2026 Cambridge IGCSE Mathematics (0580) 模拟试题及答案详解

Multiply numerator and denominator by the conjugate \(3 - \sqrt{3}\):
\[\frac{18(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{18(3 - \sqrt{3})}{3^2 - (\sqrt{3})^2}\]
\[= \frac{18(3 - \sqrt{3})}{9 - 3} = \frac{18(3 - \sqrt{3})}{6}\]
\[= 3(3 - \sqrt{3}) = 9 - 3\sqrt{3}\]

M1 for multiplying numerator and denominator by \(3 - \sqrt{3}\)
A1 for denominator correctly simplified to 6
A1 (or remaining mark fraction) for final answer \(9 - 3\sqrt{3}\)

Apply the negative index to take the reciprocal:
\[\left(\frac{125}{64}\right)^{-\frac{2}{3}} = \left(\frac{64}{125}\right)^{\frac{2}{3}}\]
Take the cube root of numerator and denominator:
\[\left(\frac{\sqrt[3]{64}}{\sqrt[3]{125}}\right)^2 = \left(\frac{4}{5}\right)^2\]
Square the result:
\[\frac{4^2}{5^2} = \frac{16}{25}\]

M1 for dealing with negative index or cube root correctly, e.g. \(\left(\frac{64}{125}\right)^{\frac{2}{3}}\) or \(\left(\frac{5}{4}\right)^{-2}\)
A1 for obtaining \(\frac{4}{5}\) before squaring
A1 (or remaining mark fraction) for \(\frac{16}{25}\) cao

Let \(x = 0.2454545\dots\)
Multiply by 10: \(10x = 2.454545\dots\)
Multiply by 1000: \(1000x = 245.454545\dots\)
Subtract the two equations:
\[1000x - 10x = 245.4545\dots - 2.4545\dots\]
\[990x = 243\]
\[x = \frac{243}{990}\]
Divide numerator and denominator by 9:
\[x = \frac{27}{110}\]

M1 for setting up equations such as \(1000x = 245.45\dots\) and \(10x = 2.45\dots\) or \(100x = 24.545\dots\)
M1 for subtraction to eliminate recurring part, e.g. \(990x = 243\) or \(99x = 24.3\)
A1 (or remaining mark fraction) for \(\frac{27}{110}\) in simplest form

Apply the power \(\frac{2}{5}\) to each component inside the bracket:
\[32^{\frac{2}{5}} = (\sqrt[5]{32})^2 = 2^2 = 4\]
\[(x^{10})^{\frac{2}{5}} = x^{10 \times \frac{2}{5}} = x^4\]
\[(y^{-15})^{\frac{2}{5}} = y^{-15 \times \frac{2}{5}} = y^{-6}\]
Combining these gives \(4x^4y^{-6}\) or \(\frac{4x^4}{y^6}\).

B1 for numerical coefficient 4
B1 for \(x^4\) or \(y^{-6}\) (or \(y^6\) in denominator)
A1 (or remaining mark fraction) for fully correct expression \(4x^4y^{-6}\) or \(\frac{4x^4}{y^6}\)

Expand the brackets:
\[(4\sqrt{3})(2\sqrt{3}) + (4\sqrt{3})(5) - 1(2\sqrt{3}) - 1(5)\]
\[= (8 \times 3) + 20\sqrt{3} - 2\sqrt{3} - 5\]
\[= 24 + 18\sqrt{3} - 5\]
\[= 19 + 18\sqrt{3}\]

M1 for correct expansion with at least 3 correct terms from \(24 + 20\sqrt{3} - 2\sqrt{3} - 5\)
A1 for correct like terms combined (e.g. 19 or \(18\sqrt{3}\))
A1 (or remaining mark fraction) for \(19 + 18\sqrt{3}\) or \(18\sqrt{3} + 19\)

Divide the numerical coefficients and subtract the powers of 10:
\[\frac{4.2}{6} = 0.7\]
\[\frac{10^7}{10^{-3}} = 10^{7 - (-3)} = 10^{10}\]
Combine and adjust into standard form \(A \times 10^n\) where \(1 \le A < 10\):
\[0.7 \times 10^{10} = 7 \times 10^9\]

M1 for \(0.7 \times 10^{10}\) or correct division step
A1 for adjusting correctly to standard form \(7 \times 10^9\)
SC1 for answer \(7000000000\) not in standard form

Convert mixed numbers to improper fractions:
\[3\frac{1}{5} = \frac{16}{5}, \quad 1\frac{7}{15} = \frac{22}{15}\]
Perform division by multiplying by the reciprocal:
\[\frac{16}{5} \div \frac{22}{15} = \frac{16}{5} \times \frac{15}{22} = \frac{16 \times 3}{22} = \frac{48}{22} = \frac{24}{11}\]
Subtract the second fraction:
\[\frac{24}{11} - \frac{5}{11} = \frac{19}{11} \quad (\text{or } 1\frac{8}{11})\]

M1 for improper fractions \(\frac{16}{5}\) and \(\frac{22}{15}\) and multiplying by reciprocal
A1 for \(\frac{24}{11}\) or equivalent fraction
A1 (or remaining mark fraction) for \(\frac{19}{11}\) or \(1\frac{8}{11}\) cao

Simplify each surd by extracting square factors:
\[\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\]
\[2\sqrt{27} = 2\sqrt{9 \times 3} = 2 \times 3\sqrt{3} = 6\sqrt{3}\]
\[\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\]
Combine the terms:
\[5\sqrt{3} - 6\sqrt{3} + 6\sqrt{3} = 5\sqrt{3}\]

M1 for simplifying at least two surds correctly to terms in \(\sqrt{3}\) (e.g. \(5\sqrt{3}\), \(6\sqrt{3}\))
A1 for all three terms correctly simplified: \(5\sqrt{3} - 6\sqrt{3} + 6\sqrt{3}\)
A1 (or remaining mark fraction) for \(5\sqrt{3}\) cao

First, find \(\vec{AB}\):
\[\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}\]

Next, find the position vector of \(P\):
\[\vec{OP} = \vec{OA} + \frac{2}{5}\vec{AB} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a}) = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\]

Since \(M\) is the midpoint of \(OB\):
\[\vec{OM} = \frac{1}{2}\mathbf{b}\]

Now find \(\vec{MP}\):
\[\vec{MP} = \vec{OP} - \vec{OM} = \left(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\right) - \frac{1}{2}\mathbf{b}\]
\[\vec{MP} = \frac{3}{5}\mathbf{a} + \left(\frac{2}{5} - \frac{1}{2}\right)\mathbf{b} = \frac{3}{5}\mathbf{a} - \frac{1}{10}\mathbf{b}\]

M1 for \(\vec{AB} = \mathbf{b} - \mathbf{a}\) or \(\vec{BA} = \mathbf{a} - \mathbf{b}\)
M1 for \(\vec{OP} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})\) or \(\vec{MP} = \vec{MO} + \vec{OA} + \frac{2}{5}\vec{AB}\)
M1 for a correct unsimplified expression for \(\vec{MP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), e.g. \(-\frac{1}{2}\mathbf{b} + \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})\)
A1 for \(\frac{3}{5}\mathbf{a} - \frac{1}{10}\mathbf{b}\) or \(\frac{6\mathbf{a} - \mathbf{b}}{10}\)

Observing the coordinates of corresponding vertices:
\((1, 2) \to (-2, -1)\)
\((3, 2) \to (-2, -3)\)
\((3, 5) \to (-5, -3)\)

Each point \((x, y)\) maps to \((-y, -x)\).
This is the standard matrix transformation \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\), which represents a reflection in the line \(y = -x\).

B1 for Reflection
B2 for in the line \(y = -x\)
(Note: award 0 marks if more than one transformation is mentioned)

(a)
Factorise the coefficient of \(x^2\) from the first two terms:
\[2x^2 - 12x + 11 = 2(x^2 - 6x) + 11\]
Complete the square inside the bracket:
\[x^2 - 6x = (x - 3)^2 - 9\]
Substitute back:
\[2[(x - 3)^2 - 9] + 11 = 2(x - 3)^2 - 18 + 11 = 2(x - 3)^2 - 7\]

(b)
From the completed square form \(y = 2(x - 3)^2 - 7\), the minimum value occurs when \(x - 3 = 0 \implies x = 3\), giving \(y = -7\).
So the turning point is \((3, -7)\).

(a)
B1 for \(a = 2\)
M1 for \((x - 3)^2\) seen in working
A1 for \(2(x - 3)^2 - 7\)

(b)
B1 FT for \((3, -7)\) from their \(a(x + p)^2 + q\)

First, find the sum vector \(\mathbf{p} + \mathbf{q}\):
\[\mathbf{p} + \mathbf{q} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} + \begin{pmatrix} k \\ 10 \end{pmatrix} = \begin{pmatrix} k + 1 \\ 8 \end{pmatrix}\]

The magnitude of this vector is given by:
\[|\mathbf{p} + \mathbf{q}| = \sqrt{(k + 1)^2 + 8^2}\]

Given that the magnitude is \(10\):
\[\sqrt{(k + 1)^2 + 64} = 10\]
Square both sides:
\[(k + 1)^2 + 64 = 100\]
\[(k + 1)^2 = 36\]
Taking square roots:
\[k + 1 = 6 \quad \text{or} \quad k + 1 = -6\]
\[k = 5 \quad \text{or} \quad k = -7\]
Since \(k\) is positive, \(k = 5\).

M1 for \(\mathbf{p} + \mathbf{q} = \begin{pmatrix} k + 1 \\ 8 \end{pmatrix}\)
M1 for \((k + 1)^2 + 8^2 = 10^2\) or \(\sqrt{(k + 1)^2 + 8^2} = 10\)
M1 for solving quadratic or taking square root: \((k + 1)^2 = 36 \implies k + 1 = \pm 6\)
A1 for \(k = 5\) only

For an enlargement with centre \(C(x_0, y_0)\) and scale factor \(k = -\frac{1}{2}\), the image of a point \(P(x, y)\) is given by:
\[P' = C + k(P - C)\]
Here \(C = (0, 3)\).

1. For \((2, 1)\):
Vector from centre: \(\begin{pmatrix} 2 - 0 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}\)
Multiply by scale factor \(-\frac{1}{2}\): \(-\frac{1}{2}\begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\)
Image vertex: \(\begin{pmatrix} 0 \\ 3 \end{pmatrix} + \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\)

2. For \((4, 1)\):
Vector from centre: \(\begin{pmatrix} 4 - 0 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}\)
Multiply by scale factor \(-\frac{1}{2}\): \(-\frac{1}{2}\begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}\)
Image vertex: \(\begin{pmatrix} 0 \\ 3 \end{pmatrix} + \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}\)

3. For \((2, 5)\):
Vector from centre: \(\begin{pmatrix} 2 - 0 \\ 5 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}\)
Multiply by scale factor \(-\frac{1}{2}\): \(-\frac{1}{2}\begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}\)
Image vertex: \(\begin{pmatrix} 0 \\ 3 \end{pmatrix} + \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\)

The vertices of triangle \(B\) are \((-1, 4)\), \((-2, 4)\) and \((-1, 2)\).

M1 for correctly calculating vectors from centre \((0, 3)\) to at least two vertices
M1 for multiplying vector displacements by \(-\frac{1}{2}\)
A1 for two correct vertices found
A1 for all three correct vertices: \((-1, 4)\), \((-2, 4)\), \((-1, 2)\)

1. Find the coordinates of the midpoint \(M\):
\[M = \left(\frac{-3 + 5}{2}, \frac{8 + (-4)}{2}\right) = (1, 2)\]

2. Find the gradient of the line \(AB\):
\[m_{AB} = \frac{-4 - 8}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2}\]

3. Find the gradient of the perpendicular line \(m_\perp\):
\[m_\perp = -\frac{1}{m_{AB}} = \frac{2}{3}\]

4. Use the point-slope formula with \(M(1, 2)\):
\[y - 2 = \frac{2}{3}(x - 1)\]
\[y = \frac{2}{3}x - \frac{2}{3} + 2\]
\[y = \frac{2}{3}x + \frac{4}{3}\]

B1 for midpoint \((1, 2)\)
M1 for gradient of \(AB = -\frac{3}{2}\)
M1 for perpendicular gradient \(m = \frac{2}{3}\)
A1 for \(y = \frac{2}{3}x + \frac{4}{3}\) or \(y = \frac{2}{3}x + 1\frac{1}{3}\)

(a)
In parallelogram \(OABC\), \(\vec{CB} = \vec{OA} = 6\mathbf{a}\) and \(\vec{AB} = \vec{OC} = 9\mathbf{c}\).

\[\vec{OB} = \vec{OA} + \vec{AB} = 6\mathbf{a} + 9\mathbf{c}\]
Since \(\vec{OY} = \frac{1}{3}\vec{OB}\):
\[\vec{OY} = \frac{1}{3}(6\mathbf{a} + 9\mathbf{c}) = 2\mathbf{a} + 3\mathbf{c}\]

For \(\vec{CX}\):
\(X\) divides \(AB\) in the ratio \(1 : 2\), so \(\vec{AX} = \frac{1}{3}\vec{AB} = \frac{1}{3}(9\mathbf{c}) = 3\mathbf{c}\).
\[\vec{OX} = \vec{OA} + \vec{AX} = 6\mathbf{a} + 3\mathbf{c}\]
\[\vec{CX} = \vec{OX} - \vec{OC} = (6\mathbf{a} + 3\mathbf{c}) - 9\mathbf{c} = 6\mathbf{a} - 6\mathbf{c}\]

(b)
Alternative relationship check:
If finding \(\vec{YX}\):
\[\vec{YX} = \vec{OX} - \vec{OY} = (6\mathbf{a} + 3\mathbf{c}) - (2\mathbf{a} + 3\mathbf{c}) = 4\mathbf{a}\]
Since \(\vec{YX} = 4\mathbf{a}\) and \(\vec{OA} = 6\mathbf{a}\), the line segment \(YX\) is parallel to \(OA\) (and \(CB\)) and has length \(\frac{2}{3}\) of \(OA\).

M1 for \(\vec{OB} = 6\mathbf{a} + 9\mathbf{c}\)
A1 for \(\vec{OY} = 2\mathbf{a} + 3\mathbf{c}\)
M1 for \(\vec{OX} = 6\mathbf{a} + 3\mathbf{c}\) or \(\vec{CX} = \vec{CO} + \vec{OA} + \vec{AX}\)
A1 for \(\vec{CX} = 6\mathbf{a} - 6\mathbf{c}\)

(a)
The speed-time graph forms a trapezium with parallel sides of length:
Top base = \(16\text{ s}\) (constant speed section)
Bottom base = \(8 + 16 + 6 = 30\text{ s}\) (total time)
Height = \(v\text{ m/s}\)

The area under the speed-time graph represents the total distance travelled:
\[\text{Distance} = \frac{1}{2} \times (16 + 30) \times v = 276\]
\[\frac{1}{2} \times 46 \times v = 276\]
\[23v = 276\]
\[v = \frac{276}{23} = 12\]

(b)
Acceleration is the gradient of the speed-time graph during the first \(8\text{ s}\):
\[\text{Acceleration} = \frac{v - 0}{8 - 0} = \frac{12}{8} = 1.5\text{ m/s}^2\]

(a)
M1 for total time \(8 + 16 + 6 = 30\)
M1 for area formula: \(\frac{1}{2}(16 + 30)v = 276\) or sum of three sections \(\frac{1}{2}(8)(v) + 16v + \frac{1}{2}(6)(v) = 276\)
A1 for \(v = 12\)

(b)
B1 FT for \(\frac{\text{their } v}{8} = 1.5\)

(a)
\(\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}\)
Since \(AX : XB = 2 : 1\), \(\vec{AX} = \frac{2}{3}\vec{AB}\).
\(\vec{AX} = \frac{2}{3}(\mathbf{b} - \mathbf{a})\) or \(-\frac{2}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}\)

(b)
\(\vec{AY} = \vec{OY} - \vec{OA} = (2\mathbf{b} - \mathbf{a}) - \mathbf{a} = 2\mathbf{b} - 2\mathbf{a} = 2(\mathbf{b} - \mathbf{a})\)
Comparing \(\vec{AY}\) and \(\vec{AX}\):
\(\vec{AY} = 2(\mathbf{b} - \mathbf{a}) = 3 \times \left(\frac{2}{3}(\mathbf{b} - \mathbf{a})\right) = 3\vec{AX}\)
Since \(\vec{AY}\) is a scalar multiple of \(\vec{AX}\) (they are parallel) and both vectors share the common point \(A\), the points \(A\), \(X\) and \(Y\) lie on a straight line.

(a)
M1 for \(\vec{AB} = \mathbf{b} - \mathbf{a}\) or for \(\frac{2}{3}\vec{AB}\) seen
A1 for \(\frac{2}{3}(\mathbf{b} - \mathbf{a})\) or \(-\frac{2}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}\)

(b)
M1 for finding \(\vec{AY} = 2\mathbf{b} - 2\mathbf{a}\) or \(\vec{XY} = \frac{4}{3}\mathbf{b} - \frac{4}{3}\mathbf{a}\)
A1 for demonstrating scalar relationship (e.g. \(\vec{AY} = 3\vec{AX}\) or \(\vec{XY} = 2\vec{AX}\)) AND stating that they share a common point (e.g. \(A\) or \(X\))

(a)
Reflection in the line \(y = -x\) maps each point \((x, y)\) to \((-y, -x)\):
- \((1, 2) \to (-2, -1)\)
- \((4, 2) \to (-2, -4)\)
- \((4, 4) \to (-4, -4)\)

(b)
Comparing coordinates of \(A\) and \(C\):
- \((1, 2) \to (-2, -4)\)
- \((4, 2) \to (-8, -4)\)
- \((4, 4) \to (-8, -8)\)
Each coordinate of triangle \(A\) is multiplied by \(-2\).
This is an enlargement with scale factor \(-2\) and centre of enlargement \((0, 0)\).

(a)
B2 for all three correct coordinates: \((-2, -1)\), \((-2, -4)\), \((-4, -4)\)
(B1 for two correct pairs of coordinates or for reflection in \(y = x\) giving \((2, 1)\), \((2, 4)\), \((4, 4)\))

(b)
B1 for Enlargement
B1 for scale factor \(-2\) and centre \((0, 0)\) or origin
(Note: Award 0 marks if more than one transformation is mentioned)

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

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

When \(x = 1\):
\(y = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0\)
First point: \((1, 0)\)

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

(c)
The stationary point with the smaller \(x\)-coordinate is \((1, 0)\).
Using the second derivative:
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18\)
At \(x = 1\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(1) - 18 = -6\).
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0\), the point \((1, 0)\) is a local maximum.

(a)
M1 for differentiating at least two terms correctly from \(2x^3 - 9x^2 + 12x - 5\)
A1 for \(6x^2 - 18x + 12\) or fully correct equivalent

(b)
M1 for setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and attempting to solve the quadratic (by factorisation or formula)
A1 for \(x = 1\) and \(x = 2\)
A1 for both pairs of coordinates correct: \((1, 0)\) and \((2, -1)\)

(c)
B1 for Maximum with valid reason (e.g. \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(1) - 18 = -6 < 0\) or gradient changes from positive to negative through \(x = 1\))

(a)
Since \(OP : PA = 3 : 1\), \(\vec{OP} = \frac{3}{4}\vec{OA} = \frac{3}{4}\mathbf{a}\).
\(M\) is the midpoint of \(AB\):
\(\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}\)
\(\vec{OM} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)

Then:
\(\vec{PM} = \vec{PO} + \vec{OM} = -\frac{3}{4}\mathbf{a} + \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = -\frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\)

(b)
\(\vec{PN} = 3\vec{PM} = 3\left(-\frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = -\frac{3}{4}\mathbf{a} + \frac{3}{2}\mathbf{b}\)

Now find \(\vec{ON}\):
\(\vec{ON} = \vec{OP} + \vec{PN} = \frac{3}{4}\mathbf{a} + \left(-\frac{3}{4}\mathbf{a} + \frac{3}{2}\mathbf{b}\right) = \frac{3}{2}\mathbf{b}\)

(c)
\(\vec{ON} = \frac{3}{2}\mathbf{b} = \frac{3}{2}\vec{OB}\).
This shows that \(\vec{ON}\) is a scalar multiple of \(\vec{OB}\) and both vectors start at the common point \(O\). Therefore, the points \(O\), \(B\) and \(N\) lie on a straight line (are collinear).

(a)
B1 for \(\vec{OP} = \frac{3}{4}\mathbf{a}\) or \(\vec{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)
M1 for \(\vec{PM} = -\vec{OP} + \vec{OM}\) or \(\vec{PA} + \vec{AM}\)
A1 for \(-\frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\) or \(\frac{1}{4}(2\mathbf{b} - \mathbf{a})\)

(b)
M1 for \(\vec{ON} = \vec{OP} + 3\vec{PM}\) substituted correctly
A1 for \(\frac{3}{2}\mathbf{b}\) or \(1.5\mathbf{b}\)

(c)
B1 for stating that \(O, B, N\) are collinear / lie on a straight line because \(\vec{ON}\) is a scalar multiple of \(\vec{OB}\) with common point \(O\) (or equivalent clear justification)

(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 \(AOB = 360^\circ - 90^\circ - 90^\circ - 44^\circ = 136^\circ\).

(b)
Angle \(ADB = \frac{1}{2} \times \text{Angle } AOB = \frac{1}{2} \times 136^\circ = 68^\circ\).
Reason: The angle subtended by an arc at the centre is twice the angle subtended by it at the circumference.

(c)
In cyclic quadrilateral \(ABCD\), opposite angles sum to \(180^\circ\):
Angle \(BCD = 180^\circ - \text{Angle } ADB = 180^\circ - 68^\circ = 112^\circ\).
Reason: Opposite angles of a cyclic quadrilateral sum to \(180^\circ\).

(a)
B1 for \(136^\circ\)
B1 for stating: tangent is perpendicular to radius / angle between tangent and radius is \(90^\circ\) AND angles in a quadrilateral add to \(360^\circ\)

(b)
B1 for \(68^\circ\) (or FT \(\frac{1}{2} \times \text{their } AOB\))
B1 for reason: angle at the centre is twice the angle at the circumference

(c)
B1 for \(112^\circ\) (or FT \(180^\circ - \text{their } ADB\))
B1 for reason: opposite angles of a cyclic quadrilateral add up to \(180^\circ\)

From the linear equation:
\(y = 2x + 1\)

Substitute \(y = 2x + 1\) into the non-linear equation:
\(x^2 + x(2x + 1) = 10\)
\(x^2 + 2x^2 + x = 10\)
\(3x^2 + x - 10 = 0\)

Factorise the quadratic:
\((3x - 5)(x + 2) = 0\)

So:
\(3x - 5 = 0 \implies x = \frac{5}{3}\)
\(x + 2 = 0 \implies x = -2\)

Now find corresponding \(y\)-values using \(y = 2x + 1\):
When \(x = -2\):
\(y = 2(-2) + 1 = -4 + 1 = -3\)

When \(x = \frac{5}{3}\):
\(y = 2\left(\frac{5}{3}\right) + 1 = \frac{10}{3} + \frac{3}{3} = \frac{13}{3}\)

The solutions are \(x = -2, y = -3\) and \(x = \frac{5}{3}, y = \frac{13}{3}\).

M1 for rearranging linear equation to make \(y\) (or \(x\)) the subject, e.g. \(y = 2x + 1\)
M1 for substituting their expression into the second equation to form an equation in one variable
A1 for correct 3-term quadratic equal to zero: \(3x^2 + x - 10 = 0\)
M1 for factorising their 3-term quadratic, e.g. \((3x - 5)(x + 2) = 0\), or correct use of quadratic formula
A1 for \(x = -2\) and \(x = \frac{5}{3}\) (or \(1\frac{2}{3}\))
A1 for \(y = -3\) and \(y = \frac{13}{3}\) (or \(4\frac{1}{3}\)) correctly paired

(a)
Rewrite \(y\) as \(y = 4x + 9x^{-1}\).
Differentiating with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 4 + 9(-1)x^{-2} = 4 - 9x^{-2} = 4 - \frac{9}{x^2}\)

(b)
Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3\):
\(4 - \frac{9}{x^2} = 3\)
\(1 = \frac{9}{x^2}\)
\(x^2 = 9\)
Since \(x > 0\), \(x = 3\).

Find the \(y\)-coordinate:
\(y = 4(3) + \frac{9}{3} = 12 + 3 = 15\)
Coordinates are \((3, 15)\).

(c)
At the minimum stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(4 - \frac{9}{x^2} = 0 \implies x^2 = \frac{9}{4} \implies x = \frac{3}{2}\) (as \(x > 0\)).

Substitute \(x = \frac{3}{2}\) into \(y\):
\(y = 4\left(\frac{3}{2}\right) + \frac{9}{\frac{3}{2}} = 6 + 6 = 12\)
Minimum value of \(y\) is 12.

(a)
M1 for derivative of form \(4 + k x^{-2}\) where \(k \ne 0\)
A1 for \(4 - 9x^{-2}\) or \(4 - \frac{9}{x^2}\)

(b)
M1 for setting their derivative equal to 3
M1 for solving to get \(x = 3\) (rejecting \(x = -3\))
A1 for \((3, 15)\)

(c)
B1 for 12

(a)
Using the formula for the area of a triangle:
\(\text{Area} = \frac{1}{2} a b \sin C\)
\(\text{Area} = \frac{1}{2} (x + 3)(2x - 1) \sin(150^\circ)\)

Since \(\sin(150^\circ) = \sin(30^\circ) = \frac{1}{2}\):
\(6.5 = \frac{1}{2}(x + 3)(2x - 1)\left(\frac{1}{2}\right)\)
\(6.5 = \frac{1}{4}(2x^2 - x + 6x - 3)\)
\(6.5 = \frac{1}{4}(2x^2 + 5x - 3)\)
Multiply by 4:
\(26 = 2x^2 + 5x - 3\)
\(2x^2 + 5x - 29 = 0\)

(b)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2, b = 5, c = -29\):
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-29)}}{2(2)}\)
\(x = \frac{-5 \pm \sqrt{25 + 232}}{4}\)
\(x = \frac{-5 \pm \sqrt{257}}{4}\)

(c)
Since \(\sqrt{257} > \sqrt{256} = 16\), \(\frac{-5 - \sqrt{257}}{4} < \frac{-5 - 16}{4} = -5.25 < 0\).
A negative value of \(x\) would make side length \(QR = 2x - 1 < 0\), which is impossible for a geometrical length. Therefore, the only valid solution is \(x = \frac{-5 + \sqrt{257}}{4}\).

(a)
M1 for substituting into area formula: \(\frac{1}{2}(x + 3)(2x - 1)\sin(150^\circ) = 6.5\)
M1 for using \(\sin(150^\circ) = \frac{1}{2}\) and expanding brackets: \(2x^2 + 5x - 3\)
A1 for completing all algebraic steps clearly to reach \(2x^2 + 5x - 29 = 0\) with no steps omitted

(b)
M1 for correct substitution into quadratic formula: \(\frac{-5 \pm \sqrt{5^2 - 4(2)(-29)}}{2(2)}\)
M1 for simplifying the discriminant: \(25 + 232 = 257\)
A1 for \(\frac{-5 \pm \sqrt{257}}{4}\)

(c)
B1 for choosing \(x = \frac{-5 + \sqrt{257}}{4}\) with correct geometric reason (side lengths must be positive / lengths cannot be negative)

(a)
Factorise 2 from the \(x\)-terms:
\(\mathrm{f}(x) = 2(x^2 - 6x) + 13\)

Complete the square inside the bracket:
\(x^2 - 6x = (x - 3)^2 - 3^2 = (x - 3)^2 - 9\)

Substitute back:
\(\mathrm{f}(x) = 2[(x - 3)^2 - 9] + 13\)
\(\mathrm{f}(x) = 2(x - 3)^2 - 18 + 13\)
\(\mathrm{f}(x) = 2(x - 3)^2 - 5\)

(b)
Since \(x \ge 3\), \((x - 3)^2 \ge 0\).
The minimum value occurs at \(x = 3\) where \(\mathrm{f}(3) = -5\).
Hence the range is \(\mathrm{f}(x) \ge -5\) (or \(y \ge -5\)).

(c)
Let \(y = 2(x - 3)^2 - 5\) for \(x \ge 3\):
\(y + 5 = 2(x - 3)^2\)
\(\frac{y + 5}{2} = (x - 3)^2\)

Take the positive square root since \(x \ge 3\):
\(x - 3 = \sqrt{\frac{y + 5}{2}}\)
\(x = 3 + \sqrt{\frac{y + 5}{2}}\)

Therefore:
\(\mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x + 5}{2}}\)

(a)
B1 for \(a = 2\)
M1 for \((x - 3)^2\)
A1 for \(2(x - 3)^2 - 5\)

(b)
B1 for \(\mathrm{f}(x) \ge -5\) or \(y \ge -5\) (accept \([-5, \infty)\))

(c)
M1 for correctly rearranging their \(y = a(x - h)^2 + k\) to make \((x - h)^2\) the subject
A1 for \(3 + \sqrt{\frac{x + 5}{2}}\) (must be in terms of \(x\) and positive square root)

Let the original value of the car be \(P\).

Using the formula for compound depreciation:
\[ V = P(1 - r)^n \]
\[ 11\,585.12 = P(1 - 0.12)^3 \]
\[ 11\,585.12 = P(0.88)^3 \]
\[ 11\,585.12 = P \times 0.681472 \]
\[ P = \frac{11\,585.12}{0.681472} = 17\,000 \]

The original value of the car is $17\,000.

M1 for \(P \times (1 - 0.12)^3 = 11\,585.12\) or \(P \times 0.88^3 = 11\,585.12\) oe
M1 for \(\frac{11\,585.12}{0.88^3}\) or \(\frac{11\,585.12}{0.681472}\)
A1 for 17000

First find the upper bound of the mass and the lower bound of the volume:
- Upper bound of mass: \(m_{\text{UB}} = 450 + 5 = 455\text{ g}\)
- Lower bound of volume: \(V_{\text{LB}} = 52 - 0.5 = 51.5\text{ cm}^3\)

To maximise density \(D = \frac{m}{V}\):
\[ D_{\text{UB}} = \frac{m_{\text{UB}}}{V_{\text{LB}}} = \frac{455}{51.5} \approx 8.834951... \]

Rounding to 3 decimal places gives \(8.835\text{ g/cm}^3\).

B1 for 455 or 51.5 seen
M1 for \(\frac{\text{their } 455}{\text{their } 51.5}\) where \(450 < \text{their } 455 \le 455\) and \(51.5 \le \text{their } 51.5 < 52\)
A1 for 8.835 (correct to 3 d.p.)

Since \(y\) is inversely proportional to \((x + 1)^2\):
\[ y = \frac{k}{(x + 1)^2} \]

Substitute \(x = 3\) and \(y = 5\):
\[ 5 = \frac{k}{(3 + 1)^2} = \frac{k}{16} \implies k = 5 \times 16 = 80 \]

Now find \(y\) when \(x = 7\):
\[ y = \frac{80}{(7 + 1)^2} = \frac{80}{64} = 1.25 \]

M1 for \(y = \frac{k}{(x + 1)^2}\) oe
M1 for substituting \(x = 3\), \(y = 5\) to find \(k = 80\)
A1 for 1.25 or \(\frac{5}{4}\)

1. Calculate the commission fee or remaining percentage in pounds:
\[ 850 \times (1 - 0.015) = 850 \times 0.985 = 837.25 \text{ pounds} \]

2. Convert the remaining pounds to Euros at the rate of £1 = €1.16:
\[ 837.25 \times 1.16 = 971.21 \text{ Euros} \]

M1 for \(850 \times (1 - 0.015)\) or \(850 - 0.015 \times 850\) oe (giving 837.25)
M1 for \(\text{their } 837.25 \times 1.16\)
A1 for 971.21

Find the lower bound for distance and upper bound for time:
- Distance: \(d_{\text{LB}} = 400 - 0.5 = 399.5\text{ m}\)
- Time: \(t_{\text{UB}} = 52.4 + 0.05 = 52.45\text{ s}\)

To minimise average speed \(s = \frac{d}{t}\):
\[ s_{\text{LB}} = \frac{d_{\text{LB}}}{t_{\text{UB}}} = \frac{399.5}{52.45} \approx 7.61677788... \]

Correct to 4 significant figures, the lower bound is \(7.617\text{ m/s}\).

B1 for 399.5 or 52.45 seen
M1 for \(\frac{\text{their } 399.5}{\text{their } 52.45}\) where \(399.5 \le \text{their } 399.5 < 400\) and \(52.4 < \text{their } 52.45 \le 52.45\)
A1 for 7.617 (correct to 4 s.f.)

Since \(P\) is directly proportional to \(\sqrt[3]{w}\):
\[ P = k\sqrt[3]{w} \]

Substitute \(w = 64\) and \(P = 18\):
\[ 18 = k\sqrt[3]{64} = k \times 4 \implies k = \frac{18}{4} = 4.5 \]

Now set \(P = 27\) to solve for \(w\):
\[ 27 = 4.5\sqrt[3]{w} \]
\[ \sqrt[3]{w} = \frac{27}{4.5} = 6 \]
\[ w = 6^3 = 216 \]

M1 for \(P = k\sqrt[3]{w}\) and substituting \(w = 64, P = 18\) to get \(k = 4.5\) oe
M1 for \(27 = 4.5\sqrt[3]{w}\) leading to \(\sqrt[3]{w} = 6\) or \(w = \left(\frac{27}{4.5}\right)^3\)
A1 for 216

1. Calculate the total value of the investment after 4 years using the compound interest formula:
\[ A = P(1 + r)^n \]
\[ A = 6500(1 + 0.032)^4 = 6500(1.032)^4 \approx 6500 \times 1.13426702 = 7372.7356... \]

2. Calculate the total interest earned by subtracting the principal:
\[ \text{Interest} = A - P = 7372.7356... - 6500 = 872.7356... \]

Rounding to the nearest cent (2 decimal places) gives $872.74.

M1 for \(6500 \times (1 + 0.032)^4\) or \(6500 \times 1.032^4\) (giving 7372.74)
M1 for \(\text{their } 7372.74 - 6500\)
A1 for 872.74

(a)
For \(x = 1\):
\(a = \dfrac{6}{1} + 1^2 - 4(1) = 6 + 1 - 4 = 3\)

For \(x = 2.5\):
\(b = \dfrac{6}{2.5} + 2.5^2 - 4(2.5) = 2.4 + 6.25 - 10 = -1.35\)

(b)
Plot all 9 points accurately and join with a smooth, continuous single curve.

(c)(i)
From the turning point on the graph, the minimum value of \(y\) occurs at \(x \approx 2.4\) where \(y \approx -1.38\) to \(-1.4\).
Acceptable range: \(-1.45 \le y \le -1.35\).

(c)(ii)
Rearrange the equation to match the graphed function:
\(\dfrac{6}{x} + x^2 - 5x + 1 = 0 \implies \dfrac{6}{x} + x^2 - 4x = x - 1\)
Draw the straight line \(y = x - 1\) on the grid through points such as \((1, 0)\), \((3, 2)\), \((5, 4)\).
Find the \(x\)-coordinates of the intersection points of the line and curve:
\(x \approx 1.2\) (range \(1.15\) to \(1.25\))
\(x = 2.0\)
\(x \approx 4.8\) (range \(4.75\) to \(4.85\))

(a) [2 marks]
B1 for \(a = 3\)
B1 for \(b = -1.35\)

(b) [2.5 marks]
M1 for at least 6 points plotted correctly
A1 for all 9 points plotted correctly
A0.5 for a smooth curve drawn through all points without discontinuity or flat apex

(c)(i) [1 mark]
B1 for value in range \([-1.45, -1.35]\)

(c)(ii) [2 marks]
M1 for drawing the straight line \(y = x - 1\) with at least two correct points
A1 for all three correct intersection values: \(x \in [1.15, 1.25]\), \(x = 2.0\), \(x \in [4.75, 4.85]\) (A0.5 for two correct)

(a) Substitute the given \(x\)-values into \(y = x^2 - \dfrac{6}{x} - 1\):
- For \(x = 1.5\): \(p = (1.5)^2 - \dfrac{6}{1.5} - 1 = 2.25 - 4 - 1 = -2.75\)
- For \(x = 2.5\): \(q = (2.5)^2 - \dfrac{6}{2.5} - 1 = 6.25 - 2.4 - 1 = 2.85\)
- For \(x = 4\): \(r = (4)^2 - \dfrac{6}{4} - 1 = 16 - 1.5 - 1 = 13.5\)

(b) Plot the 8 points \((0.8, -7.86)\), \((1, -6)\), \((1.5, -2.75)\), \((2, 0)\), \((2.5, 2.85)\), \((3, 6)\), \((3.5, 9.54)\), \((4, 13.5)\) and join them with a smooth, continuous curve.

(c) (i) Draw or read across from the horizontal line \(y = 4\) to find the \(x\)-coordinate on the curve: \(x \approx 2.67\) (accept \(2.65\) to \(2.70\)).

(ii) Rearrange \(x^2 - 2x - \dfrac{6}{x} + 2 = 0\) to match the plotted function:
\(x^2 - \dfrac{6}{x} - 1 = 2x - 3\)
Draw the straight line \(y = 2x - 3\) on the grid passing through e.g. \((1, -1)\), \((2, 1)\), \((3, 3)\).
The \(x\)-coordinates of the points of intersection are \(x \approx 1.25\) (accept \(1.20\) to \(1.30\)) and \(x \approx 2.78\) (accept \(2.75\) to \(2.85\)).

(a) B2 for all 3 values correct (B1 for 2 values correct)
(b) M1 for at least 6 points correctly plotted within 1 mm (half a small square)
A1 for all 8 points correctly plotted
A1 for a smooth, single-strand curve drawn through all points from x = 0.8 to x = 4
(c)(i) B1 for x = 2.65 to 2.70 (or FT from their curve)
(c)(ii) M1 for drawing the line y = 2x - 3 with correct intercept or gradient
A1 for both solutions x = 1.20 to 1.30 and x = 2.75 to 2.85 (or FT their intersection points)

(a) The angle \(\angle QPR = 145^\circ - 62^\circ = 83^\circ\).
Using the cosine rule in \(\triangle PQR\):
\[ QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos(\angle QPR) \]
\[ QR^2 = 48^2 + 75^2 - 2(48)(75)\cos(83^\circ) \]
\[ QR^2 = 2304 + 5625 - 7200\cos(83^\circ) = 7929 - 877.459 = 7051.541 \]
\[ QR = \sqrt{7051.541} \approx 83.973\text{ km} \approx 84.0\text{ km} \]

(b) Using the sine rule in \(\triangle PQR\) to find \(\angle PQR\):
\[ \frac{\sin(\angle PQR)}{75} = \frac{\sin(83^\circ)}{83.973} \]
\[ \sin(\angle PQR) = \frac{75 \times \sin(83^\circ)}{83.973} \approx 0.8865 \]
\[ \angle PQR = \arcsin(0.8865) \approx 62.44^\circ \]
The back-bearing of \(P\) from \(Q\) is \(062^\circ + 180^\circ = 242^\circ\).
Therefore, the bearing of \(R\) from \(Q\) is:
\[ 242^\circ - 62.44^\circ = 179.56^\circ \approx 180^\circ \text{ or } 179.6^\circ \]

(a)
M1 for \(\angle QPR = 145^\circ - 62^\circ = 83^\circ\)
M1 for correct implicit cosine rule: \(48^2 + 75^2 - 2(48)(75)\cos(83^\circ)\)
A1 for \(84.0\) or \(83.97\dots\) (accept 84)

(b)
M1 for correct sine rule setup: \(\frac{\sin(\angle PQR)}{75} = \frac{\sin(83^\circ)}{83.97}\) or cosine rule to find \(\angle PQR\)
A1 for \(\angle PQR = 62.4^\circ\) or \(62.44^\circ\)
A1 for bearing \(179.6^\circ\) or \(180^\circ\) (accept \(179.5^\circ\) to \(180^\circ\))

(a) In the right-angled triangle \(VTA\):
\[ TA = \frac{h}{\tan 28^\circ} \]
In the right-angled triangle \(VTB\):
\[ TB = \frac{h}{\tan 35^\circ} \]
Since \(A\) is due South and \(B\) is due East of \(T\), \(\angle ATB = 90^\circ\).
Using Pythagoras' theorem in \(\triangle ATB\):
\[ TA^2 + TB^2 = AB^2 \]
\[ \left(\frac{h}{\tan 28^\circ}\right)^2 + \left(\frac{h}{\tan 35^\circ}\right)^2 = 85^2 \]
\[ h^2 \left( \frac{1}{\tan^2 28^\circ} + \frac{1}{\tan^2 35^\circ} \right) = 7225 \]
\[ h^2 (3.53713 + 2.03961) = 7225 \implies h^2 (5.57674) = 7225 \]
\[ h^2 = \frac{7225}{5.57674} \approx 1295.56 \implies h \approx 35.994\text{ m} = 36.0\text{ m} \text{ (3 s.f.)} \]

(b) Since \(\angle ATB = 90^\circ\) and \(M\) is the midpoint of the hypotenuse \(AB\):
\[ TM = \frac{1}{2} AB = \frac{85}{2} = 42.5\text{ m} \]
In the right-angled triangle \(VTM\):
\[ \tan(\angle VMT) = \frac{VT}{TM} = \frac{35.994}{42.5} \approx 0.8469 \]
\[ \angle VMT = \arctan(0.8469) \approx 40.26^\circ \approx 40.3^\circ \]

(a)
M1 for expressing \(TA = \frac{h}{\tan 28^\circ}\) and \(TB = \frac{h}{\tan 35^\circ}\)
M1 for substituting into Pythagoras: \(\left(\frac{h}{\tan 28^\circ}\right)^2 + \left(\frac{h}{\tan 35^\circ}\right)^2 = 85^2\)
A1 for fully correct calculation leading to \(h = 35.99\dots\) and rounding to \(36.0\)

(b)
M1 for finding \(TM = 42.5\) (or using coordinates / Pythagoras)
M1 for \(\tan(\theta) = \frac{35.994}{42.5}\) or \(\frac{36.0}{42.5}\)
A1 for \(40.3^\circ\) or \(40.26^\circ\dots\)

(a) In \(\triangle ABC\), use the cosine rule:
\[ AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) \]
\[ AC^2 = 65^2 + 82^2 - 2(65)(82)\cos(112^\circ) \]
\[ AC^2 = 4225 + 6724 - 10660(-0.3746) = 10949 + 3993.31 = 14942.31 \]
\[ AC = \sqrt{14942.31} \approx 122.24\text{ m} \approx 122\text{ m} \]

(b) The area of \(\triangle ABC\) is:
\[ \text{Area} = \frac{1}{2} \times 65 \times 82 \times \sin(112^\circ) \approx 2470.95\text{ m}^2 \]
The shortest distance \(d\) from \(B\) to \(AC\) is the perpendicular height:
\[ \text{Area} = \frac{1}{2} \times AC \times d \]
\[ 2470.95 = \frac{1}{2} \times 122.24 \times d \]
\[ d = \frac{2 \times 2470.95}{122.24} \approx 40.43\text{ m} \approx 40.4\text{ m} \]

(a)
M1 for implicit cosine rule: \(65^2 + 82^2 - 2(65)(82)\cos(112^\circ)\)
A1 for \(14942\dots\)
A1 for \(122\) or \(122.2\dots\)

(b)
M1 for area of \(\triangle ABC = \frac{1}{2} \times 65 \times 82 \times \sin(112^\circ)\) or finding an interior angle
M1 for equating area to \(\frac{1}{2} \times AC \times d\) or using \(d = 65\sin(\angle BAC)\)
A1 for \(40.4\) or \(40.42\dots\) to \(40.43\dots\)

(a) Let \(A\) be the origin \((0, 0, 0)\).
Then \(B = (14, 0, 0)\), \(D = (0, 9, 0)\), and \(E = (0, 0, 6)\).
The centre \(M\) of the top face \(EFGH\) has coordinates \((7, 4.5, 6)\).
The distance \(AM\) is:
\[ AM = \sqrt{7^2 + 4.5^2 + 6^2} = \sqrt{49 + 20.25 + 36} = \sqrt{105.25} \approx 10.259\text{ m} \approx 10.3\text{ m} \]

(b) Let \(M'\) be the projection of \(M\) onto the base \(ABCD\), so \(M' = (7, 4.5, 0)\).
The length of \(AM'\) on the base is:
\[ AM' = \sqrt{7^2 + 4.5^2} = \sqrt{69.25} \approx 8.322\text{ m} \]
The angle \(\theta\) between \(AM\) and the horizontal base is \(\angle MAM'\) in right-angled \(\triangle AMM'\):
\[ \tan\theta = \frac{MM'}{AM'} = \frac{6}{\sqrt{69.25}} \approx 0.7210 \]
\[ \theta = \arctan(0.7210) \approx 35.79^\circ \approx 35.8^\circ \]

(a)
M1 for finding half-lengths \(7\) and \(4.5\) (or diagonal of face \(\sqrt{14^2 + 9^2}\))
M1 for \(7^2 + 4.5^2 + 6^2\) or \((\frac{1}{2}\sqrt{277})^2 + 6^2\)
A1 for \(10.3\) or \(10.25\dots\) to \(10.26\dots\)

(b)
M1 for identifying the correct right-angled triangle with vertical height \(6\) and base distance \(\sqrt{69.25}\) (or using \(\sin\theta = \frac{6}{AM}\))
M1 for \(\tan\theta = \frac{6}{\sqrt{69.25}}\) or \(\sin\theta = \frac{6}{\sqrt{105.25}}\) or \(\cos\theta = \frac{\sqrt{69.25}}{\sqrt{105.25}}\)
A1 for \(35.8^\circ\) or \(35.79^\circ\dots\)

(a) \(\angle YXZ = 134^\circ - 078^\circ = 56^\circ\).
The back-bearing of \(X\) from \(Y\) is \(078^\circ + 180^\circ = 258^\circ\).
The bearing of \(Z\) from \(Y\) is \(218^\circ\).
Therefore, \(\angle XYZ = 258^\circ - 218^\circ = 40^\circ\).

(b) In \(\triangle XYZ\), the third angle is:
\[ \angle XZY = 180^\circ - (56^\circ + 40^\circ) = 84^\circ \]
Using the sine rule:
\[ \frac{XZ}{\sin(\angle XYZ)} = \frac{XY}{\sin(\angle XZY)} \]
\[ \frac{XZ}{\sin(40^\circ)} = \frac{120}{\sin(84^\circ)} \]
\[ XZ = \frac{120 \times \sin(40^\circ)}{\sin(84^\circ)} \approx \frac{77.1345}{0.9945} \approx 77.56\text{ m} \approx 77.6\text{ m} \]

(c) The area of \(\triangle XYZ\) is:
\[ \text{Area} = \frac{1}{2} \times XY \times XZ \times \sin(\angle YXZ) \]
\[ \text{Area} = \frac{1}{2} \times 120 \times 77.56 \times \sin(56^\circ) \approx 60 \times 77.56 \times 0.8290 \approx 3858\text{ m}^2 \approx 3860\text{ m}^2 \]

(a)
B1 for \(\angle YXZ = 134^\circ - 78^\circ = 56^\circ\) correctly shown
B1 for \(\angle XYZ = 40^\circ\)

(b)
M1 for correct sine rule setup: \(\frac{XZ}{\sin 40^\circ} = \frac{120}{\sin 84^\circ}\)
A1 for \(77.6\) or \(77.55\dots\) to \(77.56\dots\)

(c)
M1 for area formula: \(\frac{1}{2} \times 120 \times (\text{their } XZ) \times \sin(56^\circ)\)
A1 for \(3860\) or \(3858\dots\)

(a) Using coordinates with \(P = (0, 0, 0)\):
\(Q = (18, 0, 0)\), \(R = (18, 25, 0)\), \(S = (0, 25, 0)\), \(T = (0, 0, 11)\).
The distance \(TR\) is:
\[ TR = \sqrt{(18 - 0)^2 + (25 - 0)^2 + (0 - 11)^2} = \sqrt{18^2 + 25^2 + 11^2} \]
\[ TR = \sqrt{324 + 625 + 121} = \sqrt{1070} \approx 32.71\text{ cm} \approx 32.7\text{ cm} \]

(b) The projection of \(T\) onto the base \(PQRS\) is \(P\).
The horizontal distance \(PR\) is:
\[ PR = \sqrt{18^2 + 25^2} = \sqrt{949} \approx 30.81\text{ cm} \]
In right-angled \(\triangle TPR\):
\[ \tan(\angle TRP) = \frac{PT}{PR} = \frac{11}{\sqrt{949}} \approx 0.3571 \]
\[ \angle TRP = \arctan(0.3571) \approx 19.65^\circ \approx 19.7^\circ \]

(c) The coordinates of \(K\) are \((0, 25, 4)\) and \(T = (0, 0, 11)\).
The horizontal distance between \(K\) and \(T\) is \(25\text{ cm}\).
The vertical difference in height is \(11 - 4 = 7\text{ cm}\).
The angle of elevation \(\phi\) is:
\[ \tan\phi = \frac{7}{25} = 0.28 \]
\[ \phi = \arctan(0.28) \approx 15.64^\circ \approx 15.6^\circ \]

(a)
M1 for \(PR = \sqrt{18^2 + 25^2} = \sqrt{949}\) (or \(30.8\dots\))
M1 for \(TR = \sqrt{949 + 11^2} = \sqrt{1070}\)
A1 for \(32.7\) or \(32.71\dots\)

(b)
M1 for \(\tan\theta = \frac{11}{\sqrt{949}}\) or \(\sin\theta = \frac{11}{\sqrt{1070}}\) or \(\cos\theta = \frac{\sqrt{949}}{\sqrt{1070}}\)
A1 for \(19.7^\circ\) or \(19.65^\circ\dots\)

(c)
B1 for \(\tan\phi = \frac{7}{25} \implies \phi = 15.6^\circ\) or \(15.64^\circ\dots\)

For the class \(120 < m \le 130\):
- Class width = \(130 - 120 = 10\)
- Frequency density \(= \frac{33}{10} = 3.3\)
- Width on histogram \(= 2\text{ cm} \implies 1\text{ unit on } m\text{-axis} = \frac{2}{10} = 0.2\text{ cm}\)
- Height on histogram \(= 6.6\text{ cm} \implies 1\text{ unit of frequency density} = \frac{6.6}{3.3} = 2\text{ cm}\)

For the class \(145 < m \le 170\):
- Class width \(= 170 - 145 = 25\)
- Width of bar \(= 25 \times 0.2 = 5\text{ cm}\)
- Frequency density \(= \frac{24}{25} = 0.96\)
- Height of bar \(= 0.96 \times 2 = 1.92\text{ cm}\)

B1: for width \(= 5\text{ cm}\)
M1: for finding frequency density \(= \frac{24}{25} = 0.96\) or for scale factor \(\frac{6.6}{3.3} = 2\) (or equivalent ratio)
A1: for height \(= 1.92\text{ cm}\)

Find midpoints of each interval:
- \(0 < d \le 5\): midpoint \(= 2.5\)
- \(5 < d \le 15\): midpoint \(= 10\)
- \(15 < d \le 25\): midpoint \(= 20\)
- \(25 < d \le 40\): midpoint \(= 32.5\)

Calculate \(\sum f x\):
\[ \sum f x = (16 \times 2.5) + (28 \times 10) + (22 \times 20) + (14 \times 32.5) \]
\[ \sum f x = 40 + 280 + 440 + 455 = 1215 \]

Calculate the estimated mean:
\[ \text{Mean} = \frac{1215}{80} = 15.1875 \]
Rounding to 3 significant figures gives \(15.2\text{ km}\).

M1: for identifying correct midpoints \(2.5, 10, 20, 32.5\) (at least 3 correct)
M1: (dep on previous M1) for \(\sum f x\) calculation \(= 16(2.5) + 28(10) + 22(20) + 14(32.5) (= 1215)\) divided by 80
A1: for \(15.2\) or \(15.1875\) or \(15.19\)

To estimate the number of visitors aged between 25 and 45 years:

1. From \(20 < x \le 30\), the interval from \(x = 25\) to \(x = 30\) has width \(30 - 25 = 5\).
\[ \text{Frequency} = 5 \times 3.6 = 18 \]

2. From \(30 < x \le 50\), the interval from \(x = 30\) to \(x = 45\) has width \(45 - 30 = 15\).
\[ \text{Frequency} = 15 \times 2.2 = 33 \]

3. Total estimated visitors:
\[ 18 + 33 = 51 \]

M1: for calculating \(5 \times 3.6 (= 18)\)
M1: for calculating \(15 \times 2.2 (= 33)\)
A1: for \(51\)

For the class \(4 < t \le 8\):
- Class width \(= 8 - 4 = 4\)
- Frequency density \(= \frac{56}{4} = 14\)
- Since height \(= 7\text{ cm}\), the scale factor is \(\frac{7}{14} = 0.5\text{ cm}\) per unit of frequency density (or \(\text{frequency density} = 2 \times \text{height}\)).

For the class \(12 < t \le 20\):
- Height \(= 2.5\text{ cm} \implies \text{frequency density} = 2.5 \times 2 = 5\)
- Class width \(= 20 - 12 = 8\)
- \(b = 5 \times 8 = 40\)

The total number of customers is 200:
\[ a + 56 + 64 + b + 20 = 200 \]
\[ a + 56 + 64 + 40 + 20 = 200 \]
\[ a + 180 = 200 \implies a = 20 \]

M1: for frequency density of \(4 < t \le 8\) as \(\frac{56}{4} = 14\) or relation \(1\text{ cm} = 2\text{ units of FD}\)
M1: for \(b = 2.5 \times 2 \times 8\) or equivalent method
A1: for \(b = 40\)
A1: for \(a = 20\)

(a)(i) \(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{45}{x}\text{ hours}\).

(a)(ii) \(\text{Time} = \frac{45}{x - 3}\text{ hours}\).

(b) \(45\text{ minutes} = \frac{45}{60} = \frac{3}{4}\text{ hours}\).

Setting up the difference in time:
\(\frac{45}{x - 3} - \frac{45}{x} = \frac{3}{4}\)

Combine the fractions over a common denominator:
\(\frac{45x - 45(x - 3)}{x(x - 3)} = \frac{3}{4}\)
\(\frac{45x - 45x + 135}{x(x - 3)} = \frac{3}{4}\)
\(\frac{135}{x^2 - 3x} = \frac{3}{4}\)

Cross-multiply:
\(3(x^2 - 3x) = 4 \times 135\)
\(3x^2 - 9x = 540\)
\(3x^2 - 9x - 540 = 0\)

Divide the entire equation by 3:
\(x^2 - 3x - 180 = 0\).

(c) Factorising the quadratic equation:
\((x - 15)(x + 12) = 0\)
\(x = 15\) or \(x = -12\)

Since speed must be positive, \(x = 15\text{ km/h}\).

(a)(i) B1: for \(\frac{45}{x}\)
(a)(ii) B1: for \(\frac{45}{x - 3}\)
(b) M1: for correct algebraic expression for time difference set equal to \(\frac{3}{4}\) or \(0.75\), e.g. \(\frac{45}{x-3} - \frac{45}{x} = \frac{3}{4}\)
A1: for correct expansion and clear steps shown reaching \(x^2 - 3x - 180 = 0\) with no steps omitted
(c) B1: for \(15\) (ignore \(-12\) if seen, must select positive value)

Multiply all terms by \((2x - 1)(x + 2)\) to eliminate denominators:
\(4(x + 2) - 3(2x - 1) = 1(2x - 1)(x + 2)\)

Expand both sides:
\(4x + 8 - 6x + 3 = 2x^2 + 4x - x - 2\)
\(-2x + 11 = 2x^2 + 3x - 2\)

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):
\(2x^2 + 3x + 2x - 2 - 11 = 0\)
\(2x^2 + 5x - 13 = 0\)

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(a = 2, b = 5, c = -13\)
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-13)}}{2(2)}\)
\(x = \frac{-5 \pm \sqrt{25 + 104}}{4}\)
\(x = \frac{-5 \pm \sqrt{129}}{4}\)

\(x = \frac{-5 + 11.3578...}{4} = 1.589... \approx 1.59\)
\(x = \frac{-5 - 11.3578...}{4} = -4.089... \approx -4.09\)

M1: for attempting common denominator or multiplying by \((2x - 1)(x + 2)\), e.g. \(4(x + 2) - 3(2x - 1)\)
M1: for expanding brackets correctly on both sides, reaching \(-2x + 11 = 2x^2 + 3x - 2\) or equivalent
A1: for correct 3-term quadratic equation in standard form: \(2x^2 + 5x - 13 = 0\)
M1: for correct substitution into quadratic formula \(\frac{-5 \pm \sqrt{5^2 - 4(2)(-13)}}{2(2)}\) or \(\frac{-5 \pm \sqrt{129}}{4}\) (FT their 3-term quadratic)
A1: for both \(1.59\) and \(-4.09\) correct to 2 decimal places

(a) The total width including the path is:
\(w + 2(1.5) = w + 3\text{ m}\).

The total length including the path is:
\((w + 3) + 2(1.5) = w + 3 + 3 = w + 6\text{ m}\).

The total combined area is:
\(\text{Area} = (w + 3)(w + 6)\)

Expanding the brackets:
\((w + 3)(w + 6) = w^2 + 6w + 3w + 18 = w^2 + 9w + 18\)

Given that the total area is \(70\text{ m}^2\):
\(w^2 + 9w + 18 = 70\)
\(w^2 + 9w + 18 - 70 = 0\)
\(w^2 + 9w - 52 = 0\).

(b) Finding factors of \(-52\) that sum to \(+9\):
\(13 \times (-4) = -52\) and \(13 + (-4) = 9\).
Thus, \(w^2 + 9w - 52 = (w + 13)(w - 4)\).

(c) Setting \((w + 13)(w - 4) = 0\) gives \(w = -13\) or \(w = 4\).
Since a physical width cannot be negative, the width of the lawn is \(4\text{ m}\).

(a) M1: for identifying outer dimensions as \(w + 3\) and \(w + 6\)
M1: for setting up area equation \((w + 3)(w + 6) = 70\)
A1: for expanding brackets correctly to \(w^2 + 9w + 18 = 70\) and concluding with \(w^2 + 9w - 52 = 0\) with all steps clearly shown
(b) B1: for \((w + 13)(w - 4)\)
(c) B1: for \(4\) (must reject \(-13\))