2025 Edexcel IAS-Level Pure Mathematics (XPM01) 模擬試題連答案詳解

For the equation to have no real roots, its discriminant must be less than zero. The quadratic equation is \(2x^2 + kx + (k + 6) = 0\), so \(a = 2\), \(b = k\), and \(c = k + 6\). The discriminant is given by \(\Delta = b^2 - 4ac = k^2 - 4(2)(k + 6) = k^2 - 8k - 48\). For no real roots, we require \(k^2 - 8k - 48 < 0\). Factorising the quadratic expression gives \((k - 12)(k + 4) < 0\). The critical values are \(k = 12\) and \(k = -4\). Since we want the expression to be less than zero, the set of values is \(-4 < k < 12\).

M1: Attempts to use the discriminant \(\Delta = b^2 - 4ac\) with \(a = 2\), \(b = k\), \(c = k + 6\) set to \(\Delta < 0\). M1: Correctly factorises or solves their three-term quadratic in \(k\) to find the critical values \(12\) and \(-4\). A1: Correct inequality \(-4 < k < 12\) (or equivalent such as \(\{k : -4 < k < 12\}\)).

First, factorise all the quadratic expressions: 1) \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\), 2) \(x^2 - 9 = (x - 3)(x + 3)\), 3) \(2x^2 + 5x + 2 = (2x + 1)(x + 2)\), and 4) \(x^2 + x - 6 = (x + 3)(x - 2)\). Now write the expression as a multiplication by inverting the second fraction: \(\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} \times \frac{(x + 3)(x - 2)}{(2x + 1)(x + 2)}\). Cancel the common terms in the numerator and denominator: \((x - 3)\) cancels, \((2x + 1)\) cancels, and \((x + 3)\) cancels. This leaves \(\frac{x - 2}{x + 2}\).

M1: Factorises at least two of the quadratic expressions correctly. M1: Correctly inverts the second fraction and attempts to cancel common factors across the expression. A1: Correct simplified fraction \(\frac{x - 2}{x + 2}\) or equivalent.

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) into the equation: \(3(1 - \cos^2 \theta) - \cos \theta - 1 = 0 \Rightarrow 3 - 3\cos^2 \theta - \cos \theta - 1 = 0 \Rightarrow 3\cos^2 \theta + \cos \theta - 2 = 0\). Let \(y = \cos \theta\): \(3y^2 + y - 2 = 0\). Factorise the quadratic: \((3y - 2)(y + 1) = 0\). So, \(\cos \theta = \frac{2}{3}\) or \(\cos \theta = -1\). For \(\cos \theta = \frac{2}{3}\), \(\theta = \arccos\left(\frac{2}{3}\right) \approx 48.2^\circ\). Another solution in the range \(0 \le \theta < 360^\circ\) is \(\theta = 360^\circ - 48.19^\circ \approx 311.8^\circ\). For \(\cos \theta = -1\), \(\theta = 180^\circ\). Thus, the solutions are \(\theta = 48.2^\circ, 180^\circ, 311.8^\circ\).

M1: Uses \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a quadratic equation in terms of \(\cos \theta\). M1: Solves the quadratic to find \(\cos \theta = \frac{2}{3}\) or \(\cos \theta = -1\). A1: All three correct values: \(\theta = 48.2^\circ, 180^\circ, 311.8^\circ\) (accept answers rounding to these values).

The formula for the area of a triangle is \(\text{Area} = \frac{1}{2} a c \sin B\). Substitute the given values into the formula: \(15 = \frac{1}{2} (x + 2)(2x - 1) \sin 30^\circ\). Since \(\sin 30^\circ = \frac{1}{2\}}, we have \)15 = \frac{1}{2} (2x^2 + 3x - 2) \left(\frac{1}{2}\right) \Rightarrow 15 = \frac{1}{4} (2x^2 + 3x - 2)\). Multiply both sides by 4: \(60 = 2x^2 + 3x - 2 \Rightarrow 2x^2 + 3x - 62 = 0\). Use the quadratic formula to solve for \(x\): \(x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-62)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 496}}{4} = \frac{-3 \pm \sqrt{505}}{4}\). Since \(x\) must be positive, \(x = \frac{-3 + \sqrt{505}}{4} \approx 4.868\). To 3 significant figures, \(x = 4.87\).

M1: Sets up the area equation using \(\frac{1}{2} a c \sin B = 15\) with correct substitutions. M1: Obtains a correct 3-term quadratic equation, e.g., \(2x^2 + 3x - 62 = 0\). A1: Solves to find the positive value \(x \approx 4.87\) (accept 4.87, or exact \(\frac{-3 + \sqrt{505}}{4}\)).

First, find the midpoint \(M\) of the line segment \(AB\): \(M = \left( \frac{-2 + 4}{2}, \frac{5 + 1}{2} \right) = (1, 3)\). Next, find the gradient of \(l_1\): \(m_1 = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}\). Since \(l_2\) is perpendicular to \(l_1\), its gradient \(m_2\) is \(m_2 = -\frac{1}{m_1} = \frac{3}{2}\). Using the point-gradient formula for \(l_2\) with point \((1, 3)\) and gradient \(\frac{3}{2}\): \(y - 3 = \frac{3}{2}(x - 1)\). Multiply both sides by 2: \(2y - 6 = 3x - 3\). Rearrange into the form \(ax + by + c = 0\): \(3x - 2y + 3 = 0\) (or \(-3x + 2y - 3 = 0\)).

M1: Correctly finds the midpoint of \(AB\) as \((1, 3)\) or find the gradient of \(AB\) as \(-\frac{2}{3}\). M1: Uses the perpendicular gradient rule \(m_2 = -1/m_1\) and attempts to write down an equation of a line using their midpoint and perpendicular gradient. A1: Correct equation in the specified form, e.g. \(3x - 2y + 3 = 0\) or equivalent integer form.

First, find the coordinates of \(P\) and \(Q\). To find \(P\) (where the line intersects the \(x\)-axis, so \(y = 0\)): \(2x - 3(0) + 12 = 0 \Rightarrow 2x = -12 \Rightarrow x = -6\, so \)P\) is \((-6, 0)\). To find \(Q\) (where the line intersects the \(y\)-axis, so \(x = 0\)): \(2(0) - 3y + 12 = 0 \Rightarrow 3y = 12 \Rightarrow y = 4\), so \(Q\) is \((0, 4)\). The length of \(PQ\) is \(PQ = \sqrt{(0 - (-6))^2 + (4 - 0)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}\). Simplify \(\sqrt{52}\): \(\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}\). Thus, \(k = 2\) and the length is \(2\sqrt{13}\).

M1: Sets \(y = 0\) to find \(P\) and \(x = 0\) to find \(Q\). M1: Correctly applies the distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to their points \(P\) and \(Q\). A1: Correctly simplifies to \(2\sqrt{13}\) (or states \(k = 2\)).

First, rewrite the equation in index form: \(y = 4x^{3/2} - 8x^{-1/2}\). Now differentiate \(y\) with respect to \(x\): \(\frac{dy}{dx} = 4 \left(\frac{3}{2}\right) x^{1/2} - 8 \left(-\frac{1}{2}\right) x^{-3/2} = 6x^{1/2} + 4x^{-3/2}\). To find the gradient at \(x = 4\), substitute \(x = 4\) into \(\frac{dy}{dx}\): \(\frac{dy}{dx} = 6(4)^{1/2} + 4(4)^{-3/2} = 6(2) + 4\left(\frac{1}{8}\right) = 12 + 0.5 = 12.5\).

M1: Expresses the terms of \(y\) as powers of \(x\), e.g. \(4x^{3/2}\) and \(-8x^{-1/2}\), and attempts to differentiate. M1: Correct differentiation of at least one term. A1: Correct final value of \(12.5\) (or \(\frac{25}{2}\) or \(12\frac{1}{2}\)).

First, write the integrand in index form: \(\int \left( 3x^2 - 4x^{-3} + 5 \right) dx\). Integrate each term by adding 1 to the power and dividing by the new power: \(\int 3x^2 dx = \frac{3x^3}{3} = x^3\), \(\int -4x^{-3} dx = \frac{-4x^{-2}}{-2} = 2x^{-2} = \frac{2}{x^2}\), and \(\int 5 dx = 5x\). Don't forget to add the constant of integration \(C\): \(\int \left( 3x^2 - \frac{4}{x^3} + 5 \right) dx = x^3 + \frac{2}{x^2} + 5x + C\).

M1: Integrates at least one term of \(x^n\) correctly to \(\frac{x^{n+1}}{n+1}\). M1: At least two terms integrated correctly. A1: Completely correct integration including the constant of integration \(+ C\), simplified to \(x^3 + \frac{2}{x^2} + 5x + C\) or \(x^3 + 2x^{-2} + 5x + C\).

Substitute \( y = 2x - 3 \) into the quadratic equation \( x^2 + y^2 = 5 \):

\( x^2 + (2x - 3)^2 = 5 \)

\( x^2 + 4x^2 - 12x + 9 = 5 \)

\( 5x^2 - 12x + 4 = 0 \)

Factorise the quadratic equation:

\( (5x - 2)(x - 2) = 0 \)

This gives \( x = 2 \) or \( x = 0.4 \).

Substitute these values back into the linear equation to find the corresponding values of \( y \):

If \( x = 2 \), then \( y = 2(2) - 3 = 1 \).

If \( x = 0.4 \), then \( y = 2(0.4) - 3 = -2.2 \).

Thus, the solutions are \( x = 2, y = 1 \) and \( x = 0.4, y = -2.2 \).

- M1: Substitutes \( y = 2x - 3 \) (or equivalent) into the quadratic equation to form a 3-term quadratic in one variable.
- M1: Attempts to solve their quadratic equation to obtain two values for \( x \) (or \( y \)).
- A1: Correct matched pairs of solutions: \( x = 2, y = 1 \) and \( x = 0.4, y = -2.2 \) (or equivalent exact fractions).

First, find the \( y \)-coordinate of the point \( P \) by substituting \( x = 3 \) into the curve's equation:

\( y = 2(3)^2 - 8(3) + 5 = 18 - 24 + 5 = -1 \)

So \( P \) is \( (3, -1) \).

Next, find the gradient function by differentiating:

\( \frac{dy}{dx} = 4x - 8 \)

Substitute \( x = 3 \) into the gradient function to find the gradient of the tangent, \( m \):

\( m = 4(3) - 8 = 4 \)

Using the straight line equation formula with \( m = 4 \) and point \( (3, -1) \):

\( y - (-1) = 4(x - 3) \)

\( y + 1 = 4x - 12 \)

\( y = 4x - 13 \)

- M1: Differentiates to find \( \frac{dy}{dx} = ax + b \) with at least one term correct (e.g. \( 4x - 8 \)).
- M1: Finds the gradient of the tangent at \( x = 3 \) and attempts to write down a straight-line equation using their \( y \)-coordinate and gradient.
- A1: Correct equation in the form \( y = mx + c \), which is \( y = 4x - 13 \).

Use the identity \( \sin^2\theta = 1 - \cos^2\theta \):

\( 3(1 - \cos^2\theta) = 2\cos\theta + 2 \)

\( 3 - 3\cos^2\theta = 2\cos\theta + 2 \)

Rearrange into a quadratic equation in \( \cos\theta \):

\( 3\cos^2\theta + 2\cos\theta - 1 = 0 \)

Factorise the quadratic expression:

\( (3\cos\theta - 1)(\cos\theta + 1) = 0 \)

This gives the solutions:

\( \cos\theta = \frac{1}{3} \) or \( \cos\theta = -1 \)

For \( \cos\theta = \frac{1}{3} \):

\( \theta = \arccos(1/3) \approx 70.5^\circ \)

\( \theta = 360^\circ - 70.53^\circ \approx 289.5^\circ \)

For \( \cos\theta = -1 \):

\( \theta = 180^\circ \)

So the solutions in the interval are \( \theta = 70.5^\circ, 180^\circ, 289.5^\circ \).

- M1: Applies the identity \( \sin^2\theta = 1 - \cos^2\theta \) to obtain a quadratic equation in \( \cos\theta \) only.
- M1: Solves their quadratic to find solutions for \( \cos\theta \) and finds at least one correct angle \( \theta \).
- A1: All three correct values of \( \theta \): \( 70.5^\circ, 180^\circ, 289.5^\circ \) (and no extra values in the range).

Find the gradient of the line \( L_1 \):

\( m_1 = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2} \)

Since the line \( L_2 \) is perpendicular to \( L_1 \), its gradient \( m_2 \) must satisfy \( m_1 \times m_2 = -1 \):

\( m_2 = -\frac{1}{-1/2} = 2 \)

Now, find the equation of \( L_2 \) using the point \( B(4, 2) \):

\( y - 2 = 2(x - 4) \)

\( y - 2 = 2x - 8 \)

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

\( 2x - y - 6 = 0 \)

- M1: Finds the gradient of \( L_1 \) and uses the perpendicular condition to determine the gradient of \( L_2 \).
- M1: Uses their perpendicular gradient with the coordinates of point \( B \) to construct the equation of the line.
- A1: Correct final equation with integer coefficients (e.g. \( 2x - y - 6 = 0 \) or any non-zero integer multiple).

Rewrite the terms with fractional and negative exponents first:

\( 4x^3 - 6x^{-1/2} + 2 \)

Integrate term by term, raising the power by 1 and dividing by the new power:

\( \int 4x^3 \text{d}x = \frac{4x^4}{4} = x^4 \)

\( \int -6x^{-1/2} \text{d}x = \frac{-6x^{1/2}}{1/2} = -12x^{1/2} \)

\( \int 2 \text{d}x = 2x \)

Add the constant of integration \( C \):

\( x^4 - 12x^{1/2} + 2x + C \)

(Note: \( 12\sqrt{x} \) is also acceptable in place of \( 12x^{1/2} \).)

- M1: Integrates at least one of the variable terms correctly (power increased by 1).
- A1: At least two terms integrated correctly and simplified.
- A1: Fully correct simplified expression including the constant of integration \( C \).

A translation by the vector \( \begin{pmatrix} 2 \\ -3 \end{pmatrix} \) corresponds to replacing \( x \) with \( x - 2 \) and subtracting \( 3 \) from the entire function.

Starting with the original equation:
\( y = (x - 2)^2 - 4(x - 2) + 7 - 3 \)

Expand the terms:
\( (x - 2)^2 = x^2 - 4x + 4 \)
\( -4(x - 2) = -4x + 8 \)

Combine the terms:
\( y = (x^2 - 4x + 4) + (-4x + 8) + 4 \)
\( y = x^2 - 8x + 16 \)

Alternatively, complete the square on \( C_1 \) first:
\( y = (x - 2)^2 + 3 \)
Applying the translation:
\( y = (x - 2 - 2)^2 + 3 - 3 \)
\( y = (x - 4)^2 = x^2 - 8x + 16 \)

- M1: Identifies the transformation correctly as \( y = \text{f}(x-2) - 3 \) or applies translation to the vertex of the completed square form.
- M1: Attempts to expand the substituted expression or simplifies the translated vertex form.
- A1: Correct final equation in the required form: \( y = x^2 - 8x + 16 \).

Equating the equations of the line and the curve:
\[ x^2 - 4x - 1 = kx - 5 \]

Rearranging into a quadratic equation in \(x\):
\[ x^2 - (4 + k)x + 4 = 0 \]

For the line and curve not to intersect, the discriminant of this quadratic equation must be negative:
\[ b^2 - 4ac < 0 \]

Here, \(a = 1\), \(b = -(4 + k)\), and \(c = 4\). Substituting these values:
\[ [-(4 + k)]^2 - 4(1)(4) < 0 \]
\[ (k + 4)^2 - 16 < 0 \]
\[ k^2 + 8k + 16 - 16 < 0 \]
\[ k^2 + 8k < 0 \]
\[ k(k + 8) < 0 \]

The critical values are \(k = 0\) and \(k = -8\).

Since we want the quadratic expression to be less than zero, the solution set is:
\[ -8 < k < 0 \]

M1: Equates the equation of the line and the curve and attempts to rearrange into a 3-term quadratic equation in \(x\) (e.g. \(x^2 - (k+4)x + 4 = 0\)).
A1: Correct simplified quadratic equation (with the coefficients of \(x\) collected).
M1: Attempts to use the discriminant condition \(b^2 - 4ac < 0\) with their coefficients to set up an inequality in \(k\).
M1: Solves the resulting quadratic inequality to find the critical values \(k = 0\) and \(k = -8\).
A1: Correct inequality: \(-8 < k < 0\) (or equivalent set notation, such as \(k \in (-8, 0)\)).

Using the trigonometric identity \(\sin^2(2x) + \cos^2(2x) = 1\), substitute \(\sin^2(2x) = 1 - \cos^2(2x)\):
\[ 2(1 - \cos^2(2x)) - \cos(2x) = 1 \]
\[ 2 - 2\cos^2(2x) - \cos(2x) = 1 \]

Rearranging into a standard quadratic form:
\[ 2\cos^2(2x) + \cos(2x) - 1 = 0 \]

Factorising the quadratic equation:
\[ (2\cos(2x) - 1)(\cos(2x) + 1) = 0 \]

This yields two possible values for \(\cos(2x)\):
\[ \cos(2x) = \frac{1}{2} \quad \text{or} \quad \cos(2x) = -1 \]

Since the interval for \(x\) is \(0 \le x \le 180^\circ\), the interval for \(2x\) is \(0 \le 2x \le 360^\circ\).

For \(\cos(2x) = \frac{1}{2}\):
\[ 2x = 60^\circ, 300^\circ \implies x = 30^\circ, 150^\circ \]

For \(\cos(2x) = -1\):
\[ 2x = 180^\circ \implies x = 90^\circ \]

Combining all solutions in the given range, we get:
\[ x = 30^\circ, 90^\circ, 150^\circ \]

M1: Applies the identity \(\sin^2(2x) = 1 - \cos^2(2x)\) to obtain an equation containing only terms of \(\cos(2x)\).
A1: Obtains the correct 3-term quadratic equation \(2\cos^2(2x) + \cos(2x) - 1 = 0\) (or equivalent).
M1: Solves their quadratic equation to find the values \(\cos(2x) = \frac{1}{2}\) and \(\cos(2x) = -1\).
A1: Identifies any two of the three correct angles \(30^\circ\), \(90^\circ\), or \(150^\circ\).
A1: Identifies all three correct solutions: \(x = 30^\circ, 90^\circ, 150^\circ\) with no extra solutions in the given range.

First, find the midpoint \(M\) of the line segment \(AB\):
\[ M = \left(\frac{-1 + 5}{2}, \frac{4 + 2}{2}\right) = (2, 3) \]

Next, find the gradient of the line \(l_1\) passing through \(A\) and \(B\):
\[ m_1 = \frac{2 - 4}{5 - (-1)} = \frac{-2}{6} = -\frac{1}{3} \]

Since \(l_2\) is perpendicular to \(l_1\), its gradient \(m_2\) is the negative reciprocal of \(m_1\):
\[ m_2 = -\frac{1}{m_1} = 3 \]

Now, use the point-gradient formula with the midpoint \(M(2, 3)\) and the perpendicular gradient \(3\) to find the equation of \(l_2\):
\[ y - 3 = 3(x - 2) \]
\[ y - 3 = 3x - 6 \]
\[ 3x - y - 3 = 0 \]

M1: Attempts to find the midpoint of the line segment \(AB\) using a correct formula.
A1: Correct midpoint coordinate \((2, 3)\).
M1: Attempts to find the gradient of \(AB\) and uses the perpendicular gradient rule \(m_1 m_2 = -1\) to calculate the gradient of \(l_2\).
M1: Uses their midpoint and their perpendicular gradient to construct a linear equation.
A1: Correct equation in the required form, e.g., \(3x - y - 3 = 0\) (or any integer multiple thereof, such as \(-3x + y + 3 = 0\)).

First, determine the \(y\)-coordinate on \(C\) at \(x = 4\):
\[ y = 2(4^2) - \frac{8}{\sqrt{4}} + 3 = 2(16) - \frac{8}{2} + 3 = 32 - 4 + 3 = 31 \]
So the coordinates of the point of tangency are \((4, 31)\).

Express the equation of \(C\) in index form:
\[ y = 2x^2 - 8x^{-1/2} + 3 \]

Differentiate with respect to \(x\) to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = 4x - 8\left(-\frac{1}{2}\right)x^{-3/2} = 4x + 4x^{-3/2} \]

Evaluate \(\frac{dy}{dx}\) at \(x = 4\) to find the gradient of the tangent, \(m\):
\[ m = 4(4) + 4(4^{-3/2}) = 16 + 4\left(\frac{1}{8}\right) = 16 + \frac{1}{2} = 16.5 = \frac{33}{2} \]

Using the point-slope formula with point \((4, 31)\) and gradient \(\frac{33}{2}\):
\[ y - 31 = \frac{33}{2}(x - 4) \]
\[ 2(y - 31) = 33(x - 4) \]
\[ 2y - 62 = 33x - 132 \]
\[ 33x - 2y - 70 = 0 \]

B1: Evaluates and finds the correct \(y\)-coordinate, \(y = 31\), when \(x = 4\).
M1: Attempts to differentiate with at least one non-zero term integrated/differentiated correctly (e.g., \(2x^2 \to 4x\) or \(-8x^{-1/2} \to 4x^{-3/2}\)).
A1: Correct derivative expression \(\frac{dy}{dx} = 4x + 4x^{-3/2}\).
M1: Substitutes \(x = 4\) into their \(\frac{dy}{dx}\) to find the gradient, and uses this gradient and their \((4, 31)\) to form the equation of the line.
A1: Correct equation in the required form, e.g., \(33x - 2y - 70 = 0\) (or any integer multiple).

To find the equation of the curve, integrate the gradient function with respect to \(x\):
\[ y = \int \left(\frac{3}{\sqrt{x}} - 2x\right) dx = \int \left(3x^{-1/2} - 2x\right) dx \]

Integrating each term:
\[ y = \frac{3x^{1/2}}{\frac{1}{2}} - \frac{2x^2}{2} + c \]
\[ y = 6\sqrt{x} - x^2 + c \]

Since the curve passes through the point \(P(9, 10)\), substitute \(x = 9\) and \(y = 10\) to calculate the constant of integration \(c\):
\[ 10 = 6\sqrt{9} - 9^2 + c \]
\[ 10 = 6(3) - 81 + c \]
\[ 10 = 18 - 81 + c \]
\[ 10 = -63 + c \]
\[ c = 73 \]

So, the equation of the curve \(C\) is:
\[ y = 6\sqrt{x} - x^2 + 73 \]

M1: Integrates at least one term of the gradient function correctly (e.g., \(\frac{3}{\sqrt{x}} \to 6x^{1/2}\) or \(-2x \to -x^2\)).
A1: Correct integration of both terms, giving \(6x^{1/2} - x^2\) (ignore presence or absence of \(c\) for this mark).
M1: Includes the constant of integration \(c\) and substitutes \(x = 9\) and \(y = 10\) to form an equation for \(c\).
A1: Correctly finds \(c = 73\).
A1: Correct final equation in the form \(y = f(x)\), i.e., \(y = 6\sqrt{x} - x^2 + 73\) (or equivalent index form, such as \(y = 6x^{0.5} - x^2 + 73\)).

(a) The total volume of the toy, \(V\), is the sum of the volume of the cylinder and the volume of the hemisphere:
\[V = \pi r^2 h + \frac{2}{3}\pi r^3\]
We are given that the volume is \(360\pi\text{ cm}^3\), so:
\[\pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi\]
Dividing through by \(\pi\):
\[r^2 h + \frac{2}{3} r^3 = 360\]
Rearranging to make \(h\) the subject:
\[r^2 h = 360 - \frac{2}{3}r^3\]
\[h = \frac{360}{r^2} - \frac{2}{3}r\]

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

Substituting our expression for \(h\) into this equation:
\[A = 3\pi r^2 + 2\pi r \left(\frac{360}{r^2} - \frac{2}{3}r\right)\]
\[A = 3\pi r^2 + \frac{720\pi}{r} - \frac{4}{3}\pi r^2\]
\[A = \left(3 - \frac{4}{3}\right)\pi r^2 + \frac{720\pi}{r}\]
\[A = \frac{5}{3}\pi r^2 + \frac{720\pi}{r}\]
This is the required expression.

(b) To find the minimum value of \(A\), we differentiate \(A\) with respect to \(r\):
\[\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{\mathrm{d}}{\mathrm{d}r}\left(\frac{5}{3}\pi r^2 + 720\pi r^{-1}\right)\]
\[\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{10}{3}\pi r - 720\pi r^{-2} = \frac{10}{3}\pi r - \frac{720\pi}{r^2}\]

At the minimum value, \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\):
\[\frac{10}{3}\pi r - \frac{720\pi}{r^2} = 0\]
\[\frac{10}{3}\pi r = \frac{720\pi}{r^2}\]
Dividing both sides by \(\pi\):
\[\frac{10}{3} r = \frac{720}{r^2}\]
\[10r^3 = 2160\]
\[r^3 = 216\]
\[r = 6\]

To find the minimum surface area, substitute \(r = 6\) back into the equation for \(A\):
\[A = \frac{5}{3}\pi (6)^2 + \frac{720\pi}{6}\]
\[A = \frac{5}{3}\pi (36) + 120\pi\]
\[A = 60\pi + 120\pi = 180\pi\]

Thus, the minimum surface area is \(180\pi\text{ cm}^2\).

(a)
- B1: Writes a correct equation for the volume of the toy in terms of \(r\) and \(h\): \(\pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi\) (or equivalent with \(\pi\) cancelled).
- M1: Rearranges their volume equation to express \(h\) in terms of \(r\).
- M1: Writes a correct formula for the total surface area \(A = \pi r^2 + 2\pi r h + 2\pi r^2\) or \(A = 3\pi r^2 + 2\pi r h\).
- A1*: Substitutes the expression for \(h\) into the surface area formula and simplifies correctly to show the given result \(A = \frac{5}{3}\pi r^2 + \frac{720\pi}{r}\). This is a given answer, so all working must be shown and correct.

(b)
- M1: Attempts to differentiate \(A\) with respect to \(r\), with at least one term differentiated correctly (power reduced by 1).
- A1: Correct derivative: \(\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{10}{3}\pi r - \frac{720\pi}{r^2}\) (or equivalent).
- M1: Sets their derivative equal to 0 and solves for \(r\) to find a non-zero value of \(r\) (or \(r^3\)).
- A1: Obtains \(r = 6\) and substitutes it into the formula for \(A\) to find the minimum value of \(180\pi\) (accept \(180\pi\) as the final answer, do not accept decimal approximations like 565 unless the exact value is also clearly shown).

Using the remainder theorem, we substitute \(x = \frac{3}{2}\) into \(f(x)\). We have \(f\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 3\left(\frac{3}{2}\right)^2 - 8\left(\frac{3}{2}\right) + 15 = 2\left(\frac{27}{8}\right) - 3\left(\frac{9}{4}\right) - 12 + 15 = \frac{27}{4} - \frac{27}{4} + 3 = 3\).

M1: Substituting \(x = \frac{3}{2}\) (or \(1.5\)) into \(f(x)\). M1: Correctly evaluating the substitution, showing at least one step of simplification. A0.5: Correct remainder of 3.

Using the laws of logarithms, we can combine the terms on the left-hand side: \(\log_3\left(\frac{x+4}{x-2}\right) = 2\). This gives \(\frac{x+4}{x-2} = 3^2 = 9\). Rearranging gives \(x + 4 = 9(x - 2)\), which simplifies to \(x + 4 = 9x - 18\). Solving for \(x\) yields \(8x = 22\), so \(x = \frac{11}{4}\).

M1: Applying subtraction law of logarithms to write as a single logarithm: \(\log_3\left(\frac{x+4}{x-2}\right) = 2\). M1: Eliminating logarithm correctly to get \(\frac{x+4}{x-2} = 9\) and attempting to solve for \(x\). A0.5: Correct answer of \(x = \frac{11}{4}\) (or \(2.75\)).

The sum of the first two terms is \(a + ar = a(1+r) = 4.5\). The sum to infinity is \(\frac{a}{1-r} = 8\), which implies \(a = 8(1-r)\). Substituting this expression for \(a\) into the first equation gives \(8(1-r)(1+r) = 4.5\), which simplifies to \(8(1-r^2) = 4.5\). This gives \(1-r^2 = \frac{4.5}{8} = \frac{9}{16}\), hence \(r^2 = \frac{7}{16}\). Taking the square root gives the possible values of \(r = \pm \frac{\sqrt{7}}{4}\).

M1: Writing two correct equations in terms of \(a\) and \(r\), e.g. \(a(1+r) = 4.5\) and \(\frac{a}{1-r} = 8\). M1: Substituting to eliminate \(a\) and attempting to solve for \(r^2\). A0.5: Finding both correct values \(r = \pm\frac{\sqrt{7}}{4}\) or exact equivalents.

We complete the square for \(x\) and \(y\): \((x - 5)^2 - 25 + (y + 3)^2 - 9 + k = 0\), which simplifies to \((x - 5)^2 + (y + 3)^2 = 34 - k\). Since the equation of a circle is \((x - a)^2 + (y - b)^2 = R^2\) where \(R\) is the radius, we have \(R^2 = 34 - k\). Since the radius is \(6\), we set \(34 - k = 6^2 = 36\), which gives \(k = -2\).

M1: Attempting to complete the square for both \(x\) and \(y\) to find the expression for \(R^2\) in terms of \(k\). M1: Equating their expression for \(R^2\) to \(36\). A0.5: Correct value of \(k = -2\).

Using the identity \(\sin^2\theta = 1 - \cos^2\theta\), we write the equation as \(4(1 - \cos^2\theta) + 5\cos\theta = 5\), which simplifies to \(4 - 4\cos^2\theta + 5\cos\theta = 5\). Rearranging gives the quadratic equation \(4\cos^2\theta - 5\cos\theta + 1 = 0\). Factoring gives \((4\cos\theta - 1)(\cos\theta - 1) = 0\), so \(\cos\theta = 1\) or \(\cos\theta = \frac{1}{4}\). From \(\cos\theta = 1\) in the given range, we get \(\theta = 0^\circ\). From \(\cos\theta = 0.25\), we get \(\theta \approx 75.5^\circ\) and \(\theta \approx 360^\circ - 75.5^\circ = 284.5^\circ\).

M1: Substituting \(\sin^2\theta = 1-\cos^2\theta\) to form a 3-term quadratic in \(\cos\theta\). M1: Solving the quadratic to find \(\cos\theta = 1\) and \(\cos\theta = 0.25\), and finding at least one valid angle. A0.5: All three angles correct: \(\theta = 0^\circ\), \(75.5^\circ\), and \(284.5^\circ\).

First, find the derivative: \(\frac{dy}{dx} = x^2 - 8x + 12\). Stationary points occur when \(\frac{dy}{dx} = 0\), so we solve \(x^2 - 8x + 12 = 0\). Factoring the quadratic gives \((x - 2)(x - 6) = 0\), which yields the solutions \(x = 2\) and \(x = 6\).

M1: Differentiating the curve's equation to find \(\frac{dy}{dx}\) (at least two terms differentiated correctly). M1: Setting their \(\frac{dy}{dx} = 0\) and attempting to solve the resulting quadratic equation. A0.5: Obtaining both correct coordinates \(x = 2\) and \(x = 6\).

First, write the integrand in index form: \(3x^{1/2} - 2x^{-2}\). Integrating term by term, we get: \(\int \left( 3x^{1/2} - 2x^{-2} \right) \mathrm{d}x = \left[ 2x^{3/2} + 2x^{-1} \right]\). Evaluating this expression from \(1\) to \(4\): at \(x = 4\), we have \(2(4)^{3/2} + \frac{2}{4} = 2(8) + 0.5 = 16.5\). At \(x = 1\), we have \(2(1)^{3/2} + \frac{2}{1} = 2 + 2 = 4\). Subtracting these gives \(16.5 - 4 = 12.5\).

M1: Integrating the terms of the expression, with at least one term integrated correctly (power increased by 1). M1: Substituting both limits \(4\) and \(1\) into their integrated function and subtracting the lower limit from the upper. A0.5: Correct evaluation of \(12.5\) (or \(\frac{25}{2}\)).

To disprove the statement by counterexample, we choose a pair of non-zero real numbers \(a\) and \(b\) of opposite sign. For instance, let \(a = -1\) and \(b = 1\). Since \(a \neq b\), this pair satisfies the conditions. Substituting these values into the expression: \(\frac{a}{b} + \frac{b}{a} = \frac{-1}{1} + \frac{1}{-1} = -1 - 1 = -2\). Since \(-2\) is not greater than \(2\), this is a valid counterexample which disproves the statement.

M1: Selecting two non-zero real values with opposite signs, such as \(a = -1\) and \(b = 1\). M1: Correctly substituting their chosen values into the expression \(\frac{a}{b} + \frac{b}{a}\) and evaluating. A0.5: Concluding clearly that the result is less than or equal to \(2\), thereby disproving the statement.

Take natural logarithms of both sides:

\(\ln(3^{2x+1}) = \ln(5^{x-1})\)

Apply the power law of logarithms:

\((2x+1)\ln 3 = (x-1)\ln 5\)

Expand the terms:

\(2x\ln 3 + \ln 3 = x\ln 5 - \ln 5\)

Rearrange to group all terms with \(x\) on one side:

\(2x\ln 3 - x\ln 5 = -\ln 5 - \ln 3\)

Factor out \(x\) and apply laws of logarithms:

\(x(2\ln 3 - \ln 5) = -(\ln 5 + \ln 3)\)

\(x(\ln 9 - \ln 5) = -\ln 15\)

\(x\ln(9/5) = \ln(1/15)\)

Divide to solve for \(x\):

\(x = \frac{\ln(1/15)}{\ln(9/5)}\)

Alternatively, this can be written as \(\frac{\ln 15}{\ln(5/9)}\).

M1: For taking logarithms of both sides and applying the power law to obtain \((2x+1)\ln 3 = (x-1)\ln 5\) (or using any other base of logarithms).
M0.5: For isolating the terms in \(x\) and factorising, leading to \(x(2\ln 3 - \ln 5) = -\ln 5 - \ln 3\) or equivalent.
A1: For the correct final answer in the required form, such as \(\frac{\ln(1/15)}{\ln(9/5)}\) or \(\frac{\ln 15}{\ln(5/9)}\).

Let the first term of the geometric series be \(a\) and the common ratio be \(r\).

Using the formula for the \(n\)-th term of a geometric series, \(u_n = a r^{n-1}\):

\(u_3 = a r^2 = 12\) (Equation 1)

\(u_6 = a r^5 = -1.5\) (Equation 2)

Divide Equation 2 by Equation 1:

\(\frac{a r^5}{a r^2} = \frac{-1.5}{12}\)

\(r^3 = -\frac{1}{8}\)

Taking the cube root of both sides gives:

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

Substitute \(r = -\frac{1}{2}\) back into Equation 1 to find \(a\):

\(a \left(-\frac{1}{2}\right)^2 = 12\)

\(\frac{a}{4} = 12 \implies a = 48\)

Since \(|r| < 1\), the series has a sum to infinity:

\(S_{\infty} = \frac{a}{1 - r} = \frac{48}{1 - (-1/2)} = \frac{48}{1.5} = 32\).

M1: Sets up the equations for both terms and attempts to find \(r\) by dividing them, obtaining \(r^3 = -1/8\) or equivalent.
M0.5: Uses their value of \(r\) to find a correct value for the first term \(a = 48\).
A1: Applies the sum to infinity formula correctly to obtain \(32\).

First, rewrite the integrand by dividing each term in the numerator by the denominator:

\(\frac{x^2 - 3}{\sqrt{x}} = \frac{x^2}{x^{1/2}} - \frac{3}{x^{1/2}} = x^{3/2} - 3x^{-1/2}\)

Now integrate each term with respect to \(x\):

\(\int (x^{3/2} - 3x^{-1/2}) \text{d}x = \frac{x^{5/2}}{5/2} - 3\left(\frac{x^{1/2}}{1/2}\right) = \frac{2}{5}x^{5/2} - 6x^{1/2}\)

Next, substitute the limits 4 and 1 into the integrated expression:

At \(x = 4\):

\(\frac{2}{5}(4^{5/2}) - 6(4^{1/2}) = \frac{2}{5}(32) - 6(2) = \frac{64}{5} - 12 = \frac{4}{5}\)

At \(x = 1\):

\(\frac{2}{5}(1^{5/2}) - 6(1^{1/2}) = \frac{2}{5} - 6 = -\frac{28}{5}\)

Subtract the lower limit value from the upper limit value:

\(\frac{4}{5} - \left(-\frac{28}{5}\right) = \frac{32}{5}\).

M1: Splits the fraction and integrates at least one term correctly by increasing the power by 1.
M0.5: Integrates both terms correctly, obtaining \(\frac{2}{5}x^{5/2} - 6x^{1/2}\).
A1: Substitutes the limits of 4 and 1 correctly, subtracts, and simplifies to obtain \(\frac{32}{5}\) (or \(6.4\)).

Since \((x - 2)\) is a factor of \(\text{f}(x)\), by the factor theorem, \(\text{f}(2) = 0\):

\(2(2)^3 + a(2)^2 + b(2) - 6 = 0\)

\(16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5\) (Equation 1)

Since dividing \(\text{f}(x)\) by \((x + 1)\) gives a remainder of \(-15\), by the remainder theorem, \(\text{f}(-1) = -15\):

\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -15\)

\(-2 + a - b - 6 = -15 \implies a - b = -7\) (Equation 2)

We now solve the simultaneous equations:

From Equation 2, \(b = a + 7\).

Substitute this into Equation 1:

\(2a + (a + 7) = -5\)

\(3a = -12 \implies a = -4\)

Find \(b\):

\(b = -4 + 7 = 3\)

Thus, \(a = -4\) and \(b = 3\).

M1: Uses the factor theorem and remainder theorem to establish two simultaneous equations in \(a\) and \(b\), namely \(2a + b = -5\) and \(a - b = -7\) (or equivalent equations).
M0.5: Attempts to solve the simultaneous equations to find at least one of the constants.
A1: Obtains \(a = -4\) and \(b = 3\).

Since the circle passes through the point \((7, -1)\), these coordinates must satisfy the circle's equation:

\(7^2 + (-1)^2 - 6(7) + 8(-1) + k = 0\)

\(49 + 1 - 42 - 8 + k = 0\)

\(50 - 50 + k = 0 \implies k = 0\)

Now, substitute \(k = 0\) back into the equation of the circle:

\(x^2 + y^2 - 6x + 8y = 0\)

Complete the square for both \(x\) and \(y\):

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

\((x - 3)^2 + (y + 4)^2 = 25\)

Comparing this to the standard circle equation \((x - h)^2 + (y - k)^2 = R^2\), we find:

\(R^2 = 25 \implies R = 5\)

So the radius of the circle is 5.

M1: Substitutes the coordinates of the point \((7, -1)\) into the circle's equation and solves to find \(k = 0\).
M0.5: Completes the square for \(x\) and \(y\) to rearrange the equation into the form \((x - a)^2 + (y - b)^2 = R^2\).
A1: Correctly identifies that the radius of the circle is \(5\).

Given the equation:

\(3 \sin(2\theta - 30^{\circ}) = 2 \cos(2\theta - 30^{\circ})\)

Divide both sides by \(3 \cos(2\theta - 30^{\circ})\):

\(\tan(2\theta - 30^{\circ}) = \frac{2}{3}\)

Let \(\phi = 2\theta - 30^{\circ}\). Since the range for \(\theta\) is \(0 \le \theta < 180^{\circ}\), the range for \(\phi\) is:

\(0 \le 2\theta < 360^{\circ} \implies -30^{\circ} \le 2\theta - 30^{\circ} < 330^{\circ}\)

Find the principal value of \(\phi\):

\(\phi = \arctan(2/3) \approx 33.69^{\circ}\)

Find the other value of \(\phi\) in the interval \([-30^{\circ}, 330^{\circ})\) by adding \(180^{\circ}\):

\(\phi = 33.69^{\circ} + 180^{\circ} = 213.69^{\circ}\)

Now set \(2\theta - 30^{\circ}\) equal to each of these values and solve for \(\theta\):

Case 1:

\(2\theta - 30^{\circ} = 33.69^{\circ} \implies 2\theta = 63.69^{\circ} \implies \theta \approx 31.8^{\circ}\)

Case 2:

\(2\theta - 30^{\circ} = 213.69^{\circ} \implies 2\theta = 243.69^{\circ} \implies \theta \approx 121.8^{\circ}\)

Both answers are within the specified range \(0 \le \theta < 180^{\circ}\).

M1: Uses the identity \(\tan x = \frac{\sin x}{\cos x}\) to simplify the equation to \(\tan(2\theta - 30^{\circ}) = 2/3\).
M0.5: Finds at least one correct value for \(2\theta - 30^{\circ}\) (e.g. \(33.7^{\circ}\) or \(213.7^{\circ}\)) or one value for \(\theta\).
A1: Finds both correct values of \(\theta\): \(31.8^{\circ}\) and \(121.8^{\circ}\) to 1 decimal place, and no others within the range.

First, use the power law of logarithms to rewrite the first term: \(2 \log_3(x-2) = \log_3(x-2)^2\). Next, apply the subtraction law of logarithms to combine the terms: \(\log_3 \frac{(x-2)^2}{x+4} = 1\). Convert the logarithmic equation into its equivalent exponential form: \(\frac{(x-2)^2}{x+4} = 3^1\). Rearrange to form a quadratic equation: \((x-2)^2 = 3(x+4) \Rightarrow x^2 - 4x + 4 = 3x + 12 \Rightarrow x^2 - 7x - 8 = 0\). Factorise the quadratic equation: \((x - 8)(x + 1) = 0\). This gives two potential solutions: \(x = 8\) and \(x = -1\). Since \(x\) must satisfy the condition \(x - 2 > 0\) (or \(x > 2\)) for \(\log_3(x-2)\) to be real and defined, the solution \(x = -1\) is rejected. Therefore, the only valid solution is \(x = 8\).

M1: For using the power law of logarithms to write \(2 \log_3(x-2)\) as \(\log_3(x-2)^2\). M1: For using the division law of logarithms to obtain \(\log_3 \frac{(x-2)^2}{x+4} = 1\). M1: For removing logarithms correctly to get \(\frac{(x-2)^2}{x+4} = 3\) and expanding to form a quadratic equation: \(x^2 - 7x - 8 = 0\). A1: For solving the quadratic equation to obtain \(x = 8\) and \(x = -1\). A0.5: For identifying that \(x = -1\) is invalid and concluding that \(x = 8\) is the only solution.

Let the first term of the geometric series be \(a\) and the common ratio be \(r\). The second term of the series is given by \(ar = 12\). The sum to infinity is given by \(\frac{a}{1-r} = 64\). From the second equation, we can express \(a\) as: \(a = 64(1-r)\). Substitute this expression for \(a\) into the first equation: \(64(1-r)r = 12 \Rightarrow 64r - 64r^2 = 12\). Rearrange this into standard quadratic form: \(64r^2 - 64r + 12 = 0\). Divide the entire equation by 4 to simplify: \(16r^2 - 16r + 3 = 0\). Factorise the quadratic equation: \((4r - 3)(4r - 1) = 0\). This yields two solutions: \(r = \frac{3}{4}\) and \(r = \frac{1}{4}\). Both values satisfy the convergence condition \(|r| < 1\).

M1: For writing down the equations \(ar = 12\) and \(\frac{a}{1-r} = 64\). M1: For substituting \(a = 64(1-r)\) into \(ar = 12\) to obtain a quadratic equation in terms of \(r\). M1: For correctly simplifying the quadratic equation to \(16r^2 - 16r + 3 = 0\) (or equivalent). A1.5: For solving the quadratic equation to obtain both correct answers \(r = \frac{1}{4}\) and \(r = \frac{3}{4}\) (0.75 marks for each).

First, find the center of the circle by completing the square for the \(x\) and \(y\) terms: \((x - 3)^2 - 9 + (y + 4)^2 - 16 - 7 = 0 \Rightarrow (x - 3)^2 + (y + 4)^2 = 32\). The center of the circle, \(C\), is \((3, -4)\). The gradient of the radius from the center \((3, -4)\) to the point \(P(7, 0)\) is: \(m_r = \frac{0 - (-4)}{7 - 3} = \frac{4}{4} = 1\). Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent is: \(m_t = -\frac{1}{1} = -1\). The equation of the tangent line through \(P(7, 0)\) with gradient \(-1\) is: \(y - 0 = -1(x - 7) \Rightarrow y = -x + 7\). Rearranging into the requested form: \(x + y - 7 = 0\).

M1: Completes the square to find the coordinates of the center of the circle, \((3, -4)\). M1: Calculates the gradient of the radius from the center to \(P(7, 0)\) as \(1\). M1: Uses the perpendicular gradient rule to find the gradient of the tangent as \(-1\). A1.5: Obtains the correct equation of the tangent in the required form: \(x + y - 7 = 0\) (accept equivalent integer multiples, e.g., \(-x - y + 7 = 0\)).

Divide both sides of the equation by \(\cos(2\theta - 10^\circ)\) to get: \(3 \tan(2\theta - 10^\circ) = 2 \Rightarrow \tan(2\theta - 10^\circ) = \frac{2}{3}\). Let \(\alpha = 2\theta - 10^\circ\). Given the range \(0 \le \theta < 180^\circ\), the interval for \(\beta = 2\theta\) is \(0 \le 2\theta < 360^\circ\), and thus the interval for \(\alpha\) is \(-10^\circ \le \alpha < 350^\circ\). Solve \(\tan \alpha = \frac{2}{3}\): The principal value is \(\alpha \approx 33.69^\circ\). Since the period of tangent is \(180^\circ\), the other value of \(\alpha\) in the range is \(\alpha = 33.69^\circ + 180^\circ = 213.69^\circ\). Now find \(\theta\) for each case: For \(2\theta - 10^\circ = 33.69^\circ \Rightarrow 2\theta = 43.69^\circ \Rightarrow \theta \approx 21.8^\circ\). For \(2\theta - 10^\circ = 213.69^\circ \Rightarrow 2\theta = 223.69^\circ \Rightarrow \theta \approx 111.8^\circ\). Thus, the solutions in the given range are \(\theta = 21.8^\circ\) and \(\theta = 111.8^\circ\).

M1: Divides by \(\cos(2\theta - 10^\circ)\) to obtain \(\tan(2\theta - 10^\circ) = \frac{2}{3}\). M1: Finds the principal value of the angle: \(2\theta - 10^\circ \approx 33.7^\circ\) (or finds one correct value of \(\theta\)). M1: Uses the periodic properties of the tangent function to find the second angle: \(2\theta - 10^\circ \approx 213.7^\circ\). A1.5: Obtains both correct answers \(\theta \approx 21.8^\circ\) and \(\theta \approx 111.8^\circ\) rounded to 1 decimal place (0.75 marks for each).

First, rewrite the equation of the curve as \(y = 4x + 9x^{-1}\). Differentiate with respect to \(x\): \(\frac{dy}{dx} = 4 - 9x^{-2} = 4 - \frac{9}{x^2}\). To find the stationary point, set \(\frac{dy}{dx} = 0\): \(4 - \frac{9}{x^2} = 0 \Rightarrow 4 = \frac{9}{x^2} \Rightarrow x^2 = \frac{9}{4}\). Since we are given \(x > 0\), we take the positive root: \(x = \frac{3}{2} = 1.5\). Find the corresponding \(y\)-coordinate: \(y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12\). So the coordinates of the stationary point are \((1.5, 12)\). To determine its nature, find the second derivative: \(\frac{d^2y}{dx^2} = \frac{d}{dx}(4 - 9x^{-2}) = 18x^{-3} = \frac{18}{x^3}\). Evaluate the second derivative at \(x = 1.5\): \(\frac{d^2y}{dx^2} = \frac{18}{(1.5)^3} = \frac{18}{3.375} = \frac{16}{3}\). Since \(\frac{d^2y}{dx^2} > 0\), the stationary point \((1.5, 12)\) is a local minimum.

M1: Differentiates \(y = 4x + 9x^{-1}\) correctly to get \(\frac{dy}{dx} = 4 - 9x^{-2}\). M1: Sets \(\frac{dy}{dx} = 0\) and solves for \(x\) (using \(x > 0\)) to find \(x = 1.5\). A1: Calculates the correct \(y\)-coordinate to find the stationary point coordinates \((1.5, 12)\). M1: Differentiates again to find \(\frac{d^2y}{dx^2} = 18x^{-3}\). A0.5: Correctly evaluates \(\frac{d^2y}{dx^2} > 0\) at \(x = 1.5\) and concludes it is a minimum.

The area \(A\) is given by the definite integral: \(A = \int_{1}^{3} (3x^2 - 4x + 2) \, dx\). Integrate each term: \(\int (3x^2 - 4x + 2) \, dx = \left[ x^3 - 2x^2 + 2x \right]_{1}^{3}\). Substitute the upper limit \(x = 3\): \(3^3 - 2(3^2) + 2(3) = 27 - 18 + 6 = 15\). Substitute the lower limit \(x = 1\): \(1^3 - 2(1^2) + 2(1) = 1 - 2 + 2 = 1\). Subtract the lower limit evaluation from the upper limit evaluation: \(A = 15 - 1 = 14\). The area of the region is 14.

M1: Attempts to integrate the function term-by-term, showing at least two correct integrated terms. A1: Obtains the fully correct integrated expression: \(x^3 - 2x^2 + 2x\). M1: Demonstrates correct substitution of both limits 3 and 1 into their integrated expression. A1.5: Correctly calculates the area of 14.

Using the factor theorem, since \((x - 2)\) is a factor of \(f(x)\), we have \(f(2) = 0\). Substitute \(x = 2\) into the equation: \(2(2^3) + a(2^2) + b(2) - 6 = 0 \Rightarrow 16 + 4a + 2b - 6 = 0 \Rightarrow 4a + 2b = -10 \Rightarrow 2a + b = -5\) (Equation 1). Using the remainder theorem, since the remainder when \(f(x)\) is divided by \((x + 1)\) is \(-12\), we have \(f(-1) = -12\). Substitute \(x = -1\) into the equation: \(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12 \Rightarrow -2 + a - b - 6 = -12 \Rightarrow a - b = -4\) (Equation 2). Solve Equation 1 and Equation 2 simultaneously: Add the two equations: \((2a + b) + (a - b) = -5 + (-4) \Rightarrow 3a = -9 \Rightarrow a = -3\). Substitute \(a = -3\) into Equation 2: \(-3 - b = -4 \Rightarrow -b = -1 \Rightarrow b = 1\). Therefore, the values of the constants are \(a = -3\) and \(b = 1\).

M1: Applies the factor theorem \(f(2) = 0\) to obtain a linear equation in \(a\) and \(b\). M1: Applies the remainder theorem \(f(-1) = -12\) to obtain a second linear equation in \(a\) and \(b\). M1: Solves the two simultaneous linear equations to find either \(a\) or \(b\). A1.5: Obtains both correct values: \(a = -3\) and \(b = 1\) (0.75 marks for each).

(a)(i) We complete the square for the equation of the circle \(C\): \(x^2 - 8x + y^2 - 6y + 16 = 0 \implies (x - 4)^2 - 16 + (y - 3)^2 - 9 + 16 = 0 \implies (x - 4)^2 + (y - 3)^2 = 9\). The coordinates of the centre of \(C\) are \((4, 3)\). (a)(ii) The radius of \(C\) is \(\sqrt{9} = 3\). (b) Method 1: Substituting \(y = kx\) into the circle equation gives \(x^2 + (kx)^2 - 8x - 6(kx) + 16 = 0 \implies (1 + k^2)x^2 - (8 + 6k)x + 16 = 0\). Since the line is tangent to the circle, this quadratic equation must have exactly one real root, so its discriminant must be zero: \((-(8 + 6k))^2 - 4(1 + k^2)(16) = 0 \implies (8 + 6k)^2 - 64(1 + k^2) = 0 \implies 64 + 96k + 36k^2 - 64 - 64k^2 = 0 \implies -28k^2 + 96k = 0 \implies 4k(24 - 7k) = 0\). This gives \(k = 0\) or \(k = \frac{24}{7}\). Method 2: The perpendicular distance from the centre of the circle \((4, 3)\) to the line \(kx - y = 0\) must equal the radius of the circle, which is 3. Using the distance formula: \(\frac{|4k - 3|}{\sqrt{k^2 + (-1)^2}} = 3 \implies |4k - 3| = 3\sqrt{k^2 + 1}\). Squaring both sides: \((4k - 3)^2 = 9(k^2 + 1) \implies 16k^2 - 24k + 9 = 9k^2 + 9 \implies 7k^2 - 24k = 0 \implies k(7k - 24) = 0\). This gives \(k = 0\) or \(k = \frac{24}{7}\).

Part (a): M1: Attempts to complete the square for both \(x\) and \(y\). Look for \((x \pm 4)^2\) and \((y \pm 3)^2\). A1: Correct centre coordinates \((4, 3)\). A1: Correct radius \(3\). Part (b): M1: Substitutes \(y = kx\) into the equation of \(C\) to form a quadratic in \(x\) OR uses the perpendicular distance formula with the centre \((4, 3)\) and the line \(kx - y = 0\). M1: Collects terms to form a quadratic in \(x\) and sets discriminant to 0 OR squares both sides of the distance equation to get a quadratic in \(k\). M1 (1.5 marks): Eliminates the constant terms to arrive at a simplified quadratic in \(k\). A1: Obtains a correct simplified quadratic, e.g., \(7k^2 - 24k = 0\) or \(28k^2 - 96k = 0\). A1: Find both \(k = 0\) and \(k = \frac{24}{7}\) (or equivalent exact fraction).