( 14 - 3n \).

M1 for identifying the common difference is \( -3 \) or writing \( -3n \). A1 for finding the constant term 14. A0.33 for the fully correct expression \( 14 - 3n \) or equivalent.

We expand the brackets by multiplying each term in the first bracket by each term in the second bracket: \( (2x - 3)(x + 5) = 2x(x) + 2x(5) - 3(x) - 3(5) \). This simplifies to: \( 2x^2 + 10x - 3x - 15 \). Combining the like terms (\( 10x - 3x = 7x \)), we get: \( 2x^2 + 7x - 15 \).

M1 for expanding to obtain at least 3 correct terms out of 4 in \( 2x^2 + 10x - 3x - 15 \). A1 for obtaining any two terms correct in the final trinomial. A1.33 for the fully correct simplified expression \( 2x^2 + 7x - 15 \).

The formula for simple interest is \(I = \frac{P \times R \times T}{100}\), where \(P\) is the principal, \(R\) is the annual interest rate, and \(T\) is the time in years. Substituting the given values: \(I = \frac{3500 \times 2.8 \times 6}{100}\). Calculating this gives \(I = 35 \times 2.8 \times 6 = 588\). Thus, the total interest earned is \(\$588\).

M1 for correct substitution into simple interest formula, e.g. \(3500 \times 0.028 \times 6\); A1 for \(588\)

First, expand each bracket: \(5(2x - 3) = 10x - 15\) and \(-3(x - 4) = -3x + 12\). Next, collect and combine like terms: \((10x - 3x) + (-15 + 12) = 7x - 3\).

M1 for expanding at least one bracket correctly to get \(10x - 15\) or \(-3x + 12\); A1 for \(7x - 3\)

Find the common difference by subtracting consecutive terms: \(18 - 11 = 7\). Since the common difference is \(7\), the \(n\)-th term is of the form \(7n + c\). Use the first term where \(n = 1\): \(7(1) + c = 11 \implies c = 4\). Therefore, the expression for the \(n\)-th term is \(7n + 4\).

M1 for finding the common difference of \(7\) or writing \(7n + c\); A1 for \(7n + 4\)

Mariam's amount after 8 years is: \(A_M = 5000 \times (1.024)^8 \approx 6049.131...\) Noah's amount after 8 years is: \(A_N = 4500 \times \left(1 + \frac{r}{100}\right)^8\) Since their final balances are equal: \(4500 \times \left(1 + \frac{r}{100}\right)^8 = 5000 \times (1.024)^8\) Dividing by 4500: \(\left(1 + \frac{r}{100}\right)^8 = \frac{5000 \times 1.024^8}{4500} \approx 1.344251\) Taking the 8th root of both sides: \(1 + \frac{r}{100} = (1.344251)^{\frac{1}{8}} \approx 1.037672\) Solving for \(r\): \(\frac{r}{100} \approx 0.037672 \implies r \approx 3.7672\) Correct to 2 decimal places, \(r = 3.77\).

[M1] for setting up Mariam's compound interest formula: \(5000 \times 1.024^8\) [M1] for setting up the equation equating both investments: \(4500 \times (1 + r/100)^8 = 5000 \times 1.024^8\) [M1] for a correct method to solve for \(r\), e.g. \(1 + r/100 = (1.34425...)^{1/8}\) [A1.17] for \(3.77\) (or any answer rounding to 3.77)

We want to solve \(\text{f}(\text{g}(x)) = 5\). Let \(y = \text{g}(x)\). Then: \(\text{f}(y) = 5 \implies \frac{2y + 3}{y - 1} = 5\) Multiplying both sides by \(y - 1\): \(2y + 3 = 5(y - 1) \implies 2y + 3 = 5y - 5\) Subtracting \(2y\) and adding 5: \(8 = 3y \implies y = \frac{8}{3}\) Since \(y = \text{g}(x)\), we have: \(3x - 5 = \frac{8}{3} \implies 3x = \frac{8}{3} + 5 = \frac{23}{3}\) Dividing by 3 gives: \(x = \frac{23}{9}\).

[M1] for a correct algebraic setup of the composite function, e.g., \(\frac{2(3x-5)+3}{(3x-5)-1} = 5\) or \(\text{f}(y)=5\) [M1] for simplifying the equation to a linear form, e.g., \(6x - 7 = 5(3x - 6)\) or \(2y + 3 = 5y - 5\) [M1] for solving to isolate \(x\) or finding \(y = 8/3\) [A1.17] for \(23/9\) (or equivalent, e.g., \(2\frac{5}{9}\))

Since the terms form a geometric sequence, the common ratio is constant: \(\frac{x + 3}{x - 1} = \frac{3x + 1}{x + 3}\) Cross-multiplying: \((x + 3)^2 = (3x + 1)(x - 1)\) Expanding both sides: \(x^2 + 6x + 9 = 3x^2 - 3x + x - 1 \implies x^2 + 6x + 9 = 3x^2 - 2x - 1\) Rearranging to form a quadratic equation: \(2x^2 - 8x - 10 = 0 \implies x^2 - 4x - 5 = 0\) Factorizing: \((x - 5)(x + 1) = 0\) This gives two possible values of \(x\): \(x = 5\) or \(x = -1\). If \(x = 5\), the first three terms are \(4\), \(8\), and \(16\), which has common ratio \(r = 2\). The 4th term is \(16 \times 2 = 32\). If \(x = -1\), the first three terms are \(-2\), \(2\), and \(-2\), which has common ratio \(r = -1\). The 4th term is \(-2 \times (-1) = 2\).

[M1] for setting up the common ratio equation: \(\frac{x+3}{x-1} = \frac{3x+1}{x+3}\) [M1] for expanding and simplifying to a standard quadratic form, e.g., \(x^2 - 4x - 5 = 0\) [M1] for solving the quadratic to find \(x = 5\) and \(x = -1\) [A1.17] for both correct 4th terms: 32 and 2

1. Find the rate of interest \(r\): \(8000 \left(1 + \frac{r}{100}\right)^6 = 9834.04\) which gives \(\left(1 + \frac{r}{100}\right)^6 = 1.229255\). Taking the 6th root: \(1 + \frac{r}{100} = 1.035\), so \(r = 3.5\). 2. Find the remaining amount after withdrawing \(\$3000\): \(9834.04 - 3000 = 6834.04\). 3. Calculate the amount after a further \(4\) years: \(6834.04 \times (1.035)^4 \approx 7842.22\). Rounding to the nearest dollar gives \(7842\).

M1 for setting up the equation \(8000(1 + r/100)^6 = 9834.04\). A1 for finding \(r = 3.5\). M1 for \((9834.04 - 3000) \times (1.035)^4\). A1 for \(7842\) (accept \(7842.22\)).

Let the \(n\)-th term be \(u_n = a n^2 + b n + c\). The first differences are \(19 - 10 = 9\) and \(32 - 19 = 13\). The second difference is \(13 - 9 = 4\). Therefore, \(2a = 4\) which gives \(a = 2\). Subtracting \(2n^2\) from the terms: for \(n=1\), \(10 - 2 = 8\); for \(n=2\), \(19 - 8 = 11\); for \(n=3\), \(32 - 18 = 14\). This linear sequence \(8, 11, 14\) has first term \(8\) and difference \(3\), so it is \(3n + 5\). Thus, the general term is \(u_n = 2n^2 + 3n + 5\). The \(20\)th term is \(u_{20} = 2(20)^2 + 3(20) + 5 = 800 + 60 + 5 = 865\).

M1 for finding the second difference of 4. M1 for setting up equations to find coefficients, leading to \(a = 2\). M1 for finding the full formula \(u_n = 2n^2 + 3n + 5\). A1 for \(865\).

The vertex form of a quadratic equation is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Substituting the vertex \((3, -8)\) gives \(y = a(x - 3)^2 - 8\). Since the curve passes through \((1, 0)\), we substitute these coordinates to solve for \(a\): \(0 = a(1 - 3)^2 - 8\) which simplifies to \(4a = 8\), so \(a = 2\). Thus, the equation of the curve is \(y = 2(x - 3)^2 - 8\). To find the \(y\)-intercept, set \(x = 0\): \(y = 2(0 - 3)^2 - 8 = 2(9) - 8 = 10\).

M1 for writing the vertex form \(y = a(x - 3)^2 - 8\). M1 for substituting \((1, 0)\) to find \(a\). A1 for \(a = 2\). A1 for \(y\)-intercept = \(10\).

Let \( n \) be the number of complete years.

The value of Adele's investment after \( n \) years is given by:
\( A = 4500 \times 1.028^n \)

The value of Bertrand's investment after \( n \) years is given by:
\( B = 4200 \times 1.031^n \)

We require the smallest integer value of \( n \) such that \( B > A \):
\( 4200 \times 1.031^n > 4500 \times 1.028^n \)

Dividing both sides by \( 4200 \times 1.028^n \):
\( \left(\frac{1.031}{1.028}\right)^n > \frac{4500}{4200} \)

\( \left(1.002918...\right)^n > 1.071428... \)

Taking natural logarithms on both sides:
\( n \ln(1.002918...) > \ln(1.071428...) \)

\( n > \frac{\ln(1.071428...)}{\ln(1.002918...)} \)

\( n > 23.68 \)

Since \( n \) must be an integer, we round up to the next whole number, which is 24.

Let's verify:
For \( n = 23 \):
\( A = 4500 \times 1.028^{23} = \$8504.18 \)
\( B = 4200 \times 1.031^{23} = \$8488.62 \) (Adele's is larger)

For \( n = 24 \):
\( A = 4500 \times 1.028^{24} = \$8742.30 \)
\( B = 4200 \times 1.031^{24} = \$8751.77 \) (Bertrand's is larger)

Thus, it takes 24 complete years.

M1 for setting up the inequality or equation \( 4200 \times 1.031^n = 4500 \times 1.028^n \) or equivalent.
M1 for a valid method to solve the exponential equation (e.g., using logarithms or systematic trial and error with at least two values of \( n \)).
A1 for obtaining a decimal value \( n \approx 23.7 \) (or showing values for both \( n=23 \) and \( n=24 \)).
A1 for the correct answer of 24 (must be an integer).

First, find the inverse function \( f^{-1}(x) \).
Let \( y = \frac{2x + 3}{x - 4} \).

Rearranging to make \( x \) the subject:
\( y(x - 4) = 2x + 3 \)
\( xy - 4y = 2x + 3 \)
\( xy - 2x = 4y + 3 \)
\( x(y - 2) = 4y + 3 \)
\( x = \frac{4y + 3}{y - 2} \)

Therefore, \( f^{-1}(x) = \frac{4x + 3}{x - 2} \).

Now set \( f(x) = f^{-1}(x) \):
\( \frac{2x + 3}{x - 4} = \frac{4x + 3}{x - 2} \)

Cross-multiply:
\( (2x + 3)(x - 2) = (4x + 3)(x - 4) \)

Expand both sides:
\( 2x^2 - 4x + 3x - 6 = 4x^2 - 16x + 3x - 12 \)
\( 2x^2 - x - 6 = 4x^2 - 13x - 12 \)

Rearrange into a standard quadratic form:
\( 2x^2 - 12x - 6 = 0 \)

Divide by 2:
\( x^2 - 6x - 3 = 0 \)

Use the quadratic formula:
\( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-3)}}{2(1)} \)
\( x = \frac{6 \pm \sqrt{36 + 12}}{2} \)
\( x = \frac{6 \pm \sqrt{48}}{2} \)
\( x = 3 \pm 2\sqrt{3} \)

The positive value of \( x \) is:
\( x = 3 + 2\sqrt{3} \approx 6.4641 \)

To 3 significant figures, \( x = 6.46 \).

M1 for a correct method to find \( f^{-1}(x) \) (e.g. swapping variables and rearranging) leading to \( f^{-1}(x) = \frac{4x+3}{x-2} \).
M1 for setting up the equation \( \frac{2x+3}{x-4} = \frac{4x+3}{x-2} \) and expanding to a quadratic equation.
A1 for obtaining the simplified quadratic equation \( x^2 - 6x - 3 = 0 \) or equivalent (e.g. \( 2x^2 - 12x - 6 = 0 \)).
A1 for \( 6.46 \) (accept \( 6.464... \)).

First, we find the formula for the \( n \)-th term, \( T_n \), of the sequence: 3, 11, 25, 45.

Let's calculate the differences:
First differences: \( 11 - 3 = 8 \), \( 25 - 11 = 14 \), \( 45 - 25 = 20 \).
Second differences: \( 14 - 8 = 6 \), \( 20 - 14 = 6 \).

Since the second differences are constant, the sequence is quadratic, of the form \( T_n = an^2 + bn + c \), where:
\( 2a = 6 \implies a = 3 \).

Subtracting \( 3n^2 \) from each term of the sequence:
For \( n = 1 \): \( 3 - 3(1)^2 = 0 \)
For \( n = 2 \): \( 11 - 3(2)^2 = -1 \)
For \( n = 3 \): \( 25 - 3(3)^2 = -2 \)
For \( n = 4 \): \( 45 - 3(4)^2 = -3 \)

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

Thus, the quadratic \( n \)-th term formula is:
\( T_n = 3n^2 - n + 1 \)

We need to find \( n \) such that \( T_n = 7451 \):
\( 3n^2 - n + 1 = 7451 \)
\( 3n^2 - n - 7450 = 0 \)

Using the quadratic formula to solve for \( n \):
\( n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-7450)}}{2(3)} \)
\( n = \frac{1 \pm \sqrt{1 + 89400}}{6} \)
\( n = \frac{1 \pm \sqrt{89401}}{6} \)

Since \( 89401 = 299^2 \):
\( n = \frac{1 \pm 299}{6} \)

Since \( n \) must be positive:
\( n = \frac{300}{6} = 50 \).

Therefore, the value of \( n \) is 50.

M1 for identifying the second difference is 6 and establishing the \( 3n^2 \) term.
M1 for finding the full \( n \)-th term formula \( 3n^2 - n + 1 \).
M1 for setting up the equation \( 3n^2 - n + 1 = 7451 \) and expressing it as a quadratic equation ready to solve.
A1 for \( n = 50 \).

First, find the total amount in Account B after 8 years of simple interest:
Account B total = \(9000 + (9000 \times 0.032 \times 8) = 9000 + 2304 = \$11304\).

Since the amount in Account A after 8 years is equal to this, we set up the compound interest formula for Account A:
\(8500 \times \left(1 + \frac{r}{100}\right)^8 = 11304\)

Solve for \(r\):
\(\left(1 + \frac{r}{100}\right)^8 = \frac{11304}{8500}\)
\(1 + \frac{r}{100} = \left(\frac{11304}{8500}\right)^{\frac{1}{8}}\)
\(1 + \frac{r}{100} \approx 1.036236\)
\(\frac{r}{100} \approx 0.036236\)
\(r \approx 3.62\%\) (to 3 significant figures).

M1 for calculating the simple interest or total value of Account B: \(9000 \times 0.032 \times 8 = 2304\) or \(11304\)
M1 for setting up the compound interest equation: \(8500 \times (1 + r/100)^8 = 11304\)
M1 for rearranging to find \(1 + r/100 = (11304/8500)^{1/8}\) or \(1.0362...\)
A1 for \(3.62\) (accept 3.62 to 3.63)

We are given \(f(g(x)) = 3\). Substitute \(g(x)\) into the function \(f(x)\):
\(f(g(x)) = \frac{2(x^2 - 2) + 5}{(x^2 - 2) - 3} = 3\)

Simplify the numerator and denominator:
\(\frac{2x^2 - 4 + 5}{x^2 - 5} = 3\)
\(\frac{2x^2 + 1}{x^2 - 5} = 3\)

Multiply both sides by \(x^2 - 5\):
\(2x^2 + 1 = 3(x^2 - 5)\)
\(2x^2 + 1 = 3x^2 - 15\)

Rearrange to solve for \(x^2\):
\(16 = x^2\)

Take the square root of both sides:
\(x = 4\) or \(x = -4\).

M1 for substituting \(g(x)\) into \(f(x)\): \(\frac{2(x^2 - 2) + 5}{x^2 - 2 - 3} = 3\)
M1 for simplifying to \(2x^2 + 1 = 3(x^2 - 5)\)
M1 for obtaining \(x^2 = 16\)
A1 for both \(4\) and \(-4\) (or \(\pm 4\))

Use the given terms to set up a system of linear equations in terms of \(a\) and \(b\):
For the 3rd term, \(T_3 = a(3)^2 + b(3) + 5 = 38\):
\(9a + 3b + 5 = 38 \implies 9a + 3b = 33 \implies 3a + b = 11\) (Equation 1)

For the 6th term, \(T_6 = a(6)^2 + b(6) + 5 = 143\):
\(36a + 6b + 5 = 143 \implies 36a + 6b = 138 \implies 6a + b = 23\) (Equation 2)

Subtract Equation 1 from Equation 2:
\((6a + b) - (3a + b) = 23 - 11\)
\(3a = 12 \implies a = 4\)

Substitute \(a = 4\) back into Equation 1:
\(3(4) + b = 11 \implies 12 + b = 11 \implies b = -1\)

So, the expression for the \(n\)-th term is:
\(T_n = 4n^2 - n + 5\)

Now, find the 10th term (\(n = 10\)):
\(T_{10} = 4(10)^2 - 10 + 5 = 400 - 10 + 5 = 395\).

M1 for setting up the two initial equations: \(9a + 3b + 5 = 38\) and \(36a + 6b + 5 = 143\) (or simplified versions)
M1 for a correct method to solve the simultaneous equations (e.g., elimination or substitution)
A1 for finding \(a = 4\) and \(b = -1\)
A1 for calculating the 10th term as \(395\)

1. Calculate the value of the investment after the first 3 years of compounding at 4.2% per year: \(A_3 = 5000 \times \left(1 + \frac{4.2}{100}\right)^3 = 5000 \times 1.042^3 = 5656.8306...\). 2. Calculate the value of the investment at the end of the next 4 years with compounding quarterly at a rate of \(r\%\): There are \(4 \times 4 = 16\) compounding periods in 4 years. \(A_7 = A_3 \times \left(1 + \frac{r}{400}\right)^{16} = 7000\). 3. Substitute the value of \(A_3\) and solve for \(r\): \(5656.8306... \times \left(1 + \frac{r}{400}\right)^{16} = 7000\) which leads to \(\left(1 + \frac{r}{400}\right)^{16} = \frac{7000}{5656.8306...} \approx 1.237442\). Taking the 16th root: \(1 + \frac{r}{400} = (1.237442)^{\frac{1}{16}} \approx 1.013349\). Solving for \(r\): \(\frac{r}{400} = 0.013349 \implies r = 400 \times 0.013349 \approx 5.3397...\). Rounding to 3 significant figures, we get \(r = 5.34\).

M1 for finding the amount after 3 years: \(5000 \times 1.042^3\) or 5657. M1 for setting up the equation for quarterly compounding: \(A_3 \times (1 + r/400)^{16} = 7000\). M1 for rearranging and solving for \(r\), e.g., \((1 + r/400) = (7000 / A_3)^{1/16}\). A1 for \(r = 5.34\) (accept 5.339 to 5.341).

1. Write down equations for \(T_3\) and \(T_6\) using the given formula. For \(n = 3\): \(T_3 = a(3)^2 + b(3) + 5 = 26 \implies 9a + 3b + 5 = 26 \implies 9a + 3b = 21 \implies 3a + b = 7\) (Equation 1). For \(n = 6\): \(T_6 = a(6)^2 + b(6) + 5 = 83 \implies 36a + 6b + 5 = 83 \implies 36a + 6b = 78 \implies 6a + b = 13\) (Equation 2). 2. Solve the simultaneous equations by subtracting Equation 1 from Equation 2: \((6a + b) - (3a + b) = 13 - 7 \implies 3a = 6 \implies a = 2\). Substitute \(a = 2\) into Equation 1: \(3(2) + b = 7 \implies 6 + b = 7 \implies b = 1\). Thus, the formula for the \(n\)-th term is \(T_n = 2n^2 + n + 5\). 3. Find the 10th term, \(T_{10}\): \(T_{10} = 2(10)^2 + 10 + 5 = 200 + 10 + 5 = 215\).

M1 for writing down two equations in terms of \(a\) and \(b\), e.g., \(9a + 3b = 21\) and \(36a + 6b = 78\). M1 for solving the simultaneous equations to find \(a\) or \(b\). A1 for obtaining \(a = 2\) and \(b = 1\). A1 for substituting \(n = 10\) to get 215.

1. Find the inverse function \(f^{-1}(x)\). Let \(y = \frac{2x + 6}{x - 1}\). Multiplying both sides by \((x - 1)\) gives \(y(x - 1) = 2x + 6 \implies yx - y = 2x + 6\). Rearranging to collect \(x\) terms: \(yx - 2x = y + 6 \implies x(y - 2) = y + 6 \implies x = \frac{y + 6}{y - 2}\). Therefore, \(f^{-1}(x) = \frac{x + 6}{x - 2}\). 2. Set \(f^{-1}(x) = g(x)\): \(\frac{x + 6}{x - 2} = x + 1\). 3. Solve the equation: \(x + 6 = (x + 1)(x - 2) \implies x + 6 = x^2 - x - 2 \implies x^2 - 2x - 8 = 0\). Factorising the quadratic equation: \((x - 4)(x + 2) = 0\). Thus, \(x = 4\) or \(x = -2\).

M1 for attempting to find \(f^{-1}(x)\) (e.g., swapping variables and making \(x\) the subject). A1 for \(f^{-1}(x) = \frac{x + 6}{x - 2}\). M1 for equating \(f^{-1}(x) = g(x)\) and reducing to a quadratic equation, e.g., \(x^2 - 2x - 8 = 0\). A1 for \(x = 4\) and \(x = -2\) (both required for final accuracy mark).

First, we calculate the amount in the savings account after 3 years:
\(A_3 = 8500 \times \left(1 + \frac{2.8}{100}\right)^3\)
\(A_3 = 8500 \times 1.028^3 = 9234.152752\)

Next, this amount is invested for another 5 years (from year 3 to year 8) at a compound interest rate of \(r\)\% per year:
\(A_8 = 9234.152752 \times \left(1 + \frac{r}{100}\right)^5 = 10651.52\)

Divide both sides by 9234.152752:
\(\left(1 + \frac{r}{100}\right)^5 = \frac{10651.52}{9234.152752} \approx 1.15349185\)

Take the fifth root of both sides:
\(1 + \frac{r}{100} = 1.15349185^{0.2} \approx 1.029000\)

\(\frac{r}{100} \approx 0.029000 \Rightarrow r \approx 2.90\)

Thus, \(r = 2.90\)% (to 2 decimal places).

M1 for \(8500 \times 1.028^3\) (or interest calculation showing 9234.15)
M1 for setting up the second stage: \((\text{their } 9234.15) \times \left(1 + \frac{r}{100}\right)^5 = 10651.52\)
M1 for rearranging to find \(\left(1 + \frac{r}{100}\right) = \left(\frac{10651.52}{\text{their } 9234.15}\right)^{0.2}\)
A1.17 for 2.90 (accept 2.9)

First, we find the formula for the \(n\)-th term of the sequence:
Let the sequence be \(u_n\). The terms are: 5, 12, 23, 38.

The first differences are:
\(12 - 5 = 7\)
\(23 - 12 = 11\)
\(38 - 23 = 15\)

The second differences are:
\(11 - 7 = 4\)
\(15 - 11 = 4\)

Since the second differences are constant, the sequence is quadratic and has the form \(u_n = an^2 + bn + c\), where \(2a = 4 \Rightarrow a = 2\).

Now, let \(u_n = 2n^2 + bn + c\):
For \(n = 1\): \(2(1)^2 + b(1) + c = 5 \Rightarrow b + c = 3\)
For \(n = 2\): \(2(2)^2 + b(2) + c = 12 \Rightarrow 2b + c = 4\)

Subtracting the first equation from the second:
\(b = 1\)

Substitute \(b = 1\) back to find \(c\):
\(1 + c = 3 \Rightarrow c = 2\)

So, the \(n\)-th term of the sequence is:
\(u_n = 2n^2 + n + 2\)

We are given that the \(n\)-th term is 530:
\(2n^2 + n + 2 = 530\)
\(2n^2 + n - 528 = 0\)

Using the quadratic formula or factorisation:
\((2n + 33)(n - 16) = 0\)

Since \(n\) must be a positive integer:
\(n = 16\)

M1 for finding second differences are 4, implying a \(2n^2\) term
M1 for finding the full expression \(2n^2 + n + 2\)
M1 for equating \(2n^2 + n + 2 = 530\) and forming a quadratic equation equal to 0
A1.17 for \(n = 16\)

To find \(x\) when \(f(g(x)) = 8\), we substitute \(g(x)\) into \(f(x)\):
\(f(g(x)) = \frac{8}{3g(x) - 2} = 8\)

Divide both sides by 8:
\(\frac{1}{3g(x) - 2} = 1\)

Multiply by \(3g(x) - 2\):

\(1 = 3g(x) - 2\)

\(3g(x) = 3 \Rightarrow g(x) = 1\)

Now substitute \(g(x) = 5^{2x - 3}\):
\(5^{2x - 3} = 1\)

Since \(5^0 = 1\):
\(2x - 3 = 0\)
\(2x = 3\)

\(x = 1.5\) (or \(\frac{3}{2}\))

M1 for setting up the composite equation \(\frac{8}{3g(x) - 2} = 8\)
M1 for simplifying to \(g(x) = 1\)
M1 for setting up \(5^{2x - 3} = 1\) and concluding \(2x - 3 = 0\)
A1.17 for 1.5 (or \(\frac{3}{2}\))

For a 3D cross of size \(n\), there is 1 central cube and 6 arms, each of length \(n\) containing \(n\) cubes. (a) For size 3: \(C = 1 + 6 \times 3 = 19\). For size 10: \(C = 1 + 6 \times 10 = 61\). (b) In general, for size \(n\), \(C = 6n + 1\). (c) Set \(6n + 1 = 121 \implies 6n = 120 \implies n = 20\).

(a) [2 marks] 1 mark for 19, 1 mark for 61. (b) [2 marks] 1 mark for finding the common difference of 6, 1 mark for the correct formula 6n + 1 (or equivalent). (c) [1.71 marks] 1 mark for setting up the equation 6n + 1 = 121, 0.71 marks for solving to get n = 20.

Each of the 6 arms of length \(n\) has 1 end face exposed and \(4n\) side faces exposed. Thus, each arm contributes \(4n + 1\) exposed faces. (a) For \(n = 1\), the surface area is \(6 \times (4(1) + 1) = 6 \times 5 = 30\). (b) For \(n = 5\), the surface area is \(6 \times (4(5) + 1) = 6 \times 21 = 126\). (c) In general, \(A = 6(4n + 1) = 24n + 6\).

(a) [2 marks] 1 mark for explaining that each of the 6 arms has 5 exposed faces, 1 mark for concluding 30. (b) [2 marks] 1 mark for substituting n = 5 into a correct method, 1 mark for 126. (c) [1.71 marks] 1 mark for identifying the slope of 24, 0.71 marks for the correct expression 24n + 6.

(a) The central block contains \(2 \times 2 \times 2 = 8\) cubes. Each of the 6 arms contains \(2 \times 2 \times n = 4n\) cubes. For \(n = 5\), each arm has \(4 \times 5 = 20\) cubes. Total cubes \(K = 8 + 6 \times 20 = 128\). (b) For \(n = 1\), each of the 6 arms is a \(2 \times 2 \times 1\) block. Its outer face has area 4, and its 4 side faces each have area 2. The exposed area of one arm is \(4 + 4 \times 2 = 12\). Since there are 6 arms, the total surface area \(S_D = 6 \times 12 = 72\). (c) For an arm of length \(n\), the outer face has area 4, and the 4 side faces each have area \(2n\). The exposed area of one arm is \(4 + 8n\). Total surface area for 6 arms is \(S_D = 6(8n + 4) = 48n + 24\).

(a) [2 marks] 1 mark for identifying 8 cubes in the center or 120 cubes in the arms, 1 mark for the total of 128 cubes. (b) [2 marks] 1 mark for identifying that each arm has an exposed area of 12, 1 mark for showing 6 x 12 = 72. (c) [1.71 marks] 1 mark for identifying the coefficient of n is 48, 0.71 marks for the correct expression 48n + 24.

The perimeters for \(n = 1, 2, 3\) are 5, 8, 11. This forms an arithmetic sequence with a common difference of 3.

(a)
(i) For \(n = 4\), the perimeter is \(11 + 3 = 14\).
(ii) For \(n = 5\), the perimeter is \(14 + 3 = 17\).

(b) The general term of an arithmetic sequence is given by \(a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
Here, \(a = 5\) and \(d = 3\).
\(P = 5 + (n-1)3\)
\(P = 5 + 3n - 3\)
\(P = 3n + 2\).

(a) B1 for 14. B1 for 17.
(b) M1 for recognizing a common difference of 3 (e.g., writing \(3n + c\)). A1 for the correct simplified expression \(3n + 2\).

The sequence of total matchsticks is 6, 11, 16, ...
This is an arithmetic sequence with a first term of 6 and a common difference of 5.

(a) For \(n = 4\), \(T = 16 + 5 = 21\).

(b) The formula is of the form \(T = 5n + c\).
Substitute \(n = 1\) and \(T = 6\):
\(6 = 5(1) + c \implies c = 1\).
So, \(T = 5n + 1\).

(c) Substitute \(T = 101\) into the formula:
\(101 = 5n + 1\)
\(100 = 5n\)
\(n = 20\).
Therefore, 20 houses can be built.

(a) B1 for 21.
(b) M1 for finding the general form \(5n + c\) or using \(a + (n-1)d\). A1 for \(T = 5n + 1\).
(c) M1 for setting up the equation \(5n + 1 = 101\) and attempting to solve. A1 for 20.

The sequence for \(D\) is 9, 16, 23, ... which increases by 7 each time.

(a) The general term is of the form \(D = 7n + c\).
Substitute \(n = 1\) and \(D = 9\):
\(9 = 7(1) + c \implies c = 2\).
So, \(D = 7n + 2\).

(b) Using the formula: For \(n = 4\), \(D = 7(4) + 2 = 30\).
Using the pattern: The next term after 23 is \(23 + 7 = 30\).
Both methods yield 30, which verifies the formula.

(c) Set \(D \le 150\):
\(7n + 2 \le 150\)
\(7n \le 148\)
\(n \le 21.14\)
So, the maximum number of columns is 21.
Using 21 columns requires \(7(21) + 2 = 149\) matchsticks.
The number of leftover matchsticks is \(150 - 149 = 1\).

(a) M1 for identifying the common difference of 7. A1 for \(D = 7n + 2\).
(b) B1 for showing that both the formula and the pattern give 30 for \(n = 4\).
(c) M1 for setting up the inequality \(7n + 2 \le 150\) and solving. A1 for 21 columns. A1 for 1 leftover.

(a) The sequence of the number of matchsticks is arithmetic with a common difference of 4.
\[ p = 13 + 4 = 17 \]

(b) The first term \(a = 5\) and the common difference \(d = 4\).
\[ M = a + (n - 1)d = 5 + 4(n - 1) = 4n + 1 \]

(c) Set the expression for \(M\) equal to 101 and solve for \(n\):
\[ 4n + 1 = 101 \]
\[ 4n = 100 \]
\[ n = 25 \]

(a) [1.71 marks]
- 1.71 marks for 17 (or B1 for identifying a common difference of 4).

(b) [2 marks]
- M1 for recognizing a linear sequence of the form \(4n + k\) (where \(k\) is any constant).
- A1 for the correct expression \(4n + 1\) (or equivalent, e.g., \(5 + 4(n-1)\)).

(c) [2 marks]
- M1 for setting their expression from part (b) equal to 101 and attempting to solve.
- A1 for 25.

(a) The sequence of triangular numbers is given by \(T_n = \frac{n(n+1)}{2}\). For \(n = 4\), \(T_4 = \frac{4 \times 5}{2} = 10\). For \(n = 5\), \(T_5 = \frac{5 \times 6}{2} = 15\).
The sum \(S_5 = 1 + 3 + 6 + 10 + 15 = 35\).

(b) We know that:
\(S_1 = T_1 = 1\)
\(S_2 = T_1 + T_2 = 1 + 3 = 4\)
\(S_3 = T_1 + T_2 + T_3 = 4 + 6 = 10\)
Using \(S_n = a n^3 + b n^2 + c n\):
For \(n = 1\): \(a + b + c = 1\) (Equation 1)
For \(n = 2\): \(8a + 4b + 2c = 4 \implies 4a + 2b + c = 2\) (Equation 2)
For \(n = 3\): \(27a + 9b + 3c = 10\) (Equation 3)

Subtracting Equation 1 from Equation 2:
\(3a + b = 1\) (Equation 4)

Subtracting 3 times Equation 1 from Equation 3:
\(24a + 6b = 7 \implies 4a + b = \frac{7}{6}\) (Equation 5)

Subtracting Equation 4 from Equation 5:
\(a = \frac{7}{6} - 1 = \frac{1}{6}\)

Substitute \(a = \frac{1}{6}\) into Equation 4:
\(3\left(\frac{1}{6}\right) + b = 1 \implies \frac{1}{2} + b = 1 \implies b = \frac{1}{2}\)

Substitute \(a = \frac{1}{6}\) and \(b = \frac{1}{2}\) into Equation 1:
\(\frac{1}{6} + \frac{1}{2} + c = 1 \implies \frac{4}{6} + c = 1 \implies c = \frac{1}{3}\).

So \(a = \frac{1}{6}\), \(b = \frac{1}{2}\), and \(c = \frac{1}{3}\).

(c) Expanding the product:
\(S_n = \frac{n(n+1)(n+2)}{6} = \frac{n(n^2 + 3n + 2)}{6} = \frac{n^3 + 3n^2 + 2n}{6} = \frac{1}{6}n^3 + \frac{1}{2}n^2 + \frac{1}{3}n\).
This matches our formula from (b).
For \(n = 15\):
\(S_{15} = \frac{15(16)(17)}{6} = 680\).

(d) Set \(S_n = 1540\):
\(\frac{n(n+1)(n+2)}{6} = 1540 \implies n(n+1)(n+2) = 9240\).
Using GDC to solve or by testing consecutive integers: \(20 \times 21 \times 22 = 9240\).
Thus, \(n = 20\).

(a) \(T_4 = 10, T_5 = 15\) [1M], \(S_5 = 35\) [1A]
(b) Set up system of equations [1M], solve system to find \(a = 1/6, b = 1/2, c = 1/3\) [2A]
(c) Expand and show equivalence [1M], find \(S_{15} = 680\) [2A]
(d) Write equation \(n(n+1)(n+2) = 9240\) [1M], solve to find \(n = 20\) [1A]

(a) When \(t = 0\), \(T(0) = 80\).
\(80 = 20 + A e^{0} \implies 80 = 20 + A \implies A = 60\).

(b) For \(t = 10\), \(T(10) = 50\).
\(50 = 20 + 60 e^{-10k} \implies 30 = 60 e^{-10k} \implies e^{-10k} = 0.5\).
Taking the natural logarithm on both sides:
\(-10k = \ln(0.5) \implies -10k = -\ln(2) \implies k = \frac{\ln(2)}{10}\).
\(k \approx 0.0693147 \approx 0.0693\) (to 3 s.f.).

(c) Substituting \(t = 25\) and \(k = 0.069315\):
\(T(25) = 20 + 60 e^{-0.069315 \times 25} = 20 + 60 e^{-1.73287} \approx 20 + 60 \times 0.17678 = 20 + 10.61 = 30.61^\circ\text{C}\).
To the nearest degree, \(T = 31^\circ\text{C}\).

(d) Set \(T(t) = 25\):
\(25 = 20 + 60 e^{-k t} \implies 5 = 60 e^{-k t} \implies e^{-k t} = \frac{1}{12}\).
\(-k t = \ln\left(\frac{1}{12}\right) = -\ln(12) \implies t = \frac{\ln(12)}{k}\).
Using \(k = 0.0693147\):
\(t = \frac{\ln(12)}{0.0693147} \approx 35.849 \approx 35.8\) minutes (or \(35.9\) minutes if using \(k = 0.0693\)).

(e) As \(t \to \infty\), \(e^{-k t} \to 0\), so \(T(t) \to 20\). The equation of the horizontal asymptote is \(T = 20\).
This represents the ambient room temperature, which is the minimum temperature the liquid will approach as it cools down.

(a) Substitute \(t=0, T=80\) and solve for \(A\) [1M]
(b) Substitute \(t=10, T=50\) and \(A=60\) [1M], take ln of both sides [1M], obtain \(k = \ln(2)/10 \approx 0.0693\) [1A]
(c) Substitute \(t=25\) and calculate [1M], round to \(31^\circ\text{C}\) [1A]
(d) Set \(T(t)=25\) and solve for \(t\) [1M], obtain \(t \approx 35.8\) or \(35.9\) [1A]
(e) Write down equation \(T = 20\) [1M], explain context as room/surrounding temperature [1A]

(a) (i) Simple interest after 5 years: \(I_A = x \times 0.04 \times 5 = 0.20x\).
(ii) Compound interest total value after 5 years: \(V_B = (10000 - x) \times (1.035)^5\).

(b) The combined total value \(V(x)\) is the sum of the total value of Account A and the total value of Account B:
Total value of Account A after 5 years = \(x + 0.20x = 1.20x\).
Total value of Account B after 5 years = \((10000 - x)(1.035)^5\).
So:
\(V(x) = 1.20x + (10000 - x)(1.035)^5\)
\(V(x) = 1.20x + 10000(1.035)^5 - x(1.035)^5\)
\(V(x) = x(1.20 - 1.035^5) + 10000(1.035)^5\).
Now calculate the values:
\(1.035^5 \approx 1.187686\)
\(1.20 - 1.187686 = 0.012314 \approx 0.01231\) (4 s.f.)
\(10000(1.035)^5 \approx 11876.86 \approx 11880\) (4 s.f.)
So, \(V(x) = 0.01231x + 11880\).

(c) Using the exact expression for \(V(x) = 11900\):
\(1.20x + (10000 - x)(1.035)^5 = 11900\)
\(x(1.20 - 1.035^5) + 10000(1.035)^5 = 11900\)
\(0.0123137x + 11876.86 = 11900\)
\(0.0123137x = 23.14\)
\(x = \frac{23.14}{0.0123137} \approx 1879.13 \approx \$1879\).
(If using rounded formula from (b): \(0.01231x + 11880 = 11900 \implies 0.01231x = 20 \implies x \approx 1625\). We accept values in the range $1625 to $1880 depending on premature rounding).

(d) The constants \(p\) and \(q\) are functions of the investment period \(t\). Specifically, \(p = (1 + 0.04t) - (1.035)^t\) and \(q = 10000(1.035)^t\). Changing the years from 5 to 20 changes the exponents and linear coefficients, resulting in completely different values for \(p\) and \(q\).

(a)(i) Write \(0.20x\) [1B]
(a)(ii) Write \((10000 - x)(1.035)^5\) [2B]
(b) Combine terms to form \(V(x)\) [1M], find \(p \approx 0.01231\) [1A], find \(q \approx 11880\) [1A]
(c) Set \(V(x) = 11900\) and solve [1M], obtain \(x \approx \$1879\) (allow $1625 due to rounded coefficients) [1A]
(d) State that \(p\) and \(q\) depend on the time parameter \(t\) or compound interest is exponential [2B]

(a) For \(n = 5\): The total number of vertices is \(V_5 = \frac{6 \times 7}{2} = 21\). The number of boundary vertices is \(B_5 = 3 \times 5 = 15\). The number of internal vertices is \(I_5 = 21 - 15 = 6\). (b) Since the boundary consists of 3 sides each of length \(n\), the number of vertices on the boundary is \(B_n = 3n\). (c) Using the given relationship: \(I_n = V_n - B_n = \left(\frac{1}{2}n^2 + \frac{3}{2}n + 1\right) - 3n = \frac{1}{2}n^2 - \frac{3}{2}n + 1\). (d) Set \(I_n = 45 \implies \frac{1}{2}n^2 - \frac{3}{2}n + 1 = 45 \implies n^2 - 3n + 2 = 90 \implies n^2 - 3n - 88 = 0 \implies (n - 11)(n + 8) = 0\). Since \(n\) must be positive, \(n = 11\).

Part (a): 3 marks total: B1 for \(V_5 = 21\), B1 for \(B_5 = 15\), B1 for \(I_5 = 6\). Part (b): 1 mark: B1 for \(3n\) (or equivalent). Part (c): 3 marks total: M1 for setting up the subtraction \(V_n - B_n\), M1 for combining \(\frac{3}{2}n - 3n = -\frac{3}{2}n\), A1 for showing the final simplified expression. Part (d): 3 marks total: M1 for setting \(I_n = 45\), M1 for formulating the quadratic equation \(n^2 - 3n - 88 = 0\) or factorising, A1 for \(n = 11\).

(a) When \(t = 0\), \(T(0) = 85 \implies 20 + B \cdot c^0 = 85 \implies 20 + B = 85 \implies B = 65\). (b)(i) When \(t = 10\), \(T(10) = 52.5 \implies 20 + 65 \cdot c^{10} = 52.5 \implies 65 \cdot c^{10} = 32.5 \implies c^{10} = 0.5 \implies c = 0.5^{1/10} = 0.5^{0.1}\). (b)(ii) \(c = 0.5^{0.1} \approx 0.933\). (c) For \(t = 25\), \(T(25) = 20 + 65 \cdot (0.5^{0.1})^{25} = 20 + 65 \cdot 0.5^{2.5} \approx 20 + 11.49 = 31.5^\circ\text{C}\). (d) To find when \(T = 60\): \(20 + 65 \cdot 0.5^{0.1t} = 60 \implies 65 \cdot 0.5^{0.1t} = 40 \implies 0.5^{0.1t} = 40/65 \implies 0.1t = \log_{0.5}(8/13) \implies t \approx 7.00\) minutes. To find when \(T = 40\): \(20 + 65 \cdot 0.5^{0.1t} = 40 \implies 65 \cdot 0.5^{0.1t} = 20 \implies 0.5^{0.1t} = 20/65 \implies 0.1t = \log_{0.5}(4/13) \implies t \approx 17.00\) minutes. Thus, the coffee is pleasant to drink between 7 and 17 minutes.

Part (a): 1 mark: B1 for substitute \(t = 0\) and show \(B = 65\). Part (b)(i): 2 marks: M1 for setting up \(20 + 65 \cdot c^{10} = 52.5\), A1 for showing \(c = 0.5^{0.1}\). Part (b)(ii): 1 mark: B1 for \(0.933\). Part (c): 2 marks: M1 for substituting \(t = 25\) into the model, A1 for \(31.5\) (accept 31.4 to 31.6). Part (d): 4 marks total: M1 for setting \(T(t) = 60\) and solving, A1 for \(t = 7\); M1 for setting \(T(t) = 40\) and solving, A1 for \(t = 17\).