2023 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Divide the numerical parts and subtract the powers of 10:
\[\frac{4.8}{1.5} = 3.2\]
\[10^7 \div 10^{-3} = 10^{7 - (-3)} = 10^{10}\]
Combining these gives \(3.2 \times 10^{10}\).

M1 for \(\frac{4.8}{1.5}\) or \(10^{7 - (-3)}\) seen or \(32\,000\,000\,000\)
A0.5 for \(3.2 \times 10^{10}\) cao

First factor out the highest common factor, \(2xy\):
\[18x^3y - 8xy^3 = 2xy(9x^2 - 4y^2)\]
Then factorise the difference of two squares:
\[9x^2 - 4y^2 = (3x - 2y)(3x + 2y)\]
Thus, the fully factorised form is \(2xy(3x - 2y)(3x + 2y)\).

M1 for \(2xy(9x^2 - 4y^2)\) or for recognising difference of two squares \((3x - 2y)(3x + 2y)\)
A0.5 for \(2xy(3x - 2y)(3x + 2y)\) oe

First calculate the arc length of the sector:
\[\text{Arc length} = \frac{144}{360} \times 2 \times \pi \times 7.5 = \frac{2}{5} \times 15\pi = 6\pi \approx 18.8496\text{ cm}\]
The perimeter includes the arc length plus two radii:
\[\text{Perimeter} = 6\pi + 2(7.5) = 6\pi + 15 \approx 33.85\text{ cm} \text{ (or } 33.8\text{ to 3 s.f.)}\]

M1 for \(\frac{144}{360} \times 2 \times \pi \times 7.5\) oe (soi by \(6\pi\) or \(18.8\dots\))
A0.5 for \(33.8\) or \(33.85\) or \(6\pi + 15\)

The magnitude of the vector is given by:
\[|\mathbf{v}| = \sqrt{k^2 + (-12)^2} = 13\]
Square both sides:
\[k^2 + 144 = 169\]
\[k^2 = 25\]
Since \(k > 0\), \(k = 5\).

M1 for \(k^2 + (-12)^2 = 13^2\) oe
A0.5 for \(5\) (do not accept \(\pm 5\))

The reduced price represents \(100\% - 15\% = 85\%\) of the original price.
\[\text{Original price} = \frac{61.20}{0.85} = 72\]

M1 for \(61.20 \div (1 - 0.15)\) or \(61.20 \div 0.85\) oe
A0.5 for \(72\) or \(72.00\)

Divide the numerical coefficients and subtract the indices for the power of 10: \((4.8 \div 1.5) \times 10^{7 - (-3)} = 3.2 \times 10^{10}\).

M1 for \(3.2 \times 10^k\) or \(k \times 10^{10}\) (where \(k \ne 3.2\)) or \(32\,000\,000\,000\). A0.5 for \(3.2 \times 10^{10}\) cao.

Identify the highest common factor of \(18x^2y\) and \(24xy^3\), which is \(6xy\). Dividing each term by \(6xy\) gives \(6xy(3x - 4y^2)\).

M1 for any partial factorisation with at least two common factors extracted (e.g. \(3xy(6x - 8y^2)\) or \(6x(3xy - 4y^3)\) or \(6y(3x^2 - 4xy^2)\)). A0.5 for \(6xy(3x - 4y^2)\) cao.

The sequence has a constant first difference of \(d = 6\), so the rule starts with \(6n\). The zeroth term is \(7 - 6 = 1\). Therefore, the \(n\)th term is \(6n + 1\).

M1 for \(6n + c\) (where \(c \ne 1\)) or \(6n\) seen. A0.5 for \(6n + 1\) oe.

Use the magnitude formula for a vector: \(|\mathbf{v}| = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\).

M1 for \(\sqrt{(-8)^2 + 15^2}\) or \(\sqrt{64 + 225}\) or \(\sqrt{289}\). A0.5 for 17 cao.

Upper bound of length = \(24.5\text{ cm}\). Upper bound of width = \(15.5\text{ cm}\). Upper bound of perimeter = \(2 \times (24.5 + 15.5) = 2 \times 40 = 80\text{ cm}\).

M1 for \(24.5\) and \(15.5\) seen or used in a perimeter calculation \(2(l + w)\). A0.5 for 80 cao.

1. Find the common denominator:
\[ \frac{5(x + 4) - 2(2x - 3)}{(2x - 3)(x + 4)} \]

2. Expand the numerator:
\[ 5(x + 4) - 2(2x - 3) = 5x + 20 - 4x + 6 \]

3. Collect like terms:
\[ 5x - 4x + 20 + 6 = x + 26 \]

4. Combine into a single fraction:
\[ \frac{x + 26}{(2x - 3)(x + 4)} \]

M1 for writing over a common denominator \((2x - 3)(x + 4)\) with at least one numerator correct
M1 for correct expansion of numerators, \(5x + 20 - 4x + 6\) soi
A1 for \(\frac{x + 26}{(2x - 3)(x + 4)}\) or \(\frac{x + 26}{2x^2 + 5x - 12}\)

1. The perimeter of a sector is given by:
\[ \text{Perimeter} = 2r + \text{arc length} \]
\[ 28.2 = 2(7.5) + \text{arc length} \]
\[ \text{Arc length} = 28.2 - 15 = 13.2\text{ cm} \]

2. Use the arc length formula to find the angle \(\theta\):
\[ \text{Arc length} = \frac{\theta}{360} \times 2\pi r \]
\[ 13.2 = \frac{\theta}{360} \times 2 \times \pi \times 7.5 \]
\[ 13.2 = \frac{15\pi\theta}{360} \]

3. Solve for \(\theta\):
\[ \theta = \frac{13.2 \times 360}{15\pi} \approx 100.8406^\circ \]

Correct to 1 decimal place, \(\theta = 100.8^\circ\).

M1 for \(28.2 - 2 \times 7.5\) or \(13.2\) soi
M1 for \(\frac{\theta}{360} \times 2 \times \pi \times 7.5 = \text{their } 13.2\) oe
A1 for \(100.8\) or \(100.84\dots\)

1. Express all terms as powers of base 3:
\[ 27 = 3^3 \implies 27^{2x - 1} = (3^3)^{2x - 1} = 3^{3(2x - 1)} = 3^{6x - 3} \]
\[ 9\sqrt{3} = 3^2 \times 3^{\frac{1}{2}} = 3^{2.5} = 3^{\frac{5}{2}} \]
\[ \frac{1}{9\sqrt{3}} = 3^{-\frac{5}{2}} = 3^{-2.5} \]

2. Equate indices:
\[ 6x - 3 = -2.5 \]
\[ 6x = 0.5 \]
\[ x = \frac{0.5}{6} = \frac{1}{12} \]

M1 for writing \(27^{2x - 1}\) as \(3^{3(2x - 1)}\) or \(3^{6x - 3}\) soi
M1 for writing \(\frac{1}{9\sqrt{3}}\) as \(3^{-2.5}\) or \(3^{-\frac{5}{2}}\) soi
A1 for \(\frac{1}{12}\) or \(0.0833\) or \(0.0833\dots\)

1. Determine the bounds for distance and time:
- Distance: \(480\text{ m}\) to nearest \(10\text{ m}\) \(\implies \text{Upper bound} = 480 + 5 = 485\text{ m}\)
- Time: \(18.4\text{ s}\) to nearest \(0.1\text{ s}\) \(\implies \text{Lower bound} = 18.4 - 0.05 = 18.35\text{ s}\)

2. Calculate the upper bound of speed:
\[ \text{Upper bound of speed} = \frac{\text{Upper bound of distance}}{\text{Lower bound of time}} = \frac{485}{18.35} \]

3. Evaluate:
\[ \frac{485}{18.35} \approx 26.4305... \]

Correct to 3 significant figures, the value is \(26.4\).

B1 for \(485\) or \(18.35\) seen
M1 for \(\frac{\text{UB of distance}}{\text{LB of time}}\) where \(480 < \text{UB} \le 485\) and \(18.35 \le \text{LB} < 18.4\)
A1 for \(26.4\) or \(26.43\dots\)

1. Find first differences:
\(11 - 3 = 8\), \(23 - 11 = 12\), \(39 - 23 = 16\), \(59 - 39 = 20\)

2. Find second differences:
\(12 - 8 = 4\), \(16 - 12 = 4\), \(20 - 16 = 4\)

Since the second difference is constant (\(4\)), the sequence is quadratic with leading term \(a n^2\):
\[ 2a = 4 \implies a = 2 \]

3. Subtract \(2n^2\) from each term:
- For \(n = 1\): \(3 - 2(1)^2 = 1\)
- For \(n = 2\): \(11 - 2(2)^2 = 3\)
- For \(n = 3\): \(23 - 2(3)^2 = 5\)
- For \(n = 4\): \(39 - 2(4)^2 = 7\)

4. Find the \(n\)th term of the linear remainder \(1, 3, 5, 7, \dots\):
\[ \text{Linear part} = 2n - 1 \]

5. Combine the parts:
\[ T_n = 2n^2 + 2n - 1 \]

M1 for second difference \(= 4\) soi, giving \(2n^2\) as the first term
M1 for subtracting \(2n^2\) from the sequence terms to obtain \(1, 3, 5, 7, \dots\) oe
A1 for \(2n^2 + 2n - 1\) oe

To subtract the algebraic fractions, find a common denominator:
\[ \frac{3(x + 4) - 2(2x - 1)}{(2x - 1)(x + 4)} \]

Expand the brackets in the numerator:
\[ 3(x + 4) = 3x + 12 \]
\[ -2(2x - 1) = -4x + 2 \]

Combine like terms in the numerator:
\[ 3x + 12 - 4x + 2 = 14 - x \]

Write the resulting fraction over the common denominator:
\[ \frac{14 - x}{(2x - 1)(x + 4)} \]

M1 for writing with a common denominator \((2x - 1)(x + 4)\) seen
M1 for correct expansion of numerators: \(3(x + 4) - 2(2x - 1)\) or \(3x + 12 - 4x + 2\) soi
A1 for \(\frac{14 - x}{(2x - 1)(x + 4)}\) or \(\frac{-x + 14}{(2x - 1)(x + 4)}\) or \(\frac{14 - x}{2x^2 + 7x - 4}\) oe

First, calculate the arc length of the sector:
\[ \text{Arc length} = \frac{140}{360} \times 2 \times \pi \times 7.5 = \frac{7}{18} \times 15\pi = \frac{35\pi}{6} \approx 18.326\text{ cm} \]

Next, calculate the total perimeter by adding the two radii:
\[ \text{Perimeter} = \text{Arc length} + 2r \]
\[ \text{Perimeter} = 18.326 + 2 \times 7.5 = 18.326 + 15 = 33.326\text{ cm} \]

Rounding to 3 significant figures gives \(33.3\text{ cm}\).

M1 for \(\frac{140}{360} \times 2 \times \pi \times 7.5\) oe
M1 for \((\text{their arc length}) + 2 \times 7.5\) oe
A1 for \(33.3\) or \(33.32\dots\) to \(33.33\) or \(\frac{35\pi}{6} + 15\)

Identify the upper bound for each measurement:
- Length is given to the nearest metre: \(\text{Upper bound of length} = 68 + 0.5 = 68.5\text{ m}\)
- Width is given to 1 decimal place (nearest \(0.1\text{ m}\)): \(\text{Upper bound of width} = 42.4 + 0.05 = 42.45\text{ m}\)

Calculate the upper bound for the area by multiplying the upper bounds:
\[ \text{Upper bound for area} = 68.5 \times 42.45 = 2907.825\text{ m}^2 \]

B1 for \(68.5\) or \(42.45\) seen
M1 for \((\text{their upper bound of length}) \times (\text{their upper bound of width})\) where \(68 < \text{UB of length} \le 68.5\) and \(42.4 < \text{UB of width} \le 42.45\)
A1 for \(2907.825\) cao

1. Multiply through by the common denominator \((x - 1)(x + 2)\):
\[2(x + 2) + 3(x - 1) = 2(x - 1)(x + 2)\]

2. Expand the brackets:
\[2x + 4 + 3x - 3 = 2(x^2 + x - 2)\]
\[5x + 1 = 2x^2 + 2x - 4\]

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

4. Solve by factorising or using the quadratic formula:
\[(2x - 5)(x + 1) = 0\]

\[2x - 5 = 0 \implies x = 2.5\]
\[x + 1 = 0 \implies x = -1\]

M1 for multiplying by common denominator to obtain \(2(x + 2) + 3(x - 1) = 2(x - 1)(x + 2)\) oe
M1 for expanding and simplifying to a correct 3-term quadratic \(2x^2 - 3x - 5 = 0\) oe
M1 for factorising \((2x - 5)(x + 1) = 0\) or correct substitution into the quadratic formula for their 3-term quadratic
A1 for \(x = -1\) and \(x = 2.5\) (or \(\frac{5}{2}\) or \(2\frac{1}{2}\))

1. Find the area of sector \(OAB\):
\[\text{Area of sector } OAB = \dfrac{75}{360} \times \pi \times 12^2 = 30\pi \approx 94.2478\text{ cm}^2\]

2. Find the area of triangle \(OAC\):
\[\text{Area of } \triangle OAC = \dfrac{1}{2} \times 12 \times 12 \times \sin(45^\circ) = 72 \times \dfrac{\sqrt{2}}{2} = 36\sqrt{2} \approx 50.9117\text{ cm}^2\]

3. Find the area of triangle \(OCB\):
\[\text{Area of } \triangle OCB = \dfrac{1}{2} \times 12 \times 12 \times \sin(30^\circ) = 72 \times 0.5 = 36\text{ cm}^2\]

4. Calculate the area of the shaded region:
\[\text{Shaded Area} = \text{Area of sector } OAB - (\text{Area of } \triangle OAC + \text{Area of } \triangle OCB)\]
\[\text{Shaded Area} = 30\pi - 36\sqrt{2} - 36 \approx 94.2478 - 50.9117 - 36 = 7.3361\text{ cm}^2\]

Rounded to 3 significant figures: \(7.34\text{ cm}^2\).

M1 for area of sector \(\frac{75}{360} \times \pi \times 12^2\) soi (\(30\pi\) or \(94.2\dots\))
M1 for area of \(\triangle OAC = \frac{1}{2} \times 12^2 \times \sin(45^\circ)\) soi (\(36\sqrt{2}\) or \(50.9\dots\))
M1 for area of \(\triangle OCB = \frac{1}{2} \times 12^2 \times \sin(30^\circ)\) soi (\(36\))
M1 (dep on previous 3 method marks) for Sector Area \(- (\text{Area } \triangle OAC + \text{Area } \triangle OCB)\)
A1 for \(7.34\) or \(7.336\dots\) (accept exact \(30\pi - 36\sqrt{2} - 36\))

(a)
\[\vec{OC} = \vec{OA} + \vec{AC} = \mathbf{a} + (2\mathbf{b} - \mathbf{a}) = 2\mathbf{b}\]

(b)
Since \(OP : PC = 3 : 1\), \(\vec{OP} = \dfrac{3}{4}\vec{OC} = \dfrac{3}{4}(2\mathbf{b}) = \dfrac{3}{2}\mathbf{b}\).

\(M\) is the midpoint of \(AB\):
\[\vec{AB} = -\mathbf{a} + \mathbf{b}\]
\[\vec{OM} = \vec{OA} + \dfrac{1}{2}\vec{AB} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\]

Now find \(\vec{PM}\):
\[\vec{PM} = \vec{PO} + \vec{OM} = -\vec{OP} + \vec{OM}\]
\[\vec{PM} = -\dfrac{3}{2}\mathbf{b} + \left(\dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\right) = \dfrac{1}{2}\mathbf{a} - \mathbf{b}\]

(a) B1 for \(2\mathbf{b}\)
(b) M1 for \(\vec{OP} = \frac{3}{2}\mathbf{b}\) or \(\vec{PO} = -\frac{3}{2}\mathbf{b}\) soi
M1 for \(\vec{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\) or any valid vector route for \(\vec{PM}\) (e.g. \(\vec{PO} + \vec{OA} + \vec{AM}\))
A1 for \(\frac{1}{2}\mathbf{a} - \mathbf{b}\) or \(\frac{1}{2}(\mathbf{a} - 2\mathbf{b})\) in simplest form

(a)
The total surface area of a closed cylinder is given by:
\[A = 2\pi r^2 + 2\pi r h\]
Given \(A = 96\pi\):
\[2\pi r^2 + 2\pi r h = 96\pi\]
Divide through by \(2\pi\):
\[r^2 + rh = 48\]
\[rh = 48 - r^2\]
\[h = \dfrac{48 - r^2}{r}\]

(b)
To find the maximum volume, differentiate \(V\) with respect to \(r\):
\[V = 48\pi r - \pi r^3\]
\[\dfrac{\mathrm{d}V}{\mathrm{d}r} = 48\pi - 3\pi r^2\]
Set the derivative equal to 0 for a stationary point:
\[48\pi - 3\pi r^2 = 0\]
\[3\pi r^2 = 48\pi\]
\[r^2 = 16\]
Since \(r > 0\), \(r = 4\).

Substitute \(r = 4\) into the volume formula:
\[V = 48\pi(4) - \pi(4)^3 = 192\pi - 64\pi = 128\pi\text{ cm}^3\]

(a) M1 for equating surface area formula to \(96\pi\): \(2\pi r^2 + 2\pi rh = 96\pi\)
A1 for complete correct algebraic rearrangement to show \(h = \frac{48 - r^2}{r}\)
(b) M1 for correct differentiation of at least one term: \(\frac{\mathrm{d}V}{\mathrm{d}r} = 48\pi - 3\pi r^2\)
M1 for setting their derivative equal to 0 and solving for \(r\)
A1 for \(r = 4\) and \(V = 128\pi\) (accept \(402\) or \(402.1\dots\))

(a)(i) \(\vec{OB} = \vec{OC} + \vec{CB} = 4\mathbf{c} + 6\mathbf{a} = 6\mathbf{a} + 4\mathbf{c}\)

(ii) \(\vec{AB} = \vec{AO} + \vec{OB} = -3\mathbf{a} + (6\mathbf{a} + 4\mathbf{c}) = 3\mathbf{a} + 4\mathbf{c}\)

(iii) \(\vec{AN} = \frac{1}{3}\vec{AB} = \frac{1}{3}(3\mathbf{a} + 4\mathbf{c}) = \mathbf{a} + \frac{4}{3}\mathbf{c}\)
\(\vec{ON} = \vec{OA} + \vec{AN} = 3\mathbf{a} + \left(\mathbf{a} + \frac{4}{3}\mathbf{c}\right) = 4\mathbf{a} + \frac{4}{3}\mathbf{c}\)

(b) \(\vec{OP} = \frac{3}{4}\vec{OC} = \frac{3}{4}(4\mathbf{c}) = 3\mathbf{c}\)
\(\vec{PN} = \vec{ON} - \vec{OP} = \left(4\mathbf{a} + \frac{4}{3}\mathbf{c}\right) - 3\mathbf{c} = 4\mathbf{a} - \frac{5}{3}\mathbf{c} = \frac{1}{3}(12\mathbf{a} - 5\mathbf{c})\)
\(\vec{OM} = \vec{OC} + \frac{1}{2}\vec{CB} = 4\mathbf{c} + 3\mathbf{a} = 3\mathbf{a} + 4\mathbf{c}\)
\(\vec{PM} = \vec{OM} - \vec{OP} = (3\mathbf{a} + 4\mathbf{c}) - 3\mathbf{c} = 3\mathbf{a} + \mathbf{c}\)
Wait, check \(\vec{NM} = \vec{OM} - \vec{ON} = (3\mathbf{a} + 4\mathbf{c}) - (4\mathbf{a} + \frac{4}{3}\mathbf{c}) = -\mathbf{a} + \frac{8}{3}\mathbf{c}\).
Let's check \(\vec{PN}\) and \(\vec{PM}\):
\(\vec{PN} = 4\mathbf{a} - \frac{5}{3}\mathbf{c}\)
\(\vec{PM} = 3\mathbf{a} + \mathbf{c}\)
Notice \(\vec{PN}\) vs \(\vec{NM}\): \(\vec{PN} = \frac{4}{3}(3\mathbf{a} + \dots)\)
Alternatively, with \(\vec{PN} = 4\mathbf{a} - \frac{5}{3}\mathbf{c}\), collinearity requires \(\vec{PN} = k \vec{PM}\). Since \(\frac{4}{3} \ne -\frac{5}{3}\), let's ensure the scalar multiples match:
\(\vec{PN} = \vec{PO} + \vec{ON} = -3\mathbf{c} + 4\mathbf{a} + \frac{4}{3}\mathbf{c} = 4\mathbf{a} - \frac{5}{3}\mathbf{c}\).
Since \(\vec{PM} = \vec{PO} + \vec{OM} = -3\mathbf{c} + (4\mathbf{c} + 3\mathbf{a}) = 3\mathbf{a} + \mathbf{c}\).
To make them collinear, \(P, N, M\) must be on a straight line: \(\vec{PN} = k \vec{NM}\).
Here \(\vec{PN} = 4\mathbf{a} - \frac{5}{3}\mathbf{c}\) and \(\vec{NM} = -\mathbf{a} + \frac{8}{3}\mathbf{c}\), summing to \(3\mathbf{a} + \mathbf{c}\).

(c) Triangles \(OAN\) and \(OAB\) share the common vertex \(O\) and their bases \(AN\) and \(AB\) lie on the same straight line.
Therefore, \(\frac{\text{Area}(\triangle OAN)}{\text{Area}(\triangle OAB)} = \frac{AN}{AB} = \frac{1}{1+2} = \frac{1}{3}\).
The ratio is \(1 : 3\).

(a)(i) B1 for \(6\mathbf{a} + 4\mathbf{c}\) oe
(a)(ii) B1 for \(3\mathbf{a} + 4\mathbf{c}\) oe
(a)(iii) M1 for \(\vec{OA} + \frac{1}{3}\vec{AB}\) oe
A1 for \(4\mathbf{a} + \frac{4}{3}\mathbf{c}\) oe
(b) B1 for \(\vec{OP} = 3\mathbf{c}\) or \(\vec{OM} = 3\mathbf{a} + 4\mathbf{c}\)
M1 for finding vector expressions for any two of \(\vec{PN}\), \(\vec{NM}\), \(\vec{PM}\)
A1 for correct simplified expressions and full conclusion stating common point and scalar multiple (or showing collinearity)
(c) M1 for identifying ratio of areas equals ratio of bases \(AN : AB\) oe
A1 for \(1 : 3\) cao

(a) Total surface area consists of the circular base \(\pi r^2\), the curved surface area of the cylinder \(2\pi r h\), and the curved surface area of the hemisphere \(2\pi r^2\).
\(\text{Total Surface Area} = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\)
Given total area \(= 180\pi\):
\(3\pi r^2 + 2\pi r h = 180\pi\)
Divide through by \(\pi\):
\(3r^2 + 2rh = 180\)
\(2rh = 180 - 3r^2\)
\(h = \frac{180 - 3r^2}{2r}\)

(b)(i) \(V = 90\pi r - \frac{5}{6}\pi r^3\)
\(\frac{\mathrm{d}V}{\mathrm{d}r} = 90\pi - 3 \times \frac{5}{6}\pi r^2 = 90\pi - \frac{5}{2}\pi r^2\)

(ii) For maximum volume, \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\):
\(90\pi - \frac{5}{2}\pi r^2 = 0\)
\(\frac{5}{2}r^2 = 90\)
\(r^2 = \frac{90 \times 2}{5} = 36\)
Since \(r > 0\), \(r = 6\).

(iii) When \(r = 6\):
\(V = 90\pi(6) - \frac{5}{6}\pi(6)^3 = 540\pi - 180\pi = 360\pi \approx 1130.97\dots\)
To the nearest integer, \(V = 1131\text{ cm}^3\).

(c) Volume of the sphere \(= \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (1.5)^3 = \frac{4}{3}\pi (3.375) = 4.5\pi\text{ cm}^3\).
Volume of one cone \(= \frac{1}{3}\pi r_c^2 h_c = \frac{1}{3}\pi (0.5)^2 (1.2) = \frac{1}{3}\pi (0.25)(1.2) = 0.1\pi\text{ cm}^3\).
Number of cones \(= \frac{4.5\pi}{0.1\pi} = 45\).

(a) M1 for \(\pi r^2 + 2\pi r h + 2\pi r^2 = 180\pi\) oe
A1 for fully correct algebraic rearrangement leading to \(h = \frac{180 - 3r^2}{2r}\) with no steps omitted
(b)(i) M1 for differentiating at least one term correctly: \(90\pi\) or \(-\frac{5}{2}\pi r^2\)
A1 for \(90\pi - \frac{5}{2}\pi r^2\) oe
(b)(ii) M1 for setting their \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\)
A1 for \(r = 6\) (ignore \(-6\))
(b)(iii) M1 for substituting their \(r = 6\) into formula for \(V\)
A1 for \(1131\) or \(360\pi\) rounded to nearest integer
(c) M1 for volume of sphere: \(\frac{4}{3}\pi (1.5)^3\) soi (\(4.5\pi\) or \(14.137...\))
M1 for volume of cone: \(\frac{1}{3}\pi (0.5)^2(1.2)\) soi (\(0.1\pi\) or \(0.3141...\))
A1 for \(45\) cao

(a) Time \(= \frac{\text{Distance}}{\text{Speed}} = \frac{48}{x}\) hours.

(b) Time \(= \frac{48}{x+4}\) hours.

(c)(i) \(\frac{48}{x} + \frac{48}{x+4} = 5\)
Multiply throughout by \(x(x+4)\):
\(48(x+4) + 48x = 5x(x+4)\)
\(48x + 192 + 48x = 5x^2 + 20x\)
\(96x + 192 = 5x^2 + 20x\)
\(5x^2 + 20x - 96x - 192 = 0\)
\(5x^2 - 76x - 192 = 0\)
Wait, let's re-verify: \(48(x+4) + 48x = 5x(x+4) \implies 96x + 192 = 5x^2 + 20x \implies 5x^2 - 76x - 192 = 0\).
Let's check the quadratic in question: \(5x^2 + 20x - 384 = 0\) comes from \(\frac{48}{x} - \frac{48}{x+4} = \dots\) or difference: if difference is \(2\) hours: \(\frac{48}{x} - \frac{48}{x+4} = 2 \implies 48(4) = 2x(x+4) \implies 192 = 2x^2 + 8x \implies x^2 + 4x - 96 = 0\).
With total time = 5 hours:
\(5x^2 - 76x - 192 = 0\).
Using quadratic formula for \(5x^2 + 20x - 384 = 0\):
\(x = \frac{-20 \pm \sqrt{20^2 - 4(5)(-384)}}{2(5)} = \frac{-20 \pm \sqrt{400 + 7680}}{10} = \frac{-20 \pm \sqrt{8080}}{10}\)
\(\sqrt{8080} \approx 89.8888\)
\(x = \frac{-20 + 89.8888}{10} = 6.98888... \approx 6.99\) or \(x = \frac{-20 - 89.8888}{10} = -10.9888... \approx -10.99\).
For (c)(iii): Outward time \(= \frac{48}{6.98888} \approx 6.8680\text{ hours} = 6\text{ hours} + 0.8680 \times 60\text{ minutes} = 6\text{ hours } 52.08\dots\text{ minutes} \approx 6\text{ hours } 52\text{ minutes}\) (or \(6\text{ hours } 53\text{ minutes}\) using rounded \(x\)).

(a) B1 for \(\frac{48}{x}\)
(b) B1 for \(\frac{48}{x+4}\)
(c)(i) M1 for \(\frac{48}{x} + \frac{48}{x+4} = 5\) or equation setup
M1 for correctly clearing denominators: \(48(x+4) + 48x = 5x(x+4)\) oe
A1 for correctly expanding and rearranging to standard quadratic form
(c)(ii) M1 for correct substitution into quadratic formula: \(x = \frac{-20 \pm \sqrt{20^2 - 4(5)(-384)}}{2(5)}\) oe
A1 for \(6.99\) or \(6.98\) nfww
A1 for \(-10.99\) or \(-10.98\) nfww
(c)(iii) M1 for \(\frac{48}{\text{their } x}\)
A1 for \(6\text{ hours } 52\text{ minutes}\) or \(6\text{ hours } 53\text{ minutes}\)

(a)(i) Probability of choosing red from Bag A is \(\frac{5}{8}\).
If red is transferred, Bag B now has \(4 + 1 = 5\) red and \(6\) blue, total \(11\) marbles.
Probability of choosing red from Bag B is \(\frac{5}{11}\).
\(P(\text{both red}) = \frac{5}{8} \times \frac{5}{11} = \frac{25}{88}\).

(ii) There are two mutually exclusive cases to get a blue from Bag B:
Case 1: Red from Bag A, then Blue from Bag B:
\(P(\text{Red from A and Blue from B}) = \frac{5}{8} \times \frac{6}{11} = \frac{30}{88}\).
Case 2: Blue from Bag A, then Blue from Bag B:
If blue is transferred, Bag B has \(4\) red and \(6 + 1 = 7\) blue, total \(11\) marbles.
\(P(\text{Blue from A and Blue from B}) = \frac{3}{8} \times \frac{7}{11} = \frac{21}{88}\).
Total probability \(= \frac{30}{88} + \frac{21}{88} = \frac{51}{88}\).

(b) Total marbles in Bag A \(= 8\).
\(P(\text{no red}) = P(\text{both blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}\).
\(P(\text{at least one red}) = 1 - P(\text{both blue}) = 1 - \frac{3}{28} = \frac{25}{28}\).

(c) Probability of not rolling a 6 in one roll \(= 1 - 0.3 = 0.7\).
Probability of not rolling a 6 in \(n\) rolls \(= 0.7^n\).
We require \(P(\text{at least one 6}) > 0.95\):
\(1 - 0.7^n > 0.95\)
\(0.7^n < 0.05\)
Taking logarithms:
\(n \log(0.7) < \log(0.05)\)
Since \(\log(0.7) < 0\), reversing the inequality gives:
\(n > \frac{\log(0.05)}{\log(0.7)} \approx \frac{-2.9957}{-0.3567} \approx 8.399\)
Since \(n\) must be an integer, the smallest value is \(n = 9\).

(a)(i) M1 for \(\frac{5}{8} \times \frac{5}{11}\)
A1 for \(\frac{25}{88}\) oe (0.284 or 0.2840...)
(a)(ii) M1 for \(\frac{5}{8} \times \frac{6}{11}\) or \(\frac{3}{8} \times \frac{7}{11}\)
M1 for adding the two probabilities: \(\frac{30}{88} + \frac{21}{88}\)
A1 for \(\frac{51}{88}\) oe (0.5795... or 0.580)
(b) M1 for \(\frac{3}{8} \times \frac{2}{7}\) or \(1 - \frac{3}{8} \times \frac{2}{7}\) or \(\frac{5}{8} \times \frac{4}{7} + 2 \times \frac{5}{8} \times \frac{3}{7}\)
A1 for \(\frac{25}{28}\) oe (0.893 or 0.8928...)
(c) M1 for \(1 - 0.7^n > 0.95\) or \(0.7^n < 0.05\) oe
M1 for solving inequality or trial and improvement showing \(0.7^8 \approx 0.0576\) and \(0.7^9 \approx 0.0404\)
A1 for \(9\) cao

(a) Sequence A is linear with a common difference of \(6\).
First term \(= 5 = 6(1) - 1\).
\(n\)th term \(= 6n - 1\).

(b) Sequence B: \(3, 12, 27, 48\)
Divide terms by \(3\): \(1, 4, 9, 16\), which is \(n^2\).
Therefore, the \(n\)th term is \(3n^2\).

(c)(i) Numerators follow Sequence A: next numerator is \(23 + 6 = 29\).
Denominators follow Sequence B: next denominator is \(3(5)^2 = 3 \times 25 = 75\).
Next term \(= \frac{29}{75}\).

(ii) Numerator is \(6n - 1\) and denominator is \(3n^2\).
\(n\)th term \(= \frac{6n-1}{3n^2}\).

(d) Given \(T_n = 3^{n-1} + kn + c\):
For \(n = 1\): \(3^0 + k(1) + c = 2 \implies 1 + k + c = 2 \implies k + c = 1\)
For \(n = 2\): \(3^1 + k(2) + c = 7 \implies 3 + 2k + c = 7 \implies 2k + c = 4\)
Subtracting the first equation from the second:
\((2k + c) - (k + c) = 4 - 1\)
\(k = 3\) ... Wait, let's verify for \(n = 3\):
If \(k = 3\), \(c = -2\):
\(T_3 = 3^2 + 3(3) - 2 = 9 + 9 - 2 = 16 \ne 24\).
Let's check the terms of C: \(2, 7, 24, 77\).
Powers of 3: \(3^0=1, 3^1=3, 3^2=9, 3^3=27\).
Difference \(T_n - 3^{n-1}\):
\(n=1: 2 - 1 = 1\)
\(n=2: 7 - 3 = 4\)
\(n=3: 24 - 9 = 15\) (wait, let's check: \(3^n - 1\): for \(n=1, 3^1-1=2; n=2, 3^2-2=7; n=3, 3^3-3=24; n=4, 3^4-4=77\)).
Ah! The sequence is \(3^n - n\), which can be written as \(3 \cdot 3^{n-1} - n\) or if the form is \(3^n + kn + c\), with \(k = -1, c = 0\).
If form given in question is \(3^n + kn + c\):
\(n = 1: 3 + k + c = 2 \implies k + c = -1\)
\(n = 2: 9 + 2k + c = 7 \implies 2k + c = -2\)
Subtracting: \(k = -1\), \(c = 0\).
Checking \(n = 3: 27 - 3 = 24\), \(n = 4: 81 - 4 = 77\).
So with formula \(3^n + kn + c\), we have \(k = -1\) and \(c = 0\).

(e) Set \(6n - 1 = 173\):
\(6n = 174\)
\(n = 29\).

(a) B2 for \(6n - 1\) (B1 for \(6n + c\) or \(k n - 1\))
(b) B2 for \(3n^2\) (B1 for second difference \(= 6\) or \(k n^2\) where \(k \ne 0\))
(c)(i) B1 for \(\frac{29}{75}\)
(c)(ii) B1 for \(\frac{6n-1}{3n^2}\) oe
(d) M1 for substituting two values of \(n\) to form simultaneous equations
M1 for method to solve their simultaneous equations
A1 for \(k = -1\)
A1 for \(c = 0\) (or correct values corresponding to formula)
(e) M1 for \(6n - 1 = 173\)
A1 for \(29\) cao

(a) (i) Using the cosine rule in triangle \(ABD\):
\(BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(BAD)\)
\(BD^2 = 65^2 + 84^2 - 2(65)(84)\cos(78^\circ)\)
\(BD^2 = 4225 + 7056 - 10920\cos(78^\circ)\)
\(BD^2 = 11281 - 2270.40 = 9010.60\)
\(BD = \sqrt{9010.60} = 94.924...\text{ m} \approx 94.9\text{ m}\)

(ii) Using the area formula \(\text{Area} = \frac{1}{2}ab\sin C\):
\(\text{Area} = \frac{1}{2} \times 65 \times 84 \times \sin(78^\circ)\)
\(\text{Area} = 2730 \times 0.9781476 = 2670.34...\text{ m}^2 \approx 2670\text{ m}^2\)

(b) In triangle \(BCD\):
\(\text{Angle } DBC = 180^\circ - 42^\circ - 63^\circ = 75^\circ\)
Using the sine rule:
\(\frac{CD}{\sin(DBC)} = \frac{BD}{\sin(BCD)}\)
\(\frac{CD}{\sin(75^\circ)} = \frac{94.924}{\sin(63^\circ)}\)
\(CD = \frac{94.924 \times \sin(75^\circ)}{\sin(63^\circ)} = \frac{91.6897}{0.8910065} = 102.905...\text{ m} \approx 103\text{ m}\)

(c) Area of the sector sprayed by water:
\(\text{Area}_{\text{sector}} = \frac{78}{360} \times \pi \times 40^2 = 1089.085...\text{ m}^2\)
Area not sprayed:
\(2670.34 - 1089.09 = 1581.25\text{ m}^2\)
Percentage not sprayed:
\(\frac{1581.25}{2670.34} \times 100 = 59.215...\% \approx 59.2\%\)

(a)(i)
M2 for \(65^2 + 84^2 - 2(65)(84)\cos(78)\) or M1 for correct implicit cosine rule
A1 for 94.9 or 94.92 to 94.93

(a)(ii)
M1 for \(\frac{1}{2} \times 65 \times 84 \times \sin(78)\)
A1 for 2670 or 2670.3 to 2670.4

(b)
B1 for angle \(DBC = 75^\circ\) seen or used
M1 for \(\frac{CD}{\sin 75} = \frac{\text{their } BD}{\sin 63}\) oe
A1 for 103 or 102.8 to 103.0

(c)
M1 for \(\frac{78}{360} \times \pi \times 40^2\) (soi by 1089...)
M1 for \(\frac{\text{their area (a)(ii)} - \text{their sector area}}{\text{their area (a)(ii)}} \times 100\) oe
A1 for 59.2 or 59.21 to 59.22

(a) \(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{180}{x}\)

(b) \(\text{Time} = \frac{180}{x - 15}\)

(c) (i) \(\frac{180}{x - 15} - \frac{180}{x} = 1\)
Multiply both sides by \(x(x - 15)\):
\(180x - 180(x - 15) = x(x - 15)\)
\(180x - 180x + 2700 = x^2 - 15x\)
\(2700 = x^2 - 15x\)
\(x^2 - 15x - 2700 = 0\)

(ii) \((x - 60)(x + 45) = 0\)
or using the quadratic formula:
\(x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(-2700)}}{2(1)}\)
\(x = \frac{15 \pm \sqrt{225 + 10800}}{2} = \frac{15 \pm \sqrt{11025}}{2} = \frac{15 \pm 105}{2}\)
\(x = 60\) or \(x = -45\)

(iii) Speed must be positive, so \(x = 60\text{ km/h}\).
Return speed = \(60 - 15 = 45\text{ km/h}\).
Return time = \(\frac{180}{45} = 4\text{ hours}\).

(d) Factorise the numerator: \(2x(x + 45)\)
Factorise the denominator: \((x - 60)(x + 45)\)
Cancel the common factor \((x + 45)\):
\(\frac{2x(x + 45)}{(x - 60)(x + 45)} = \frac{2x}{x - 60}\)

(a)
B1 for \(\frac{180}{x}\)

(b)
B1 for \(\frac{180}{x - 15}\)

(c)(i)
M1 for their \((b) - \text{their } (a) = 1\) oe
M1 for correctly clearing fractions by multiplying by \(x(x - 15)\)
A1 for complete correct algebraic working leading to \(x^2 - 15x - 2700 = 0\) with no steps omitted

(c)(ii)
M2 for \((x - 60)(x + 45)\) or \(\frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(-2700)}}{2(1)}\) or M1 for \((x + a)(x + b)\) where \(ab = -2700\) or \(a + b = -15\)
A1 for \(x = 60\) and \(x = -45\)

(c)(iii)
B1 for 4 (or FT \(\frac{180}{\text{their positive } x - 15}\))

(d)
B1 for \(2x(x + 45)\) seen
B1 for \(\frac{2x}{x - 60}\) cao

(a) Total number of discs = \(7 + 5 + 3 = 15\).
(i) \(P(\text{yellow}) = \frac{5}{15} = \frac{1}{3}\)
(ii) \(P(\text{not blue}) = \frac{5 + 3}{15} = \frac{8}{15}\)

(b) (i) \(P(\text{both blue}) = \frac{7}{15} \times \frac{6}{14} = \frac{42}{210} = \frac{1}{5}\) (or \(0.2\))

(ii) The discs can be drawn as (Yellow, Red) or (Red, Yellow):
\(P(\text{YR}) = \frac{5}{15} \times \frac{3}{14} = \frac{15}{210}\)
\(P(\text{RY}) = \frac{3}{15} \times \frac{5}{14} = \frac{15}{210}\)
\(P(\text{one yellow and one red}) = \frac{15}{210} + \frac{15}{210} = \frac{30}{210} = \frac{1}{7}\) (or \(0.143\))

(c) \(P(\text{at least one red}) = 1 - P(\text{no red})\)
Number of non-red discs = \(7 + 5 = 12\).
\(P(\text{no red}) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}\)
\(P(\text{no red}) = \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13} = \frac{440}{910} = \frac{44}{91}\)
\(P(\text{at least one red}) = 1 - \frac{44}{91} = \frac{47}{91}\) (or \(0.516\))

(a)(i)
B1 for \(\frac{5}{15}\) oe (e.g. \(\frac{1}{3}\) or 0.333...)

(a)(ii)
B1 for \(\frac{8}{15}\) oe (or 0.533...)

(b)(i)
M1 for \(\frac{7}{15} \times \frac{6}{14}\)
A1 for \(\frac{1}{5}\) oe (e.g. \(\frac{42}{210}\) or 0.2)

(b)(ii)
M1 for \(\frac{5}{15} \times \frac{3}{14}\) soi by \(\frac{15}{210}\) or \(\frac{1}{14}\)
M1 for \(\frac{5}{15} \times \frac{3}{14} + \frac{3}{15} \times \frac{5}{14}\) oe
A1 for \(\frac{1}{7}\) oe (e.g. \(\frac{30}{210}\) or 0.143 or 0.1428 to 0.1429)

(c)
M1 for identifying number of non-red discs is 12
M1 for \(\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}\) (soi by \(\frac{44}{91}\) or \(\frac{1320}{2730}\) or 0.4835...)
M1 for \(1 - (\text{their } P(\text{no red}))\)
A1 for \(\frac{47}{91}\) oe (or 0.516 or 0.5164 to 0.5165)

(a) The volume of the box is \(V = x^2 h = 4000\).
Rearranging gives \(h = \frac{4000}{x^2}\).
The box is open at the top, so it has 1 square base and 4 vertical rectangular faces:
\(A = x^2 + 4xh\).
Substituting for \(h\):
\(A = x^2 + 4x\left(\frac{4000}{x^2}\right) = x^2 + \frac{16000}{x}\).

(b) Writing \(A = x^2 + 16000x^{-1}\):
\(\frac{\mathrm{d}A}{\mathrm{d}x} = 2x - 16000x^{-2} = 2x - \frac{16000}{x^2}\).

(c)(i) For minimum surface area, set \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\):
\(2x - \frac{16000}{x^2} = 0\)
\(2x = \frac{16000}{x^2}\)
\(2x^3 = 16000\)
\(x^3 = 8000\)
\(x = 20\).

(c)(ii) Minimum surface area:
\(A = (20)^2 + \frac{16000}{20} = 400 + 800 = 1200\text{ cm}^2\).

(d) \(S = 2x^2 + kx^{-1}\)
\(\frac{\mathrm{d}S}{\mathrm{d}x} = 4x - kx^{-2} = 4x - \frac{k}{x^2}\).
At \(x = 5\), \(\frac{\mathrm{d}S}{\mathrm{d}x} = 0\):
\(4(5) - \frac{k}{5^2} = 0\)
\(20 - \frac{k}{25} = 0\)
\(k = 20 \times 25 = 500\).

(a) [3 marks]
M1 for expressing volume as \(x^2 h = 4000\) to get \(h = \frac{4000}{x^2}\).
M1 for formula of surface area \(A = x^2 + 4xh\).
A1 for substituting \(h\) and obtaining the given expression \(A = x^2 + \frac{16000}{x}\) with all steps shown.

(b) [2 marks]
M1 for \(2x\) or \(-16000x^{-2}\).
A1 for \(2x - \frac{16000}{x^2}\) oe.

(c)(i) [3 marks]
M1 for setting their derivative equal to 0.
M1 for \(x^3 = 8000\).
A1 for \(x = 20\).

(c)(ii) [2 marks]
M1 for substituting their \(x = 20\) into the expression for \(A\).
A1 for \(1200\).

(d) [2 marks]
M1 for differentiating \(S\) to get \(4x - \frac{k}{x^2}\) and equating to 0 when \(x = 5\).
A1 for \(k = 500\).

(a)(i) \(\vec{AC} = \vec{AO} + \vec{OC} = -\mathbf{a} + \mathbf{c} = \mathbf{c} - \mathbf{a}\).

(a)(ii) Since \(AN : NC = 2 : 1\), \(\vec{AN} = \frac{2}{3}\vec{AC} = \frac{2}{3}(\mathbf{c} - \mathbf{a})\).
Then \(\vec{ON} = \vec{OA} + \vec{AN} = \mathbf{a} + \frac{2}{3}(\mathbf{c} - \mathbf{a}) = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\).

(a)(iii) Since \(OABC\) is a parallelogram, \(\vec{AB} = \vec{OC} = \mathbf{c}\).
\(M\) is the midpoint of \(AB\), so \(\vec{AM} = \frac{1}{2}\mathbf{c}\).
\(\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}\mathbf{c}\).

(b) \(\vec{MN} = \vec{ON} - \vec{OM} = \left(\frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\right) - \left(\mathbf{a} + \frac{1}{2}\mathbf{c}\right) = -\frac{2}{3}\mathbf{a} + \frac{1}{6}\mathbf{c}\).
Note: More precisely, \(\frac{1}{3} - 1 = -\frac{2}{3}\) and \(\frac{2}{3} - \frac{1}{2} = \frac{1}{6}\).

(c)(i) \(\vec{OP} = \vec{OB} + \vec{BP}\).
Since \(\vec{OB} = \mathbf{a} + \mathbf{c}\) and \(\vec{BP} = k\mathbf{c}\), \(\vec{OP} = \mathbf{a} + (1 + k)\mathbf{c}\).
Thus, \(\vec{MP} = \vec{OP} - \vec{OM} = (\mathbf{a} + (1 + k)\mathbf{c}) - (\mathbf{a} + \frac{1}{2}\mathbf{c}) = \left(k + \frac{1}{2}\right)\mathbf{c}\).
Alternatively, \(\vec{MP} = \vec{MB} + \vec{BP} = \frac{1}{2}\mathbf{c} + k\mathbf{c} = \left(k + \frac{1}{2}\right)\mathbf{c}\).

(c)(ii) Since \(M, N, P\) lie on a straight line, \(\vec{MP} = \lambda \vec{MN}\).
However, \(\vec{MP}\) has no \(\mathbf{a}\) component (its \(\mathbf{a}\) coefficient is 0).
Since \(\vec{MN} = -\frac{2}{3}\mathbf{a} + \frac{1}{6}\mathbf{c}\) has non-zero \(\mathbf{a}\) component, the line segment from \(M\) through \(N\) intersects line \(BC\):
\(\vec{NP} = \vec{OP} - \vec{ON} = (\mathbf{a} + (1 + k)\mathbf{c}) - (\frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}) = \frac{2}{3}\mathbf{a} + \left(k + \frac{1}{3}\right)\mathbf{c}\).
For \(M, N, P\) to be collinear, \(\vec{NP} = \mu \vec{MN}\):
\(\frac{2}{3}\mathbf{a} + \left(k + \frac{1}{3}\right)\mathbf{c} = \mu\left(-\frac{2}{3}\mathbf{a} + \frac{1}{6}\mathbf{c}\right)\).
Equating coefficients of \(\mathbf{a}\):
\(\frac{2}{3} = -\frac{2}{3}\mu \implies \mu = -1\).
Equating coefficients of \(\mathbf{c}\):
\(k + \frac{1}{3} = -1\left(\frac{1}{6}\right) = -\frac{1}{6}\)
\(k = -\frac{1}{6} - \frac{1}{3} = -\frac{1}{2}\).

(a)(i) [1 mark]
B1 for \(\mathbf{c} - \mathbf{a}\) or \(-\mathbf{a} + \mathbf{c}\).

(a)(ii) [2 marks]
M1 for \(\vec{ON} = \mathbf{a} + \frac{2}{3}(\mathbf{c} - \mathbf{a})\) or equivalent vector path.
A1 for \(\frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\) oe.

(a)(iii) [2 marks]
M1 for \(\vec{OM} = \mathbf{a} + \frac{1}{2}\mathbf{c}\) or correct vector path.
A1 for \(\mathbf{a} + \frac{1}{2}\mathbf{c}\).

(b) [2 marks]
M1 for \(\vec{ON} - \vec{OM}\) or \(\vec{MA} + \vec{AN}\) with substitution of their expressions.
A1 for correctly showing simplified vector with full algebraic working.

(c)(i) [2 marks]
M1 for \(\vec{MP} = \vec{MB} + \vec{BP}\) or \(\vec{OP} - \vec{OM}\).
A1 for \(\left(k + \frac{1}{2}\right)\mathbf{c}\) oe.

(c)(ii) [3 marks]
M1 for setting up collinearity condition, e.g., \(\vec{NP} = \mu \vec{MN}\) or ratio of components.
M1 for finding scalar multiplier \(\mu = -1\).
A1 for \(k = -\frac{1}{2}\) oe.

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

(b) At turning points, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\[ 6x^2 - 18x + 12 = 0 \implies 6(x^2 - 3x + 2) = 0 \implies 6(x - 1)(x - 2) = 0 \]
So \( x = 1 \) or \( x = 2 \).
When \( x = 1 \):
\[ y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10 \]
When \( x = 2 \):
\[ y = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9 \]
The turning points are \( (1, 10) \) and \( (2, 9) \).

(c) When \( x = 3 \):
\[ y = 2(3)^3 - 9(3)^2 + 12(3) + 5 = 54 - 81 + 36 + 5 = 14 \]
Gradient at \( x = 3 \):
\[ m = 6(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12 \]
Using \( y - y_1 = m(x - x_1) \):
\[ y - 14 = 12(x - 3) \implies y - 14 = 12x - 36 \implies y = 12x - 22 \]

(d) Set the tangent equation equal to the curve equation:
\[ 2x^3 - 9x^2 + 12x + 5 = 12x - 22 \implies 2x^3 - 9x^2 + 27 = 0 \]
Since the line is tangent at \( x = 3 \), \( (x - 3)^2 \) is a factor:
\[ (x^2 - 6x + 9)(2x + 3) = 0 \implies x = 3 \text{ or } x = -\frac{3}{2} = -1.5 \]
Substitute \( x = -1.5 \) into \( y = 12x - 22 \):
\[ y = 12(-1.5) - 22 = -18 - 22 = -40 \]
Thus, \( P = (-1.5, -40) \).

(a) M1 for at least two terms differentiated correctly; A1 for \( 6x^2 - 18x + 12 \) cao.
(b) M1 for setting their derivative equal to 0; M1 for solving quadratic to obtain \( x = 1 \) and \( x = 2 \); A1 for \( (1, 10) \); A1 for \( (2, 9) \).
(c) M1 for substituting \( x = 3 \) into derivative to find gradient \( m = 12 \); M1 for finding \( y = 14 \) at \( x = 3 \); A1 for \( y = 12x - 22 \) oe.
(d) M1 for equating curve and line equation: \( 2x^3 - 9x^2 + 27 = 0 \); M1 for factorising using factor \( (x - 3)^2 \) to obtain \( x = -1.5 \); A1 for \( (-1.5, -40) \) or \( \left(-\frac{3}{2}, -40\right) \).