2024 Edexcel IGCSE Mathematics (Specification B) Practice Paper with Answers

First, expand the two parts of the expression separately:

\((2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12\)

\(-2x(x - 1) = -2x^2 + 2x\)

Now, combine the expanded parts:

\((2x^2 + 5x - 12) + (-2x^2 + 2x) = 2x^2 - 2x^2 + 5x + 2x - 12 = 7x - 12\)

M1: for expanding either part correctly (allow 1 sign or arithmetic error), e.g., \(2x^2 + 5x - 12\) or \(-2x^2 + 2x\).
A1: for \(7x - 12\) (or equivalent).

Rationalise the denominator of the fraction \(\frac{2}{\sqrt{3} - 1}\) by multiplying the numerator and denominator by \(\sqrt{3} + 1\):

\(\frac{2}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{2(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}\)

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

Substitute this back into the original expression:

\((\sqrt{3} + 1) - \sqrt{3} = 1\)

Since 1 is an integer, the proof is complete.

M1: for attempting to rationalise the denominator by multiplying by \(\frac{\sqrt{3} + 1}{\sqrt{3} + 1}\).
A1: for fully correct working showing the final result simplifies to \(1\).

Multiply both sides of the inequality by 6 (the lowest common multiple of 2 and 3) to clear the denominators:

\(3(3x - 5) > 2(5x + 1)\)

Expand both sides:

\(9x - 15 > 10x + 2\)

Rearrange to isolate \(x\):

\(-15 - 2 > 10x - 9x\)

\(-17 > x\)

This can be written as \(x < -17\).

M1: for clearing denominators correctly, e.g., \(3(3x - 5) > 2(5x + 1)\).
M1: for isolating the \(x\) terms on one side and constant terms on the other side of an inequality or equation (e.g., \(9x - 10x > 2 + 15\)).
A1: for \(x < -17\) or \(-17 > x\).

A 15% reduction means the sale price is \(100\% - 15\% = 85\%\) of the normal price.

Let the normal price be \(P\).

\(0.85 \times P = 323\)

\(P = \frac{323}{0.85} = 380\)

The normal price of the bicycle is £380.

M1: for equating 85% to 323, e.g., \(0.85 \times P = 323\) or \(85\% = 323\).
M1: for a complete method to find the normal price, e.g., \(\frac{323}{85} \times 100\).
A1: for \(380\) (accept £380).

Find the highest common factor (HCF) of the two terms:

- The HCF of 12 and 18 is 6.
- The HCF of \(a\) and \(a^2\) is \(a\).
- The HCF of \(x^2\) and \(x\) is \(x\).

So, the HCF is \(6ax\).

Factorise \(6ax\) out of the expression:

\(12ax^2 - 18a^2x = 6ax(2x - 3a)\)

M1: for taking out a common factor of at least \(2ax\), \(3ax\), \(6x\), or \(6a\) with a correct remaining expression in parentheses.
A1: for \(6ax(2x - 3a)\).

Let \(x = 0.2\dot{3}\dot{6} = 0.2363636...\)

Multiply both sides by 10:
\(10x = 2.363636...\) (Equation 1)

Multiply both sides by 1000:
\(1000x = 236.363636...\) (Equation 2)

Subtract Equation 1 from Equation 2:
\(1000x - 10x = 236.363636... - 2.363636...\)
\(990x = 234\)

Solve for \(x\):
\(x = \frac{234}{990}\)

Simplify the fraction by dividing numerator and denominator by their greatest common divisor, 18:
\(\frac{234 \div 18}{990 \div 18} = \frac{13}{55}\)

M1: for setting up two equations that when subtracted eliminate the recurring part, e.g., \(1000x = 236.36...\) and \(10x = 2.36...\).
M1: for a correct subtraction step resulting in an equation like \(990x = 234\) or equivalent.
A1: for \(\frac{13}{55}\) (no marks awarded if no working is shown).

Multiply both sides by \((2 - t)\) to clear the fraction:

\(v(2 - t) = u + t\)

Expand the brackets:

\(2v - vt = u + t\)

Collect all terms containing \(t\) on one side and the other terms on the opposite side:

\(2v - u = t + vt\)

Factorise \(t\) on the right side:

\(2v - u = t(1 + v)\)

Divide both sides by \((1 + v)\) to make \(t\) the subject:

\(t = \frac{2v - u}{1 + v}\)

M1: for removing the fraction to get \(v(2 - t) = u + t\) or \(2v - vt = u + t\).
M1: for isolating the \(t\) terms on one side and factorising, e.g., \(2v - u = t(1 + v)\).
A1: for \(t = \frac{2v - u}{1 + v}\) (or equivalent, e.g., \(t = \frac{u - 2v}{-1 - v}\)).

The value of the car depreciates by 8% per year, so the multiplier for each year is \(1 - 0.08 = 0.92\).

From the start of 2021 to the start of 2023 is exactly 2 years.

Value at start of 2023:
\(18500 \times 0.92^2 = 18500 \times 0.8464 = 15658.40\)

Rounding to the nearest £100:
\(15658.40 \approx 15700\)

M1: for a correct calculation of the value after 2 years, e.g., \(18500 \times 0.92^2\) or evaluating \(18500 \times 0.92 = 17020\) and then \(17020 \times 0.92 = 15658.4\).
A1: for \(15700\) (accept £15,700).

First, express all terms with a base of 3: \( 9^{x+2} = (3^2)^{x+2} = 3^{2x+4} \) and \( \frac{1}{27} = 3^{-3} \). The equation becomes \( 3^{2x-1} \times 3^{2x+4} = 3^{-3} \). Using the laws of indices, we add the exponents on the left-hand side: \( 3^{(2x-1) + (2x+4)} = 3^{-3} \), which simplifies to \( 3^{4x+3} = 3^{-3} \). Equating the exponents gives \( 4x + 3 = -3 \). Subtracting 3 from both sides gives \( 4x = -6 \), and dividing by 4 gives \( x = -1.5 \).

M1: for expressing 9 as \( 3^2 \) or \( \frac{1}{27} \) as \( 3^{-3} \) (or equivalent base 3 representation). M1: for forming and simplifying a correct linear equation in \( x \), e.g., \( 2x - 1 + 2(x + 2) = -3 \). A1: for \( x = -1.5 \) or \( x = -\frac{3}{2} \).

Let \( x = 0.2363636... \). Multiply both sides by 10 to shift the non-recurring part: \( 10x = 2.363636... \). Multiply both sides of the original equation by 1000 to shift one full repeating block: \( 1000x = 236.363636... \). Subtracting the two equations: \( 1000x - 10x = 236.363636... - 2.363636... \), which simplifies to \( 990x = 234 \). Solving for \( x \) gives \( x = \frac{234}{990} \). Dividing both the numerator and denominator by their greatest common divisor, 18, gives \( x = \frac{13}{55} \).

M1: for a complete method to subtract two correct equations to eliminate the recurring part, e.g., showing \( 1000x - 10x = 236.\dot{3}\dot{6} - 2.\dot{3}\dot{6} \) or establishing \( 990x = 234 \). A1: for \( \frac{13}{55} \) (must be in simplest form).

First, multiply both sides by \( 5 - y \) to clear the fraction: \( x(5 - y) = 3y + 2 \). Expand the bracket on the left-hand side: \( 5x - xy = 3y + 2 \). Group all terms containing \( y \) on one side and the remaining terms on the other side: \( 5x - 2 = 3y + xy \). Factorise \( y \) on the right-hand side: \( 5x - 2 = y(3 + x) \). Finally, divide both sides by \( 3 + x \) to isolate \( y \): \( y = \frac{5x - 2}{x + 3} \).

M1: for multiplying both sides by \( 5 - y \) to clear the fraction. M1: for expanding the brackets and isolating the terms containing \( y \) on one side of the equation. A1: for \( y = \frac{5x - 2}{x + 3} \) (or equivalent, such as \( y = \frac{2 - 5x}{-x - 3} \)).

The length of \( 8.4\text{ cm} \) is measured correct to the nearest millimetre (which is \( 0.1\text{ cm} \)). The lower bound for the length of a single rod is \( 8.4 - 0.05 = 8.35\text{ cm} \). To find the lower bound of the total length of 15 rods placed end-to-end, we multiply the lower bound of one rod by 15: \( 15 \times 8.35 = 125.25\text{ cm} \).

M1: for identifying the lower bound of a single rod as \( 8.35 \) (or for calculating \( 15 \times L \) where \( 8.3 \le L < 8.4 \)). A1: for \( 125.25 \).

First, use the gradient formula to find the value of \(p\): \(\text{Gradient} = \frac{8 - 2}{p - 3} = \frac{3}{4}\). This simplifies to \(\frac{6}{p - 3} = \frac{3}{4}\). Multiplying both sides by \(4(p - 3)\) gives \(24 = 3(p - 3)\), which simplifies to \(24 = 3p - 9\). Solving this yields \(33 = 3p \implies p = 11\). Now, find the coordinates of the midpoint of \(AB\): \(\text{Midpoint } M = \left(\frac{3 + p}{2}, \frac{2 + 8}{2}\right) = \left(\frac{3 + 11}{2}, \frac{10}{2}\right) = (7, 5)\). Since the midpoint lies on the line \(y = 2x + c\), substitute the coordinates \(x = 7\) and \(y = 5\) into the equation: \(5 = 2(7) + c \implies 5 = 14 + c \implies c = -9\).

M1: Correct substitution into the gradient formula and setting equal to \(\frac{3}{4}\) to find \(p = 11\). M1: Correctly finding the midpoint coordinates \((7, 5)\) using their value of \(p\). M1: Substituting their midpoint coordinates into \(y = 2x + c\) to find \(c\). A1: Correct final value of \(c = -9\).

Let the five positive integers in ascending order be \(x_1, x_2, x_3, x_4, x_5\). Since the median is 5, the middle value is \(x_3 = 5\). Since the mean is 6, the sum of all five integers is \(x_1 + x_2 + x_3 + x_4 + x_5 = 5 \times 6 = 30\). Substituting \(x_3 = 5\) gives \(x_1 + x_2 + 5 + x_4 + x_5 = 30 \implies x_1 + x_2 + x_4 + x_5 = 25\). To maximize the largest number \(x_5\), we must minimize the other terms \(x_1, x_2\), and \(x_4\). Since 5 is the unique mode, 5 must appear at least twice, so we can set \(x_4 = 5\), giving the set \(\{x_1, x_2, 5, 5, x_5\}\). To prevent other numbers from repeating (keeping 5 as the unique mode), we must choose distinct positive integers for \(x_1\) and \(x_2\). To minimize these, we set \(x_1 = 1\) and \(x_2 = 2\). Substituting these values into the sum gives: \(1 + 2 + 5 + 5 + x_5 = 30 \implies 13 + x_5 = 30 \implies x_5 = 17\). Since the set \(\{1, 2, 5, 5, 17\}\) satisfies all conditions, the maximum possible value is 17.

M1: Using the mean to find that the sum of the five integers is 30. M1: Identifying that the median is 5 and establishing that 5 must appear at least twice to be the unique mode. M1: Minimizing the remaining elements by choosing \(x_1 = 1\) and \(x_2 = 2\) with \(x_3 = 5\) and \(x_4 = 5\). A1: Correct maximum value of 17.

First, find the gradient of line \(L_1\): \(m_1 = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}\). Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) is the negative reciprocal: \(m_2 = -\frac{1}{-1/2} = 2\). Now, write the equation of line \(L_2\) using the point \(C(1, -3)\): \(y - (-3) = 2(x - 1) \implies y + 3 = 2x - 2 \implies y = 2x - 5\). The line intersects the \(y\)-axis when \(x = 0\), which gives \(y = -5\). Thus, the coordinates are \((0, -5)\).

M1: Finding the gradient of \(L_1\) as \(-\frac{1}{2}\). M1: Finding the perpendicular gradient of \(L_2\) as \(2\). M1: Finding the equation of line \(L_2\) as \(y = 2x - 5\) or substituting \(x = 0\) into their linear equation. A1: Correct coordinates \((0, -5)\).

Identify the mid-interval values (\(x_m\)) for each class: 15, 25, 35, and 45. Find the sum of the products of midpoints and frequencies: \(\sum f x_m = (15 \times 3) + (25 \times 5) + (35 \times f) + (45 \times 4) = 45 + 125 + 35f + 180 = 350 + 35f\). The total frequency is \(\sum f = 3 + 5 + f + 4 = 12 + f\). Set up the equation for the estimated mean: \(\frac{350 + 35f}{12 + f} = 33 \implies 350 + 35f = 33(12 + f) \implies 350 + 35f = 396 + 33f\). Rearranging gives \(2f = 46 \implies f = 23\).

M1: Finding the correct midpoints (15, 25, 35, 45). M1: Expressing the total sum of products as \(350 + 35f\) and the total frequency as \(12 + f\). M1: Setting up the equation \(\frac{350 + 35f}{12 + f} = 33\) and attempting to solve. A1: Correct value of \(f = 23\).

The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by \(\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\). Substituting the coordinates \(A(1, 1)\), \(B(5, 3)\), and \(C(3, k)\): \(10 = \frac{1}{2} |1(3 - k) + 5(k - 1) + 3(1 - 3)| \implies 20 = |3 - k + 5k - 5 - 6| \implies 20 = |4k - 8|\). Since \(k > 3\), the term \(4k - 8\) is positive, so \(4k - 8 = 20 \implies 4k = 28 \implies k = 7\).

M1: Substituting the coordinates correctly into the triangle area formula. M1: Simplifying the expression inside the absolute value to obtain \(|4k - 8|\). M1: Resolving the absolute value using the condition \(k > 3\) to write \(4k - 8 = 20\). A1: Correct value of \(k = 7\).

First, calculate the total monthly salary for Branch A: \(\text{Total A} = 40 \times 2500 = 100,000\). The total monthly salary for Branch B is \(\text{Total B} = n \times 3100 = 3100n\). The total number of employees is \(40 + n\). The combined mean is \(\frac{100,000 + 3100n}{40 + n} = 2860\). Multiplying both sides by \(40 + n\) yields: \(100,000 + 3100n = 2860(40 + n) \implies 100,000 + 3100n = 114,400 + 2860n\). Grouping like terms: \(3100n - 2860n = 114,400 - 100,000 \implies 240n = 14,400 \implies n = 60\).

M1: Correct expression or value for total salary of Branch A (100,000). M1: Setting up the equation for the combined mean: \(\frac{100,000 + 3100n}{40+n} = 2860\). M1: Rearranging the equation to solve for \(n\) (e.g., \(240n = 14,400\)). A1: Correct value of \(n = 60\).

The tangent to the circle at point \(B\) is perpendicular to the radius (or diameter) \(AB\). First, find the gradient of \(AB\): \(m_{AB} = \frac{12 - 4}{5 - (-1)} = \frac{8}{6} = \frac{4}{3}\). The gradient of the tangent, which is perpendicular, is \(m_t = -\frac{1}{4/3} = -\frac{3}{4}\). Using the point-slope formula with point \(B(5, 12)\): \(y - 12 = -\frac{3}{4}(x - 5)\). Clear the fraction by multiplying both sides by 4: \(4(y - 12) = -3(x - 5) \implies 4y - 48 = -3x + 15\). Rearranging this into the form \(ax + by + c = 0\) gives \(3x + 4y - 63 = 0\).

M1: Correctly calculating the gradient of the diameter \(AB\) as \(\frac{4}{3}\). M1: Finding the perpendicular gradient of the tangent as \(-\frac{3}{4}\). M1: Substituting point \(B(5, 12)\) and their perpendicular gradient into a linear equation formula. A1: Correct final equation in the required form: \(3x + 4y - 63 = 0\) (or any non-zero integer multiple).

For the first class interval \(10 < w \le 15\), the actual class width is \(15 - 10 = 5\) units, which is represented by a bar width of 2 cm. The height of this bar is 6 cm, so its drawn area is \(\text{Area} = 2 \text{ cm} \times 6 \text{ cm} = 12 \text{ cm}^2\). This area represents a frequency of 30, meaning \(1 \text{ cm}^2\) represents \(\frac{30}{12} = 2.5\) frequency units. For the second class interval \(15 < w \le 25\), the actual class width is \(25 - 15 = 10\) units. Since 5 units of width is represented by 2 cm, 10 units of width is represented by \(4 \text{ cm}\). The frequency is 48, so the required drawn area is \(\text{Area} = \frac{48}{2.5} = 19.2 \text{ cm}^2\). Since \(\text{Area} = \text{Width} \times \text{Height}\), we have \(19.2 = 4 \times \text{Height} \implies \text{Height} = 4.8 \text{ cm}\).

M1: Calculating the area of the first bar as \(12 \text{ cm}^2\) and finding the frequency-to-area scale factor of \(2.5\). M1: Finding the drawn width of the second bar as \(4 \text{ cm}\). M1: Calculating the required drawn area of the second bar as \(19.2 \text{ cm}^2\) or setting up an equivalent frequency density proportion. A1: Correct height of \(4.8 \text{ cm}\).

First, find the midpoint \(M\) of the line segment \(AB\):
\(M = \left(\frac{2 + 8}{2}, \frac{5 + (-3)}{2}\right) = (5, 1)\).

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

The boundary line is the perpendicular bisector of \(AB\). Therefore, its gradient \(m\) is the negative reciprocal of \(m_{AB}\):
\(m = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}\).

Now, use the point-gradient form of a straight line with point \((5, 1)\) and gradient \(\frac{3}{4}\):
\(y - 1 = \frac{3}{4}(x - 5)\)

Multiply the entire equation by 4 to clear the fraction:
\(4(y - 1) = 3(x - 5)\)
\(4y - 4 = 3x - 15\)

Rearrange into the form \(ax + by = c\) with \(a > 0\):
\(3x - 4y = 11\).

M1: For correctly calculating the coordinates of the midpoint of \(AB\), \((5, 1)\).
M1: For finding the gradient of \(AB\) (\(-\frac{4}{3}\)) and using the perpendicular gradient rule \(m_1 m_2 = -1\) to get the perpendicular gradient as \(\frac{3}{4}\).
M1: For substituting their midpoint and perpendicular gradient into a linear equation formula, e.g., \(y - 1 = \frac{3}{4}(x - 5)\).
A1: For the final simplified equation in the required form: \(3x - 4y = 11\).

First, calculate the total monthly salary of the 12 technicians:
\(\text{Total salary of technicians} = 12 \times 2400 = 28800\)

Next, express the total monthly salary of the \(n\) managers in terms of \(n\):
\(\text{Total salary of managers} = n \times 3800 = 3800n\)

The total number of employees is \(12 + n\). The combined mean salary is given as £2960, which leads to the equation:
\(\frac{28800 + 3800n}{12 + n} = 2960\)

Multiply both sides by \(12 + n\) to solve for \(n\):
\(28800 + 3800n = 2960(12 + n)\)
\(28800 + 3800n = 35520 + 2960n\)

Rearrange the equation to isolate terms in \(n\):
\(3800n - 2960n = 35520 - 28800\)
\(840n = 6720\)

Solve for \(n\):
\(n = \frac{6720}{840} = 8\).

M1: For calculating the total salary of the technicians, i.e., \(12 \times 2400 = 28800\), or writing an algebraic expression for the total salary of managers, \(3800n\).
M1: For setting up the correct equation for the combined mean: \(\frac{28800 + 3800n}{12 + n} = 2960\).
M1: For expanding and rearranging the terms to form a linear equation in \(n\), e.g., \(840n = 6720\).
A1: For the correct answer \(n = 8\).

First, we express \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Since \(AP : PB = 2 : 3\), we have \(\overrightarrow{AP} = \frac{2}{5}\overrightarrow{AB}\). Thus, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a}) = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\). Since \(X\) lies on \(OP\), we can write: \(\overrightarrow{OX} = k\overrightarrow{OP} = k\left(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\right) = \frac{3}{5}k\mathbf{a} + \frac{2}{5}k\mathbf{b}\). Next, we express \(\overrightarrow{OX}\) using the line \(AQ\). Since \(OQ : QB = 3 : 1\), we have \(\overrightarrow{OQ} = \frac{3}{4}\mathbf{b}\). Since \(X\) lies on the line \(AQ\), there exists a scalar \(m\) such that \(\overrightarrow{AX} = m\overrightarrow{AQ}\). \(\overrightarrow{AQ} = \overrightarrow{OQ} - \overrightarrow{OA} = \frac{3}{4}\mathbf{b} - \mathbf{a}\). Thus, \(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + m\left(\frac{3}{4}\mathbf{b} - \mathbf{a}\right) = (1 - m)\mathbf{a} + \frac{3}{4}m\mathbf{b}\). We now equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from the two expressions for \(\overrightarrow{OX}\): 1) \(\frac{3}{5}k = 1 - m\) and 2) \(\frac{2}{5}k = \frac{3}{4}m \implies m = \frac{8}{15}k\). Substituting \(m = \frac{8}{15}k\) into (1) gives: \(\frac{3}{5}k = 1 - \frac{8}{15}k\). Multiplying the entire equation by 15: \(9k = 15 - 8k \implies 17k = 15 \implies k = \frac{15}{17}\).

M1: Attempt to find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) (e.g., \(\mathbf{a} + \frac{2}{5}(\mathbf{b}-\mathbf{a})\)).
A1: Correct expression \(\overrightarrow{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\).
M1: Set up expression for \(\overrightarrow{OX}\) along \(AQ\) (e.g., \((1-m)\mathbf{a} + \frac{3}{4}m\mathbf{b}\)).
M1: Equate coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations.
M1: Solve the simultaneous equations for \(k\).
A1: Correct value \(k = \frac{15}{17}\).

The total surface area \(A\) of a closed cylinder is given by: \(A = 2\pi r^2 + 2\pi r h\). We are given \(A = 150\pi\), so: \(2\pi r^2 + 2\pi r h = 150\pi \implies r^2 + rh = 75\). Rearranging for \(h\): \(h = \frac{75 - r^2}{r}\). The volume \(V\) of the cylinder is: \(V = \pi r^2 h\). Substituting the expression for \(h\): \(V = \pi r^2 \left(\frac{75 - r^2}{r}\right) = 75\pi r - \pi r^3\). To find the maximum volume, differentiate \(V\) with respect to \(r\): \(\frac{\mathrm{d}V}{\mathrm{d}r} = 75\pi - 3\pi r^2\). Setting this to 0 for a stationary point: \(75\pi - 3\pi r^2 = 0 \implies 3r^2 = 75 \implies r^2 = 25 \implies r = 5\) (since \(r > 0\)). The second derivative is \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\). At \(r = 5\), \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -30\pi < 0\), confirming a maximum. Substituting \(r = 5\) into the volume equation gives: \(V = 75\pi(5) - \pi(5)^3 = 375\pi - 125\pi = 250\pi\).

M1: Use the surface area formula \(2\pi r^2 + 2\pi r h = 150\pi\) to express \(h\) in terms of \(r\).
M1: Substitute \(h\) into the volume formula to obtain \(V = 75\pi r - \pi r^3\).
M1: Differentiate the volume expression with respect to \(r\) to find \(\frac{\mathrm{d}V}{\mathrm{d}r}\).
A1: Set \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\) and find \(r = 5\).
M1: Use second derivative to confirm the value is a maximum.
A1: Correct maximum volume \(250\pi\).

Since \(M\) is the midpoint of \(OA\), \(\overrightarrow{OM} = \frac{1}{2}\mathbf{a}\). Since \(ON : NB = 2 : 1\), \(\overrightarrow{ON} = \frac{2}{3}\mathbf{b}\). Since \(X\) lies on the line \(AN\), we can express \(\overrightarrow{OX}\) as: \(\overrightarrow{OX} = (1 - \lambda)\overrightarrow{OA} + \lambda\overrightarrow{ON} = (1 - \lambda)\mathbf{a} + \frac{2}{3}\lambda\mathbf{b}\). Since \(X\) lies on the line \(BM\), we can express \(\overrightarrow{OX}\) as: \(\overrightarrow{OX} = (1 - \mu)\overrightarrow{OB} + \mu\overrightarrow{OM} = \frac{1}{2}\mu\mathbf{a} + (1 - \mu)\mathbf{b}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): From the \(\mathbf{a}\) coefficients: \(1 - \lambda = \frac{1}{2}\mu \implies ̅\mu = 2(1 - \lambda)\). From the \(\mathbf{b}\) coefficients: \(\frac{2}{3}\lambda = 1 - \mu\). Substituting \(\mu\): \(\frac{2}{3}\lambda = 1 - 2(1 - \lambda) \implies \frac{2}{3}\lambda = 2\lambda - 1 \implies 1 = \frac{4}{3}\lambda \implies \lambda = \frac{3}{4}\). Substituting \(\lambda = \frac{3}{4}\) back into our first expression for \(\overrightarrow{OX}\): \(\overrightarrow{OX} = \left(1 - \frac{3}{4}\right)\mathbf{a} + \frac{2}{3}\left(\frac{3}{4}\right)\mathbf{b} = \frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

M1: State \(\overrightarrow{OM} = \frac{1}{2}\mathbf{a}\) and \(\overrightarrow{ON} = \frac{2}{3}\mathbf{b}\).
M1: Express \(\overrightarrow{OX}\) along \(AN\) using a parameter \(\lambda\).
M1: Express \(\overrightarrow{OX}\) along \(BM\) using a parameter \(\mu\).
M1: Set up simultaneous equations by equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\).
A1: Solve to find \(\lambda = \frac{3}{4}\) (or \(\mu = \frac{1}{2}\)).
A1: Fully complete proof to show \(\overrightarrow{OX} = \frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

First, when \(x = 4\), \(y = 4^3 - 6(4^2) + 9(4) + 5 = 64 - 96 + 36 + 5 = 9\). Thus \(P\) is \((4, 9)\). Next, we find the gradient of the tangent: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12x + 9\). At \(x = 4\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3(16) - 12(4) + 9 = 9\). The equation of the tangent line is: \(y - 9 = 9(x - 4) \implies y = 9x - 27\). To find where the tangent intersects the curve again, set them equal: \(x^3 - 6x^2 + 9x + 5 = 9x - 27 \implies x^3 - 6x^2 + 32 = 0\). Since \(x = 4\) is a point of tangency, \((x - 4)^2 = x^2 - 8x + 16\) must be a factor of the cubic. Factoring yields: \((x - 4)^2(x + 2) = 0\). Thus, the other point of intersection \(Q\) occurs at \(x = -2\). Substitute \(x = -2\) into the tangent equation to get \(y = 9(-2) - 27 = -45\). Hence, the coordinates of \(Q\) are \((-2, -45)\).

M1: Find the y-coordinate of \(P\) to obtain \((4, 9)\).
M1: Differentiate the curve equation to find the gradient function \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Determine the tangent equation is \(y = 9x - 27\).
M1: Set the curve equal to the tangent to form the equation \(x^3 - 6x^2 + 32 = 0\).
M1: Factorize using \((x - 4)^2\) as a factor to find the other root \(x = -2\).
A1: Obtain the correct coordinates of \(Q\) as \((-2, -45)\).

Let \(h\) be the depth of the water and \(r\) be the radius of the water level. Since the semi-vertical angle is \(30^\circ\), we have \(\tan(30^\circ) = \frac{r}{h} \implies r = \frac{h}{\sqrt{3}}\). The volume \(V\) of water is \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{\pi h^3}{9}\). Differentiating \(V\) with respect to \(h\) gives \(\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{\pi h^2}{3}\). We are given that \(\frac{\mathrm{d}V}{\mathrm{d}t} = 8\). Using the chain rule: \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}h}{\mathrm{d}V} \times \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{3}{\pi h^2} \times 8 = \frac{24}{\pi h^2}\). Substituting \(h = 5\): \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{24}{25\pi}\text{ cm s}^{-1}\).

M1: Use the semi-vertical angle to relate \(r\) and \(h\) via \(r = h \tan(30^\circ)\).
M1: Substitute into the volume formula of a cone to find \(V = \frac{\pi h^3}{9}\).
M1: Correctly find \(\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{\pi h^2}{3}\).
M1: Set up the chain rule expression relating \(\frac{\mathrm{d}h}{\mathrm{d}t}\), \(\frac{\mathrm{d}V}{\mathrm{d}t}\), and \(\frac{\mathrm{d}V}{\mathrm{d}h}\).
M1: Substitute the given values \(\frac{\mathrm{d}V}{\mathrm{d}t} = 8\) and \(h = 5\) into the rate expression.
A1: Correct exact rate of \(\frac{24}{25\pi}\).

1. **Identify the boundary for being closer to \(AB\) than to \(CD\)**:
Since \(ABCD\) is a rectangle of height \(BC = 8\text{ cm}\), the locus of points equidistant from \(AB\) and \(CD\) is a horizontal line parallel to \(AB\) at a distance of \(4\text{ cm}\) below \(AB\).
Points closer to \(AB\) than to \(CD\) lie in the upper half of the rectangle.
The area of this upper half is:
\(\text{Area}_{\text{half}} = 12 \times 4 = 48\text{ cm}^2\).

2. **Identify the boundary for being more than \(5\text{ cm}\) from \(A\)**:
This boundary is a circle of radius \(5\text{ cm}\) centered at \(A\).
We need to subtract the area of the region inside this circle that lies within the upper half of the rectangle.

3. **Set up a coordinate system to find the intersection**:
Let \(A\) be the origin \((0,0)\).
The side \(AB\) lies along the positive x-axis, so the upper half of the rectangle is defined by \(0 \le x \le 12\) and \(-4 \le y \le 0\).
The circle is defined by \(x^2 + y^2 = 25\).
The intersection of the circle with the boundary line \(y = -4\) occurs where:
\(x^2 + (-4)^2 = 25 \implies x^2 + 16 = 25 \implies x^2 = 9 \implies x = 3\) (since \(x \ge 0\)).
So the intersection point is \((3, -4)\).

4. **Calculate the subtracted area**:
The area inside the circle and the upper half of the rectangle can be split into two parts:
- A right-angled triangle with vertices \((0,0)\), \((0,-4)\), and \((3,-4)\).
Its area is:
\(\text{Area}_{\text{triangle}} = \frac{1}{2} \times 3 \times 4 = 6\text{ cm}^2\).
- A circular sector from the line connecting \((0,0)\) to \((3,-4)\) up to the line \(AB\) (the x-axis).
The angle \(\theta\) of this sector satisfies:
\(\cos\theta = \frac{3}{5} \implies \theta = \arccos(0.6) \approx 0.9273\text{ radians}\) (or \(53.13^\circ\)).
The area of this sector is:
\(\text{Area}_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 5^2 \times 0.9273 \approx 11.591\text{ cm}^2\).

The total area to be subtracted is:
\(\text{Area}_{\text{subtracted}} = 6 + 11.591 = 17.591\text{ cm}^2\).

5. **Calculate the area of region \(R\)**:
\(\text{Area}(R) = \text{Area}_{\text{half}} - \text{Area}_{\text{subtracted}} = 48 - 17.591 = 30.409\text{ cm}^2\).
To 3 significant figures, this is \(30.4\text{ cm}^2\).

- **M1**: For identifying the dividing line at a distance of \(4\text{ cm}\) from \(AB\) and calculating the area of the upper half of the rectangle as \(12 \times 4 = 48\text{ cm}^2\).
- **M1**: For using Pythagoras' theorem to find the horizontal distance of the intersection point of the circle with the dividing line: \(x^2 + (-4)^2 = 5^2 \implies x = 3\).
- **M1**: For partitioning the subtracted area into a triangle of area \(\frac{1}{2} \times 3 \times 4 = 6\text{ cm}^2\) and a circular sector.
- **M1**: For calculating the angle of the sector as \(\theta = \arccos(0.6) \approx 0.927\text{ rad}\) (or \(53.1^\circ\)) and finding the sector area as \(\approx 11.6\text{ cm}^2\).
- **M1**: For summing the subtraction areas to get \(\approx 17.6\text{ cm}^2\) and subtracting from \(48\).
- **A1**: For the correct final area of \(30.4\text{ cm}^2\) (accept answers in the range \(30.4\) to \(30.5\)).

(a) Express the distances \(AP\) and \(BP\) using the distance formula:
\(AP = \sqrt{(x - 1)^2 + (y - 2)^2}\)
\(BP = \sqrt{(x - 4)^2 + (y - 5)^2}\)

Given \(AP = \sqrt{2}BP\), square both sides to obtain:
\(AP^2 = 2BP^2\)

Substitute the coordinate expressions:
\((x - 1)^2 + (y - 2)^2 = 2\left[(x - 4)^2 + (y - 5)^2\right]\)

Expand the squares on both sides:
\(x^2 - 2x + 1 + y^2 - 4y + 4 = 2\left[x^2 - 8x + 16 + y^2 - 10y + 25\right]\)
\(x^2 + y^2 - 2x - 4y + 5 = 2\left[x^2 + y^2 - 16x - 20y + 41\right]\)
\(x^2 + y^2 - 2x - 4y + 5 = 2x^2 + 2y^2 - 32x - 40y + 82\)

Rearrange all terms to the right side to set the equation to 0:
\(0 = (2x^2 - x^2) + (2y^2 - y^2) - 32x + 2x - 40y + 4y + 82 - 5\)
\(x^2 + y^2 - 14x - 16y + 77 = 0\) (as required).

(b) Complete the square for both the \(x\) and \(y\) terms of the equation:
\(x^2 - 14x = (x - 7)^2 - 49\)
\(y^2 - 16y = (y - 8)^2 - 64\)

Substitute these back into the circle equation:
\((x - 7)^2 - 49 + (y - 8)^2 - 64 + 77 = 0\)
\((x - 7)^2 + (y - 8)^2 - 36 = 0\)
\((x - 7)^2 + (y - 8)^2 = 36\)

From the standard form \((x - h)^2 + (y - k)^2 = r^2\):
- The coordinates of the centre are \((7, 8)\).
- The radius is \(\sqrt{36} = 6\).

**Part (a) [3.5 Marks]**:
- **M1**: For expressing \(AP^2\) and \(BP^2\) in terms of coordinates and stating \(AP^2 = 2BP^2\).
- **M1**: For substituting the terms correctly to obtain \((x - 1)^2 + (y - 2)^2 = 2[(x - 4)^2 + (y - 5)^2]\).
- **M1**: For expanding both sides correctly to get \(x^2 + y^2 - 2x - 4y + 5 = 2x^2 + 2y^2 - 32x - 40y + 82\) (allow up to one sign/expansion error).
- **A1**: For fully correct algebraic rearrangement to show the given equation \(x^2 + y^2 - 14x - 16y + 77 = 0\).

**Part (b) [2 Marks]**:
- **M1**: For attempting to complete the square on the given equation, resulting in \((x - 7)^2 + (y - 8)^2 = C\) where \(C\) is a constant.
- **A1**: For stating the correct centre of \((7, 8)\) and the correct radius of \(6\) (accept \(\sqrt{36}\)).

For (a): Since \(f^{-1}(5) = 3\), we have \(f(3) = 5\). Substituting \(x = 3\) into the expression for \(f(x)\): \(\frac{3(3) + a}{3 - 1} = 5 \implies \frac{9 + a}{2} = 5 \implies 9 + a = 10 \implies a = 1\).

For (b): Using \(a = 1\), the function is \(f(x) = \frac{3x + 1}{x - 1}\). Substituting \(x = 2\): \(f(2) = \frac{3(2) + 1}{2 - 1} = 7\). Since \(g(f(2)) = 16\), we have \(g(7) = 16\). Substituting into \(g(x)\): \(7^2 - 2b(7) + b^2 = 16 \implies 49 - 14b + b^2 = 16 \implies b^2 - 14b + 33 = 0\). Factoring the quadratic: \((b - 3)(b - 11) = 0\), which gives \(b = 3\) or \(b = 11\).

For (c): The larger value of \(b\) is 11. Substituting this into \(g(x)\): \(g(x) = x^2 - 22x + 121\). The equation \(g(x) = k\) becomes \(x^2 - 22x + 121 - k = 0\). For this quadratic equation to have no real roots, its discriminant \(\Delta\) must be less than zero: \(\Delta = (-22)^2 - 4(1)(121 - k) < 0 \implies 484 - 484 + 4k < 0 \implies 4k < 0 \implies k < 0\).

(a) M1 for using the relationship \(f^{-1}(5) = 3 \implies f(3) = 5\). M1 for setting up the equation \(\frac{9+a}{2} = 5\). A1 for \(a = 1\).
(b) M1 for finding \(f(2) = 7\). M1 for substituting \(f(2)\) into \(g(x)\) and setting equal to 16. M1 for setting up the quadratic equation \(b^2 - 14b + 33 = 0\). A1 for finding both values \(b = 3\) and \(b = 11\).
(c) M1 for setting up the discriminant \(\Delta < 0\) for \(x^2 - 22x + 121 - k = 0\). A1 for obtaining \(4k < 0\). A0.5 for the final inequality \(k < 0\).

For (a): The derivative of the curve is \(\frac{dy}{dx} = 4x + p\). Since the line \(y = 7x - 9\) is a tangent to the curve at \(P(2, 5)\), the gradient of the curve at \(x = 2\) is equal to the gradient of the line, which is 7. Thus, \(4(2) + p = 7 \implies 8 + p = 7 \implies p = -1\). Since \(P(2, 5)\) lies on the curve, we substitute \(x = 2\) and \(y = 5\) into \(y = 2x^2 + px + q\): \(5 = 2(2)^2 + (-1)(2) + q \implies 5 = 8 - 2 + q \implies 5 = 6 + q \implies q = -1\).

For (b): The gradient of the tangent at \(P\) is 7, so the gradient of the normal at \(P\) is \(-\frac{1}{7}\). The equation of the normal is: \(y - 5 = -\frac{1}{7}(x - 2) \implies 7(y - 5) = -(x - 2) \implies 7y - 35 = -x + 2 \implies x + 7y - 37 = 0\).

For (c): To find the other intersection point \(Q\), we set the equation of the normal equal to the curve: \(2x^2 - x - 1 = -\frac{1}{7}x + \frac{37}{7}\). Multiplying the entire equation by 7 gives: \(14x^2 - 7x - 7 = -x + 37 \implies 14x^2 - 6x - 44 = 0 \implies 7x^2 - 3x - 22 = 0\). Since \(x = 2\) is a known root, we can factor the quadratic as: \((x - 2)(7x + 11) = 0\). This gives the other root \(x = -\frac{11}{7}\). Substituting \(x = -\frac{11}{7}\) into the normal equation: \(y = -\frac{1}{7}\left(-\frac{11}{7}\right) + \frac{37}{7} = \frac{11}{49} + \frac{259}{49} = \frac{270}{49}\). Thus, the coordinates of \(Q\) are \(\left(-\frac{11}{7}, \frac{270}{49}\right)\).

(a) M1 for finding \(\frac{dy}{dx} = 4x + p\). M1 for setting \(4(2) + p = 7\) and solving for \(p\). A1 for \(p = -1\). A1 for substituting \(P(2, 5)\) and finding \(q = -1\).
(b) M1 for finding the gradient of the normal as \(-\frac{1}{7}\). M1 for utilizing the point-slope form \(y - y_1 = m(x - x_1)\). A0.5 for the form \(x + 7y - 37 = 0\).
(c) M1 for setting up the equation \(2x^2 - x - 1 = -\frac{1}{7}x + \frac{37}{7}\). M1 for simplifying to a quadratic equation, e.g., \(7x^2 - 3x - 22 = 0\). A1 for finding the x-coordinate \(x = -11/7\). A1 for finding the y-coordinate \(y = 270/49\).

For (a): Let \(y = \frac{4x + 24}{x + 1}\). Rearranging to make \(x\) the subject: \(y(x + 1) = 4x + 24 \implies yx + y = 4x + 24 \implies yx - 4x = 24 - y \implies x(y - 4) = 24 - y \implies x = \frac{24 - y}{y - 4}\). Thus, the inverse function is \(f^{-1}(x) = \frac{24 - x}{x - 4}\).

For (b): First, find the expression for the composite function \(fg(x)\): \(fg(x) = f(2x - 3) = \frac{4(2x - 3) + 24}{(2x - 3) + 1} = \frac{8x - 12 + 24}{2x - 2} = \frac{8x + 12}{2x - 2} = \frac{4x + 6}{x - 1}\). We set \(fg(x) = x\): \(\frac{4x + 6}{x - 1} = x \implies 4x + 6 = x(x - 1) \implies 4x + 6 = x^2 - x \implies x^2 - 5x - 6 = 0\). Factoring the quadratic: \((x - 6)(x + 1) = 0\). This gives \(x = 6\) or \(x = -1\).

For (c): The function \(f(g(x))\) is undefined when the denominator of \(f(g(x))\) is zero. Since \(f(u)\) requires \(u \neq -1\), we must exclude values of \(x\) where \(g(x) = -1\). Setting \(g(x) = -1 \implies 2x - 3 = -1 \implies 2x = 2 \implies x = 1\). Thus, \(x = 1\) must be excluded from the domain of \(fg\).

(a) M1 for writing \(y = \frac{4x+24}{x+1}\) and expanding to \(yx + y = 4x + 24\). M1 for collecting terms in \(x\) and factoring. A1 for \(f^{-1}(x) = \frac{24 - x}{x - 4}\).
(b) M1 for substituting \(g(x)\) into \(f(x)\) and simplifying to \(fg(x) = \frac{4x+6}{x-1}\). M1 for setting \(fg(x) = x\) and forming a quadratic equation. M1 for solving the quadratic equation \(x^2 - 5x - 6 = 0\). A1 for \(x = 6\) and \(x = -1\).
(c) M1 for recognizing that \(g(x) \neq -1\) or analyzing the denominator of the simplified composite function. A1.5 for \(x = 1\).

For (a): To find the intersection points, set the equations equal: \(2x^2 - 3x + 4 = kx + 2 \implies 2x^2 - (k + 3)x + 2 = 0\). For the line to be a tangent to the curve, this quadratic equation must have exactly one real root, meaning its discriminant \(\Delta\) must equal zero: \(\Delta = [-(k + 3)]^2 - 4(2)(2) = 0 \implies (k + 3)^2 - 16 = 0 \implies (k + 3)^2 = 16\). Taking the square root of both sides: \(k + 3 = \pm 4\). Thus, \(k = 4 - 3 = 1\) or \(k = -4 - 3 = -7\).

For (b): When \(k = 1\), the quadratic equation is: \(2x^2 - 4x + 2 = 0 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1\). Substituting \(x = 1\) into the tangent line \(y = (1)x + 2\) gives \(y = 3\). So, the first point of contact is \((1, 3)\). When \(k = -7\), the quadratic equation is: \(2x^2 + 4x + 2 = 0 \implies x^2 + 2x + 1 = 0 \implies (x + 1)^2 = 0 \implies x = -1\). Substituting \(x = -1\) into the tangent line \(y = -7x + 2\) gives \(y = 9\). So, the second point of contact is \((-1, 9)\).

For (c): The two tangent lines are \(y = x + 2\) and \(y = -7x + 2\). At their intersection: \(x + 2 = -7x + 2 \implies 8x = 0 \implies x = 0\). Substituting \(x = 0\) back into either line equation gives \(y = 2\). Therefore, the intersection point of the two tangents is \((0, 2)\).

(a) M1 for equating the curve and line to get \(2x^2 - (k+3)x + 2 = 0\). M1 for writing down the expression for the discriminant \(\Delta = (k+3)^2 - 16\). M1 for setting \(\Delta = 0\) and solving. A1 for showing clearly that \(k = 1\) and \(k = -7\).
(b) M1 for substituting \(k = 1\) and finding \(x = 1\). A1 for point \((1, 3)\). M1 for substituting \(k = -7\) and finding \(x = -1\). A0.5 for point \((-1, 9)\).
(c) M1 for equating the two tangent lines. A1 for the coordinates of the intersection point \((0, 2)\).

For (a): We are given \(u = 2^x\) and \(v = 3^y\). The first equation is \(2^x + 3^y = 17\), which directly becomes \(u + v = 17\). The second equation is \(2^{x-1} + 3^{y+1} = 31\). Using the laws of indices: \(2^{x-1} = \frac{2^x}{2} = \frac{u}{2}\), and \(3^{y+1} = 3 \cdot 3^y = 3v\). Substituting these into the second equation: \(\frac{u}{2} + 3v = 31\). Multiplying the entire equation by 2 gives: \(u + 6v = 62\). (Showed)

For (b): We have the system:
1) \(u + v = 17\)
2) \(u + 6v = 62\)
Subtracting equation 1 from equation 2: \((u + 6v) - (u + v) = 62 - 17 \implies 5v = 45 \implies v = 9\). Substituting \(v = 9\) back into equation 1: \(u + 9 = 17 \implies u = 8\). Thus, the solution is \(u = 8\) and \(v = 9\).

For (c): From our substitutions: \(2^x = u \implies 2^x = 8 \implies 2^x = 2^3 \implies x = 3\). And \(3^y = v \implies 3^y = 9 \implies 3^y = 3^2 \implies y = 2\). Therefore, the solutions are \(x = 3\) and \(y = 2\).

(a) M1 for substituting \(u = 2^x\) and \(v = 3^y\) into the first equation to get \(u+v=17\). M1 for expressing \(2^{x-1}\) as \(u/2\) or \(3^{y+1}\) as \(3v\). A1 for showing the final form \(u + 6v = 62\) clearly.
(b) M1 for a valid method to eliminate one variable (substitution or elimination). A1 for \(v = 9\). A1.5 for \(u = 8\).
(c) M1 for setting up \(2^x = 8\) and \(3^y = 9\). A1 for \(x = 3\). A1 for \(y = 2\).

For (a): To find where the graph crosses the x-axis, set \(y = 0\): \(\frac{2x - 5}{x + 3} = 0 \implies 2x - 5 = 0 \implies x = 2.5\). So the point is \((2.5, 0)\). To find where it crosses the y-axis, set \(x = 0\): \(y = \frac{2(0) - 5}{0 + 3} = -\frac{5}{3}\). So the point is \(\left(0, -\frac{5}{3}\right)\).

For (b): Let \(y = \frac{2x - 5}{x + 3}\). Rearranging to make \(x\) the subject: \(y(x + 3) = 2x - 5 \implies yx + 3y = 2x - 5 \implies yx - 2x = -3y - 5 \implies x(y - 2) = -(3y + 5) \implies x = \frac{3y + 5}{2 - y}\). Thus, \(f^{-1}(x) = \frac{3x + 5}{2 - x}\).

For (c): To solve \(\frac{2x - 5}{x + 3} > 3\), we can multiply both sides of the inequality by the positive quantity \((x + 3)^2\) (since \(x \neq -3\)): \((2x - 5)(x + 3) > 3(x + 3)^2 \implies (x + 3)[(2x - 5) - 3(x + 3)] > 0 \implies (x + 3)[2x - 5 - 3x - 9] > 0 \implies (x + 3)(-x - 14) > 0 \implies -(x + 3)(x + 14) > 0 \implies (x + 3)(x + 14) < 0\). The roots of the expression are \(x = -3\) and \(x = -14\). For the product to be less than zero, the variable \(x\) must lie between these roots: \(-14 < x < -3\).

(a) M1 for setting \(y = 0\) and solving for \(x\) or setting \(x = 0\) and solving for \(y\). A1 for \((2.5, 0)\). A0.5 for \((0, -5/3)\).
(b) M1 for writing \(y(x+3) = 2x-5\) and expanding. M1 for grouping \(x\) terms together and factoring. A1 for \(f^{-1}(x) = \frac{3x + 5}{2 - x}\).
(c) M1 for multiplying by \((x+3)^2\) to maintain inequality sign or considering two cases based on the sign of \(x+3\). M1 for simplifying to a quadratic inequality, e.g., \((x+3)(x+14) < 0\). A1 for identifying the key boundary values \(-14\) and \(-3\). A1 for the correct interval \(-14 < x < -3\).

(a) Let \(y = \frac{2x+3}{x-1}\). Then \(y(x-1) = 2x+3 \Rightarrow xy - y = 2x+3 \Rightarrow xy - 2x = y+3 \Rightarrow x(y-2) = y+3 \Rightarrow x = \frac{y+3}{y-2}\). Thus, \(f^{-1}(x) = \frac{x+3}{x-2}\). (b) We want to solve \(g(f(x)) = 4\). Let \(y = f(x)\), so \(g(y) = 4 \Rightarrow y^2 - 3y = 4 \Rightarrow y^2 - 3y - 4 = 0 \Rightarrow (y-4)(y+1) = 0\), which gives \(y = 4\) or \(y = -1\). If \(f(x) = 4 \Rightarrow \frac{2x+3}{x-1} = 4 \Rightarrow 2x+3 = 4x-4 \Rightarrow 2x = 7 \Rightarrow x = 3.5\). If \(f(x) = -1 \Rightarrow \frac{2x+3}{x-1} = -1 \Rightarrow 2x+3 = -x+1 \Rightarrow 3x = -2 \Rightarrow x = -\frac{2}{3}\). (c) \(fg(x) = f(g(x)) = f(x^2 - 3x) = \frac{2(x^2-3x)+3}{(x^2-3x)-1} = \frac{2x^2 - 6x + 3}{x^2 - 3x - 1}\). (d) For the composite function \(fg\) to be defined, the output of \(g(x)\) must be in the domain of \(f(x)\). Since the domain of \(f\) excludes \(1\), we must exclude values where \(g(x) = 1 \Rightarrow x^2 - 3x = 1 \Rightarrow x^2 - 3x - 1 = 0\). Using the quadratic formula: \(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}\).

(a) M1: Setting up equation \(y = \frac{2x+3}{x-1}\) and multiplying by \(x-1\). M1: Rearranging to isolate \(x\) on one side. A1: Correct expression for \(f^{-1}(x)\). (b) M1: Setting up \(y^2-3y-4=0\) and solving for \(y\). A1: Obtaining \(y = 4\) and \(y = -1\). M1: Setting up \(f(x) = 4\) and solving for \(x\). M1: Setting up \(f(x) = -1\) and solving for \(x\). A1: Both correct values \(x = 3.5\) and \(x = -\frac{2}{3}\). (c) M1: Substituting \(g(x)\) into \(f(x)\). A1: Simplifying to the required expression. (d) M1: Setting the denominator of \(fg(x)\) to zero: \(x^2 - 3x - 1 = 0\). M1: Using the quadratic formula correctly. A1: Obtaining \(x = \frac{3 \pm \sqrt{13}}{2}\).

(a) Since \(OP : PA = 2 : 1\), \(\vec{OP} = \frac{2}{3}\mathbf{a}\). Therefore, \(\vec{BP} = \vec{OP} - \vec{OB} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). Since \(Q\) is the midpoint of \(AB\), \(\vec{OQ} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\). (b) Since \(X\) lies on \(OQ\), \(\vec{OX} = \mu \vec{OQ} = \frac{1}{2}\mu \mathbf{a} + \frac{1}{2}\mu \mathbf{b}\). Since \(X\) lies on \(BP\), \(\vec{OX} = \vec{OB} + \lambda \vec{BP} = \mathbf{b} + \lambda(\frac{2}{3}\mathbf{a} - \mathbf{b}) = \frac{2}{3}\lambda \mathbf{a} + (1-\lambda)\mathbf{b}\). Equating components of \(\mathbf{a}\) and \(\mathbf{b}\): (1) \(\frac{1}{2}\mu = \frac{2}{3}\lambda \Rightarrow \mu = \frac{4}{3}\lambda\). (2) \(\frac{1}{2}\mu = 1-\lambda\). Substituting (1) into (2): \(\frac{2}{3}\lambda = 1-\lambda \Rightarrow \frac{5}{3}\lambda = 1 \Rightarrow \lambda = \frac{3}{5}\). Then \mu = \frac{4}{5}\). Thus, \(\vec{OX} = \frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\). (c) Since \(A, X, R\) are collinear, \(\vec{AX} = k \vec{AR}\). \(\vec{AX} = \vec{OX} - \vec{OA} = \frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b} - \mathbf{a} = -\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\). Since \(R\) lies on \(OB\), \(\vec{OR} = c \mathbf{b}\) for some scalar \(c\). Then \(\vec{AR} = \vec{OR} - \vec{OA} = c\mathbf{b} - \mathbf{a}\). Substituting these into \(\vec{AX} = k \vec{AR}\): \(-\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b} = k(c\mathbf{b} - \mathbf{a}) = -k\mathbf{a} + kc\mathbf{b}\). Equating coefficients of \(\mathbf{a}\): \(k = \frac{3}{5}\). Equating coefficients of \(\mathbf{b}\): \(kc = \frac{2}{5} \Rightarrow \frac{3}{5}c = \frac{2}{5} \Rightarrow c = \frac{2}{3}\). Thus, \(\vec{OR} = \frac{2}{3}\mathbf{b}\). This means \(R\) divides \(OB\) in the ratio \(2 : 1\), so \(OR : RB = 2 : 1\).

(a) B1: Correct vector for \(\vec{BP}\). B1: Correct vector for \(\vec{OQ}\). (b) M1: Writing \(\vec{OX}\) as a scalar multiple of \(\vec{OQ}\). M1: Writing \(\vec{OX}\) using a vector path along \(BP\). M1: Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form two simultaneous equations. M1: Solving the simultaneous equations for the parameters. A1: Correct values for parameters (e.g., \(\lambda = 0.6\), \(\mu = 0.8\)). A1: Correct final vector expression for \(\vec{OX}\). (c) M1: Writing an expression for \(\vec{AX}\). M1: Writing \(\vec{OR} = c\mathbf{b}\) and forming an expression for \(\vec{AR}\). M1: Using collinearity to set up \(\vec{AX} = k \vec{AR}\). A1: Finding \(c = \frac{2}{3}\). A1: Correct ratio \(OR : RB = 2 : 1\).

(a) Using Pythagoras' theorem on the rectangular base \(ABCD\): \(AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}\). (b) The center \(O\) of the base is the midpoint of \(AC\), so \(AO = \frac{13}{2} = 6.5\text{ cm}\). In the right-angled triangle \(VOA\), the angle \(\angle VAO = 60^\circ\). Thus, \(\tan(60^\circ) = \frac{VO}{AO} \Rightarrow VO = 6.5 \times \tan(60^\circ) = 6.5 \sqrt{3} \approx 11.2583... \approx 11.3\text{ cm}\) (to 3 s.f.). (c) Let \(N\) be the midpoint of \(BC\). The line \(ON\) is parallel to \(AB\) and its length is \(ON = \frac{1}{2} AB = 6\text{ cm}\). The angle between the face \(VBC\) and the base \(ABCD\) is \(\angle VNO\). In the right-angled triangle \(VON\): \(\tan(\angle VNO) = \frac{VO}{ON} = \frac{6.5\sqrt{3}}{6} \approx 1.87639...\). Therefore, \(\angle VNO = \arctan(1.87639) \approx 61.947^\circ \approx 61.9^\circ\). (d) Since the pyramid is symmetrical, the slant edges \(VB\) and \(VC\) have the same length as \(VA\). In the right-angled triangle \(VOA\), \(VA = \frac{AO}{\cos(60^\circ)} = \frac{6.5}{0.5} = 13\text{ cm}\). Thus, \(VB = VC = 13\text{ cm}\). In the isosceles triangle \(VBC\), let \(N\) be the midpoint of \(BC\). Then \(VN\) is perpendicular to \(BC\) and bisects \(\angle BVC\). \(BN = 2.5\text{ cm}\). In the right-angled triangle \(VNB\): \(\sin(\angle BVN) = \frac{BN}{VB} = \frac{2.5}{13}\). Therefore, \(\angle BVN = \arcsin\left(\frac{2.5}{13}\right) \approx 11.086^\circ\). Hence, \(\angle BVC = 2 \times \angle BVN = 2 \times 11.086^\circ \approx 22.17^\circ \approx 22.2^\circ\).

(a) M1: Using Pythagoras' theorem for \(AC = \sqrt{12^2 + 5^2}\). A1: Correct answer of \(13\). (b) M1: Finding \(AO = 6.5\). M1: Using tangent ratio \(\tan(60^\circ) = \frac{VO}{6.5}\). A1: Show key steps to arrive at \(11.3\) (e.g., \(6.5\sqrt{3} = 11.258...\)). (c) M1: Identifying the angle \(\angle VNO\) and noting \(ON = 6\). M1: Using \(\tan(\angle VNO) = \frac{VO}{ON}\) with their \(VO\). A1: Correct evaluation of ratio. A1: Correct angle \(61.9^\circ\) (accept \(61.9\)). (d) M1: Calculating slant edge length \(VB = 13\). M1: Using half-angle trigonometry in triangle \(VBC\) or the cosine rule on triangle \(VBC\). M1: Setting up correct equation, e.g., \(\sin(\theta/2) = 2.5 / 13\) or \(5^2 = 13^2 + 13^2 - 2(13)(13)\cos(\theta)\). A1: Correct angle of \(22.2^\circ\) (accept \(22.2\)).