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To express the fraction in the required form, we can split it into two separate terms:

\[\frac{3\sqrt{x} - 4x^2}{2x\sqrt{x}} = \frac{3\sqrt{x}}{2x\sqrt{x}} - \frac{4x^2}{2x\sqrt{x}}\]

First, we express the denominator and numerator terms using laws of indices:
- \(\sqrt{x} = x^{\frac{1}{2}}\)
- \(2x\sqrt{x} = 2x^1 \cdot x^{\frac{1}{2}} = 2x^{\frac{3}{2}}\)

Now substitute these into the split terms:

\[\frac{3x^{\frac{1}{2}}}{2x^{\frac{3}{2}}} - \frac{4x^2}{2x^{\frac{3}{2}}}\]

Apply the division law of indices \(\frac{x^a}{x^b} = x^{a-b}\):

- For the first term:
\[\frac{3}{2} x^{\frac{1}{2} - \frac{3}{2}} = \frac{3}{2} x^{-1}\]

- For the second term:
\[-2 x^{2 - \frac{3}{2}} = -2 x^{\frac{1}{2}}\]

Combining the terms gives:

\[\frac{3}{2} x^{-1} - 2 x^{\frac{1}{2}}\]

Comparing this with \(A x^p + B x^q\), we obtain:
- \(A = \frac{3}{2}\) (or \(1.5\)), \(p = -1\)
- \(B = -2\), \(q = \frac{1}{2}\) (or \(0.5\))

(Alternatively, \(A = -2\), \(p = \frac{1}{2}\), \(B = \frac{3}{2}\), \(q = -1\))

M1: Attempts to write \(2x\sqrt{x}\) as \(2x^{\frac{3}{2}}\) or splits the fraction into two terms with a common denominator of \(2x\sqrt{x}\) (or \(2x^{\frac{3}{2}}\)).
M1: Attempts to subtract the indices for at least one of the terms, e.g., showing \(x^{\frac{1}{2} - \frac{3}{2}}\) or \(x^{2 - \frac{3}{2}}\).
A1: One term is fully simplified and correct, either \(\frac{3}{2}x^{-1}\) (or \(1.5x^{-1}\)) OR \(-2x^{\frac{1}{2}}\) (or \(-2\sqrt{x}\)).
A1: Fully correct constants identified: \(A = 1.5\), \(p = -1\), \(B = -2\), \(q = 0.5\) (or vice versa).

(a) The length of the rectangle is given by:
\[ \text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{12 + 4\sqrt{3}}{3 - \sqrt{3}} \]

To rationalise the denominator, multiply the numerator and the denominator by the conjugate of the denominator, \( 3 + \sqrt{3} \):
\[ \text{Length} = \frac{(12 + 4\sqrt{3})(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} \]

Expand the numerator:
\[ (12 + 4\sqrt{3})(3 + \sqrt{3}) = 12(3) + 12\sqrt{3} + 4\sqrt{3}(3) + 4\sqrt{3}(\sqrt{3}) \]
\[ = 36 + 12\sqrt{3} + 12\sqrt{3} + 12 = 48 + 24\sqrt{3} \]

Expand the denominator:
\[ (3 - \sqrt{3})(3 + \sqrt{3}) = 9 - 3 = 6 \]

Substitute these back into the expression for length:
\[ \text{Length} = \frac{48 + 24\sqrt{3}}{6} = 8 + 4\sqrt{3} \]

So, \( a = 8 \) and \( b = 4 \).

(b) The perimeter of the rectangle is given by:
\[ \text{Perimeter} = 2(\text{Length} + \text{Width}) \]
\[ \text{Perimeter} = 2((8 + 4\sqrt{3}) + (3 - \sqrt{3})) \]
\[ = 2(11 + 3\sqrt{3}) \]
\[ = 22 + 6\sqrt{3} \]

So, \( c = 22 \) and \( d = 6 \).

**Part (a) [4 Marks]**
* **M1**: Sets up a correct expression for the length, \( \frac{12 + 4\sqrt{3}}{3 - \sqrt{3}} \), and attempts to multiply the numerator and denominator by \( 3 + \sqrt{3} \).
* **M1**: Expands the numerator to get 4 terms (or 3 terms if simplified directly), with at least 3 correct terms: e.g., \( 36 + 12\sqrt{3} + 12\sqrt{3} + 12 \).
* **A1**: Correctly simplifies the numerator to \( 48 + 24\sqrt{3} \) and the denominator to \( 6 \).
* **A1**: Fully correct simplified answer: \( 8 + 4\sqrt{3} \) (or clearly states \( a = 8 \) and \( b = 4 \)).

**Part (b) [2 Marks]**
* **M1**: Attempts to find the perimeter using \( 2(\text{length} + \text{width}) \) with their length of the form \( a + b\sqrt{3} \).
* **A1**: Correct final answer of \( 22 + 6\sqrt{3} \) (or clearly states \( c = 22 \) and \( d = 6 \)).

(a) Curve \(C_1\) is a cubic with a negative \(x^3\) coefficient. It crosses the \(x\)-axis at \((-2, 0)\), \((0, 0)\), and \((3, 0)\). Curve \(C_2\) is a reciprocal curve translated vertically upwards by 2 units. Its horizontal asymptote is \(y = 2\) and its vertical asymptote is \(x = 0\). It crosses the \(x\)-axis where \(0 = \frac{4}{x} + 2 \implies x = -2\), so its only intercept with the axes is at \((-2, 0)\). Thus, both curves intersect at the point \((-2, 0)\) on the negative \(x\)-axis. (b) To find the points of intersection, we set the equations equal to each other: \(x(x+2)(3-x) = \frac{4}{x} + 2\). Since \(x \neq 0\), we multiply both sides by \(x\) to obtain: \(x^2(x+2)(3-x) = 4 + 2x\). Expanding the left-hand side: \(x^2(3x - x^2 + 6 - 2x) = x^2(-x^2 + x + 6) = -x^4 + x^3 + 6x^2\). Equating this to the right-hand side gives: \(-x^4 + x^3 + 6x^2 = 2x + 4\). Rearranging all terms to the right-hand side gives: \(0 = x^4 - x^3 - 6x^2 + 2x + 4\), which is \(x^4 - x^3 - 6x^2 + 2x + 4 = 0\) as required. (c) Substituting \(x = 1\) into the equation of \(C_1\) gives: \(y = 1(1+2)(3-1) = 1(3)(2) = 6\). Alternatively, substituting \(x = 1\) into the equation of \(C_2\) gives: \(y = \frac{4}{1} + 2 = 6\). Thus, the \(y\)-coordinate is 6.

(a) M1: For drawing a negative cubic shape (starts top-left, ends bottom-right) passing through the origin. A1: For correctly identifying and labeling the \(x\)-intercepts of \(C_1\) at \((-2, 0)\) and \((3, 0)\). M1: For drawing a reciprocal curve in the first and third quadrants (relative to asymptotes) with a horizontal asymptote of \(y = a\) where \(a > 0\). A1: For correctly labeling the horizontal asymptote as \(y = 2\) and showing both curves intersecting at their common \(x\)-intercept \((-2, 0)\). (b) M1: For equating both equations and multiplying by \(x\) to get \(x^2(x+2)(3-x) = 4 + 2x\), then expanding to a quartic form. A1*: For showing full and correct algebraic steps to arrive at the given equation \(x^4 - x^3 - 6x^2 + 2x + 4 = 0\) with no errors. (c) B1: For \(y = 6\).

(a) Applying the cosine rule to triangle \(ABC\):

\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]

Substituting the given values:

\[(\sqrt{39})^2 = x^2 + (x+2)^2 - 2(x)(x+2)\cos(60^\circ)\]

Since \(\cos(60^\circ) = \frac{1}{2}\):

\[39 = x^2 + (x^2 + 4x + 4) - 2x(x+2)\left(\frac{1}{2}\right)\]

\[39 = 2x^2 + 4x + 4 - (x^2 + 2x)\]

\[39 = x^2 + 2x + 4\]

\[x^2 + 2x - 35 = 0\quad\text{(as required)}\]

(b) Solving the quadratic equation:

\[x^2 + 2x - 35 = 0 \implies (x+7)(x-5) = 0\]

This gives \(x = -7\) or \(x = 5\).

Since the length \(x\) must be positive, we reject \(x = -7\).

Therefore, \(x = 5\).

(c) The formula for the area of a triangle is:

\[\text{Area} = \frac{1}{2}(AB)(BC)\sin(\angle ABC)\]

Substituting \(x = 5\):

\[AB = 5\text{ cm}\]

\[BC = 5 + 2 = 7\text{ cm}\]

\[\text{Area} = \frac{1}{2} \times 5 \times 7 \times \sin(60^\circ)\]

Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):

\[\text{Area} = \frac{1}{2} \times 35 \times \frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{4}\text{ cm}^2\]

(a)
* M1: Attempts to use the cosine rule with the given sides and angle, leading to an equation in \(x\). E.g., \((\sqrt{39})^2 = x^2 + (x+2)^2 - 2x(x+2)\cos(60^\circ)\).
* M1: Correctly expands \((x+2)^2\) to \(x^2+4x+4\), substitutes \(\cos(60^\circ) = \frac{1}{2}\), and simplifies to an intermediate quadratic stage.
* A1*: Fully correct derivation leading to the given quadratic equation \(x^2 + 2x - 35 = 0\) with no errors seen.

(b)
* B1: Identifies \(x = 5\) as the only valid solution and rejects the negative root.

(c)
* M1: Attempts to use the area formula \(\frac{1}{2}ab\sin C\) with their positive value of \(x\) and angle \(60^\circ\).
* M1: Employs the exact value of \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).
* A1: Obtains the correct exact area \(\frac{35\sqrt{3}}{4}\) (or \(8.75\sqrt{3}\)).

(a) Applying the cosine rule to triangle \(ABC\):

\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]

Substituting the given values:

\[(\sqrt{39})^2 = x^2 + (x+2)^2 - 2(x)(x+2)\cos(60^\circ)\]

Since \(\cos(60^\circ) = \frac{1}{2}\):

\[39 = x^2 + (x^2 + 4x + 4) - 2x(x+2)\left(\frac{1}{2}\right)\]

\[39 = 2x^2 + 4x + 4 - (x^2 + 2x)\]

\[39 = x^2 + 2x + 4\]

\[x^2 + 2x - 35 = 0\quad\text{(as required)}\]

(b) Solving the quadratic equation:

\[x^2 + 2x - 35 = 0 \implies (x+7)(x-5) = 0\]

This gives \(x = -7\) or \(x = 5\).

Since the length \(x\) must be positive, we reject \(x = -7\).

Therefore, \(x = 5\).

(c) The formula for the area of a triangle is:

\[\text{Area} = \frac{1}{2}(AB)(BC)\sin(\angle ABC)\]

Substituting \(x = 5\):

\[AB = 5\text{ cm}\]

\[BC = 5 + 2 = 7\text{ cm}\]

\[\text{Area} = \frac{1}{2} \times 5 \times 7 \times \sin(60^\circ)\]

Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):

\[\text{Area} = \frac{1}{2} \times 35 \times \frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{4}\text{ cm}^2\]

(a)
* M1: Attempts to use the cosine rule with the given sides and angle, leading to an equation in \(x\). E.g., \((\sqrt{39})^2 = x^2 + (x+2)^2 - 2x(x+2)\cos(60^\circ)\).
* M1: Correctly expands \((x+2)^2\) to \(x^2+4x+4\), substitutes \(\cos(60^\circ) = \frac{1}{2}\), and simplifies to an intermediate quadratic stage.
* A1*: Fully correct derivation leading to the given quadratic equation \(x^2 + 2x - 35 = 0\) with no errors seen.

(b)
* B1: Identifies \(x = 5\) as the only valid solution and rejects the negative root.

(c)
* M1: Attempts to use the area formula \(\frac{1}{2}ab\sin C\) with their positive value of \(x\) and angle \(60^\circ\).
* M1: Employs the exact value of \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).
* A1: Obtains the correct exact area \(\frac{35\sqrt{3}}{4}\) (or \(8.75\sqrt{3}\)).

To find the set of possible values of \( k \) for which the equation has no real roots, we must use the discriminant condition \( b^2 - 4ac < 0 \).

Identify the coefficients from the quadratic equation:
\[ a = k - 2 \]
\[ b = 2k \]
\[ c = 3k - 4 \]

Substitute these into the discriminant expression:
\[ b^2 - 4ac = (2k)^2 - 4(k - 2)(3k - 4) \]
\[ = 4k^2 - 4(3k^2 - 4k - 6k + 8) \]
\[ = 4k^2 - 4(3k^2 - 10k + 8) \]
\[ = 4k^2 - 12k^2 + 40k - 32 \]
\[ = -8k^2 + 40k - 32 \]

Since the equation has no real roots, we set the discriminant to be less than zero:
\[ -8k^2 + 40k - 32 < 0 \]

Divide the entire inequality by \(-8\) and reverse the inequality sign:
\[ k^2 - 5k + 4 > 0 \]

Factorise the quadratic expression to find the critical values:
\[ (k - 1)(k - 4) > 0 \]

This gives the critical values as \( k = 1 \) and \( k = 4 \).

Since the inequality is of the form \( > 0 \), we choose the regions outside of the critical values:
\[ k < 1 \text{ or } k > 4 \]

In set notation, this can be written as:
\[ \{k : k < 1\} \cup \{k : k > 4\} \]

**M1**: Attempts to use the discriminant \( b^2 - 4ac \) with \( a = k - 2 \), \( b = 2k \), and \( c = 3k - 4 \). Condone sign slips in substitution.
**A1**: Obtains a correct, simplified expression for the discriminant: \( -8k^2 + 40k - 32 \) (or equivalent).
**M1**: Sets their discriminant \( < 0 \) (for no real roots) and attempts to find the critical values of their 3-term quadratic expression by factorising, completing the square, or using the quadratic formula.
**A1**: Correct critical values of \( k = 1 \) and \( k = 4 \).
**M1**: Selects the outside region for their critical values (i.e. \( k < \text{lower value} \) or \( k > \text{upper value} \)).
**A1**: Correct final range: \( k < 1 \) or \( k > 4 \) (or equivalent set notation. Do not accept \( 1 > k > 4 \)).

(a) Since the point \(P(4, y_P)\) lies on the normal line, its coordinates must satisfy the equation of the normal:
\(2(4) + 11y_P - 41 = 0\)
\(8 + 11y_P - 41 = 0\)
\(11y_P = 33 \implies y_P = 3\) (as required).

(b) Rearranging the equation of the normal to find its gradient:
\(2x + 11y - 41 = 0 \implies 11y = -2x + 41 \implies y = -\frac{2}{11}x + \frac{41}{11}\)
So the gradient of the normal is \(-\frac{2}{11}\).
Since the tangent and the normal are perpendicular, the gradient of the tangent at \(P\) is:
\(f'(4) = -\frac{1}{-\frac{2}{11}} = \frac{11}{2}\) (or \(5.5\)).

(c) (i) Using \(f'(x) = 3\sqrt{x} - \frac{k}{x^2}\) at \(x = 4\):
\(f'(4) = 3\sqrt{4} - \frac{k}{4^2} = \frac{11}{2}\)
\(3(2) - \frac{k}{16} = 5.5\)
\(6 - \frac{k}{16} = 5.5\)
\(\frac{k}{16} = 0.5\)
\(k = 8\) (as required).

(ii) To find \(f(x)\), we integrate \(f'(x)\) with respect to \(x\):
\(f'(x) = 3x^{1/2} - 8x^{-2}\)
\(f(x) = \int (3x^{1/2} - 8x^{-2}) \, \mathrm{d}x = \frac{3x^{3/2}}{\frac{3}{2}} - \frac{8x^{-1}}{-1} + C\)
\(f(x) = 2x^{3/2} + \frac{8}{x} + C\)
Since the curve passes through \(P(4, 3)\), substitute \(x = 4\) and \(y = 3\):
\(3 = 2(4)^{3/2} + \frac{8}{4} + C\)
\(3 = 2(8) + 2 + C\)
\(3 = 18 + C \implies C = -15\)
Therefore, the equation of the curve is:
\(f(x) = 2x^{3/2} + \frac{8}{x} - 15\)

(a)
M1: Substituting \(x = 4\) and \(y = y_P\) (or \(y\)) into the given normal equation and attempting to solve for \(y_P\).
A1: Correctly showing that \(y_P = 3\) with no errors shown.

(b)
M1: Identifies the gradient of the normal line (e.g., \(-\frac{2}{11}\)) and uses the perpendicular rule \(m_{\text{tangent}} = -\frac{1}{m_{\text{normal}}}\).
A1: Correct value of \(f'(4) = \frac{11}{2}\) (or \(5.5\)).

(c)(i)
M1: Sets up the equation \(3\sqrt{4} - \frac{k}{4^2} = \text{their } f'(4)\) and attempts to solve for \(k\).
A1*: Shows clearly with intermediate steps that \(k = 8\) (must see at least \(6 - \frac{k}{16} = 5.5\) or equivalent).

(c)(ii)
M1: Integrates the expression \(3x^{1/2} - 8x^{-2}\) with at least one power increased by 1.
A1: Correct integration of at least one term, e.g., \(2x^{3/2}\) or \(8x^{-1}\) (constant of integration not required for this mark).
M1: Substitutes \(x = 4\) and \(y = 3\) into their integrated expression containing a constant of integration \(C\) and attempts to find \(C\).
A1: Correct final equation: \(f(x) = 2x^{3/2} + \frac{8}{x} - 15\) (accept equivalent forms, e.g., \(y = 2x\sqrt{x} + \frac{8}{x} - 15\)).

\((a)\) The perimeter \(P\) of a sector with radius \(r\) and angle \(\theta\) is given by: \(P = 2r + r\theta\). Since the perimeter is \(26\text{ cm}\), we have: \(2r + r\theta = 26 \implies r\theta = 26 - 2r\). The area \(A\) of a sector is given by: \(A = \frac{1}{2}r^2\theta = \frac{1}{2}r(r\theta)\). Substituting \(r\theta = 26 - 2r\) into this formula gives: \(A = \frac{1}{2}r(26 - 2r) = 13r - r^2\) (as required). \((b)\) Given that the area of the sector is \(36\text{ cm}^2\), we set: \(13r - r^2 = 36 \implies r^2 - 13r + 36 = 0\). Factoring this quadratic equation gives: \((r - 4)(r - 9) = 0\). Thus, the possible values of \(r\) are \(r = 4\) and \(r = 9\). To find the corresponding values of \(\theta\), we substitute these back into the perimeter equation \(2r + r\theta = 26\): For \(r = 4\): \(2(4) + 4\theta = 26 \implies 4\theta = 18 \implies \theta = 4.5\) radians. For \(r = 9\): \(2(9) + 9\theta = 26 \implies 9\theta = 8 \implies \theta = \frac{8}{9}\) radians (or approximately \(0.889\) radians).

Part (a): [3 Marks] M1: Attempts to write an equation for the perimeter of the sector, e.g., \(2r + r\theta = 26\). M1: Rearranges the perimeter equation to express \(r\theta\) or \(\theta\) in terms of \(r\) and substitutes into the area formula \(A = \frac{1}{2}r^2\theta\). A1*: Fully correct proof with no errors leading to \(A = 13r - r^2\). Part (b): [4 Marks] M1: Sets \(13r - r^2 = 36\) and forms a 3-term quadratic equation, e.g., \(r^2 - 13r + 36 = 0\). A1: Solves the quadratic to find both \(r = 4\) and \(r = 9\). M1: Uses their values of \(r\) to find at least one corresponding value of \(\theta\). A1: Both pairs of values correct: \(r = 4\), \(\theta = 4.5\) and \(r = 9\), \(\theta = \frac{8}{9}\) (or \(0.889\) to 3 s.f.).

(a) At the intersection with the \(y\)-axis, \(x = 0\).
Substituting \(x = 0\) and \(y = \frac{\sqrt{3}}{2}\) into the equation of the curve:
\[\sin(0 + \alpha) = \frac{\sqrt{3}}{2} \implies \sin\alpha = \frac{\sqrt{3}}{2}\]
Since \(0 < \alpha < 90^\circ\), we have \(\alpha = 60^\circ\).

(b) The equation of the curve is \(y = \sin(x + 60^\circ)\).
The minimum value of the sine function is \(-1\), which occurs when the angle is \(270^\circ\) (for the given domain).
Setting the argument of the sine function equal to \(270^\circ\):
\[x + 60^\circ = 270^\circ \implies x = 210^\circ\]
Thus, the coordinates of the minimum point \(P\) are \((210^\circ, -1)\).

(c) The curve intersects the \(x\)-axis when \(y = 0\):
\[\sin(x + 60^\circ) = 0\]
Since \(0 \le x \le 360^\circ\), the range for the angle is \(60^\circ \le x + 60^\circ \le 420^\circ\).
Within this range, \(\sin\theta = 0\) when \(\theta = 180^\circ\) and \(\theta = 360^\circ\).
Therefore:
\[x + 60^\circ = 180^\circ \implies x = 120^\circ\]
\[x + 60^\circ = 360^\circ \implies x = 300^\circ\]
The coordinates of the points of intersection are \((120^\circ, 0)\) and \((300^\circ, 0)\).

(a)
**M1**: Sets \(x = 0\) and \(y = \frac{\sqrt{3}}{2}\) to obtain \(\sin\alpha = \frac{\sqrt{3}}{2}\).
**A1**: \(\alpha = 60^\circ\) (or just \(60\)).

(b)
**M1**: Realises that the minimum \(y\)-coordinate is \(-1\) and sets \(x + \alpha = 270^\circ\) or shifts the point \((270^\circ, -1)\) to the left by their value of \(\alpha\).
**A1**: Correct coordinates \((210^\circ, -1)\). Accept \(x = 210^\circ, y = -1\).

(c)
**M1**: Sets \(\sin(x + 60^\circ) = 0\) and finds at least one value for \(x + 60^\circ\) (e.g. \(180^\circ\) or \(360^\circ\)), or shifts the standard intercepts of \(y = \sin x\) (at \(180^\circ\) and \(360^\circ\)) to the left by their value of \(\alpha\).
**A1**: Both \((120^\circ, 0)\) and \((300^\circ, 0)\). Accept \(x = 120^\circ, 300^\circ\).

**(a)**

To complete the square for \( f(x) = -2x^2 + 12x - 5 \):
First, factor out \( -2 \) from the terms containing \( x \):
\[ f(x) = -2(x^2 - 6x) - 5 \]

Now, complete the square inside the bracket:
\[ x^2 - 6x = (x - 3)^2 - 9 \]

Substitute this back into the expression:
\[ f(x) = -2\left[(x - 3)^2 - 9\right] - 5 \]
\[ f(x) = -2(x - 3)^2 + 18 - 5 \]
\[ f(x) = -2(x - 3)^2 + 13 \]

Thus, \( a = -2 \), \( b = -3 \), and \( c = 13 \).

**(b)**

Using the completed square form \( f(x) = -2(x-3)^2 + 13 \), the maximum value of \( f(x) \) is \( 13 \), which occurs when \( x - 3 = 0 \implies x = 3 \).
Therefore, the coordinates of the maximum point are \( (3, 13) \).

**(c)**

To find the intersection of the line \( y = 2x + k \) and the curve \( y = -2x^2 + 12x - 5 \), we set their equations equal to each other:
\[ 2x + k = -2x^2 + 12x - 5 \]

Rearranging into a standard quadratic equation form:
\[ 2x^2 - 10x + (k + 5) = 0 \]

For the line and the curve to not intersect, this quadratic equation must have no real roots. Therefore, the discriminant must be negative (\( b^2 - 4ac < 0 \)):
\[ (-10)^2 - 4(2)(k + 5) < 0 \]
\[ 100 - 8(k + 5) < 0 \]
\[ 100 - 8k - 40 < 0 \]
\[ 60 < 8k \]
\[ k > 7.5 \quad \left(\text{or } k > \frac{15}{2}\right) \]

**(d)**

When \( k = -3 \), substitute this into the quadratic equation obtained in part (c):
\[ 2x^2 - 10x + (-3 + 5) = 0 \]
\[ 2x^2 - 10x + 2 = 0 \]

Divide the entire equation by 2:
\[ x^2 - 5x + 1 = 0 \]

Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(1)}}{2(1)} \]
\[ x = \frac{5 \pm \sqrt{21}}{2} \]

To express this in the form \( p \pm \sqrt{q} \):
\[ x = \frac{5}{2} \pm \sqrt{\frac{21}{4}} \]
(Alternatively, in decimal form: \( x = 2.5 \pm \sqrt{5.25} \))

**(a)**
* **M1**: Attempts to factor out \( -2 \) from the first two terms or attempts to set up an identity like \( a(x+b)^2+c = ax^2 + 2abx + ab^2 + c \) and equates coefficients.
* **M1**: Attempts to complete the square inside the bracket, achieving \( (x-3)^2 - 9 \) or equivalent.
* **A1**: Correct expression: \( -2(x-3)^2 + 13 \). (Allow \( a = -2, b = -3, c = 13 \) written explicitly).

**(b)**
* **B1**: Correct coordinates \( (3, 13) \). Must be written as coordinates, not just \( x = 3, y = 13 \).

**(c)**
* **M1**: Equates the line and the curve and collects terms on one side to form a 3-term quadratic equation in \( x \).
* **A1**: Correct quadratic equation: \( 2x^2 - 10x + (k + 5) = 0 \) (or equivalent).
* **M1**: Attempts to use the discriminant condition \( b^2 - 4ac < 0 \) with their coefficients to form an inequality in \( k \).
* **A1**: Correct set of values: \( k > 7.5 \) (or \( k > \frac{15}{2} \)).

**(d)**
* **M1**: Substitutes \( k = -3 \) into their quadratic equation and attempts to solve for \( x \) using completing the square or the quadratic formula.
* **A1**: Correct \( x \)-coordinates in the form \( p \pm \sqrt{q} \), e.g., \( x = \frac{5}{2} \pm \sqrt{\frac{21}{4}} \) or \( x = 2.5 \pm \sqrt{5.25} \).

To prove the statement by exhaustion, we must test all possible integer values of \(n\) in the given interval \(2 \le n \le 6\). The integers to test are \(n = 2, 3, 4, 5, 6\). For \(n = 2\): \(2^2 + 2 = 6\), which is not divisible by 4. For \(n = 3\): \(3^2 + 2 = 11\), which is not divisible by 4. For \(n = 4\): \(4^2 + 2 = 18\), which is not divisible by 4. For \(n = 5\): \(5^2 + 2 = 27\), which is not divisible by 4. For \(n = 6\): \(6^2 + 2 = 38\), which is not divisible by 4. Since all possible cases have been tested and none of the resulting values are divisible by 4, the statement has been proven by exhaustion.

M1: Attempts to substitute at least three of the integers from the set \(\{2, 3, 4, 5, 6\}\) into \(n^2 + 2\). A1: Correctly calculates all five values: 6, 11, 18, 27, and 38. A1: Complete proof with all 5 cases correctly evaluated and a concluding remark stating that none of the values are divisible by 4.

To prove the statement by exhaustion, we must test all possible integer values of \(n\) in the given interval \(2 \le n \le 6\). The integers to test are \(n = 2, 3, 4, 5, 6\). For \(n = 2\): \(2^2 + 2 = 6\), which is not divisible by 4. For \(n = 3\): \(3^2 + 2 = 11\), which is not divisible by 4. For \(n = 4\): \(4^2 + 2 = 18\), which is not divisible by 4. For \(n = 5\): \(5^2 + 2 = 27\), which is not divisible by 4. For \(n = 6\): \(6^2 + 2 = 38\), which is not divisible by 4. Since all possible cases have been tested and none of the resulting values are divisible by 4, the statement has been proven by exhaustion.

M1: Attempts to substitute at least three of the integers from the set \(\{2, 3, 4, 5, 6\}\) into \(n^2 + 2\). A1: Correctly calculates all five values: 6, 11, 18, 27, and 38. A1: Complete proof with all 5 cases correctly evaluated and a concluding remark stating that none of the values are divisible by 4.

We use the trigonometric identity \(\sin^2(2\theta) + \cos^2(2\theta) = 1\) to substitute for \(\sin^2(2\theta)\): \(3(1 - \cos^2(2\theta)) - 5\cos(2\theta) - 1 = 0\). Expanding and simplifying this gives: \(3 - 3\cos^2(2\theta) - 5\cos(2\theta) - 1 = 0\), which simplifies to the quadratic equation: \(3\cos^2(2\theta) + 5\cos(2\theta) - 2 = 0\). We can solve this quadratic equation by factoring: \((3\cos(2\theta) - 1)(\cos(2\theta) + 2) = 0\). This gives two possible solutions: \(\cos(2\theta) = \frac{1}{3}\) or \(\cos(2\theta) = -2\). Since \(\cos(2\theta)\) must lie in the range \([-1, 1]\), the equation \(\cos(2\theta) = -2\) has no real solutions. Now we solve \(\cos(2\theta) = \frac{1}{3}\) for the interval \(0 \le \theta < 360^\circ\), which means we need to find values of \(2\theta\) in the interval \(0 \le 2\theta < 720^\circ\). The principal value is: \(2\theta = \arccos\left(\frac{1}{3}\right) \approx 70.53^\circ\). Using the cosine graph or ASTC diagram, the other solution in the first cycle is: \(2\theta = 360^\circ - 70.53^\circ = 289.47^\circ\). To find the solutions in the second cycle, we add \(360^\circ\) to our first two solutions: \(2\theta = 70.53^\circ + 360^\circ = 430.53^\circ\) and \(2\theta = 289.47^\circ + 360^\circ = 649.47^\circ\). Dividing each value by 2 gives the values of \(\theta\): \(\theta \approx 35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\) (each rounded to one decimal place).

M1: Uses \(\sin^2(2\theta) = 1 - \cos^2(2\theta)\) to form an equation in \(\cos(2\theta)\) only. A1: Obtains the correct quadratic equation \(3\cos^2(2\theta) + 5\cos(2\theta) - 2 = 0\) (or equivalent). M1: Solves their quadratic to obtain \(\cos(2\theta) = \frac{1}{3}\) (condoning failure to explicitly reject \(\cos(2\theta) = -2\)). M1: Finds at least one correct value of \(2\theta\) (approx \(70.5^\circ\) or \(289.5^\circ\)) or one correct value of \(\theta\) (approx \(35.3^\circ\) or \(144.7^\circ\)). M1: Employs a correct method to find at least two values for \(2\theta\) in the range \(0 \le 2\theta < 720^\circ\) (e.g. \(360^\circ - \text{PV}\) or adding \(360^\circ\)). A1: Obtains any two of the correct solutions for \(\theta\) (approx \(35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\)). A1: Obtains all four correct solutions: \(\theta = 35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\) and no extras in the range.

We use the trigonometric identity \(\sin^2(2\theta) + \cos^2(2\theta) = 1\) to substitute for \(\sin^2(2\theta)\): \(3(1 - \cos^2(2\theta)) - 5\cos(2\theta) - 1 = 0\). Expanding and simplifying this gives: \(3 - 3\cos^2(2\theta) - 5\cos(2\theta) - 1 = 0\), which simplifies to the quadratic equation: \(3\cos^2(2\theta) + 5\cos(2\theta) - 2 = 0\). We can solve this quadratic equation by factoring: \((3\cos(2\theta) - 1)(\cos(2\theta) + 2) = 0\). This gives two possible solutions: \(\cos(2\theta) = \frac{1}{3}\) or \(\cos(2\theta) = -2\). Since \(\cos(2\theta)\) must lie in the range \([-1, 1]\), the equation \(\cos(2\theta) = -2\) has no real solutions. Now we solve \(\cos(2\theta) = \frac{1}{3}\) for the interval \(0 \le \theta < 360^\circ\), which means we need to find values of \(2\theta\) in the interval \(0 \le 2\theta < 720^\circ\). The principal value is: \(2\theta = \arccos\left(\frac{1}{3}\right) \approx 70.53^\circ\). Using the cosine graph or ASTC diagram, the other solution in the first cycle is: \(2\theta = 360^\circ - 70.53^\circ = 289.47^\circ\). To find the solutions in the second cycle, we add \(360^\circ\) to our first two solutions: \(2\theta = 70.53^\circ + 360^\circ = 430.53^\circ\) and \(2\theta = 289.47^\circ + 360^\circ = 649.47^\circ\). Dividing each value by 2 gives the values of \(\theta\): \(\theta \approx 35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\) (each rounded to one decimal place).

M1: Uses \(\sin^2(2\theta) = 1 - \cos^2(2\theta)\) to form an equation in \(\cos(2\theta)\) only. A1: Obtains the correct quadratic equation \(3\cos^2(2\theta) + 5\cos(2\theta) - 2 = 0\) (or equivalent). M1: Solves their quadratic to obtain \(\cos(2\theta) = \frac{1}{3}\) (condoning failure to explicitly reject \(\cos(2\theta) = -2\)). M1: Finds at least one correct value of \(2\theta\) (approx \(70.5^\circ\) or \(289.5^\circ\)) or one correct value of \(\theta\) (approx \(35.3^\circ\) or \(144.7^\circ\)). M1: Employs a correct method to find at least two values for \(2\theta\) in the range \(0 \le 2\theta < 720^\circ\) (e.g. \(360^\circ - \text{PV}\) or adding \(360^\circ\)). A1: Obtains any two of the correct solutions for \(\theta\) (approx \(35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\)). A1: Obtains all four correct solutions: \(\theta = 35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\) and no extras in the range.

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

(a) Since \((x - 3)\) is a factor of \(\mathrm{f}(x)\), by the factor theorem, \(\mathrm{f}(3) = 0\).
Substituting \(x = 3\) into \(\mathrm{f}(x)\):
\(2(3)^3 + p(3)^2 + q(3) - 12 = 0\)
\(2(27) + 9p + 3q - 12 = 0\)
\(54 + 9p + 3q - 12 = 0\)
\(9p + 3q + 42 = 0\)
Dividing by 3:
\(3p + q + 14 = 0 \implies 3p + q = -14\) (as required).

(b) Since the remainder is \(-30\) when \(\mathrm{f}(x)\) is divided by \((x + 2)\), by the remainder theorem, \(\mathrm{f}(-2) = -30\).
Substituting \(x = -2\) into \(\mathrm{f}(x)\):
\(2(-2)^3 + p(-2)^2 + q(-2) - 12 = -30\)
\(2(-8) + 4p - 2q - 12 = -30\)
\(-16 + 4p - 2q - 12 = -30\)
\(4p - 2q - 28 = -30\)
\(4p - 2q = -2\)
Dividing by 2:
\(2p - q = -1\)
We now solve the simultaneous equations:
1) \(3p + q = -14\)
2) \(2p - q = -1\)
Adding these two equations:
\((3p + 2p) + (q - q) = -14 - 1\)
\(5p = -15 \implies p = -3\)
Substituting \(p = -3\) into equation (2):
\(2(-3) - q = -1 \implies -6 - q = -1 \implies q = -5\).
So, \(p = -3\) and \(q = -5\).

(c) From parts (a) and (b), we have:
\(\mathrm{f}(x) = 2x^3 - 3x^2 - 5x - 12\)
Since \((x - 3)\) is a factor, we can write:
\(2x^3 - 3x^2 - 5x - 12 = (x - 3)(2x^2 + kx + 4)\)
Comparing the \(x^2\) coefficients:
\(-3 = k - 6 \implies k = 3\)
Thus, the quadratic factor is \(2x^2 + 3x + 4\).
Checking the discriminant of \(2x^2 + 3x + 4\):
\(b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23 < 0\), which has no real roots.
Thus, \(\mathrm{f}(x)\) is factorised completely as:
\(\mathrm{f}(x) = (x - 3)(2x^2 + 3x + 4)\)

Part (a):
M1: Attempts to use the factor theorem by setting \(\mathrm{f}(3) = 0\) to obtain an equation in \(p\) and \(q\).
A1*: Fully correct proof with no errors or omissions. Must show at least one intermediate simplification step before reaching \(3p + q = -14\).

Part (b):
M1: Attempts to use the remainder theorem by setting \(\mathrm{f}(-2) = -30\) to obtain an equation in \(p\) and \(q\).
A1: Obtains a correct simplified equation in \(p\) and \(q\), e.g., \(4p - 2q = -2\) or \(2p - q = -1\).
M1: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
A1: Correct values of \(p = -3\) and \(q = -5\).

Part (c):
M1: Attempts to find the quadratic factor by division, grid method, or equating coefficients. Must obtain a quadratic of the form \(2x^2 + kx + c\) where \(c\) is a non-zero constant (usually 4).
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 4\).
A1: Fully factorised form \((x - 3)(2x^2 + 3x + 4)\). (Note: no need to prove that the quadratic cannot be factorised further to obtain this mark, but the final expression must be correct).

Using the change of base formula, we can write:

\[ \log_9 (x + 8) = \frac{\log_3 (x + 8)}{\log_3 9} \]

Since \(\log_3 9 = 2\), this becomes:

\[ \log_9 (x + 8) = \frac{1}{2} \log_3 (x + 8) \]

Substituting this back into the original equation gives:

\[ \log_3 (x + 2) - \frac{1}{2} \log_3 (x + 8) = \frac{1}{2} \]

Multiply the entire equation by 2 to clear the fraction:

\[ 2 \log_3 (x + 2) - \log_3 (x + 8) = 1 \]

Using the power law of logarithms, \(2 \log_3 (x + 2) = \log_3 (x + 2)^2\):

\[ \log_3 (x + 2)^2 - \log_3 (x + 8) = 1 \]

Using the subtraction law (quotient law) of logarithms:

\[ \log_3 \left( \frac{(x + 2)^2}{x + 8} \right) = 1 \]

Rewrite the logarithmic equation in exponential form:

\[ \frac{(x + 2)^2}{x + 8} = 3^1 \]

\[ (x + 2)^2 = 3(x + 8) \]

Expand both sides:

\[ x^2 + 4x + 4 = 3x + 24 \]

Rearrange into a standard quadratic form:

\[ x^2 + x - 20 = 0 \]

Factorise the quadratic:

\[ (x + 5)(x - 4) = 0 \]

This gives two potential solutions:

\[ x = -5 \quad \text{or} \quad x = 4 \]

We must check the validity of these solutions in the original logarithmic terms:
- For \(\log_3(x+2)\) to be defined, we require \(x + 2 > 0 \implies x > -2\).
- If \(x = -5\), then \(x + 2 = -3\), so \(\log_3(-3)\) is undefined. Thus, \(x = -5\) is rejected as a solution.
- If \(x = 4\), then both \(x + 2 = 6 > 0\) and \(x + 8 = 12 > 0\), so the terms are well-defined.

Therefore, the only valid solution is:

\[ x = 4 \]

- **M1**: Applies the change of base rule to express \(\log_9 (x+8)\) in terms of base 3: \(\log_9 (x+8) = \frac{\log_3 (x+8)}{\log_3 9} = \frac{1}{2}\log_3 (x+8)\).
- **A1**: Correctly forms an equation in base 3 only, e.g., \(\log_3 (x+2) - \frac{1}{2}\log_3 (x+8) = \frac{1}{2}\) or \(2\log_3 (x+2) - \log_3 (x+8) = 1\).
- **M1**: Applies logarithm laws correctly to combine the terms on the left-hand side into a single logarithm, e.g., obtaining \(\log_3 \left(\frac{(x+2)^2}{x+8}\right)\).
- **M1**: Eliminates logarithms correctly to form a quadratic equation, e.g. \(\frac{(x+2)^2}{x+8} = 3^1\), leading to \(x^2 + x - 20 = 0\).
- **A1**: Solves the quadratic equation to find both \(x = 4\) and \(x = -5\).
- **A1**: Selects \(x = 4\) as the only valid solution and rejects \(x = -5\) with a valid reason (e.g., stating that \(x+2 > 0\) or that logarithms of negative numbers are undefined).

(a) The width of each interval is \(h = 3\). Using the trapezium rule: \(\text{Area} \approx \frac{h}{2} [ d_0 + d_4 + 2(d_1 + d_2 + d_3) ]\). Substituting the values: \(\text{Area} \approx \frac{3}{2} [ 0 + 0 + 2(1.8 + 2.5 + 2.1) ] = 1.5 [ 2(6.4) ] = 1.5 \times 12.8 = 19.2\text{ m}^2\). (b) The speed of the water is \(0.6\text{ m s}^{-1}\), so the distance the water travels in one minute (60 seconds) is \(0.6 \times 60 = 36\text{ m}\). The estimated volume of water flowing past the cross-section in one minute is \(\text{Volume} = \text{Area} \times \text{distance} = 19.2 \times 36 = 691.2\text{ m}^3\).

Part (a): M1: Identifies \(h = 3\). M1: Applies the correct structure of the trapezium rule: \(\frac{h}{2} [ y_0 + y_4 + 2(y_1 + y_2 + y_3) ]\) with their value of \(h\) and at most one slip in the \(y\)-values. A1: Correctly substituted expression: \(\frac{3}{2} [ 0 + 0 + 2(1.8 + 2.5 + 2.1) ]\). A1: \(19.2\) or exact equivalent. Part (b): M1: Multiplies their area from part (a) by \(0.6 \times 60\). A1ft: \(691.2\) (accept \(691\)), follow through on their area from part (a).

**(a) (i) & (ii)**

To find the center and the radius of the circle, we complete the square for both the \(x\) and \(y\) terms:

\[x^2 - 6x + y^2 + 8y - 15 = 0\]

\[(x - 3)^2 - 9 + (y + 4)^2 - 16 - 15 = 0\]

\[(x - 3)^2 + (y + 4)^2 - 40 = 0\]

\[(x - 3)^2 + (y + 4)^2 = 40\]

Comparing this to the standard equation of a circle \((x - a)^2 + (y - b)^2 = r^2\):
* The coordinates of the center are \((3, -4)\).
* The radius is \(\sqrt{40} = 2\sqrt{10}\).

**(b)**

The point \(P\) is \((5, 2)\) and the center of the circle is \(A(3, -4)\).

First, find the gradient of the radius \(AP\):

\[m_{\text{radius}} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-4)}{5 - 3} = \frac{6}{2} = 3\]

The tangent at \(P\) is perpendicular to the radius \(AP\). Therefore, the gradient of the tangent, \(m_{\text{tangent}}\), is:

\[m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{1}{3}\]

Using the point-slope form of a line equation with the point \(P(5, 2)\):

\[y - y_1 = m(x - x_1)\]

\[y - 2 = -\frac{1}{3}(x - 5)\]

Multiply the entire equation by 3 to eliminate the fraction:

\[3(y - 2) = -(x - 5)\]

\[3y - 6 = -x + 5\]

Rearranging into the form \(ax + by + c = 0\):

\[x + 3y - 11 = 0\]

**Part (a)**
* **M1**: Attempts to complete the square for both \(x\) and \(y\). Look for \((x \pm 3)^2\) and \((y \pm 4)^2\).
* **A1**: Correct coordinates of the center: \((3, -4)\).
* **A1**: Correct exact radius: \(\sqrt{40}\) or \(2\sqrt{10}\).

**Part (b)**
* **M1**: Attempts to find the gradient of the radius \(AP\) using their center \(A\) and the point \(P(5, 2)\).
* **A1ft**: Correct gradient for their center. If the center is \((3, -4)\), the gradient must be \(3\).
* **M1**: Uses the perpendicular gradient rule \(m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}\) to find the gradient of the tangent.
* **M1**: Attempts to find the equation of the tangent line passing through \(P(5, 2)\) with their perpendicular gradient.
* **A1**: Correct equation in the form \(ax + by + c = 0\) with integer coefficients, e.g., \(x + 3y - 11 = 0\) or any non-zero integer multiple.

(a) Let \(a\) be the number of seats in the first row and \(d\) be the common difference. Using the formula for the \(n\)th term of an arithmetic sequence, \(u_n = a + (n - 1)d\), we have: \(u_5 = a + 4d = 26\) (Equation 1). Using the formula for the sum of the first \(n\) terms, \(S_n = \frac{n}{2}[2a + (n - 1)d]\), we have: \(S_{12} = \frac{12}{2}[2a + 11d] = 366\), which simplifies to \(6(2a + 11d) = 366 \implies 2a + 11d = 61\) (Equation 2). Multiplying Equation 1 by 2 gives \(2a + 8d = 52\) (Equation 3). Subtracting Equation 3 from Equation 2 gives \(3d = 9 \implies d = 3\). Substituting \(d = 3\) back into Equation 1 gives \(a + 4(3) = 26 \implies a = 14\). So there are 14 seats in the first row. (b) Using the sum formula for \(S_{24}\) with \(a = 14\) and \(d = 3\): \(S_{24} = \frac{24}{2}[2(14) + (24 - 1)(3)] = 12[28 + 23(3)] = 12[28 + 69] = 12[97] = 1164\). Thus, the total number of seats in the theatre is 1164.

(a) M1: Attempts to write an equation for the 5th row using \(a + 4d = 26\). M1: Attempts to write an equation for the sum of the first 12 rows, e.g., \(\frac{12}{2}(2a+11d) = 366\) or \(2a+11d = 61\). dM1: Solves their two linear equations simultaneously to find a value for \(a\) or \(d\). Dependent on at least one of the previous M marks. A1: Correctly finds \(a = 14\) (also accept \(d = 3\) shown as part of the working). (b) M1: Attempts to use the sum formula for 24 terms with their values of \(a\) and \(d\). A1ft: Correct substitution of their \(a\) and \(d\) into \(\frac{24}{2}[2a + 23d]\). A1: Correct final answer of 1164.

(a) First, we find the gradient of the curve by differentiating \(y = x^2 - 4x + 7\):

\[\frac{\text{d}y}{\text{d}x} = 2x - 4\]

At the point \(P(4, 7)\), the gradient of the tangent is:

\[m = 2(4) - 4 = 4\]

Using the equation of a straight line with gradient \(m = 4\) passing through \((4, 7)\):

\[y - 7 = 4(x - 4)\]
\[y - 7 = 4x - 16\]
\[y = 4x - 9\]

(b) To find the intersection points of the curve \(C\) and the tangent, we equate their expressions:

\[x^2 - 4x + 7 = 4x - 9\]
\[x^2 - 8x + 16 = 0\]
\[(x - 4)^2 = 0\]

This quadratic equation has a single repeated root at \(x = 4\). Since there are no other real solutions, the tangent does not intersect the curve at any other point.

(c) The area of the region \(R\) is given by the integral of the upper curve minus the lower line from \(x = 0\) to \(x = 4\):

\[\text{Area} = \int_{0}^{4} \left( (x^2 - 4x + 7) - (4x - 9) \right) \text{d}x\]
\[\text{Area} = \int_{0}^{4} (x^2 - 8x + 16) \text{d}x\]

Integrating each term:

\[\text{Area} = \left[ \frac{1}{3}x^3 - 4x^2 + 16x \right]_{0}^{4}\]

Substituting the upper limit \(x = 4\):

\[\left( \frac{1}{3}(4)^3 - 4(4)^2 + 16(4) \right) = \frac{64}{3} - 64 + 64 = \frac{64}{3}\]

Substituting the lower limit \(x = 0\):

\[0\]

Subtracting the limits gives:

\[\text{Area} = \frac{64}{3} - 0 = \frac{64}{3}\]

(a)
- M1: Attempts to differentiate the equation of the curve \(y = x^2 - 4x + 7\) with at least one term differentiated correctly.
- A1: Correct derivative \(\frac{\text{d}y}{\text{d}x} = 2x - 4\).
- M1: Substitutes \(x = 4\) into their derivative to find the gradient and attempts the equation of the tangent using \(y - 7 = m(x - 4)\).
- A1: Correct equation in the form \(y = 4x - 9\).

(b)
- M1: Equates the curve and their tangent equation to form a quadratic equation in \(x\).
- A1: Factores to show \((x-4)^2 = 0\) (or equivalent justification using discriminant \(b^2 - 4ac = 0\)) and concludes that \(x = 4\) is the only solution, hence no other intersection point.

(c)
- M1: Attempts to integrate the curve equation \(x^2 - 4x + 7\) with at least one term integrated correctly (power increased by 1).
- A1: Correct integrated expression: \(\frac{1}{3}x^3 - 2x^2 + 7x\).
- M1: Integrates the line equation to get \(2x^2 - 9x\) (or integrates the combined function \(x^2 - 8x + 16\) to get \(\frac{1}{3}x^3 - 4x^2 + 16x\)).
- A1: Correct combined integrated expression: \(\frac{1}{3}x^3 - 4x^2 + 16x\) (or separate expressions correctly integrated).
- M1: Substitutes both limits of \(0\) and \(4\) into their integrated expressions and subtracts the lower limit value from the upper limit value.
- A1: Correct exact area of \(\frac{64}{3}\) or \(21\frac{1}{3}\).

(a) The sum of the first two terms is given by:
\(a + ar = 15 \implies a(1+r) = 15\)

The sum to infinity is given by:
\(\frac{a}{1-r} = 27 \implies a = 27(1-r)\)

Substituting the expression for \(a\) into the first equation:
\(27(1-r)(1+r) = 15\)
\(27(1-r^2) = 15\)
\(1-r^2 = \frac{15}{27} = \frac{5}{9}\)
\(r^2 = 1 - \frac{5}{9} = \frac{4}{9}\)

Since \(r > 0\), we take the positive square root:
\(r = \frac{2}{3}\) (as required).

(b) First, find the first term \(a\):
\(a = 27\left(1 - \frac{2}{3}\right) = 9\)

The 5th term is given by:
\(u_5 = ar^4 = 9 \left(\frac{2}{3}\right)^4 = 9 \times \frac{16}{81} = \frac{16}{9}\)

(c) The sum of the first \(n\) terms is:
\(S_n = \frac{a(1-r^n)}{1-r} = \frac{9\left(1 - (2/3)^n\right)}{1 - 2/3} = 27\left(1 - \left(\frac{2}{3}\right)^n\right)\)

We are given:
\(S_{\infty} - S_n < 0.1\)
\(27 - 27\left(1 - \left(\frac{2}{3}\right)^n\right) < 0.1\)
\(27\left(\frac{2}{3}\right)^n < 0.1\)
\(\left(\frac{2}{3}\right)^n < \frac{0.1}{27} = \frac{1}{270}\)

Taking natural logarithms (or common logarithms) of both sides:
\(n \ln\left(\frac{2}{3}\right) < \ln\left(\frac{1}{270}\right)\)

Since \(\ln\left(\frac{2}{3}\right) < 0\), we reverse the inequality when dividing:
\(n > \frac{\ln\left(\frac{1}{270}\right)}{\ln\left(\frac{2}{3}\right)}\)
\(n > 13.807...\)

Since \(n\) must be an integer, the smallest value is \(n = 14\).

(a)
- M1: Attempts to write down equations for the sum of the first two terms and the sum to infinity, i.e., \(a(1+r) = 15\) and \(\frac{a}{1-r} = 27\).
- M1: Substitutes \(a = 27(1-r)\) into \(a(1+r) = 15\) to obtain a quadratic equation in \(r\).
- A1: Obtains \(r^2 = \frac{4}{9}\) or equivalent simplified form such as \(27r^2 = 12\).
- A1*: Reaches \(r = \frac{2}{3}\) with convincing working, explicitly stating or showing why the negative root is rejected (e.g., citing \(r > 0\)).

(b)
- M1: Correct method to find \(a\) (e.g. \(a = 27(1-2/3) = 9\)) and attempts to find the 5th term using \(u_5 = ar^4\).
- A1: \(\frac{16}{9}\) or \(1 \frac{7}{9}\).

(c)
- M1: Formulates a correct expression for \(S_n\) or \(S_{\infty} - S_n\) in terms of \(n\).
- M1: Sets up the correct inequality \(27\left(\frac{2}{3}\right)^n < 0.1\) or equivalent equation.
- M1: Divides by 27 to get \(\left(\frac{2}{3}\right)^n < \frac{1}{270}\) (or equivalent).
- M1: Correct use of logarithms to solve an equation or inequality of the form \(p^n < q\) where \(0 < p < 1\).
- A1: Obtains a critical value of \(n \approx 13.8\) (or \(13.8\) to \(13.9\)).
- A1: Deduces \(n = 14\) (must be an integer, and must follow from correct work).