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First, rationalise the denominator of the expression:
\[\frac{4 + 3\sqrt{5}}{2 + \sqrt{5}} = \frac{(4 + 3\sqrt{5})(2 - \sqrt{5})}{(2 + \sqrt{5})(2 - \sqrt{5})}\]

Expand the numerator:
\[(4 + 3\sqrt{5})(2 - \sqrt{5}) = 8 - 4\sqrt{5} + 6\sqrt{5} - 3(5) = -7 + 2\sqrt{5}\]

Expand the denominator:
\[(2 + \sqrt{5})(2 - \sqrt{5}) = 4 - 5 = -1\]

Thus, the expression simplifies to:
\[\frac{-7 + 2\sqrt{5}}{-1} = 7 - 2\sqrt{5}\]
This gives \(a = 7\) and \(b = -2\).

Now substitute this back into the original equation:
\[\sqrt{5}(x - 3) = 7 - 2\sqrt{5} - x\]

Expand the left-hand side:
\[x\sqrt{5} - 3\sqrt{5} = 7 - 2\sqrt{5} - x\]

Rearrange to collect all terms in \(x\) on one side:
\[x\sqrt{5} + x = 7 - 2\sqrt{5} + 3\sqrt{5}\]
\[x(\sqrt{5} + 1) = 7 + \sqrt{5}\]

Solve for \(x\):
\[x = \frac{7 + \sqrt{5}}{\sqrt{5} + 1}\]

Rationalise this fraction by multiplying the numerator and denominator by \(\sqrt{5} - 1\):
\[x = \frac{(7 + \sqrt{5})(\sqrt{5} - 1)}{(\sqrt{5} + 1)(\sqrt{5} - 1)}\]
\[x = \frac{7\sqrt{5} - 7 + 5 - \sqrt{5}}{5 - 1}\]
\[x = \frac{-2 + 6\sqrt{5}}{4}\]

Simplify the fraction by dividing numerator and denominator by 2:
\[x = \frac{-1 + 3\sqrt{5}}{2}\]

**M1**: Multiply the numerator and denominator of \(\frac{4 + 3\sqrt{5}}{2 + \sqrt{5}}\) by \(2 - \sqrt{5}\) (or equivalent method).
**A1**: Obtain \(7 - 2\sqrt{5}\) (accept \(a = 7, b = -2\)).
**M1**: Substitute their simplified expression, expand the brackets, collect the terms in \(x\), and attempt to isolate \(x\) to the form \(x = \frac{A + B\sqrt{5}}{C + D\sqrt{5}}\).
**A1**: Correct final simplified answer in the form \(\frac{p + q\sqrt{5}}{r}\), i.e., \(\frac{-1 + 3\sqrt{5}}{2}\) or equivalent.

(a) To express \( 3x^2 + 12x + 17 \) in the form \( a(x+b)^2 + c \):
Factor out the coefficient of \( x^2 \) from the terms in \( x \):
\( 3(x^2 + 4x) + 17 \)
Complete the square inside the bracket:
\( 3[(x + 2)^2 - 4] + 17 \)
Expand the outer bracket:
\( 3(x + 2)^2 - 12 + 17 \)
Simplify:
\( 3(x + 2)^2 + 5 \)
So, \( a = 3 \), \( b = 2 \), and \( c = 5 \).

(b) The completed square form is \( y = 3(x + 2)^2 + 5 \).
Since \( 3(x+2)^2 \ge 0 \) for all real \( x \), the minimum value of \( y \) occurs when \( x + 2 = 0 \), which gives \( x = -2 \).
At \( x = -2 \), \( y = 5 \).
Thus, the coordinates of the minimum point are \( (-2, 5) \).

(c) Let \( g(x) = \frac{20}{3x^2 + 12x + 17} = \frac{20}{3(x+2)^2 + 5} \).
To maximize the fraction, we must minimize the denominator.
From part (b), the minimum value of the denominator \( 3(x+2)^2 + 5 \) is \( 5 \), which occurs at \( x = -2 \).
Therefore, the maximum value of \( g(x) \) is:
\( \frac{20}{5} = 4 \)
This maximum value occurs at \( x = -2 \).

(a)
M1: Attempts to factor out 3 from the first two terms or starts to complete the square, e.g., showing \( 3(x^2 + 4x) \) or achieving \( 3(x + 2)^2 + k \).
A1: For getting \( 3(x+2)^2 \).
A1: For the fully correct expression \( 3(x + 2)^2 + 5 \) (or stating \( a=3, b=2, c=5 \)).

(b)
B1: For the correct coordinates \( (-2, 5) \). Accept written as \( x = -2, y = 5 \).

(c)
M1: Realises that the maximum of the expression occurs when the denominator is minimized, and substitutes their minimum value from part (a) or (b) into the denominator.
A1: Correct maximum value of \( 4 \).
A1: Correct value of \( x = -2 \).

(a) First, we find the first derivative of \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = 6x^2 + 2px + q \]

Since the tangent to the curve at \(x = 0\) has gradient \(-8\):
\[ \left. \frac{dy}{dx} \right|_{x=0} = -8 \implies q = -8 \]

Since the curve has a stationary point at \(x = 1\):
\[ \left. \frac{dy}{dx} \right|_{x=1} = 0 \implies 6(1)^2 + 2p(1) + q = 0 \]

Substitute \(q = -8\):
\[ 6 + 2p - 8 = 0 \implies 2p = 2 \implies p = 1 \]

Let \(f(x) = 2x^3 + px^2 + qx + r = 2x^3 + x^2 - 8x + r\). Since \((x - 2)\) is a factor of \(f(x)\), by the Factor Theorem:
\[ f(2) = 0 \implies 2(2)^3 + (2)^2 - 8(2) + r = 0 \]
\[ 16 + 4 - 16 + r = 0 \implies r = -4 \]

Thus, \(p = 1\), \(q = -8\), and \(r = -4\).

(b) The equation of the curve is \(y = 2x^3 + x^2 - 8x - 4\), and its derivative is:
\[ \frac{dy}{dx} = 6x^2 + 2x - 8 \]

To find the stationary points, set \(\frac{dy}{dx} = 0\):
\[ 6x^2 + 2x - 8 = 0 \implies 3x^2 + x - 4 = 0 \]
\[ (3x + 4)(x - 1) = 0 \]

This gives \(x = 1\) (the given stationary point) and \(x = -\frac{4}{3}\) for the other stationary point.

Now find the corresponding \(y\)-coordinate by substituting \(x = -\frac{4}{3}\) into the equation of \(C\):
\[ y = 2\left(-\frac{4}{3}\right)^3 + \left(-\frac{4}{3}\right)^2 - 8\left(-\frac{4}{3}\right) - 4 \]
\[ y = 2\left(-\frac{64}{27}\right) + \frac{16}{9} + \frac{32}{3} - 4 \]
\[ y = -\frac{128}{27} + \frac{48}{27} + \frac{288}{27} - \frac{108}{27} \]
\[ y = \frac{100}{27} \]

So the coordinates of the other stationary point are \(\left(-\frac{4}{3}, \frac{100}{27}\right)\).

**Part (a)**
* **M1**: Differentiate \(y\) to obtain an expression of the form \(Ax^2 + Bpx + q\).
* **A1**: Correctly state \(q = -8\).
* **M1**: Set their \(\frac{dy}{dx} = 0\) at \(x = 1\) and substitute their \(q\) to find \(p\). (Correctly finds \(p = 1\))
* **M1**: Apply the Factor Theorem \(f(2) = 0\) with their \(p\) and \(q\) to form an equation in \(r\).
* **A1**: Correctly find \(r = -4\).

**Part (b)**
* **M1**: Set their \(\frac{dy}{dx} = 0\) to obtain a 3-term quadratic equation in \(x\) and solve for the other value of \(x\) (must get \(x = -\frac{4}{3}\) or equivalent).
* **M1**: Substitute their other \(x\)-value into \(y\) to find the \(y\)-coordinate.
* **A1**: Correct coordinates \(\left(-\frac{4}{3}, \frac{100}{27}\right)\) or exact equivalent.

**(a)**

The gradient of the line \(L_1\) passing through \(A(1, 4)\) and \(B(7, -2)\) is:
\[m = \frac{-2 - 4}{7 - 1} = \frac{-6}{6} = -1\]

Using the point-slope form with point \(A(1, 4)\):
\[y - 4 = -1(x - 1)\]
\[y - 4 = -x + 1\]
\[y = -x + 5\]

**(b)**

The midpoint \(M\) of \(AB\) is:
\[M = \left(\frac{1 + 7}{2}, \frac{4 + (-2)}{2}\right) = (4, 1)\]

Since \(L_2\) is perpendicular to \(L_1\), the gradient of \(L_2\) is:
\[m' = -\frac{1}{m} = -\frac{1}{-1} = 1\]

Using the point-slope form with the midpoint \(M(4, 1)\):
\[y - 1 = 1(x - 4)\]
\[y - 1 = x - 4\]
\[y - x + 3 = 0\]

This is in the form \(py + qx + r = 0\), where \(p = 1\), \(q = -1\), and \(r = 3\).

**(c)**

To find the coordinates of \(C\), the intersection of \(L_2\) with the \(x\)-axis, set \(y = 0\) in the equation of \(L_2\):
\[0 - x + 3 = 0 \implies x = 3\]

So, \(C\) has coordinates \((3, 0)\).

We can find the area of triangle \(ABC\) using the coordinates of its vertices \(A(1, 4)\), \(B(7, -2)\), and \(C(3, 0)\):

Using the shoelace formula:
\[\text{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|\]
\[\text{Area} = \frac{1}{2} |1(-2 - 0) + 7(0 - 4) + 3(4 - (-2))|\]
\[\text{Area} = \frac{1}{2} |-2 - 28 + 18| = \frac{1}{2} |-12| = 6\]

Alternatively, using the base and height:
The length of the base \(AB\) is:
\[AB = \sqrt{(7 - 1)^2 + (-2 - 4)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{72} = 6\sqrt{2}\]

The height is the distance from \(C(3, 0)\) to the midpoint \(M(4, 1)\):
\[MC = \sqrt{(4 - 3)^2 + (1 - 0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\]

\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6\sqrt{2} \times \sqrt{2} = 6\]

**(a)**
* **M1**: Attempts to find the gradient of \(L_1\), showing correct substitution of coordinates into the gradient formula.
* **A1**: Correct equation of the line, e.g., \(y = -x + 5\) or \(y + x - 5 = 0\).

**(b)**
* **M1**: Finds the midpoint \(M\) of \(AB\) with at least one coordinate correct.
* **M1**: Uses the perpendicular gradient condition \(m_1 m_2 = -1\) to find the gradient of \(L_2\).
* **M1**: Attempts to find the equation of \(L_2\) using their midpoint and perpendicular gradient.
* **A1**: Correct equation in the required form, e.g., \(y - x + 3 = 0\) (accept any equivalent integer multiple, e.g., \(x - y - 3 = 0\)).

**(c)**
* **B1**: Finds the correct coordinates of \(C\), which is \((3, 0)\).
* **M1**: A complete and valid method to find the area of triangle \(ABC\), either by the coordinate area formula (shoelace method) or by calculating \(\frac{1}{2} \times AB \times MC\).
* **A1**: Correct area of \(6\).

The volume of revolution is given by \(V = \pi \int_{0}^{1} y^2 \text{d}x\). Here, \(y^2 = (2x\text{e}^{-x})^2 = 4x^2\text{e}^{-2x}\). So, \(V = \pi \int_{0}^{1} 4x^2\text{e}^{-2x} \text{d}x\). We find the integral using integration by parts. Let \(u = x^2\) and \(\text{d}v = \text{e}^{-2x} \text{d}x\), then \(\text{d}u = 2x \text{d}x\) and \(v = -\frac{1}{2}\text{e}^{-2x}\). Thus, \(\int x^2\text{e}^{-2x} \text{d}x = -\frac{1}{2}x^2\text{e}^{-2x} + \int x\text{e}^{-2x} \text{d}x\). For the second integral, let \(u = x\) and \(\text{d}v = \text{e}^{-2x} \text{d}x\), so \(\text{d}u = \text{d}x\) and \(v = -\frac{1}{2}\text{e}^{-2x}\). This gives \(\int x\text{e}^{-2x} \text{d}x = -\frac{1}{2}x\text{e}^{-2x} - \frac{1}{4}\text{e}^{-2x}\). Combining these, we have \(\int 4x^2\text{e}^{-2x} \text{d}x = 4\left[ -\frac{1}{2}x^2\text{e}^{-2x} - \frac{1}{2}x\text{e}^{-2x} - \frac{1}{4}\text{e}^{-2x} \right] = \left[ -\text{e}^{-2x}(2x^2 + 2x + 1) \right]\). Evaluating between \(0\) and \(1\): at \(x = 1\), the expression is \(-\text{e}^{-2}(2(1)^2 + 2(1) + 1) = -5\text{e}^{-2}\); at \(x = 0\), the expression is \(-\text{e}^{0}(1) = -1\). Subtracting the lower limit value from the upper limit value gives \(-5\text{e}^{-2} - (-1) = 1 - 5\text{e}^{-2}\). Multiplying by \(\pi\), the volume is \(\pi \left( 1 - \frac{5}{\text{e}^2} \right)\).

M1: Identifies the correct volume formula \(V = \pi \int y^2 \text{d}x\) and writes down \(y^2 = 4x^2\text{e}^{-2x}\). A1: Correctly identifies the limits of integration from \(0\) to \(1\) and sets up the integral \(\pi \int_{0}^{1} 4x^2\text{e}^{-2x} \text{d}x\). M1: Applies integration by parts to \(\int x^2\text{e}^{-2x} \text{d}x\) once. A1: Obtains the correct first-stage integration expression. M1: Applies integration by parts a second time to \(\int x\text{e}^{-2x} \text{d}x\). A1: Obtains the fully integrated expression \(\left[ -\text{e}^{-2x}(2x^2 + 2x + 1) \right]\) or equivalent. M1: Correctly substitutes the limits of \(0\) and \(1\) into their integrated expression. A1: Obtains the final exact volume \(\pi \left( 1 - \frac{5}{\text{e}^2} \right)\), identifying \(a=1\) and \(b=5\).

**(a)**

First, find the length of the diagonal \(AC\) of the rectangular base \(ABCD\) using Pythagoras' Theorem:
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = 12^2 + 16^2 = 144 + 256 = 400\)
\(AC = \sqrt{400} = 20\text{ cm}\)

Since \(O\) is the center of the base, the length \(AO\) is half of the diagonal:
\(AO = \frac{1}{2}AC = \frac{1}{2} \times 20 = 10\text{ cm}\)

Now, in the right-angled triangle \(VAO\), the height of the pyramid \(VO\) is perpendicular to the base, so:
\(VO^2 = VA^2 - AO^2\)
\(VO^2 = 26^2 - 10^2 = 676 - 100 = 576\)
\(VO = \sqrt{576} = 24\text{ cm}\) (as required)

---

**(b)**

Let \(M\) be the midpoint of the side \(BC\).
The line \(VM\) is perpendicular to \(BC\), and the line \(OM\) is perpendicular to \(BC\).
The angle between the plane \(VBC\) and the base \(ABCD\) is the angle \(\angle VMO\).

Since \(O\) is the center of the base and \(ABCD\) is a rectangle:
\(OM = \frac{1}{2}AB = \frac{12}{2} = 6\text{ cm}\)

In the right-angled triangle \(VOM\):
\(\tan(\angle VMO) = \frac{VO}{OM} = \frac{24}{6} = 4\)

\(\angle VMO = \arctan(4) \approx 75.96375^\circ\)

Rounded to 1 decimal place, the angle is \(76.0^\circ\).

---

**(c)**

Let \(N\) be the midpoint of \(AB\) and \(P\) be the midpoint of \(CD\).
The angle between the plane \(VAB\) and the plane \(VCD\) is the angle \(\angle NVP\).

In the triangle \(VNP\):
* \(NP = BC = 16\text{ cm}\)
* \(O\) is the midpoint of \(NP\), so \(ON = OP = 8\text{ cm}\)
* \(VO = 24\text{ cm}\) is perpendicular to \(NP\) at \(O\)

In the right-angled triangle \(VON\):
\(\tan(\angle OVN) = \frac{ON}{VO} = \frac{8}{24} = \frac{1}{3}\)
\(\angle OVN = \arctan\left(\frac{1}{3}\right) \approx 18.4349^\circ\)

By symmetry, \(\angle NVP = 2 \times \angle OVN\):
\(\angle NVP = 2 \times 18.4349^\circ \approx 36.8699^\circ\)

*Alternative method using the Cosine Rule:*
First, find the slant height \(VN\) (and \(VP\)):
\(VN^2 = VO^2 + ON^2 = 24^2 + 8^2 = 576 + 64 = 640\)

In triangle \(VNP\):
\(\cos(\angle NVP) = \frac{VN^2 + VP^2 - NP^2}{2 \times VN \times VP}\)
\(\cos(\angle NVP) = \frac{640 + 640 - 16^2}{2 \times \sqrt{640} \times \sqrt{640}} = \frac{1280 - 256}{1280} = \frac{1024}{1280} = 0.8\)
\(\angle NVP = \arccos(0.8) \approx 36.8699^\circ\)

Rounded to 1 decimal place, the angle is \(36.9^\circ\).

**(a)**
* **M1**: Correct use of Pythagoras' theorem to find the diagonal of the base \(AC\) or half-diagonal \(AO\) (e.g., \(AC = \sqrt{12^2 + 16^2} = 20\) or \(AO^2 = 6^2 + 8^2\)).
* **M1**: Correct application of Pythagoras' theorem in the vertical triangle to express \(VO\) (e.g., \(VO = \sqrt{26^2 - 10^2}\)).
* **A1**: Fully correct working shown to arrive at \(VO = 24\) (c.s.o.).

**(b)**
* **M1**: Identification of the required angle as \(\angle VMO\) where \(M\) is the midpoint of \(BC\).
* **M1**: Finding the horizontal distance \(OM = 6\) or the slant height \(VM = \sqrt{26^2 - 8^2} = \sqrt{612} = 6\sqrt{17}\).
* **M1**: Applying a correct trigonometric ratio (e.g., \(\tan(\theta) = \frac{24}{6}\) or \(\cos(\theta) = \frac{6}{\sqrt{612}}\)).
* **A1**: Correct angle of \(76.0^\circ\) (must be given to 1 decimal place as requested).

**(c)**
* **M1**: Identification of the required angle as \(\angle NVP\) where \(N\) and \(P\) are the midpoints of \(AB\) and \(CD\) respectively.
* **M1**: Correctly finding the length of \(VN\) (or \(VP\)) as \(\sqrt{24^2 + 8^2} = \sqrt{640}\) or identifying the half-angle trigonometric ratio \(\tan(\angle OVN) = \frac{8}{24}\).
* **M1**: A complete and correct method to calculate the angle, e.g., using the Cosine Rule \(\cos(\angle NVP) = \frac{640+640-256}{2 \times 640}\) or doubling the half-angle \(2 \times \arctan\left(\frac{1}{3}\right)\).
* **A1**: Correct angle of \(36.9^\circ\) (must be given to 1 decimal place as requested).

(a) Since \(u_2 = \log_9 x\), we can use the change of base formula:

\(u_2 = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \frac{1}{2}\log_3 x\).

The common ratio \(r\) of a geometric series is given by:

\(r = \frac{u_2}{a} = \frac{\frac{1}{2}\log_3 x}{\log_3 x} = \frac{1}{2}\) (since \(x > 1\), \(\log_3 x \neq 0\)).

(b) The sum to infinity of a geometric series is given by:

\(S_{\infty} = \frac{a}{1 - r}\)

Substituting \(a = \log_3 x\) and \(r = \frac{1}{2}\):

\(S_{\infty} = \frac{\log_3 x}{1 - \frac{1}{2}} = 2\log_3 x = \log_3 (x^2)\)

We are given that:

\(S_{\infty} = \log_3 (10x - 9)\)

Therefore:

\(\log_3 (x^2) = \log_3 (10x - 9)\)

\(x^2 = 10x - 9\)

\(x^2 - 10x + 9 = 0\)

\((x - 9)(x - 1) = 0\)

Since \(x > 1\), we must have \(x = 9\).

(c) When \(x = 9\):

\(a = \log_3 9 = 2\)

\(r = \frac{1}{2}\)

We want to find the sum of the first 5 terms, \(S_5\):

\(S_5 = \frac{a(1 - r^5)}{1 - r}\)

\(S_5 = \frac{2\left(1 - \left(\frac{1}{2}\right)^5\right)}{1 - \frac{1}{2}} = \frac{2\left(1 - \frac{1}{32}\right)}{\frac{1}{2}} = 4 \times \frac{31}{32} = \frac{31}{8}\)

(a)
M1: Applies the change of base formula to express \(\log_9 x\) in terms of base 3, e.g., \(\frac{\log_3 x}{\log_3 9}\) or \(\frac{1}{2}\log_3 x\).
A1: Divides the second term by the first term to show \(r = \frac{1}{2}\) clearly with no errors.

(b)
M1: Recalls the formula for the sum to infinity \(S_{\infty} = \frac{a}{1-r}\) and substitutes their values to get \(2\log_3 x\).
M1: Uses the power law of logarithms to write \(2\log_3 x = \log_3 (x^2)\) and equates this to \(\log_3 (10x - 9)\) to form a quadratic equation.
A1: Solves the quadratic equation to get \(x = 9\) and \(x = 1\).
A1: Correctly rejects \(x = 1\) due to the condition \(x > 1\) to give the final answer \(x = 9\).

(c)
M1: Evaluates the first term to find \(a = \log_3 9 = 2\).
M1: Uses the sum of a geometric series formula \(S_n = \frac{a(1-r^n)}{1-r}\) with \(n = 5\), \(a = 2\), and \(r = \frac{1}{2}\).
A1: Obtains the correct simplified fraction \(\frac{31}{8}\) (accept \(3\frac{7}{8}\) or \(3.875\)).

(a) Let the height of the cylinder be \( h \text{ cm} \).
The volume of a cylinder is given by:
\( V = \pi r^2 h \)

Given that \( V = 128\pi \):
\( \pi r^2 h = 128\pi \implies h = \frac{128}{r^2} \)

The total surface area \( A \) of a solid cylinder is:
\( A = 2\pi r^2 + 2\pi r h \)

Substituting \( h = \frac{128}{r^2} \) into the expression for \( A \):
\( A = 2\pi r^2 + 2\pi r \left( \frac{128}{r^2} \right) \)
\( A = 2\pi r^2 + \frac{256\pi}{r} \) (as required)

(b) Differentiating \( A \) with respect to \( r \):
\( A = 2\pi r^2 + 256\pi r^{-1} \)
\( \frac{\text{d}A}{\text{d}r} = 4\pi r - 256\pi r^{-2} = 4\pi r - \frac{256\pi}{r^2} \)

When \( r = 8 \):
\( \frac{\text{d}A}{\text{d}r} = 4\pi(8) - \frac{256\pi}{8^2} \)
\( \frac{\text{d}A}{\text{d}r} = 32\pi - \frac{256\pi}{64} = 32\pi - 4\pi = 28\pi \)

(c) Using the small changes approximation:
\( \delta A \approx \frac{\text{d}A}{\text{d}r} \times \delta r \)

Here, \( \delta r = p \) and at \( r = 8 \), \( \frac{\text{d}A}{\text{d}r} = 28\pi \).

Therefore,
\( \delta A \approx 28\pi p \)

(a)
- M1: Attempts to write an expression for volume in terms of \( r \) and \( h \) and sets it equal to \( 128\pi \) to make \( h \) the subject.
- M1: Writes the formula for the total surface area of a cylinder and substitutes their expression for \( h \).
- A1: Fully correct algebraic steps leading to the given show-that formula \( A = 2\pi r^2 + \frac{256\pi}{r} \).

(b)
- M1: Attempts to differentiate the given expression for \( A \) with at least one term correctly differentiated.
- A1: Correctly evaluates \( \frac{\text{d}A}{\text{d}r} = 28\pi \) when \( r = 8 \).

(c)
- M1: Uses the approximation formula \( \delta A \approx \frac{\text{d}A}{\text{d}r} \delta r \) with their value from part (b).
- M1: Identifies \( \delta r = p \).
- A1: Obtains the correct approximate change of \( 28\pi p \) (or equivalent form like \( 28 p \pi \)).

Part (a):
Using the binomial expansion formula:
\((1 + Y)^n = 1 + nY + \frac{n(n-1)}{2!}Y^2 + \frac{n(n-1)(n-2)}{3!}Y^3 + \dots\)
With \(n = -\frac{1}{3}\) and \(Y = 3x\):
\((1 + 3x)^{-\frac{1}{3}} = 1 + \left(-\frac{1}{3}\right)(3x) + \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2}(3x)^2 + \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\left(-\frac{7}{3}\right)}{6}(3x)^3 + \dots\)
\((1 + 3x)^{-\frac{1}{3}} = 1 - x + \frac{4/9}{2}(9x^2) + \frac{-28/27}{6}(27x^3) + \dots\)
\((1 + 3x)^{-\frac{1}{3}} = 1 - x + 2x^2 - \frac{14}{3}x^3\)

This expansion is valid for \(|3x| < 1\), which simplifies to \(|x| < \frac{1}{3}\) (or \(-\frac{1}{3} < x < \frac{1}{3}\)).

Part (b):
We want to find the expansion of \(\frac{\text{d}}{\text{d}x}\left[x(1 + 3x)^{-\frac{1}{3}}\right]\).
Using the result from part (a):
\(x(1 + 3x)^{-\frac{1}{3}} \approx x\left(1 - x + 2x^2 - \frac{14}{3}x^3\right) = x - x^2 + 2x^3 - \frac{14}{3}x^4\)
Now, differentiating term-by-term with respect to \(x\):
\(\frac{\text{d}}{\text{d}x}\left[x(1 + 3x)^{-\frac{1}{3}}\right] \approx \frac{\text{d}}{\text{d}x}\left(x - x^2 + 2x^3 - \frac{14}{3}x^4\right) = 1 - 2x + 6x^2 - \frac{56}{3}x^3\)

Part (c):
Using the expansion from part (a):
\(\int_{0}^{0.1} (1 + 3x)^{-\frac{1}{3}} \text{d}x \approx \int_{0}^{0.1} \left(1 - x + 2x^2 - \frac{14}{3}x^3\right) \text{d}x\)
\(= \left[ x - \frac{x^2}{2} + \frac{2x^3}{3} - \frac{7x^4}{6} \right]_{0}^{0.1}\)
\(= \left( 0.1 - \frac{(0.1)^2}{2} + \frac{2(0.1)^3}{3} - \frac{7(0.1)^4}{6} \right) - 0\)
\(= 0.1 - 0.005 + \frac{0.002}{3} - \frac{0.0007}{6}\)
\(= 0.095 + 0.00066667 - 0.00011667\)
\(= 0.09555\) (to 5 decimal places)

Part (a):
- M1: Attempt binomial expansion with \(n = -1/3\) and term in \(x\), showing the structure of at least three terms.
- A1: Correctly simplified second and third terms: \(-x + 2x^2\).
- A1: Correctly simplified fourth term: \(-\frac{14}{3}x^3\).
- B1: States correct validity range, e.g., \(|x| < \frac{1}{3}\) or \(-\frac{1}{3} < x < \frac{1}{3}\).

Part (b):
- M1: Multiplies expansion from part (a) by \(x\) to obtain a polynomial of the form \(x - ax^2 + bx^3 - cx^4\).
- M1: Differentiates their polynomial term-by-term with respect to \(x\).
- A1: Correct simplified expansion: \(1 - 2x + 6x^2 - \frac{56}{3}x^3\) (allow ft from incorrect coefficients in part a, if differentiated correctly).

Part (c):
- M1: Integrates their binomial expansion from part (a) term-by-term (at least 3 terms integrated correctly).
- M1: Correctly substitutes the limits \(0.1\) and \(0\) into their integrated expression.
- A1: Correct final answer of \(0.09555\) (to 5 decimal places).

**(a)**
Using the compound angle formula:
\(\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x\)

Apply double-angle formulas \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\):
\(\cos 3x = (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x\)
\(\cos 3x = 2\cos^3 x - \cos x - 2\sin^2 x \cos x\)

Use the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\):
\(\cos 3x = 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x\)
\(\cos 3x = 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x\)
\(\cos 3x = 4\cos^3 x - 3\cos x\) (as required)

**(b)**
From (a):
\(\cos 3x = 4\cos^3 x - 3\cos x \implies 4\cos^3 x = \cos 3x + 3\cos x\)
\(\cos^3 x = \frac{1}{4}(\cos 3x + 3\cos x)\)

**(c)**
\(\int \cos^3 x \text{ d}x = \int \frac{1}{4}(\cos 3x + 3\cos x) \text{ d}x\)
\(= \frac{1}{4} \left( \frac{1}{3}\sin 3x + 3\sin x \right) + C\)
\(= \frac{1}{12}\sin 3x + \frac{3}{4}\sin x + C\)

**(d)**
Given equation:
\(8\cos^3 x - 6\cos x = \sqrt{3}\)
\(2(4\cos^3 x - 3\cos x) = \sqrt{3}\)

Substitute \(\cos 3x\):
\(2\cos 3x = \sqrt{3} \implies \cos 3x = \frac{\sqrt{3}}{2}\)

Since \(0 \le x \le \pi\), we have \(0 \le 3x \le 3\pi\).
The solutions for \(\cos 3x = \frac{\sqrt{3}}{2}\) in this interval:
\(3x = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6}\)
\(x = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}\)

**(e)**
Using the volume of revolution formula:
\(V = \pi \int_{a}^{b} y^2 \text{ d}x\)
Given \(y = 3\cos^{3/2} x\), so \(y^2 = 9\cos^3 x\)
\(V = \pi \int_{0}^{\pi/4} 9\cos^3 x \text{ d}x = 9\pi \int_{0}^{\pi/4} \cos^3 x \text{ d}x\)

Using the integral from part (c):
\(V = 9\pi \left[ \frac{1}{12}\sin 3x + \frac{3}{4}\sin x \right]_{0}^{\pi/4}\)

Evaluate at upper limit \(x = \frac{\pi}{4}\):
\(\frac{1}{12}\sin\left(\frac{3\pi}{4}\right) + \frac{3}{4}\sin\left(\frac{\pi}{4}\right) = \frac{1}{12}\left(\frac{\sqrt{2}}{2}\right) + \frac{3}{4}\left(\frac{\sqrt{2}}{2}\right) = \frac{10}{24}\sqrt{2} = \frac{5\sqrt{2}}{12}\)

Evaluate at lower limit \(x = 0\):
\(\frac{1}{12}\sin 0 + \frac{3}{4}\sin 0 = 0\)

Therefore:
\(V = 9\pi \left( \frac{5\sqrt{2}}{12} \right) = \frac{45\sqrt{2}\pi}{12} = \frac{15\sqrt{2}\pi}{4}\)

**Part (a) [4 Marks]**
- **M1**: Uses compound angle formula \(\cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x\) (or equivalent).
- **M1**: Substitutes double-angle formulas \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\).
- **M1**: Uses \(\sin^2 x = 1 - \cos^2 x\) to express the entire equation in terms of \(\cos x\).
- **A1**: Fully correct algebraic steps leading to the given identity: \(\cos 3x = 4\cos^3 x - 3\cos x\).

**Part (b) [1 Mark]**
- **B1**: Rearranges the formula correctly to find \(\cos^3 x = \frac{1}{4}(\cos 3x + 3\cos x)\) or equivalent.

**Part (c) [3 Marks]**
- **M1**: Attempts to integrate \(\cos 3x\) and \(\cos x\) (must see \(\pm \sin 3x\) and \(\pm \sin x\) terms).
- **A1**: Correct integration of both terms: \(\int \cos 3x \text{ d}x = \frac{1}{3}\sin 3x\) and \(\int \cos x \text{ d}x = \sin x\).
- **A1**: Correct expression including constant of integration: \(\frac{1}{12}\sin 3x + \frac{3}{4}\sin x + C\).

**Part (d) [4 Marks]**
- **M1**: Recognizes \(8\cos^3 x - 6\cos x = 2\cos 3x\) and writes \(2\cos 3x = \sqrt{3}\) or equivalent.
- **M1**: Finds at least one correct value of \(3x\) (e.g., \(3x = \frac{\pi}{6}\)) or \(x\) (e.g., \(x = \frac{\pi}{18}\)).
- **A1**: Finds two correct values of \(x\) from \(x = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}\).
- **A1**: Finds all three correct values of \(x\) and no extras in the range \(0 \le x \le \pi\).

**Part (e) [5 Marks]**
- **M1**: Identifies the formula for the volume of revolution \(V = \pi \int y^2 \text{ d}x\) and writes down \(\pi \int 9\cos^3 x \text{ d}x\).
- **M1**: Uses the integration result from (c) to write \([A\sin 3x + B\sin x]_0^{\pi/4}\).
- **M1**: Substitutes limits \(0\) and \(\frac{\pi}{4}\) into their integrated expression.
- **A1**: Correctly evaluates \(\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}\) and \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\).
- **A1**: Obtains the exact volume \(\frac{15\sqrt{2}\pi}{4}\) (or exact equivalent).

First, express the term \(\log_4 x\) in base 2 using the change of base formula: \(\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}\). Substitute this into the first equation to get \(\frac{1}{2}\log_2 x + \log_2 y = 4\). Multiplying the entire equation by 2 gives \(\log_2 x + 2\log_2 y = 8\). Using the laws of logarithms, this can be written as \(\log_2(xy^2) = 8\), which in exponential form is \(xy^2 = 2^8 = 256\) (Equation 1). Now, remove the logarithm from the second equation: \(\log_2(x - 2y) = 3\) gives \(x - 2y = 2^3 = 8\), which simplifies to \(x = 2y + 8\) (Equation 2). Substitute Equation 2 into Equation 1 to obtain a cubic equation in terms of \(y\): \((2y + 8)y^2 = 256\), which simplifies to \(2y^3 + 8y^2 - 256 = 0\), and dividing by 2 yields \(y^3 + 4y^2 - 128 = 0\). By the factor theorem, test integer factors of 128 to find a real root: let \(f(y) = y^3 + 4y^2 - 128\), and checking \(y = 4\) gives \(f(4) = 4^3 + 4(4^2) - 128 = 64 + 64 - 128 = 0\). Thus, \(y = 4\) is a root and \((y - 4)\) is a factor. Factorising the cubic gives \((y - 4)(y^2 + 8y + 32) = 0\). For the quadratic part \(y^2 + 8y + 32 = 0\), the discriminant is \(b^2 - 4ac = 8^2 - 4(1)(32) = 64 - 128 = -64\). Since the discriminant is negative, there are no other real roots. Therefore, \(y = 4\) is the only real solution. Substituting \(y = 4\) back into Equation 2 gives \(x = 2(4) + 8 = 16\). Checking both values in the original equations confirms they are valid since \(x > 0\), \(y > 0\), and \(x - 2y > 0\). Thus, the final solution is \(x = 16, y = 4\).

- **M1**: For using the change of base formula to write \(\log_4 x\) as \(\frac{\log_2 x}{2}\). - **M1**: For applying laws of logarithms to the first equation to form a single logarithm, such as \(\log_2(x^{1/2}y) = 4\) or \(\log_2(xy^2) = 8\). - **A1**: For the correct index equation \(xy^2 = 256\) (or equivalent). - **B1**: For correctly removing the logarithm from the second equation to get \(x - 2y = 8\). - **M1**: For substituting \(x = 2y + 8\) (or equivalent) into their first equation to form a cubic equation in \(y\) (or a polynomial in \(x\)), e.g. \(y^3 + 4y^2 - 128 = 0\). - **M1**: For attempting to solve the cubic equation by identifying \(y = 4\) as a root (e.g., showing \(f(4) = 0\) or factorising as \((y-4)(y^2+8y+32)=0\)). - **A1**: For finding the correct value \(y = 4\) and showing there are no other real roots. - **A1**: For finding the correct value \(x = 16\).

For the quadratic equation to have two distinct real roots, its discriminant must be strictly greater than zero: \(b^2 - 4ac > 0\). From the given equation, we identify: \(a = 1\), \(b = k + 3\), and \(c = 2k + 3\). Substituting these into the discriminant expression: \(\Delta = (k + 3)^2 - 4(1)(2k + 3)\). Expanding the terms: \(\Delta = k^2 + 6k + 9 - 8k - 12 = k^2 - 2k - 3\). We require: \(k^2 - 2k - 3 > 0\). Factorising the quadratic inequality: \((k - 3)(k + 1) > 0\). The critical values are \(k = 3\) and \(k = -1\). Since we require the expression to be strictly positive, the solution lies outside the interval between the critical values. Thus, the set of possible values is \(k < -1\) or \(k > 3\).

M1: Attempts to use the discriminant \(b^2 - 4ac\) with correct substitution of \(a, b, c\). A1: Simplifies the discriminant to get the correct three-term quadratic expression \(k^2 - 2k - 3\). M1: Sets their discriminant expression \(> 0\) and finds the correct critical values \(k = 3\) and \(k = -1\). M1: Identifies the correct regions (outside the critical values) for a 'greater than' inequality. A1: Correct final inequalities: \(k < -1\) or \(k > 3\) (or equivalent set notation).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) Using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} a b \sin C\). Here, \(\text{Area} = \frac{1}{2} (AB)(BC) \sin(ABC)\), so \(15\sqrt{3} = \frac{1}{2} (x+3)(2x-1) \sin(120^\circ)\). Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\), we have \(15\sqrt{3} = \frac{1}{2} (2x^2 + 5x - 3) \left(\frac{\sqrt{3}}{2}\right)\), which simplifies to \(15\sqrt{3} = \frac{\sqrt{3}}{4} (2x^2 + 5x - 3)\). Dividing both sides by \(\sqrt{3}\) and multiplying by 4 gives \(60 = 2x^2 + 5x - 3\), which simplifies to the quadratic equation \(2x^2 + 5x - 63 = 0\). Factorizing this gives \((2x - 9)(x + 7) = 0\). Since the side lengths of the triangle must be positive, \(x\) must be positive, which gives \(x = 4.5\). (b) Substituting \(x = 4.5\) into the expressions for the side lengths, we get \(AB = 4.5 + 3 = 7.5\text{ cm}\) and \(BC = 2(4.5) - 1 = 8\text{ cm}\). Using the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). Since \(\cos(120^\circ) = -0.5\), we have \(AC^2 = 56.25 + 64 - 120(-0.5) = 120.25 + 60 = 180.25\). Therefore, \(AC = \sqrt{180.25} \approx 13.4257\text{ cm}\). To 3 significant figures, \(AC = 13.4\text{ cm}\).

Part (a): - M1: Uses the area formula to write an equation in terms of \(x\): \(\frac{1}{2}(x+3)(2x-1)\sin(120^\circ) = 15\sqrt{3}\). - M1: Rearranges and simplifies to a standard quadratic equation: \(2x^2 + 5x - 63 = 0\). - M1: Solves the quadratic equation by factorizing to \((2x - 9)(x + 7) = 0\) or by using the quadratic formula. - A1: Identifies \(x = 4.5\) as the only valid solution and rejects the negative root. Part (b): - M1: Finds \(AB = 7.5\) and \(BC = 8\) and substitutes them into the Cosine Rule: \(AC^2 = 7.5^2 + 8^2 - 2(7.5)(8)\cos(120^\circ)\). - A1: Obtains a correct intermediate value, e.g., \(AC^2 = 180.25\). - A1: For the final answer \(AC = 13.4\) (3 s.f.).

(a) The velocity \(v\) of the particle at time \(t\) is found by differentiating the displacement with respect to \(t\):
\(v = \frac{ds}{dt} = 3t^2 - 12t + 9\)

When \(P\) is instantaneously at rest, \(v = 0\):
\(3t^2 - 12t + 9 = 0\)
\(3(t^2 - 4t + 3) = 0\)
\(3(t - 1)(t - 3) = 0\)

So, \(t = 1\) or \(t = 3\).

(b) When the velocity is \(45\text{ m/s}\):
\(3t^2 - 12t + 9 = 45\)
\(3t^2 - 12t - 36 = 0\)
\(t^2 - 4t - 12 = 0\)

Factorising the quadratic equation:
\((t - 6)(t + 2) = 0\)

Since \(t \ge 0\), we take \(t = 6\).

The acceleration \(a\) of \(P\) at time \(t\) is found by differentiating the velocity with respect to \(t\):
\(a = \frac{dv}{dt} = 6t - 12\)

At \(t = 6\):
\(a = 6(6) - 12 = 24\text{ m/s}^2\).

(a)
M1: For differentiating \(s\) with respect to \(t\) to find an expression for \(v\) (at least two terms correct).
M1: For setting their \(v = 0\) and attempting to solve the quadratic equation.
A1: For both correct values \(t = 1\) and \(t = 3\).

(b)
M1: For setting their expression for \(v\) equal to 45 and attempting to solve the resulting quadratic equation.
A1: For finding the correct positive solution \(t = 6\) (accept rejection of \(t = -2\) shown or implied).
M1: For differentiating their \(v\) to obtain an expression for the acceleration \(a\).
A1: For substituting their positive \(t\) value into their expression for \(a\) to obtain \(24\) (or \(24\text{ m/s}^2\)).

(a) Writing down the inequalities from the given conditions:
1. The total number of items must not exceed 15: \(x + y \le 15\).
2. The number of items of Type B must be at least 3: \(y \ge 3\).
3. The number of items of Type B must be at most twice Type A: \(y \le 2x\).
4. Total time constraint: \(2x + 3y \le 36\).

(b) To find the vertices of the feasible region \(R\), we find the intersections of the boundary lines:
- Intersection of \(y = 3\) and \(y = 2x\):
\(y = 3 \implies 2x = 3 \implies x = 1.5\). Vertex 1 is \((1.5, 3)\).
- Intersection of \(y = 3\) and \(x + y = 15\):
\(y = 3 \implies x = 12\). Vertex 2 is \((12, 3)\).
- Intersection of \(x + y = 15\) and \(2x + 3y = 36\):
Multiplying the first equation by 2 gives \(2x + 2y = 30\).
Subtracting this from the second equation gives \(y = 6\).
Then \(x = 15 - 6 = 9\). Vertex 3 is \((9, 6)\).
- Intersection of \(2x + 3y = 36\) and \(y = 2x\):
\(2x + 3(2x) = 36 \implies 8x = 36 \implies x = 4.5\).
Then \(y = 2(4.5) = 9\). Vertex 4 is \((4.5, 9)\).

Thus, the four vertices of \(R\) are \((1.5, 3)\), \((12, 3)\), \((9, 6)\), and \((4.5, 9)\).

(c) The objective profit function is \(P = 15x + 25y\).
Since \(x\) and \(y\) must be integers, we test integer coordinate points within the feasible region \(R\) near the upper boundaries where the profit is likely to be maximized:
- At the integer vertex \((9, 6)\):
\(P = 15(9) + 25(6) = 135 + 150 = 285\).
- Near the non-integer vertex \((4.5, 9)\):
The maximum possible integer value for \(y\) is 8 (since if \(y=9\), the only feasible \(x\) is 4.5).
If \(y = 8\), the boundaries require:
\(y \le 2x \implies 8 \le 2x \implies x \ge 4\).
\(2x + 3y \le 36 \implies 2x + 24 \le 36 \implies 2x \le 12 \implies x \le 6\).
\(x + y \le 15 \implies x + 8 \le 15 \implies x \le 7\).
So the possible integer values of \(x\) when \(y = 8\) are \(x = 4, 5, 6\).
Testing these points:
At \((4, 8)\): \(P = 15(4) + 25(8) = 60 + 200 = 260\).
At \((5, 8)\): \(P = 15(5) + 25(8) = 75 + 200 = 275\).
At \((6, 8)\): \(P = 15(6) + 25(8) = 90 + 200 = 290\).
- If \(y = 7\), the maximum possible integer value of \(x\) is 7 (since \(2x + 3(7) \le 36 \implies 2x \le 15 \implies x \le 7.5\)):
At \((7, 7)\): \(P = 15(7) + 25(7) = 105 + 175 = 280\).
Comparing these results, the maximum weekly profit is 290, which occurs when \(x = 6\) and \(y = 8\).

Part (a):
- M1: For writing down at least three of the four inequalities correctly (accept strict inequalities for this mark).
- A1: For all four inequalities correctly written (must include non-strict inequalities: \(x+y \le 15\), \(y \ge 3\), \(y \le 2x\), and \(2x+3y \le 36\)).

Part (b):
- M1: For attempting to solve simultaneous equations to find the intersection of any two boundary lines.
- A1: For finding any two correct vertices.
- A1: For finding all four correct vertices: \((1.5, 3)\), \((12, 3)\), \((9, 6)\), and \((4.5, 9)\).

Part (c):
- B1: For establishing the profit function \(P = 15x + 25y\).
- M1: For testing at least two candidate integer coordinates close to the upper boundary lines of the feasible region (e.g., testing \((9, 6)\) and \((6, 8)\)).
- A1: For correctly identifying the maximum weekly profit as 290, achieved with \(x = 6\) items of Type A and \(y = 8\) items of Type B.

At \(x = 1\), \(y = 8\sqrt{1} - 2(1) = 6\), so the coordinates of \(P\) are \((1, 6)\). (a) Differentiating the equation of the curve with respect to \(x\) gives \(\frac{dy}{dx} = 4x^{-1/2} - 2\). At \(P\) where \(x = 1\), the gradient of the tangent is \(\frac{dy}{dx} = 4(1)^{-1/2} - 2 = 2\). Therefore, the gradient of the normal to \(C\) at \(P\) is \(m = -\frac{1}{2}\). Using the equation of a straight line: \(y - 6 = -\frac{1}{2}(x - 1)\), which simplifies to \(2y - 12 = -x + 1\), or \(x + 2y - 13 = 0\). (b) The normal crosses the \(x\)-axis when \(y = 0\). Substituting this into the normal equation gives \(x + 2(0) - 13 = 0\), which leads to \(x = 13\). Thus, the coordinates of \(Q\) are \((13, 0)\). (c) The curve crosses the positive \(x\)-axis when \(y = 0\). Setting \(8\sqrt{x} - 2x = 0\) gives \(2\sqrt{x}(4 - \sqrt{x}) = 0\). Since \(x > 0\) for the positive \(x\)-axis, we have \(4 - \sqrt{x} = 0\), which gives \(x = 16\). The coordinates of \(S\) are \((16, 0)\). (d) The region \(R\) is bounded by the curve from \(x = 0\) to \(x = 1\), and the normal line from \(x = 1\) to \(x = 13\). Total Area \(= \int_{0}^{1} (8x^{1/2} - 2x) \, dx + \text{Area of triangle } PQQ'\) (where \(Q'\) is the point \((1,0)\)). The area under the curve is: \(\int_{0}^{1} (8x^{1/2} - 2x) \, dx = \left[ \frac{16}{3}x^{3/2} - x^2 \right]_{0}^{1} = \left(\frac{16}{3} - 1\right) - 0 = \frac{13}{3}\). The area of the triangle with base from \(x = 1\) to \(x = 13\) (length \(12\)) and height \(6\) is: \(\text{Area} = \frac{1}{2} \times 12 \times 6 = 36\). Therefore, the total exact area of \(R\) is \(\frac{13}{3} + 36 = \frac{121}{3}\) (or \(40\frac{1}{3}\)).

(a) M1: Attempt to differentiate \(y = 8x^{1/2} - 2x\) with at least one term correct. A1: Correct derivative \(\frac{dy}{dx} = 4x^{-1/2} - 2\). M1: Substitute \(x = 1\) into their derivative to find the tangent gradient, and obtain the negative reciprocal. M1: Find the \(y\)-coordinate of \(P\) as \(6\) and use their normal gradient to set up a linear equation. A1: Correct equation in the form \(ax + by + c = 0\) with integer coefficients, e.g., \(x + 2y - 13 = 0\). (b) M1: Substitute \(y = 0\) into their normal equation and solve for \(x\). A1: \((13, 0)\) or \(x = 13, y = 0\). (c) M1: Set \(y = 0\) in the curve equation and attempt to solve for \(x > 0\). A1: \((16, 0)\) or \(x = 16, y = 0\). (d) M1: Realise that the area is split into two parts, split at \(x = 1\). M1: Attempt to integrate the curve expression (increasing at least one fractional/integer power by 1). A1: Correct integration of the curve: \(\frac{16}{3}x^{3/2} - x^2\). M1: Correct substitute of limits \(0\) and \(1\) to find the first area as \(\frac{13}{3}\). M1: Correct calculation of the triangular area under the normal line as \(36\) (either by integration or geometry). A1: Correct final exact area of \(\frac{121}{3}\) (or \(40\frac{1}{3}\)).

(a) The volume of a right circular cylinder is given by the formula:
\[V = \pi r^2 h\]

Since both \(r\) and \(h\) are functions of time \(t\), we differentiate with respect to \(t\) using the product rule:
\[\frac{\text{d}V}{\text{d}t} = \pi \left[ \frac{\text{d}}{\text{d}t}(r^2) \cdot h + r^2 \cdot \frac{\text{d}h}{\text{d}t} \right]\]

Using the chain rule, \(\frac{\text{d}}{\text{d}t}(r^2) = 2r \frac{\text{d}r}{\text{d}t}\). Substituting this back into our expression:
\[\frac{\text{d}V}{\text{d}t} = \pi \left( 2r \frac{\text{d}r}{\text{d}t} h + r^2 \frac{\text{d}h}{\text{d}t} \right)\]

Factoring out \(r\) from both terms inside the bracket:
\[\frac{\text{d}V}{\text{d}t} = \pi r \left( 2h \frac{\text{d}r}{\text{d}t} + r \frac{\text{d}h}{\text{d}t} \right) \quad \text{(as required)}\]

(b) From the given information, at this particular instant:
\[r = 6\]
\[\frac{\text{d}r}{\text{d}t} = 0.5\]
\[h = 10\]
\[\frac{\text{d}h}{\text{d}t} = -0.8 \quad \text{(negative because the height is decreasing)}\]

Substituting these values into the formula derived in part (a):
\[\frac{\text{d}V}{\text{d}t} = \pi (6) \left( 2(10)(0.5) + (6)(-0.8) \right)\]
\[\frac{\text{d}V}{\text{d}t} = 6\pi \left( 10 - 4.8 \right)\]
\[\frac{\text{d}V}{\text{d}t} = 6\pi (5.2) = 31.2\pi\text{ cm}^3\text{/s}\]

**Part (a)**
* **M1**: For attempting to use the product rule on \(V = \pi r^2 h\) with respect to \(t\).
* **M1**: For correctly applying the chain rule to find the derivative of \(r^2\) as \(2r \frac{\text{d}r}{\text{d}t}\).
* **A1**: For obtaining the correct differentiated expression and factoring out \(r\) to show the given result exactly as shown.

**Part (b)**
* **B1**: For correctly identifying that \(\frac{\text{d}h}{\text{d}t} = -0.8\) (must include the negative sign).
* **M1**: For substituting \(r = 6\), \(h = 10\), \(\frac{\text{d}r}{\text{d}t} = 0.5\), and their \(\frac{\text{d}h}{\text{d}t}\) into the formula from part (a).
* **A1**: For simplifying to \(6\pi(10 - 4.8)\) or an equivalent exact expression.
* **A1**: For the final exact answer of \(31.2\pi\) (or \(\frac{156}{5}\pi\)).

(a) The volume of the cylinder is \(V_{\text{cylinder}} = \pi r^2 h\).
The volume of the hemisphere is \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\).
The total volume of the solid is given by:
\[V = \pi r^2 h + \frac{2}{3}\pi r^3\]
Since \(V = 360\pi\):
\[\pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi\]
Dividing both sides by \(\pi\):
\[r^2 h + \frac{2}{3}r^3 = 360 \implies h = \frac{360}{r^2} - \frac{2}{3}r\]

The total surface area \(A\) of the solid consists of the curved surface area of the hemisphere, the curved surface area of the cylinder, and the flat circular base:
\[A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\]

Substituting \(h = \frac{360}{r^2} - \frac{2}{3}r\) into the surface area equation:
\[A = 3\pi r^2 + 2\pi r \left(\frac{360}{r^2} - \frac{2}{3}r\right)\]
\[A = 3\pi r^2 + \frac{720\pi}{r} - \frac{4}{3}\pi r^2\]
\[A = \frac{5}{3}\pi r^2 + \frac{720\pi}{r} \quad \text{(as required)}\]

(b) To find the stationary value, we differentiate \(A\) with respect to \(r\):
\[\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{10}{3}\pi r - \frac{720\pi}{r^2}\]

Setting \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\):
\[\frac{10}{3}\pi r = \frac{720\pi}{r^2}\]
\[r^3 = 216\]
\[r = 6\]

To justify that \(r = 6\) gives a minimum, we find the second derivative:
\[\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10}{3}\pi + \frac{1440\pi}{r^3}\]

Substituting \(r = 6\):
\[\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10}{3}\pi + \frac{1440\pi}{216} = 10\pi\]
Since \(10\pi > 0\), the surface area \(A\) is a minimum at \(r = 6\).

(c) Substituting \(r = 6\) into the expression for \(A\):
\[A = \frac{5}{3}\pi (6)^2 + \frac{720\pi}{6} = 60\pi + 120\pi = 180\pi\]

(a)
- **M1**: Writes an expression for the total volume \(V = \pi r^2 h + \frac{2}{3}\pi r^3\).
- **A1**: Sets \(V = 360\pi\) and successfully expresses \(h\) in terms of \(r\): \(h = \frac{360}{r^2} - \frac{2}{3}r\).
- **M1**: Writes the total surface area formula \(A = 3\pi r^2 + 2\pi r h\) and substitutes their expression for \(h\).
- **A1\***: Fully simplifies without errors to obtain \(A = \frac{5}{3}\pi r^2 + \frac{720\pi}{r}\).

(b)
- **M1**: Standard differentiation of \(A\) with respect to \(r\) resulting in at least one term correct.
- **A1**: Correct derivative: \(\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{10}{3}\pi r - \frac{720\pi}{r^2}\).
- **M1**: Sets their \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\) and solves for \(r\).
- **A1**: Obtains \(r = 6\) and justifies minimum using the second derivative \(\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = 10\pi > 0\) (or an acceptable first derivative sign test).

(c)
- **B1**: Substitutes \(r = 6\) to find the minimum area as \(180\pi\).

(a) The volume of the cylinder is \(V_{\text{cylinder}} = \pi r^2 h\).
The volume of the hemisphere is \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\).
The total volume of the solid is given by:
\[V = \pi r^2 h + \frac{2}{3}\pi r^3\]
Since \(V = 360\pi\):
\[\pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi\]
Dividing both sides by \(\pi\):
\[r^2 h + \frac{2}{3}r^3 = 360 \implies h = \frac{360}{r^2} - \frac{2}{3}r\]

The total surface area \(A\) of the solid consists of the curved surface area of the hemisphere, the curved surface area of the cylinder, and the flat circular base:
\[A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\]

Substituting \(h = \frac{360}{r^2} - \frac{2}{3}r\) into the surface area equation:
\[A = 3\pi r^2 + 2\pi r \left(\frac{360}{r^2} - \frac{2}{3}r\right)\]
\[A = 3\pi r^2 + \frac{720\pi}{r} - \frac{4}{3}\pi r^2\]
\[A = \frac{5}{3}\pi r^2 + \frac{720\pi}{r} \quad \text{(as required)}\]

(b) To find the stationary value, we differentiate \(A\) with respect to \(r\):
\[\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{10}{3}\pi r - \frac{720\pi}{r^2}\]

Setting \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\):
\[\frac{10}{3}\pi r = \frac{720\pi}{r^2}\]
\[r^3 = 216\]
\[r = 6\]

To justify that \(r = 6\) gives a minimum, we find the second derivative:
\[\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10}{3}\pi + \frac{1440\pi}{r^3}\]

Substituting \(r = 6\):
\[\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10}{3}\pi + \frac{1440\pi}{216} = 10\pi\]
Since \(10\pi > 0\), the surface area \(A\) is a minimum at \(r = 6\).

(c) Substituting \(r = 6\) into the expression for \(A\):
\[A = \frac{5}{3}\pi (6)^2 + \frac{720\pi}{6} = 60\pi + 120\pi = 180\pi\]

(a)
- **M1**: Writes an expression for the total volume \(V = \pi r^2 h + \frac{2}{3}\pi r^3\).
- **A1**: Sets \(V = 360\pi\) and successfully expresses \(h\) in terms of \(r\): \(h = \frac{360}{r^2} - \frac{2}{3}r\).
- **M1**: Writes the total surface area formula \(A = 3\pi r^2 + 2\pi r h\) and substitutes their expression for \(h\).
- **A1\***: Fully simplifies without errors to obtain \(A = \frac{5}{3}\pi r^2 + \frac{720\pi}{r}\).

(b)
- **M1**: Standard differentiation of \(A\) with respect to \(r\) resulting in at least one term correct.
- **A1**: Correct derivative: \(\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{10}{3}\pi r - \frac{720\pi}{r^2}\).
- **M1**: Sets their \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\) and solves for \(r\).
- **A1**: Obtains \(r = 6\) and justifies minimum using the second derivative \(\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = 10\pi > 0\) (or an acceptable first derivative sign test).

(c)
- **B1**: Substitutes \(r = 6\) to find the minimum area as \(180\pi\).

(a) Using the formula for the sum of an arithmetic series \(S_n = \frac{n}{2}[2a + (n-1)d]\):

\(S_k = \frac{k}{2}[2a + (k-1)d] = ak + \frac{1}{2}dk^2 - \frac{1}{2}dk\)

\(S_{2k} = \frac{2k}{2}[2a + (2k-1)d] = 2ak + 2dk^2 - dk\)

Substitute these expressions into the given relation \(S_{2k} = 4S_k - 6k\):

\(2ak + 2dk^2 - dk = 4\left(ak + \frac{1}{2}dk^2 - \frac{1}{2}dk\right) - 6k\)

\(2ak + 2dk^2 - dk = 4ak + 2dk^2 - 2dk - 6k\)

Subtract \(2dk^2\) from both sides:

\(2ak - dk = 4ak - 2dk - 6k\)

Rearrange the terms to one side:

\(2ak - dk - 6k = 0\)

\(k(2a - d - 6) = 0\)

Since \(k\) is a positive integer, \(k \ne 0\), we can divide by \(k\):

\(2a - d - 6 = 0 \implies 2a - d = 6\) (as required).

(b) Using the given sum \(S_{15} = 495\):

\(S_{15} = \frac{15}{2}[2a + 14d] = 15(a + 7d) = 495\)

Divide by 15:

\(a + 7d = 33\)

From part (a), we have \(d = 2a - 6\). Substitute this into the equation:

\(a + 7(2a - 6) = 33\)

\(a + 14a - 42 = 33\)

\(15a = 75 \implies a = 5\)

Using \(d = 2a - 6\):

\(d = 2(5) - 6 = 4\)

Thus, \(a = 5\) and \(d = 4\).

(c) The sum of the first \(p\) terms is:

\(S_p = \frac{p}{2}[2(5) + (p-1)4] = \frac{p}{2}[10 + 4p - 4] = \frac{p}{2}[4p + 6] = p(2p + 3) = 2p^2 + 3p\)

We require \(S_p > 2000\):

\(2p^2 + 3p > 2000 \implies 2p^2 + 3p - 2000 > 0\)

Solving the critical quadratic equation \(2p^2 + 3p - 2000 = 0\):

\(p = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2000)}}{4} = \frac{-3 \pm \sqrt{16009}}{4}\)

Since \(p > 0\), we choose the positive root:

\(p \approx \frac{-3 + 126.53}{4} \approx 30.88\)

We test the adjacent integers:

For \(p = 30\): \(S_{30} = 2(30)^2 + 3(30) = 1890 < 2000\).

For \(p = 31\): \(S_{31} = 2(31)^2 + 3(31) = 2015 > 2000\).

Thus, the smallest integer value of \(p\) is 31.

Part (a):
- M1: For attempting to write algebraic expressions for \(S_k\) and \(S_{2k}\) using the arithmetic series sum formula.
- M1: For substituting these expressions into \(S_{2k} = 4S_k - 6k\) and attempting to simplify the resulting equation to linear terms in \(k\).
- A1: For a complete and mathematically sound proof showing \(2a - d = 6\) with clear intermediate steps.

Part (b):
- M1: For using \(S_{15} = 495\) to obtain the linear equation \(a + 7d = 33\) (or equivalent).
- M1: For solving the system of two linear equations simultaneously to find a value for either \(a\) or \(d\).
- A1: For both correct values: \(a = 5\) and \(d = 4\).

Part (c):
- M1: For establishing an expression for \(S_p = 2p^2 + 3p\) using their values of \(a\) and \(d\).
- M1: For setting up the inequality \(2p^2 + 3p > 2000\) and attempting to find the critical value using the quadratic formula.
- A1: For finding the critical decimal boundary \(p \approx 30.88\) or showing evaluations for \(p = 30\) and \(p = 31\).
- A1: For identifying the correct integer value \(p = 31\) with sufficient supporting evidence.

(a) Let \(M\) be the midpoint of the arc \(AB\). By symmetry, the line of symmetry of the sector passes through \(O\), \(C\), and \(M\). Since the angle \(AOB = 2\theta\), we have \(\angle AOM = \theta\). The circle touches the arc \(AB\) at \(M\), so the distance from \(O\) to \(M\) is the radius of the sector, \(OM = r\). The radius of the inscribed circle is \(CM = R\). Since \(C\) lies on the line segment \(OM\), the distance from \(O\) to \(C\) is \(OC = OM - CM = r - R\). The circle touches the radius \(OA\) at \(T_1\). Therefore, the radius \(CT_1\) is perpendicular to \(OA\), which means \(\angle OT_1 C = 90^\circ\). In the right-angled triangle \(OT_1 C\), we have \(\sin(\angle T_1 O C) = \frac{CT_1}{OC}\), which gives \(\sin\theta = \frac{R}{r - R}\). Multiplying both sides by \(r - R\) gives \((r - R)\sin\theta = R\). Expanding and rearranging: \(r\sin\theta - R\sin\theta = R \implies r\sin\theta = R(1 + \sin\theta) \implies R = \frac{r \sin\theta}{1 + \sin\theta}\). (b) Given \(r = 15\) and \(\theta = \frac{\pi}{6}\), we find \(R = \frac{15 \sin(\frac{\pi}{6})}{1 + \sin(\frac{\pi}{6})} = \frac{15 \times 0.5}{1 + 0.5} = \frac{7.5}{1.5} = 5\) cm. The area of the region \(R_1\) is given by \(\text{Area}(R_1) = \text{Area of quadrilateral } OT_1 C T_2 - \text{Area of sector } C T_1 T_2\). The quadrilateral consists of two congruent right-angled triangles, \(OT_1 C\) and \(OT_2 C\). In \(\triangle OT_1 C\), \(OT_1 = \frac{CT_1}{\tan\theta} = \frac{5}{\tan(\frac{\pi}{6})} = 5\sqrt{3}\) cm. The area of \(\triangle OT_1 C\) is \(\frac{1}{2} \times OT_1 \times CT_1 = \frac{1}{2} \times 5\sqrt{3} \times 5 = \frac{25\sqrt{3}}{2}\) cm\(^2\). Thus, the area of quadrilateral \(OT_1 C T_2\) is \(2 \times \frac{25\sqrt{3}}{2} = 25\sqrt{3}\) cm\(^2\). The angles of quadrilateral \(OT_1 C T_2\) sum to \(2\pi\) radians, so \(\angle T_1 C T_2 = 2\pi - (\angle T_1 O T_2 + \angle O T_1 C + \angle C T_2 O) = 2\pi - (\frac{\pi}{3} + \frac{\pi}{2} + \frac{\pi}{2}) = \frac{2\pi}{3}\) radians. The area of sector \(C T_1 T_2\) with radius \(R = 5\) and angle \(\frac{2\pi}{3}\) is \(\frac{1}{2} R^2 \theta_C = \frac{1}{2} \times 5^2 \times \frac{2\pi}{3} = \frac{25\pi}{3}\) cm\(^2\). Therefore, \(\text{Area}(R_1) = 25\sqrt{3} - \frac{25\pi}{3}\) cm\(^2\). This gives \(a = 25\) and \(b = -\frac{25}{3}\). (c) The boundary of region \(R_1\) consists of the straight segments \(OT_1\) and \(OT_2\), and the minor arc \(T_1 T_2\). The lengths of the straight segments are \(OT_1 = OT_2 = 5\sqrt{3}\) cm, so their sum is \(10\sqrt{3}\) cm. The length of the minor arc \(T_1 T_2\) is \(\text{Arc length} = R \times \angle T_1 C T_2 = 5 \times \frac{2\pi}{3} = \frac{10\pi}{3}\) cm. Therefore, the total perimeter is \(10\sqrt{3} + \frac{10\pi}{3}\) cm. This gives \(c = 10\) and \(d = \frac{10}{3}\).

Part (a): M1: Identifies \(OC = r - R\) or equivalent from symmetry. M1: Sets up a correct trigonometric ratio in triangle \(OT_1 C\), e.g., \(\sin\theta = \frac{R}{r - R}\). M1: Rearranges the equation to isolate \(R\), showing clear algebraic steps. A1: Correctly shows the given formula for \(R\) with no errors. Part (b): B1: Calculates \(R = 5\). M1: Identifies the correct strategy to find the area of \(R_1\) by subtracting the sector area from the quadrilateral area. A1: Calculates the correct area of quadrilateral \(OT_1 C T_2 = 25\sqrt{3}\). A1: Calculates the correct angle \(\angle T_1 C T_2 = \frac{2\pi}{3}\) and the correct sector area \(\frac{25\pi}{3}\). A1: Obtains the correct area expression \(25\sqrt{3} - \frac{25}{3}\pi\), with \(a = 25\) and \(b = -\frac{25}{3}\). Part (c): M1: Identifies the boundary components of the perimeter of \(R_1\). A1: Correctly calculates \(OT_1 + OT_2 = 10\sqrt{3}\). A1: Correctly calculates the arc length as \(\frac{10\pi}{3}\). A1: Obtains the correct perimeter expression \(10\sqrt{3} + \frac{10\pi}{3}\), with \(c = 10\) and \(d = \frac{10}{3}\).

From the given equation \(2x^2 - 5x + 4 = 0\), we have \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\). (a)(i) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\). (a)(ii) \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{5}{8}\). (b) For the new equation, the roots are \(\gamma = \frac{\alpha^2}{\beta}\) and \(\delta = \frac{\beta^2}{\alpha}\). The sum of the new roots is: \(\gamma + \delta = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{5/8}{2} = \frac{5}{16}\). The product of the new roots is: \(\gamma\delta = \left(\frac{\alpha^2}{\beta}\right)\left(\frac{\beta^2}{\alpha}\right) = \alpha\beta = 2\). The quadratic equation is given by \(x^2 - (\text{sum})x + \text{product} = 0\), which is \(x^2 - \frac{5}{16}x + 2 = 0\). Multiplying by 16 to obtain integer coefficients gives: \(16x^2 - 5x + 32 = 0\).

(a)(i) M1: For attempting to expand \((\alpha+\beta)^2 - 2\alpha\beta\) and substituting correct values of sum and product. A1: For obtaining \(\frac{9}{4}\) (or 2.25). (a)(ii) M1: For using a correct identity for \(\alpha^3 + \beta^3\) and substituting values. A1: For obtaining \(\frac{5}{8}\) (or 0.625). (b) B1: For finding the product of the new roots is 2. M1: For setting up the sum of the new roots as \(\frac{\alpha^3 + \beta^3}{\alpha\beta}\). A1: For finding the sum of the new roots is \(\frac{5}{16}\). M1: For writing the new equation in the form \(x^2 - (\text{sum})x + \text{product} = 0\). M1: For multiplying by 16 to obtain integer coefficients. A1: For the correct equation \(16x^2 - 5x + 32 = 0\) (or any equivalent equation with integer coefficients, including \(= 0\)).

(a) Since \(AD : DB = 4 : 1\), we have \(\vec{AD} = \frac{4}{5}\vec{AB}\). Thus, \(\vec{OD} = \vec{OA} + \vec{AD} = \mathbf{a} + \frac{4}{5}(\mathbf{b} - \mathbf{a}) = \frac{1}{5}\mathbf{a} + \frac{4}{5}\mathbf{b}\). (b) Since \(OC : CB = 3 : 2\), we have \(\vec{OC} = \frac{3}{5}\mathbf{b}\). Therefore, \(\vec{AC} = \vec{OC} - \vec{OA} = \frac{3}{5}\mathbf{b} - \mathbf{a}\). (c) Since \(X\) lies on the line \(OD\), we can write \(\vec{OX} = \mu \vec{OD} = \mu \left(\frac{1}{5}\mathbf{a} + \frac{4}{5}\mathbf{b}\right) = \frac{\mu}{5}\mathbf{a} + \frac{4\mu}{5}\mathbf{b}\) for some scalar \(\mu\). Since \(X\) lies on the line \(AC\), we can write \(\vec{OX} = \vec{OA} + \lambda \vec{AC} = \mathbf{a} + \lambda \left(\frac{3}{5}\mathbf{b} - \mathbf{a}\right) = (1 - \lambda)\mathbf{a} + \frac{3\lambda}{5}\mathbf{b}\) for some scalar \(\lambda\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) gives: \(\frac{\mu}{5} = 1 - \lambda\) and \(\frac{4\mu}{5} = \frac{3\lambda}{5}\). From the second equation, we get \(4\mu = 3\lambda \implies \mu = \frac{3}{4}\lambda\). Substituting this into the first equation: \(\frac{3}{20}\lambda = 1 - \lambda \implies \frac{23}{20}\lambda = 1 \implies \lambda = \frac{20}{23}\). Thus, \(\mu = \frac{3}{4} \left(\frac{20}{23}\right) = \frac{15}{23}\). Substituting \(\mu\) back into the expression for \(\vec{OX}\): \(\vec{OX} = \frac{15}{23}\left(\frac{1}{5}\mathbf{a} + \frac{4}{5}\mathbf{b}\right) = \frac{3}{23}\mathbf{a} + \frac{12}{23}\mathbf{b}\). (d) Since \(\vec{OX} = \frac{15}{23}\vec{OD}\), we have \(OX = \frac{15}{23}OD\), which means \(XD = \frac{8}{23}OD\). Therefore, the ratio \(OX : XD = 15 : 8\).

(a) M1: For attempting to write \(\vec{OD} = \vec{OA} + \frac{4}{5}\vec{AB}\) or equivalent. A1: Correct expression \(\vec{OD} = \frac{1}{5}\mathbf{a} + \frac{4}{5}\mathbf{b}\) (or equivalent). (b) B1: Correct expression \(\vec{AC} = \frac{3}{5}\mathbf{b} - \mathbf{a}\) (or equivalent). (c) M1: For expressing \(\vec{OX}\) in the form \(\mu\vec{OD}\) or \((1-\lambda)\mathbf{a} + \lambda\vec{OC}\). M1: For equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to obtain two simultaneous equations. M1: For solving the simultaneous equations to find the value of one parameter. A1: Correct value of either \(\lambda = \frac{20}{23}\) or \(\mu = \frac{15}{23}\). A1: Correct final expression \(\vec{OX} = \frac{3}{23}\mathbf{a} + \frac{12}{23}\mathbf{b}\). (d) B1: Correct ratio \(15 : 8\) (or equivalent).