2026 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Simplify each surd into multiples of \(\sqrt{3}\):
\(\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\)
\(2\sqrt{27} = 2\sqrt{9 \times 3} = 2 \times 3\sqrt{3} = 6\sqrt{3}\)
\(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\)

Combine the terms:
\(4\sqrt{3} - 6\sqrt{3} + 5\sqrt{3} = (4 - 6 + 5)\sqrt{3} = 3\sqrt{3}\)

M1 for at least two of \(4\sqrt{3}\), \(6\sqrt{3}\), \(5\sqrt{3}\) seen (or simplifying \(\sqrt{48}=4\sqrt{3}\), \(\sqrt{27}=3\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\))
A1 for \(3\sqrt{3}\)

1. Find the arc length:
\(\text{Arc length} = \frac{80}{360} \times 2 \times \pi \times 9 = \frac{2}{9} \times 18\pi = 4\pi\text{ cm}\)

2. Find the total perimeter by adding the two radii:
\(\text{Perimeter} = \text{arc length} + 2r = 4\pi + 2(9) = 18 + 4\pi\text{ cm}\)

M1 for \(\frac{80}{360} \times 2 \times \pi \times 9\) oe
A1 for arc length \(= 4\pi\)
A1 for \(18 + 4\pi\) or \(4\pi + 18\)

Convert both mixed numbers to improper fractions:
\(1\frac{5}{7} = \frac{12}{7}\)
\(2\frac{2}{5} = \frac{12}{5}\)

Divide by multiplying by the reciprocal:
\(\frac{12}{7} \div \frac{12}{5} = \frac{12}{7} \times \frac{5}{12} = \frac{5}{7}\)

M1 for converting to improper fractions \(\frac{12}{7}\) and \(\frac{12}{5}\) and writing \(\frac{12}{7} \times \frac{5}{12}\)
A1 for \(\frac{5}{7}\) cao

Find the exterior angle of the regular polygon:
\(\text{Exterior angle} = 180^\circ - 156^\circ = 24^\circ\)

The sum of the exterior angles of any convex polygon is \(360^\circ\):
\(n = \frac{360^\circ}{24^\circ} = 15\)

M1 for \(180 - 156 (= 24)\) or \(\frac{(n-2) \times 180}{n} = 156\) oe
A1 for \(15\)

Apply index laws:
\(\left(\frac{125}{64}\right)^{-\frac{2}{3}} = \left(\frac{64}{125}\right)^{\frac{2}{3}}\)

Take the cube root and then square:
\(\sqrt[3]{\frac{64}{125}} = \frac{4}{5}\)
\(\left(\frac{4}{5}\right)^2 = \frac{16}{25}\)

M1 for dealing with negative power or cube root, e.g. \(\left(\frac{64}{125}\right)^{\frac{2}{3}}\) or \(\left(\frac{5}{4}\right)^{-2}\) or \(\frac{4}{5}\) seen
A1 for \(\frac{16}{25}\)

The area of a trapezium is given by \(A = \frac{1}{2}(a + b)h\).

Substitute the given expressions:
\(51 = \frac{1}{2} \big((x + 3) + (2x - 1)\big) \times 6\)
\(51 = 3(3x + 2)\)
\(51 = 9x + 6\)
\(9x = 45\)
\(x = 5\)

M1 for \(\frac{1}{2}\big((x+3) + (2x-1)\big) \times 6 = 51\) oe
M1 for simplifying correctly to \(9x + 6 = 51\) or \(3x + 2 = 17\)
A1 for \(5\)

1. In triangle \(ABX\), angle \(AXB\) and angle \(BXC\) lie on a straight line:
\(\text{Angle } AXB = 180^\circ - 82^\circ = 98^\circ\)

2. Angles in triangle \(ABX\) sum to \(180^\circ\):
\(\text{Angle } ABD = 180^\circ - (98^\circ + 38^\circ) = 180^\circ - 136^\circ = 44^\circ\)

3. Angles in the same segment subtended by arc \(AD\) are equal:
\(\text{Angle } ACD = \text{angle } ABD = 44^\circ\)

M1 for angle \(AXB = 180 - 82 (= 98^\circ)\) or using exterior angle \(82^\circ = 38^\circ + \text{angle } ABD\)
M1 for angle \(ABD = 44^\circ\)
A1 for \(44^\circ\)

Divide the numerical coefficients and powers of 10 separately:
\(\frac{4.8 \times 10^7}{8 \times 10^{-3}} = \left(\frac{4.8}{8}\right) \times 10^{7 - (-3)}\)
\(= 0.6 \times 10^{10}\)

Convert to standard form \(A \times 10^n\) where \(1 \le A < 10\):
\(0.6 \times 10^{10} = 6 \times 10^9\)

M1 for \(0.6 \times 10^{10}\) oe or \(\frac{4.8}{8} \times 10^{10}\)
A1 for \(6 \times 10^9\)

1. Evaluate the expression inside the brackets:
\[ \frac{5}{6} - \frac{1}{3} = \frac{5}{6} - \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]

2. Convert the mixed number to an improper fraction:
\[ 2\frac{1}{4} = \frac{9}{4} \]

3. Divide the fractions by multiplying by the reciprocal:
\[ \frac{9}{4} \div \frac{1}{2} = \frac{9}{4} \times \frac{2}{1} = \frac{18}{4} = \frac{9}{2} \]

4. Convert the result to a mixed number:
\[ \frac{9}{2} = 4\frac{1}{2} \]

M1 for a correct method to subtract the fractions inside the bracket, e.g. \( \frac{5}{6} - \frac{2}{6} = \frac{3}{6} \) oe, or for converting \( 2\frac{1}{4} \) to \( \frac{9}{4} \)
M1 for multiplying by the reciprocal of their bracket, e.g. \( \frac{9}{4} \times \frac{2}{1} \) or \( \frac{9}{4} \times \frac{6}{3} \) oe
A1 for \( 4\frac{1}{2} \) (cao)

1. The sum of the exterior angles of any convex polygon is \( 360^\circ \).

2. Set up the equation:
\[ x + 2x + (x + 20) + (2x - 10) + 50 = 360 \]

3. Simplify the left-hand side:
\[ 6x + 60 = 360 \]

4. Solve for \( x \):
\[ 6x = 300 \]
\[ x = 50 \]

M1 for equating the sum of the five given exterior angles to 360: \( x + 2x + (x + 20) + (2x - 10) + 50 = 360 \) oe
M1 for simplifying to the form \( 6x + 60 = 360 \) or \( 6x = 300 \)
A1 for 50

1. Factorise the numerator by grouping or inspection:
$$2x^2 - 5x - 12 = 2x^2 - 8x + 3x - 12 = 2x(x - 4) + 3(x - 4) = (2x + 3)(x - 4)$$

2. Factorise the denominator using the difference of two squares:
$$4x^2 - 9 = (2x - 3)(2x + 3)$$

3. Cancel the common factor of \((2x + 3)\):
$$\frac{(2x + 3)(x - 4)}{(2x - 3)(2x + 3)} = \frac{x - 4}{2x - 3}$$

M2 for \((2x + 3)(x - 4)\) (or M1 for \((2x + a)(x + b)\) where \(ab = -12\) or \(2b + a = -5\))
M1 for \((2x - 3)(2x + 3)\)
A1 for \(\frac{x - 4}{2x - 3}\)

1. Find the midpoint of \(AB\):
$$\text{Midpoint} = \left(\frac{1 + 7}{2}, \frac{-3 + 5}{2}\right) = (4, 1)$$

2. Calculate the gradient of the line segment \(AB\):
$$m_{AB} = \frac{5 - (-3)}{7 - 1} = \frac{8}{6} = \frac{4}{3}$$

3. Determine the gradient of the perpendicular bisector:
$$m_{\perp} = -\frac{1}{m_{AB}} = -\frac{3}{4}$$

4. Find the equation using the point \((4, 1)\):
$$y - y_1 = m(x - x_1) \implies y - 1 = -\frac{3}{4}(x - 4)$$
$$y - 1 = -\frac{3}{4}x + 3 \implies y = -\frac{3}{4}x + 4$$

B1 for midpoint \((4, 1)\)
M1 for gradient of \(AB = \frac{8}{6} = \frac{4}{3}\)
M1 for perpendicular gradient \(-\frac{3}{4}\) (FT their gradient)
A1 for \(y = -\frac{3}{4}x + 4\) or \(y = -0.75x + 4\)

1. Use the sine rule:
$$\frac{AC}{\sin(\angle ABC)} = \frac{AB}{\sin(\angle ACB)}$$
$$\frac{AC}{\sin 45^\circ} = \frac{6}{\sin 30^\circ}$$

2. Substitute the exact trigonometric values:
$$\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 30^\circ = \frac{1}{2}$$

3. Solve for \(AC\):
$$AC = \frac{6 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 6\sqrt{2}\text{ cm}$$

M1 for applying sine rule correctly: \(\frac{AC}{\sin 45^\circ} = \frac{6}{\sin 30^\circ}\)
B1 for both exact values stated: \(\sin 45^\circ = \frac{\sqrt{2}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\)
A1 for \(6\sqrt{2}\)

(a) Completing the square:
$$x^2 - 8x + 11 = (x - 4)^2 - 4^2 + 11 = (x - 4)^2 - 16 + 11 = (x - 4)^2 - 5$$

(b) Using the completed square form:
$$(x - 4)^2 - 5 = 0$$
$$(x - 4)^2 = 5$$
$$x - 4 = \pm\sqrt{5}$$
$$x = 4 \pm \sqrt{5}$$

(a) M1 for \((x - 4)^2 + k\)
A1 for \((x - 4)^2 - 5\)
(b) M1 for \((x - 4)^2 = 5\) or \(x - 4 = \pm\sqrt{5}\) (FT their \(p\) and \(q\))
A1 for \(x = 4 \pm \sqrt{5}\)

(a) Let \(y = 5 - 2x\):
$$2x = 5 - y \implies x = \frac{5 - y}{2}$$
$$f^{-1}(x) = \frac{5 - x}{2}$$

(b) First evaluate \(f(2)\):
$$f(2) = 5 - 2(2) = 1$$
Then evaluate \(g(1)\):
$$g(1) = \frac{3(1) + 1}{4} = \frac{4}{4} = 1$$

(c) Set \(g(x) = f(x)\):
$$\frac{3x + 1}{4} = 5 - 2x$$
$$3x + 1 = 4(5 - 2x)$$
$$3x + 1 = 20 - 8x$$
$$11x = 19 \implies x = \frac{19}{11}$$

(a) B1 for \(f^{-1}(x) = \frac{5 - x}{2}\) or \(2.5 - 0.5x\)
(b) B1 for \(1\)
(c) M1 for correct clearance of fraction: \(3x + 1 = 4(5 - 2x)\)
A1 for \(x = \frac{19}{11}\) or \(1\frac{8}{11}\)

1. Rearrange to find \(\sin x\):
$$2\sin x = -\sqrt{3} \implies \sin x = -\frac{\sqrt{3}}{2}$$

2. Find the reference angle:
$$\sin \alpha = \frac{\sqrt{3}}{2} \implies \alpha = 60^\circ$$

3. Since \(\sin x\) is negative, solutions lie in the 3rd and 4th quadrants:
$$\text{Quadrant III: } x = 180^\circ + 60^\circ = 240^\circ$$
$$\text{Quadrant IV: } x = 360^\circ - 60^\circ = 300^\circ$$

M1 for \(\sin x = -\frac{\sqrt{3}}{2}\)
B1 for reference angle \(60^\circ\) or one correct angle (\(240^\circ\) or \(300^\circ\))
A1 for both \(x = 240^\circ\) and \(x = 300^\circ\) and no extra solutions in range

1. Express \(y\) in terms of \(x\) from the linear equation:
$$y = x - 3$$

2. Substitute into the second equation:
$$x^2 + (x - 3)^2 = 29$$
$$x^2 + (x^2 - 6x + 9) = 29$$
$$2x^2 - 6x - 20 = 0$$

3. Simplify the quadratic equation by dividing by 2:
$$x^2 - 3x - 10 = 0$$

4. Factorise:
$$(x - 5)(x + 2) = 0$$
So \(x = 5\) or \(x = -2\).

5. Find the corresponding \(y\)-values:
When \(x = 5\), \(y = 5 - 3 = 2\).
When \(x = -2\), \(y = -2 - 3 = -5\).

M1 for correct substitution of linear expression into non-linear: \(x^2 + (x - 3)^2 = 29\) or \((y + 3)^2 + y^2 = 29\)
M1 for simplifying to a 3-term quadratic: \(2x^2 - 6x - 20 = 0\) or \(x^2 - 3x - 10 = 0\)
A1 for finding both correct values of one variable: \(x = 5, x = -2\) (or \(y = 2, y = -5\))
A1 for both complete pairs: \(x = 5, y = 2\) and \(x = -2, y = -5\)

(a) Differentiating term by term:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2(3x^2) - 9(2x) + 12(1) - 0 = 6x^2 - 18x + 12$$

(b) At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
$$6x^2 - 18x + 12 = 0$$
$$6(x^2 - 3x + 2) = 0$$
$$6(x - 1)(x - 2) = 0$$
So \(x = 1\) or \(x = 2\).

Substitute the \(x\)-values back into the curve equation:
When \(x = 1\):
$$y = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0$$
Stationary point: \((1, 0)\).

When \(x = 2\):
$$y = 2(2)^3 - 9(2)^2 + 12(2) - 5 = 16 - 36 + 24 - 5 = -1$$
Stationary point: \((2, -1)\).

(a) B1 for any two correct non-zero differentiated terms, B1 for fully correct derivative \(6x^2 - 18x + 12\)
(b) M1 for setting their derivative \(= 0\) and solving for \(x\) (obtaining \(x = 1\) and \(x = 2\))
A1 for both coordinates \((1, 0)\) and \((2, -1)\)

1. Find the midpoint, \(M\), of line segment \(AB\):
\[ M = \left(\frac{-2 + 6}{2}, \frac{7 + 1}{2}\right) = (2, 4) \]

2. Find the gradient of line \(AB\):
\[ m_{AB} = \frac{1 - 7}{6 - (-2)} = \frac{-6}{8} = -\frac{3}{4} \]

3. The perpendicular gradient \(m_{\perp}\) is the negative reciprocal of \(m_{AB}\):
\[ m_{\perp} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3} \]

4. Use the point-slope form with midpoint \((2, 4)\):
\[ y - 4 = \frac{4}{3}(x - 2) \]
\[ y = \frac{4}{3}x - \frac{8}{3} + 4 \]
\[ y = \frac{4}{3}x + \frac{4}{3} \]

M1: for finding the midpoint of \(AB\) as \((2, 4)\)
M1: for finding the gradient of \(AB\) as \(-\frac{3}{4}\) oe
M1: for using the perpendicular gradient \(m_{\perp} = \frac{4}{3}\) with their midpoint in \(y - y_1 = m(x - x_1)\) or \(y = mx + c\)
A1: for \(y = \frac{4}{3}x + \frac{4}{3}\) oe (e.g. \(y = 1\frac{1}{3}x + 1\frac{1}{3}\))

(a) Use the area formula \(\text{Area} = \frac{1}{2}ab\sin C\):
\[ \text{Area} = \frac{1}{2} \times 5 \times 8 \times \sin 60^\circ \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[ \text{Area} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2 \]

(b) Use the cosine rule \(b^2 = a^2 + c^2 - 2ac\cos B\):
\[ AC^2 = 5^2 + 8^2 - 2(5)(8)\cos 60^\circ \]
Since \(\cos 60^\circ = \frac{1}{2}\):
\[ AC^2 = 25 + 64 - 80\left(\frac{1}{2}\right) \]
\[ AC^2 = 89 - 40 = 49 \]
\[ AC = \sqrt{49} = 7\text{ cm} \]

(a)
M1: for \(\frac{1}{2} \times 5 \times 8 \times \sin 60^\circ\) or stating \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
A1: for \(10\sqrt{3}\)

(b)
M1: for \(5^2 + 8^2 - 2(5)(8)\cos 60^\circ\) with \(\cos 60^\circ = \frac{1}{2}\) substituted correctly
A1: for \(7\)

(a) \(\frac{\mathrm{d}y}{\mathrm{d}x} = x^2 - 4x + 3\).
Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(x^2 - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0\), so \(x = 1\) or \(x = 3\).
When \(x = 1\), \(y = \frac{1}{3}(1)^3 - 2(1)^2 + 3(1) + 4 = \frac{1}{3} - 2 + 3 + 4 = 5\frac{1}{3} = \frac{16}{3}\).
When \(x = 3\), \(y = \frac{1}{3}(27) - 2(9) + 3(3) + 4 = 9 - 18 + 9 + 4 = 4\).
The stationary points are \((1, \frac{16}{3})\) and \((3, 4)\).

(b) \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2x - 4\).
At \(x = 1\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(1) - 4 = -2 < 0\), hence \((1, \frac{16}{3})\) is a maximum.
At \(x = 3\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(3) - 4 = 2 > 0\), hence \((3, 4)\) is a minimum.

M1: Correct derivative \(x^2 - 4x + 3\)
M1: Equating derivative to 0 and solving to find \(x = 1, x = 3\)
A1: Correct \(y\)-values \(\frac{16}{3}\) and \(4\)
M1: Method to determine nature (e.g. correct second derivative \(2x - 4\) or sign testing of gradient)
A1: Correct identification of maximum at \((1, \frac{16}{3})\) and minimum at \((3, 4)\) with full justification

Multiply the entire equation by \((2x - 1)(x + 2)\):
\(3(x + 2) - 2(2x - 1) = 1(2x - 1)(x + 2)\)
Expand both sides:
\(3x + 6 - 4x + 2 = 2x^2 + 4x - x - 2\)
\(-x + 8 = 2x^2 + 3x - 2\)
Rearrange to form a quadratic equation:
\(2x^2 + 4x - 10 = 0\)
Divide by 2:
\(x^2 + 2x - 5 = 0\)
Apply the quadratic formula or complete the square:
\((x + 1)^2 - 1 - 5 = 0 \implies (x + 1)^2 = 6\)
\(x + 1 = \pm \sqrt{6} \implies x = -1 \pm \sqrt{6}\).

M1: Multiplies by common denominator to obtain \(3(x + 2) - 2(2x - 1) = (2x - 1)(x + 2)\)
M1: Correct expansion of all brackets
M1: Rearranges into standard quadratic form \(2x^2 + 4x - 10 = 0\) or \(x^2 + 2x - 5 = 0\)
M1: Applies quadratic formula or completes the square correctly on their 3-term quadratic
A1: Correct exact final answer \(-1 \pm \sqrt{6}\) (or \(x = -1 + \sqrt{6}\), \(x = -1 - \sqrt{6}\))

(a) \(\text{Area} = \frac{1}{2} a c \sin B = \frac{1}{2}(2\sqrt{3} - 1)(2\sqrt{3} + 1)\sin 60^\circ\).
\((2\sqrt{3} - 1)(2\sqrt{3} + 1) = (2\sqrt{3})^2 - 1^2 = 12 - 1 = 11\).
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\(\text{Area} = \frac{1}{2} \times 11 \times \frac{\sqrt{3}}{2} = \frac{11\sqrt{3}}{4}\text{ cm}^2\).

(b) Using the cosine rule:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos 60^\circ\)
\(AB^2 = (2\sqrt{3} - 1)^2 = 12 - 4\sqrt{3} + 1 = 13 - 4\sqrt{3}\)
\(BC^2 = (2\sqrt{3} + 1)^2 = 12 + 4\sqrt{3} + 1 = 13 + 4\sqrt{3}\)
\(AB^2 + BC^2 = (13 - 4\sqrt{3}) + (13 + 4\sqrt{3}) = 26\)
\(2(AB)(BC)\cos 60^\circ = 2(11)\left(\frac{1}{2}\right) = 11\)
\(AC^2 = 26 - 11 = 15\)
\(AC = \sqrt{15}\text{ cm}\).

M1 (part a): Uses \(\frac{1}{2} a c \sin 60^\circ\) with \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
A1 (part a): Expands \((2\sqrt{3}-1)(2\sqrt{3}+1) = 11\) correctly to reach \(\frac{11\sqrt{3}}{4}\)
M1 (part b): Correct substitution into cosine rule \(AC^2 = (2\sqrt{3}-1)^2 + (2\sqrt{3}+1)^2 - 2(11)\cos 60^\circ\)
M1 (part b): Correct expansions of both brackets and substitution of \(\cos 60^\circ = \frac{1}{2}\)
A1 (part b): \(\sqrt{15}\) correctly derived with all intermediate steps shown

(a) \(\vec{AB} = \mathbf{b} - \mathbf{a}\).
\(\vec{OP} = \vec{OA} + \frac{2}{5}\vec{AB} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a}) = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\).
\(\vec{OQ} = \frac{3}{5}\mathbf{b}\), so \(\vec{AQ} = \vec{AO} + \vec{OQ} = -\mathbf{a} + \frac{3}{5}\mathbf{b}\).

(b) \(\vec{OX} = \lambda \vec{OP} = \frac{3}{5}\lambda \mathbf{a} + \frac{2}{5}\lambda \mathbf{b}\).
Also, \(\vec{OX} = \vec{OA} + \vec{AX} = \mathbf{a} + \mu \vec{AQ} = \mathbf{a} + \mu(-\mathbf{a} + \frac{3}{5}\mathbf{b}) = (1 - \mu)\mathbf{a} + \frac{3}{5}\mu \mathbf{b}\).
Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
(1) \(\frac{3}{5}\lambda = 1 - \mu \implies 3\lambda + 5\mu = 5\)
(2) \(\frac{2}{5}\lambda = \frac{3}{5}\mu \implies 2\lambda = 3\mu \implies \lambda = \frac{3}{2}\mu\)
Substitute into (1):
\(3\left(\frac{3}{2}\mu\right) + 5\mu = 5 \implies \frac{9}{2}\mu + \frac{10}{2}\mu = 5 \implies \frac{19}{2}\mu = 5 \implies \mu = \frac{10}{19}\).
\(\lambda = \frac{3}{2}\left(\frac{10}{19}\right) = \frac{15}{19}\).

B1: Correct \(\vec{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\)
B1: Correct \(\vec{AQ} = -\mathbf{a} + \frac{3}{5}\mathbf{b}\)
M1: Sets up two vector paths for \(\vec{OX}\) in terms of \(\lambda\) and \(\mu\)
M1: Equates coefficients to form two simultaneous equations in \(\lambda\) and \(\mu\)
A1: Correct values \(\lambda = \frac{15}{19}\) and \(\mu = \frac{10}{19}\)

(a) Let \(y = \frac{2x + 3}{x - 4}\).
\(y(x - 4) = 2x + 3 \implies yx - 4y = 2x + 3\)
\(yx - 2x = 4y + 3 \implies x(y - 2) = 4y + 3\)
\(x = \frac{4y + 3}{y - 2}\)
So \(\mathrm{f}^{-1}(x) = \frac{4x + 3}{x - 2}\).

(b) \(\mathrm{g}(\mathrm{f}(x)) = 3\left(\frac{2x + 3}{x - 4}\right) - 1 = 8\)
\(3\left(\frac{2x + 3}{x - 4}\right) = 9 \implies \frac{2x + 3}{x - 4} = 3\)
\(2x + 3 = 3(x - 4) \implies 2x + 3 = 3x - 12\)
\(x = 15\).

(c) \(\frac{2x + 3}{x - 4} = \frac{4x + 3}{x - 2}\)
\((2x + 3)(x - 2) = (4x + 3)(x - 4)\)
\(2x^2 - x - 6 = 4x^2 - 13x - 12\)
\(2x^2 - 12x - 6 = 0 \implies x^2 - 6x - 3 = 0\)
Using the quadratic formula:
\(x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-3)}}{2} = \frac{6 \pm \sqrt{36 + 12}}{2} = \frac{6 \pm \sqrt{48}}{2} = \frac{6 \pm 4\sqrt{3}}{2} = 3 \pm 2\sqrt{3}\).

M1 (part a): Correct rearrangement steps to isolate \(x\): \(yx - 2x = 4y + 3\)
A1 (part a): \(\mathrm{f}^{-1}(x) = \frac{4x + 3}{x - 2}\)
M1 (part b): Equating \(3\mathrm{f}(x) - 1 = 8\) and solving to obtain \(x = 15\)
M1 (part c): Equates \(\mathrm{f}(x) = \mathrm{f}^{-1}(x)\) and simplifies to \(x^2 - 6x - 3 = 0\) (or \(2x^2 - 12x - 6 = 0\))
A1 (part c): Fully simplified exact answers \(3 \pm 2\sqrt{3}\)

(a) Total surface area of a cone = \(\pi r^2 + \pi r l\).
With \(l = 3r\):
\(\text{Total surface area} = \pi r^2 + \pi r(3r) = \pi r^2 + 3\pi r^2 = 4\pi r^2\).

(b) Surface area of a sphere = \(4\pi R^2\).
Equating surface areas: \(4\pi R^2 = 4\pi r^2 \implies R^2 = r^2 \implies R = r\) (since \(R, r > 0\)).

(c) Height of the cone, \(h = \sqrt{l^2 - r^2} = \sqrt{(3r)^2 - r^2} = \sqrt{9r^2 - r^2} = \sqrt{8r^2} = 2\sqrt{2}r\).
\(\text{Volume of cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2\sqrt{2}r) = \frac{2\sqrt{2}}{3}\pi r^3\).
\(\text{Volume of sphere} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi r^3\).
Ratio \(\text{Volume of cone} : \text{Volume of sphere} = \frac{2\sqrt{2}}{3}\pi r^3 : \frac{4}{3}\pi r^3 = 2\sqrt{2} : 4 = \sqrt{2} : 2\).

B1 (part a): Correct addition of base area and curved surface area: \(\pi r^2 + \pi r(3r) = 4\pi r^2\)
B1 (part b): Equating \(4\pi R^2 = 4\pi r^2\) and concluding \(R = r\)
M1 (part c): Finds vertical height of cone \(h = \sqrt{8}r = 2\sqrt{2}r\)
M1 (part c): Writes correct volume formulas: \(\frac{1}{3}\pi r^2(2\sqrt{2}r)\) and \(\frac{4}{3}\pi r^3\)
A1 (part c): Simplifies ratio to exact form \(\sqrt{2} : 2\)

(a) \(\angle OAT = 90^\circ\) and \(\angle OBT = 90^\circ\) because the angle between a tangent and a radius is \(90^\circ\).
In quadrilateral \(OATB\), the sum of angles is \(360^\circ\):
\(\angle AOB = 360^\circ - (90^\circ + 90^\circ + 44^\circ) = 180^\circ - 44^\circ = 136^\circ\).

(b) \(\angle ADB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 136^\circ = 68^\circ\) because the angle subtended at the centre is twice the angle subtended at the circumference.

(c)(i) \(\angle ABC = 90^\circ\) because the angle in a semicircle is \(90^\circ\).

(c)(ii) In the isosceles triangle \(OAB\) (where \(OA = OB\) are radii):
\(\angle OAB = \frac{180^\circ - 136^\circ}{2} = \frac{44^\circ}{2} = 22^\circ\).
Since \(O\) lies on the line segment \(AC\), \(\angle CAB = \angle OAB = 22^\circ\).

B1: \(136^\circ\) with valid reason (tangent-radius is \(90^\circ\) / angles in quadrilateral sum to \(360^\circ\))
B1: \(68^\circ\) with valid reason (angle at centre is twice angle at circumference)
B1: \(\angle ABC = 90^\circ\) (angle in semicircle is \(90^\circ\))
M1: Method to find \(\angle CAB\), e.g. using isosceles triangle \(OAB\): \(\frac{180 - 136}{2}\) or \(90 - 68\)
A1: \(\angle CAB = 22^\circ\)

(a) Midpoint of \( AB = \left(\frac{-3 + 5}{2}, \frac{5 + 1}{2}\right) = (1, 3) \).
Gradient of \( AB = \frac{1 - 5}{5 - (-3)} = \frac{-4}{8} = -\frac{1}{2} \).
Gradient of the perpendicular bisector = \(-\frac{1}{-\frac{1}{2}} = 2\).
Equation of the line passing through \((1, 3)\) with gradient 2:
\(y - 3 = 2(x - 1) \implies y - 3 = 2x - 2 \implies 2x - y + 1 = 0\).

(b) We have the system of linear equations:
(1) \(y = 2x + 1\)
(2) \(2x + y = 9\)
Substitute (1) into (2):
\(2x + (2x + 1) = 9 \implies 4x + 1 = 9 \implies 4x = 8 \implies x = 2\).
\(y = 2(2) + 1 = 5\).
Thus, \(P = (2, 5)\).

(c) Distance from origin \(O(0, 0)\) to \(P(2, 5)\):
\(OP = \sqrt{(2 - 0)^2 + (5 - 0)^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}\).

M1 (part a): Calculates midpoint \((1, 3)\) and gradient \(-\frac{1}{2}\)
M1 (part a): Uses perpendicular gradient \(2\) with midpoint in straight line equation
A1 (part a): \(2x - y + 1 = 0\) (or any integer multiple equivalent such as \(-2x + y - 1 = 0\))
M1 (part b): Solves simultaneous linear equations to find \(P(2, 5)\)
A1 (part c): Correct exact distance \(\sqrt{29}\)

Using the compound decay formula:
\(V = P\left(1 - \frac{r}{100}\right)^n\)
\(V = 18\,500 \times (1 - 0.045)^6\)
\(V = 18\,500 \times (0.955)^6\)
\(V = 18\,500 \times 0.7584168... = 14\,030.71...\)

Rounding to the nearest dollar gives $14\,031.

M1 for \(18\,500 \times (1 - 0.045)^6\) or \(18\,500 \times 0.955^6\) oe
A1 for answer in range \([14\,030.7, 14\,031]\)
B1 for rounding correctly to the nearest integer (14\,031) dep on at least M1 scored

Use the formula for the area of a non-right-angled triangle:
\(\text{Area} = \frac{1}{2}ab\sin(C)\)
\(\text{Area} = \frac{1}{2} \times 7.4 \times 9.8 \times \sin(53^\circ)\)
\(\text{Area} = 36.26 \times \sin(53^\circ) = 28.9585...\text{ cm}^2\)

Rounding to 3 significant figures gives \(29.0\text{ cm}^2\).

M2 for \(\frac{1}{2} \times 7.4 \times 9.8 \times \sin(53^\circ)\) (or M1 for \(\frac{1}{2} \times 7.4 \times 9.8 \times \sin(\dots)\))
A1 for 29.0 or 28.95 to 28.96

First, calculate the arc length of the sector:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{132}{360} \times 2 \times \pi \times 8.5 = 19.5826...\text{ cm}\)

Then, add the two radii to find the total perimeter:
\(\text{Perimeter} = \text{Arc length} + 2r\)
\(\text{Perimeter} = 19.5826... + 2 \times 8.5 = 19.5826... + 17 = 36.5826...\text{ cm}\)

Rounding to 3 significant figures gives \(36.6\text{ cm}\).

M1 for \(\frac{132}{360} \times 2 \times \pi \times 8.5\) oe (arc length \(= 19.58...\))
M1 for arc length \(+ 2 \times 8.5\)
A1 for 36.6 or 36.58 to 36.59

Find the linear scale factor:
\(k = \frac{21}{14} = 1.5\)

The volume scale factor is \(k^3 = 1.5^3 = 3.375\).

Calculate the volume of the larger vase:
\(\text{Volume} = 680 \times 3.375 = 2295\text{ cm}^3\).

M1 for linear scale factor \(\frac{21}{14}\) or \(\frac{14}{21}\) oe
M1 for volume scale factor \(\left(\frac{21}{14}\right)^3\) or \(1.5^3\) (3.375)
A1 for 2295

Rearrange the equation of line \(L\) into the form \(y = mx + c\):
\(y = \frac{5}{2}x - 4\)

So the gradient of line \(L\) is \(m_1 = \frac{5}{2} = 2.5\).

For perpendicular lines, \(m_1 \times m_2 = -1\):
\(m_2 = -\frac{1}{2.5} = -\frac{2}{5} = -0.4\).

M1 for finding the gradient of \(L\) as \(\frac{5}{2}\) or 2.5
M1 for using \(m_1 \times m_2 = -1\) or \(-\frac{1}{m}\)
A1 for \(-\frac{2}{5}\) or \(-0.4\)

Find the upper bounds for the measurements:
Upper bound for length \(= 3.4 + 0.05 = 3.45\text{ m}\)
Upper bound for width \(= 2.15 + 0.005 = 2.155\text{ m}\)

Calculate the upper bound for the area:
\(\text{Upper Bound for Area} = 3.45 \times 2.155 = 7.43475\text{ m}^2\)

Rounding to 3 significant figures gives \(7.43\text{ m}^2\).

B1 for length upper bound \(3.45\)
B1 for width upper bound \(2.155\)
M1 for multiplying their upper bound length by their upper bound width
A1 for 7.43 or exact 7.43475

Use the formula for the surface area of a sphere:
\(A = 4\pi r^2\)
\(A = 4 \times \pi \times 4.2^2\)
\(A = 4 \times \pi \times 17.64 = 70.56\pi\)
\(A = 221.6707...\text{ cm}^2\)

Rounding to 3 significant figures gives \(222\text{ cm}^2\).

M1 for substitution of \(r = 4.2\) into \(4\pi r^2\)
A1 for answer in range \([221.6, 222]\) or \(70.56\pi\)

In the right-angled triangle \(PQR\), with right angle at \(Q\):
\(\tan(\angle PRQ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{PQ}{QR}\)
\(\tan(\angle PRQ) = \frac{18.6}{12.4} = 1.5\)
\(\angle PRQ = \arctan(1.5) = 56.3099...^\circ\)

Rounding to 1 decimal place gives \(56.3^\circ\).

M1 for \(\tan(\theta) = \frac{18.6}{12.4}\) oe
M1 for \(\arctan(1.5)\) oe
A1 for 56.3 or 56.30 to 56.31

(a) Using the cosine rule on triangle \(PQR\):
\[PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)\]
\[PR^2 = 74^2 + 115^2 - 2(74)(115)\cos(64^\circ)\]
\[PR^2 = 5476 + 13225 - 17020 \times 0.438371...\]
\[PR^2 = 18701 - 7461.077...\]
\[PR^2 = 11239.92...\]
\[PR = \sqrt{11239.92...} \approx 106.018...\text{ m}\]
To 3 significant figures, \(PR = 106\text{ m}\).

(b) Area of triangle \(PQR\):
\[\text{Area} = \frac{1}{2} \times 74 \times 115 \times \sin(64^\circ) = 4255 \times 0.898794... \approx 3824.37\text{ m}^2\]
The shortest distance \(h\) is the perpendicular height from \(Q\) to \(PR\):
\[\text{Area} = \frac{1}{2} \times PR \times h\]
\[3824.37 = \frac{1}{2} \times 106.018 \times h\]
\[h = \frac{2 \times 3824.37}{106.018} \approx 72.146...\text{ m}\]
To 3 significant figures, \(h = 72.1\text{ m}\).

(a)
M1 for correct implicit cosine rule statement: \(74^2 + 115^2 - 2(74)(115)\cos(64^\circ)\)
A1 for \(11239.9...\) or \(11240\)
A1 for 106 or 106.0...

(b)
M1 for \(0.5 \times 74 \times 115 \times \sin(64^\circ)\) or \(0.5 \times 106.0... \times h = \text{Area}\)
A1 for 72.1 or 72.14...

(a)
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2(3)x^2 - 9(2)x + 12(1) - 0 = 6x^2 - 18x + 12\]

(b) At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[6x^2 - 18x + 12 = 0\]
\[6(x^2 - 3x + 2) = 0\]
\[6(x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2\]

When \(x = 1\):
\[y = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0\]
Stationary point is \((1, 0)\).

When \(x = 2\):
\[y = 2(2)^3 - 9(2)^2 + 12(2) - 5 = 16 - 36 + 24 - 5 = -1\]
Stationary point is \((2, -1)\).

To determine the nature, evaluate the second derivative \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18\):
- At \(x = 1\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(1) - 18 = -6 < 0\), so \((1, 0)\) is a local maximum.
- At \(x = 2\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(2) - 18 = 6 > 0\), so \((2, -1)\) is a local minimum.

(a)
B2 for \(6x^2 - 18x + 12\) (B1 for two correct terms)

(b)
M1 for setting their derivative \(= 0\) and solving to find \(x = 1, 2\)
A1 for correct points \((1, 0)\) and \((2, -1)\)
B1 dep on M1 for correct determination of maximum and minimum with valid mathematical justification (e.g. correct second derivative test or evaluating gradient/values on either side)

(a) Find the midpoints \(x\) for each class interval:
- \(80 < m \le 110\): midpoint \(= 95\)
- \(110 < m \le 130\): midpoint \(= 120\)
- \(130 < m \le 150\): midpoint \(= 140\)
- \(150 < m \le 180\): midpoint \(= 165\)
- \(180 < m \le 200\): midpoint \(= 190\)

Calculate \(\sum fx\):
\[\sum fx = (18 \times 95) + (32 \times 120) + (44 \times 140) + (21 \times 165) + (5 \times 190)\]
\[\sum fx = 1710 + 3840 + 6160 + 3465 + 950 = 16125\]
\[\text{Estimated mean} = \frac{16125}{120} = 134.375\text{ g}\]
To 3 significant figures, mean \(= 134\text{ g}\).

(b) Frequency density for \(110 < m \le 130\):
\[\text{Class width} = 130 - 110 = 20\]
\[\text{Frequency density} = \frac{32}{20} = 1.6\]
Given bar height \(= 4.8\text{ cm}\), the scale factor is \(\frac{4.8}{1.6} = 3\text{ cm per unit of frequency density}\).

Frequency density for \(150 < m \le 180\):
\[\text{Class width} = 180 - 150 = 30\]
\[\text{Frequency density} = \frac{21}{30} = 0.7\]
\[\text{Height of bar} = 0.7 \times 3 = 2.1\text{ cm}\]

(a)
M1 for using correct midpoints (at least 3 correct midpoints seen)
M1 (dep on previous M1) for \(\sum fx / 120\) using their midpoints
A1 for 134.375 or 134

(b)
M1 for frequency density \(21 / 30 = 0.7\) or scale relationship \(\frac{4.8}{(32/20)}\)
A1 for 2.1

(a) In the right-angled triangle \(ABC\):
\[AC^2 = AB^2 + BC^2\]
\[AC = \sqrt{18^2 + 14^2} = \sqrt{324 + 196} = \sqrt{520}\]
\[AC = 22.8035...\text{ cm}\]
Rounding to 1 decimal place gives \(AC = 22.8\text{ cm}\).

(b) \(M\) is the midpoint of diagonal \(AC\), so:
\[AM = \frac{1}{2} AC = \frac{\sqrt{520}}{2} = \sqrt{130} \approx 11.40175\text{ cm}\]
Triangle \(VMA\) is right-angled at \(M\). The angle between \(VA\) and the base is \(\angle VAM\):
\[\tan(\angle VAM) = \frac{VM}{AM} = \frac{22}{\sqrt{130}} = \frac{22}{11.40175...} \approx 1.92953...\]
\[\angle VAM = \tan^{-1}(1.92953...) \approx 62.602...^\circ\]
To 1 decimal place, the angle is \(62.6^\circ\).

(a)
M1 for \(18^2 + 14^2\) or \(\sqrt{520}\)
A1 for \(22.803...\) shown to at least 4 significant figures before rounding to 22.8

(b)
M1 for identifying base length \(AM = 0.5 \times 22.803... = 11.40...\)
M1 for \(\tan(\theta) = \frac{22}{11.40...}\) or equivalent correct trig ratio
A1 for 62.6 or 62.60...

(a) Let \(y = \frac{3x + 4}{x - 2}\):
\[y(x - 2) = 3x + 4\]
\[yx - 2y = 3x + 4\]
\[yx - 3x = 2y + 4\]
\[x(y - 3) = 2y + 4\]
\[x = \frac{2y + 4}{y - 3}\]
Therefore,
\[\mathrm{g}^{-1}(x) = \frac{2x + 4}{x - 3}\]

(b)
\[\mathrm{f}(\mathrm{h}(x)) = \mathrm{f}(2^x) = 5 - 2(2^x)\]
Set equal to \(-27\):
\[5 - 2(2^x) = -27\]
\[-2(2^x) = -32\]
\[2^x = 16\]
Since \(16 = 2^4\):
\[x = 4\]

(a)
M1 for multiplying by \((x - 2)\) to remove fraction: \(y(x - 2) = 3x + 4\)
M1 for isolating terms in \(x\): \(x(y - 3) = 2y + 4\) or \(y(x - 3) = 2x + 4\)
A1 for \(\frac{2x+4}{x-3}\) (or equivalent form \(\frac{-2x-4}{3-x}\))

(b)
M1 for setting \(5 - 2(2^x) = -27\) or \(2(2^x) = 32\) or \(2^{x+1} = 32\)
A1 for \(x = 4\)

(a) Gradient of \(AB\):
\[m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 7}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4} = -0.75\]

(b) Midpoint \(M\) of \(AB\):
\[M = \left(\frac{-3 + 5}{2}, \frac{7 + 1}{2}\right) = \left(\frac{2}{2}, \frac{8}{2}\right) = (1, 4)\]

Gradient of the perpendicular line \(m_{\perp}\):
\[m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}\]

Using equation \(y - y_1 = m(x - x_1)\) with point \((1, 4)\):
\[y - 4 = \frac{4}{3}(x - 1)\]
\[y - 4 = \frac{4}{3}x - \frac{4}{3}\]
\[y = \frac{4}{3}x - \frac{4}{3} + \frac{12}{3}\]
\[y = \frac{4}{3}x + \frac{8}{3}\]

(a)
M1 for \(\frac{1 - 7}{5 - (-3)}\) or \(\frac{7 - 1}{-3 - 5}\)
A1 for \(-\frac{3}{4}\) or \(-0.75\)

(b)
B1 for midpoint \((1, 4)\)
M1 for gradient \(m = -\frac{1}{\text{their } (a)} = \frac{4}{3}\)
A1 for \(y = \frac{4}{3}x + \frac{8}{3}\) (or \(y = 1.33x + 2.67\) with recurring decimals / equivalent fractional form)

(a)
\[\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\]

(b) Since \(AP : PB = 2 : 3\), \(\vec{AP} = \frac{2}{5}\vec{AB}\):
\[\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})\]
\[\vec{OP} = \mathbf{a} + \frac{2}{5}\mathbf{b} - \frac{2}{5}\mathbf{a} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\]

(c)
\[\vec{QP} = \vec{QO} + \vec{OP} = -\vec{OQ} + \vec{OP}\]
\[\vec{QP} = -\frac{2}{5}\mathbf{b} + \left(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\right) = \frac{3}{5}\mathbf{a}\]
Geometric relationship: \(\vec{QP}\) is a scalar multiple of \(\vec{OA}\) (\(\vec{QP} = \frac{3}{5}\vec{OA}\)), which means \(QP\) is parallel to \(OA\) (and \(QP = \frac{3}{5} OA\)).

(a)
B1 for \(\mathbf{b} - \mathbf{a}\)

(b)
M1 for \(\mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})\) or \(\mathbf{b} - \frac{3}{5}(\mathbf{b} - \mathbf{a})\)
A1 for \(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\)

(c)
B1 for \(\vec{QP} = \frac{3}{5}\mathbf{a}\)
B1 for stating that \(QP\) is parallel to \(OA\)

(a) The volume of the container with a square base of side \(x\) and height \(h\) is:
\[ V = x^2 h = 32 \implies h = \frac{32}{x^2} \]
The container is open at the top, so it has 1 square base and 4 vertical rectangular faces:
\[ A = x^2 + 4(xh) = x^2 + 4x \left(\frac{32}{x^2}\right) = x^2 + \frac{128}{x} \]

(b) Differentiating \(A = x^2 + 128x^{-1}\) with respect to \(x\):
\[ \frac{\mathrm{d}A}{\mathrm{d}x} = 2x - 128x^{-2} = 2x - \frac{128}{x^2} \]

(c) For a stationary value (minimum surface area), set \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\):
\[ 2x - \frac{128}{x^2} = 0 \implies 2x = \frac{128}{x^2} \implies 2x^3 = 128 \implies x^3 = 64 \implies x = 4 \]

(d) Substituting \(x = 4\) into the area formula:
\[ A = 4^2 + \frac{128}{4} = 16 + 32 = 48\text{ m}^2 \]

(a) M1 for expressing \(h\) in terms of \(x\) as \(h = \frac{32}{x^2}\) or \(x^2h = 32\)
A1 for complete unsimplified area expression \(x^2 + 4xh\) and substituting \(h\) to show \(A = x^2 + \frac{128}{x}\)

(b) B1 for \(2x\)
B1 for \(-128x^{-2}\) or \(-\frac{128}{x^2}\)

(c) M1 for setting their derivative \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\) and rearranging for \(x^3\)
A1 for \(x = 4\)

(d) B0.5 for \(48\) (or \(48\text{ m}^2\))

(a) Volume of the cone \(= \frac{1}{3}\pi r^2 h\).
Volume of the hemisphere \(= \frac{2}{3}\pi r^3\).
Setting them equal:
\[ \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3 \]
Dividing both sides by \(\frac{1}{3}\pi r^2\):
\[ h = 2r \]

(b)(i) When \(r = 4.5\text{ cm}\), \(h = 2(4.5) = 9\text{ cm}\).
Volume of the hemisphere \(= \frac{2}{3}\pi (4.5)^3 = 60.75\pi \approx 190.852\text{ cm}^3\).
Total volume \(= 2 \times 60.75\pi = 121.5\pi \approx 381.7035\text{ cm}^3\).
To 3 significant figures, \(\text{Volume} = 382\text{ cm}^3\).

(b)(ii) The slant height \(l\) of the cone is found using Pythagoras' theorem:
\[ l = \sqrt{r^2 + h^2} = \sqrt{4.5^2 + 9^2} = \sqrt{20.25 + 81} = \sqrt{101.25} \approx 10.0623\text{ cm} \]
The total external surface area consists of the curved surface area of the cone and the curved surface area of the hemisphere:
\[ \text{Curved surface of cone} = \pi r l = \pi \times 4.5 \times \sqrt{101.25} \approx 142.247\text{ cm}^2 \]
\[ \text{Curved surface of hemisphere} = 2\pi r^2 = 2\pi \times (4.5)^2 = 40.5\pi \approx 127.2345\text{ cm}^2 \]
\[ \text{Total surface area} = 142.247 + 127.2345 = 269.4815\text{ cm}^2 \]
To 3 significant figures, \(\text{Surface Area} = 270\text{ cm}^2\).

(a) M1 for equating \(\frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3\)
A1 for completing division clearly to arrive at \(h = 2r\)

(b)(i) M1 for \(\frac{4}{3}\pi (4.5)^3\) or \(\frac{1}{3}\pi(4.5)^2(9) + \frac{2}{3}\pi(4.5)^3\)
A1 for \(382\) or \(381.7\dots\)

(b)(ii) M1 for finding slant height \(l = \sqrt{4.5^2 + 9^2} = \sqrt{101.25}\) or \(10.06\dots\)
M1 for \(\pi \times 4.5 \times l + 2 \pi (4.5)^2\)
A0.5 for \(270\) or \(269.5\)

(a) Let \(y = 3x - 7\).
\(y + 7 = 3x \implies x = \frac{y + 7}{3}\).
Therefore, \(f^{-1}(x) = \frac{x + 7}{3}\).

(b) \(gf(x) = g(f(x)) = g(3x - 7) = \frac{6}{(3x - 7) + 1} = \frac{6}{3x - 6}\).
Factorising the denominator: \(\frac{6}{3(x - 2)} = \frac{2}{x - 2}\).
Wait, check arithmetic: \(\frac{6}{3(x - 2)} = \frac{2}{x - 2}\).

(c) \(h(x) = 2^{x+3} = 128\).
Since \(128 = 2^7\):
\(x + 3 = 7 \implies x = 4\).

(d) \(f(x) = g(x) \implies 3x - 7 = \frac{6}{x + 1}\).
Multiplying by \((x + 1)\):
\((3x - 7)(x + 1) = 6\)
\(3x^2 + 3x - 7x - 7 = 6\)
\(3x^2 - 4x - 13 = 0\)
Applying the quadratic formula:
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-13)}}{2(3)} = \frac{4 \pm \sqrt{16 + 156}}{6} = \frac{4 \pm \sqrt{172}}{6} \]
\(\sqrt{172} \approx 13.11488\)
\[ x_1 = \frac{4 + 13.11488}{6} = \frac{17.11488}{6} \approx 2.8525 \to 2.85 \]
\[ x_2 = \frac{4 - 13.11488}{6} = \frac{-9.11488}{6} \approx -1.5191 \to -1.52 \]

(a) B1 for \(\frac{x + 7}{3}\) or \(\frac{1}{3}(x + 7)\)

(b) M1 for \(\frac{6}{(3x-7)+1}\)
A1 for \(\frac{2}{x - 2}\) or \(\frac{6}{3x - 6}\)

(c) B1 for \(4\)

(d) M1 for forming quadratic \(3x^2 - 4x - 13 = 0\)
M1 for applying quadratic formula \(\frac{4 \pm \sqrt{(-4)^2 - 4(3)(-13)}}{2(3)}\)
A1.5 for \(x = 2.85\) and \(x = -1.52\) (awarded 0.75 for each correct value)

(a) Using Pythagoras' theorem on the right-angled triangle \(ABC\):
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20\text{ cm} \]

(b) Since \(M\) is the centre of the rectangle, \(AM = \frac{1}{2} AC = \frac{20}{2} = 10\text{ cm}\).
In the right-angled triangle \(VMA\):
\[ VM = \sqrt{VA^2 - AM^2} = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24\text{ cm} \]

(c) In right-angled triangle \(VMA\), the angle between edge \(VA\) and the base is \(\angle VAM\):
\[ \cos(\angle VAM) = \frac{AM}{VA} = \frac{10}{26} \]
\[ \angle VAM = \cos^{-1}\left(\frac{10}{26}\right) = \cos^{-1}(0.3846) \approx 67.380\dots^\circ \]
To 1 decimal place, the angle is \(67.4^\circ\).

(d) Let \(N\) be the midpoint of \(BC\). Then \(MN = \frac{1}{2} AB = \frac{16}{2} = 8\text{ cm}\).
The angle between the face \(VBC\) and the base \(ABCD\) is \(\angle VNM\):
\[ \tan(\angle VNM) = \frac{VM}{MN} = \frac{24}{8} = 3 \]
\[ \angle VNM = \tan^{-1}(3) \approx 71.565\dots^\circ \]
To 1 decimal place, the angle is \(71.6^\circ\).

(a) B1 for \(\sqrt{16^2 + 12^2} = \sqrt{400} = 20\)

(b) M1 for \(AM = 10\) and \(VM = \sqrt{26^2 - 10^2}\)
A1 for \(24\)

(c) M1 for \(\cos(\angle VAM) = \frac{10}{26}\) or \(\sin(\angle VAM) = \frac{24}{26}\) or \(\tan(\angle VAM) = \frac{24}{10}\)
A1 for \(67.4^\circ\) or \(67.38\dots^\circ\)

(d) M1 for identifying \(MN = 8\) and using \(\tan(\theta) = \frac{24}{8}\)
A1.5 for \(71.6^\circ\) or \(71.56\dots^\circ\)

(a) Differentiating term by term:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x + 12 \]

(b) Stationary points occur when \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 6x^2 - 18x + 12 = 0 \implies x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0 \]
So \(x = 1\) and \(x = 2\).
When \(x = 1\), \(y = 2(1)^3 - 9(1)^2 + 12(1) + 1 = 2 - 9 + 12 + 1 = 6\).
When \(x = 2\), \(y = 2(8) - 9(4) + 12(2) + 1 = 16 - 36 + 24 + 1 = 5\).
The coordinates are \((1, 6)\) and \((2, 5)\).

(c) Method 1: Second derivative
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18 \]
At \(x = 1\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(1) - 18 = -6 < 0\), so \((1, 6)\) is a local maximum.
At \(x = 2\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(2) - 18 = 6 > 0\), so \((2, 5)\) is a local minimum.

(d) At \(x = 3\):
\(y = 2(3)^3 - 9(3)^2 + 12(3) + 1 = 54 - 81 + 36 + 1 = 10\).
Gradient \(m = \frac{\mathrm{d}y}{\mathrm{d}x} = 6(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12\).
Equation of the tangent: \(y - 10 = 12(x - 3) \implies y = 12x - 36 + 10 \implies y = 12x - 26\).

(a) B1 for \(6x^2 - 18x + 12\)

(b) M1 for solving their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\)
A1 for both points \((1, 6)\) and \((2, 5)\)

(c) M1 for evaluating second derivative (or gradient test on either side) at \(x = 1\) and \(x = 2\)
A1 for correct deduction with working: \((1, 6)\) is a maximum and \((2, 5)\) is a minimum

(d) M1 for calculating \(y = 10\) and \(m = 12\) at \(x = 3\)
A0.5 for \(y = 12x - 26\)

(a) The volume of the conical funnel is:
\[ V_1 = \frac{1}{3}\pi R^2 H = \frac{1}{3}\pi (10)^2 (24) = 800\pi \approx 2513.27\text{ cm}^3 \]
To 3 significant figures, \(V_1 = 2510\text{ cm}^3\).

(b)(i) The water forms a similar cone to the full funnel.
Using the scale factor of lengths:
\[ \frac{r}{10} = \frac{18}{24} \implies r = 10 \times \frac{3}{4} = 7.5\text{ cm} \]

(b)(ii) Volume of the water cone:
\[ V_2 = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (7.5)^2 (18) = \frac{1}{3}\pi (56.25)(18) = 337.5\pi \approx 1060.2875\text{ cm}^3 \]
Alternatively, using the volume ratio:
\[ V_2 = \left(\frac{18}{24}\right)^3 \times 800\pi = \left(\frac{3}{4}\right)^3 \times 800\pi = \frac{27}{64} \times 800\pi = 337.5\pi \approx 1060\text{ cm}^3 \]
To 3 significant figures, \(V_2 = 1060\text{ cm}^3\).

(c) Time taken to drain:
\[ t = \frac{V_2}{\text{rate}} = \frac{337.5\pi}{15} = 22.5\pi \approx 70.6858\dots\text{ seconds} \]
To 1 decimal place, \(t = 70.7\text{ seconds}\).

(a) M1 for \(\frac{1}{3}\pi (10)^2(24)\)
A1 for \(2510\) or \(2513\dots\) or \(800\pi\)

(b)(i) B1 for \(7.5\)

(b)(ii) M1 for \(\frac{1}{3}\pi(7.5)^2(18)\) or \(\left(\frac{18}{24}\right)^3 \times \text{their}(a)\)
A1 for \(1060\) or \(1060.28\dots\) or \(337.5\pi\)

(c) M1 for \(\frac{\text{their } 1060}{15}\)
A1.5 for \(70.7\) (accept \(70.68\dots\) to \(70.7\))