2025 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

To find the equation of the curve, we integrate the gradient function with respect to \(x\):

\(y = \int \left(4(2x+1)^{-\frac{1}{2}} - 3\right) dx\)

Using the rule for integrating \((ax+b)^n\):

\(y = 4 \cdot \frac{(2x+1)^{\frac{1}{2}}}{\frac{1}{2} \times 2} - 3x + c\)

\(y = 4(2x+1)^{\frac{1}{2}} - 3x + c\)

We are given that the curve passes through \((4, 5)\). Substitute \(x = 4\) and \(y = 5\) into the equation to find \(c\):

\(5 = 4(2(4)+1)^{\frac{1}{2}} - 3(4) + c\)

\(5 = 4(9)^{\frac{1}{2}} - 12 + c\)

\(5 = 4(3) - 12 + c\)

\(5 = 12 - 12 + c \implies c = 5\)

Therefore, the equation of the curve is:

\(y = 4\sqrt{2x+1} - 3x + 5\)

M1: Attempt to integrate to obtain a term of the form \(k(2x+1)^{\frac{1}{2}}\).
A1: Obtain correct integrated expression \(4(2x+1)^{\frac{1}{2}} - 3x\) (condone absence of \(+ c\)).
M1: Substitute \(x = 4, y = 5\) into an integrated expression containing \(+ c\) to find \(c\).
A1: Obtain the correct equation \(y = 4\sqrt{2x+1} - 3x + 5\) or equivalent.

To find the equation of the curve, we integrate the gradient function with respect to \(x\):

\(y = \int \left(4(2x+1)^{-\frac{1}{2}} - 3\right) dx\)

Using the rule for integrating \((ax+b)^n\):

\(y = 4 \cdot \frac{(2x+1)^{\frac{1}{2}}}{\frac{1}{2} \times 2} - 3x + c\)

\(y = 4(2x+1)^{\frac{1}{2}} - 3x + c\)

We are given that the curve passes through \((4, 5)\). Substitute \(x = 4\) and \(y = 5\) into the equation to find \(c\):

\(5 = 4(2(4)+1)^{\frac{1}{2}} - 3(4) + c\)

\(5 = 4(9)^{\frac{1}{2}} - 12 + c\)

\(5 = 4(3) - 12 + c\)

\(5 = 12 - 12 + c \implies c = 5\)

Therefore, the equation of the curve is:

\(y = 4\sqrt{2x+1} - 3x + 5\)

M1: Attempt to integrate to obtain a term of the form \(k(2x+1)^{\frac{1}{2}}\).
A1: Obtain correct integrated expression \(4(2x+1)^{\frac{1}{2}} - 3x\) (condone absence of \(+ c\)).
M1: Substitute \(x = 4, y = 5\) into an integrated expression containing \(+ c\) to find \(c\).
A1: Obtain the correct equation \(y = 4\sqrt{2x+1} - 3x + 5\) or equivalent.

We start with the equation: \( 8\sin^2 \theta + 2\cos \theta - 7 = 0 \). Using the trigonometric identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we substitute to get: \( 8(1 - \cos^2 \theta) + 2\cos \theta - 7 = 0 \). Expanding the terms gives: \( 8 - 8\cos^2 \theta + 2\cos \theta - 7 = 0 \), which simplifies to: \( -8\cos^2 \theta + 2\cos \theta + 1 = 0 \). Multiplying by \(-1\), we get the quadratic equation: \( 8\cos^2 \theta - 2\cos \theta - 1 = 0 \). Let \( y = \cos \theta \). The equation becomes \( 8y^2 - 2y - 1 = 0 \). Factoring the quadratic expression, we get: \( (4y + 1)(2y - 1) = 0 \). This yields two possible values for \( y \): \( y = -\frac{1}{4} \) or \( y = \frac{1}{2} \). Therefore, \( \cos \theta = \frac{1}{2} \) or \( \cos \theta = -\frac{1}{4} \). For \( \cos \theta = \frac{1}{2} \) in the range \( 0^\circ \le \theta \le 360^\circ \), we have: \( \theta = 60^\circ \) and \( \theta = 360^\circ - 60^\circ = 300^\circ \). For \( \cos \theta = -\frac{1}{4} \), the basic angle \( \alpha \) is given by: \( \alpha = \cos^{-1}\left(\frac{1}{4}\right) \approx 75.52^\circ \). Since the cosine value is negative, \( \theta \) must lie in the second or third quadrant: \( \theta = 180^\circ - 75.52^\circ \approx 104.5^\circ \) (to 1 decimal place) and \( \theta = 180^\circ + 75.52^\circ \approx 255.5^\circ \) (to 1 decimal place). Thus, the full set of solutions is \( \theta = 60^\circ, 104.5^\circ, 255.5^\circ, 300^\circ \).

**M1**: For using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to form a quadratic equation in terms of \( \cos \theta \). **A1**: For obtaining the correct simplified quadratic equation \( 8\cos^2 \theta - 2\cos \theta - 1 = 0 \) (or equivalent). **M1**: For solving their quadratic equation to find two values for \( \cos \theta \) (must be a valid method for solving a quadratic, e.g., factoring or formula). **A1**: For obtaining both \( \theta = 60^\circ \) and \( \theta = 300^\circ \) (both required for this mark). **A1**: For obtaining both \( \theta = 104.5^\circ \) and \( \theta = 255.5^\circ \) (both required, accept \( 104.5 \) and \( 255.5 \) to 1 d.p.). Note: Deduct 1 mark (from the final A marks) for any extra solutions within the range, or if answers are not rounded to 1 decimal place where appropriate.

The volume of revolution \(V\) about the \(x\)-axis is given by the formula:
\[V = \pi \int_{a}^{b} y^2 \, dx\]

First, square the expression for \(y\):
\[y^2 = \left( \frac{3}{(2x+1)^{3/2}} \right)^2 = \frac{9}{(2x+1)^3} = 9(2x+1)^{-3}\]

Next, integrate this expression with respect to \(x\):
\[\int 9(2x+1)^{-3} \, dx = 9 \left[ \frac{(2x+1)^{-2}}{-2 \times 2} \right] = -\frac{9}{4}(2x+1)^{-2} = -\frac{9}{4(2x+1)^2}\]

Now, evaluate this integral between the limits \(x = 0\) and \(x = 4\):
\[V = \pi \left[ -\frac{9}{4(2x+1)^2} \right]_{0}^{4}\]
\[V = \pi \left( -\frac{9}{4(2(4)+1)^2} - \left( -\frac{9}{4(2(0)+1)^2} \right) \right)\]
\[V = \pi \left( -\frac{9}{4(9)^2} + \frac{9}{4(1)^2} \right)\]
\[V = \pi \left( -\frac{9}{324} + \frac{9}{4} \right)\]
\[V = \pi \left( -\frac{1}{36} + \frac{81}{36} \right)\]
\[V = \frac{80}{36}\pi = \frac{20}{9}\pi\]

M1: For attempting to find the volume of revolution using \(V = \pi \int y^2 \, dx\) with an attempt to square \(y\).
A1: For obtaining the correct integrand, \(9(2x+1)^{-3}\), with or without \(\pi\).
M1: For integrating to the form \(k(2x+1)^{-2}\) where \(k\) is a constant.
A1: For the correct integrated expression \(-\frac{9}{4}(2x+1)^{-2}\).
A1: For substituting the limits \(0\) and \(4\) correctly into their integrated expression and obtaining \(\frac{20}{9}\pi\) (or exact equivalent).

**(a)**
Using the formulae for the area and the arc length of a sector:
\[ \text{Area} = \frac{1}{2}r^2\theta = 48 \]
\[ \text{Arc length} = r\theta = 12 \]

Dividing the area equation by the arc length equation to eliminate \(\theta\):
\[ \frac{\frac{1}{2}r^2\theta}{r\theta} = \frac{48}{12} \]
\[ \frac{1}{2}r = 4 \implies r = 8 \]

Substituting \(r = 8\) back into the arc length equation:
\[ 8\theta = 12 \implies \theta = 1.5 \text{ radians} \]

**(b)**
The perimeter of the shaded region consists of the arc \(AB\), the line segment \(AC\), and the line segment \(BC\).
We are given that the length of the arc \(AB\) is \(12\text{ cm}\).

In the right-angled triangle \(OCA\), the angle \(COA = \theta = 1.5\text{ rad}\) and the hypotenuse \(OA = r = 8\text{ cm}\):
\[ AC = OA \sin\theta = 8 \sin(1.5) \approx 7.97996\text{ cm} \]
\[ OC = OA \cos\theta = 8 \cos(1.5) \approx 0.56590\text{ cm} \]

Since \(C\) lies on the radius \(OB\) of length \(8\text{ cm}\):
\[ BC = OB - OC = 8 - 8 \cos(1.5) \approx 7.43410\text{ cm} \]

Thus, the total perimeter \(P\) is:
\[ P = \text{arc } AB + AC + BC \]
\[ P = 12 + 8 \sin(1.5) + (8 - 8 \cos(1.5)) \]
\[ P \approx 12 + 7.97996 + 7.43410 = 27.41406\text{ cm} \]

To 3 significant figures, the perimeter of the shaded region is \(27.4\text{ cm}\).

**(a)**
* **M1**: For attempting to eliminate one variable from \( \frac{1}{2}r^2\theta = 48 \) and \( r\theta = 12 \) (e.g., substituting \(\theta = \frac{12}{r}\)).
* **A1**: For obtaining both correct values: \( r = 8 \) and \( \theta = 1.5 \).

**(b)**
* **M1**: For finding \( AC = r \sin\theta \) using their \( r \) and \( \theta \).
* **M1**: For finding \( BC = r - r \cos\theta \) using their \( r \) and \( \theta \).
* **M1**: For summing the three boundaries: \( 12 + AC + BC \).
* **A1**: For obtaining \( 27.4 \) (accept \( 27.41 \) or \( 27.4\text{ cm} \)).

**(a)**

For Scheme A, the deposits form an arithmetic progression (AP) with:
- First term \(a = 150\)
- Common difference \(d = 5\)
- Number of terms \(n = 48\)

The sum of the first \(n\) terms of an AP is given by:
\[S_n = \frac{n}{2} [2a + (n-1)d]\]

Substituting the values:
\[S_{48} = \frac{48}{2} [2(150) + (48-1)(5)]\]
\[S_{48} = 24 [300 + 47(5)]\]
\[S_{48} = 24 [300 + 235]\]
\[S_{48} = 24 [535] = 12840\]

Thus, the total amount saved under Scheme A is **$12,840**.

**(b)**

For Scheme B, the deposits form a geometric progression (GP) with:
- First term \(a = X\)
- Common ratio \(r = 1.02\)
- Number of terms \(n = 48\)

The sum of the first \(n\) terms of a GP is given by:
\[S_n = \frac{a(r^n - 1)}{r - 1}\]

Substituting the values:
\[S_{48} = \frac{X(1.02^{48} - 1)}{1.02 - 1} = \frac{X(1.02^{48} - 1)}{0.02}\]

We are given that the total amount saved under Scheme B is equal to the total amount saved under Scheme A:
\[\frac{X(1.02^{48} - 1)}{0.02} = 12840\]

Multiply both sides by \(0.02\):
\[X(1.02^{48} - 1) = 256.8\]

Calculate \(1.02^{48}\):
\[1.02^{48} \approx 2.58707038\]

Substitute and solve for \(X\):
\[X(2.58707038 - 1) = 256.8\]
\[1.58707038 X = 256.8\]
\[X = \frac{256.8}{1.58707038} \approx 161.80757\]

Rounding to 2 decimal places, we get:
\[X = 161.81\]

**Part (a)**
- **M1**: For attempting to use the sum formula for an AP with \(a = 150\), \(d = 5\), and \(n = 48\).
- **A1**: For a correct unsimplified expression, e.g., \(\frac{48}{2}[300 + 47 \times 5]\).
- **A1**: For obtaining the correct total of \(12840\).

**Part (b)**
- **M1**: For identifying the monthly deposits as a geometric progression and attempting to use the GP sum formula with \(a = X\), \(r = 1.02\), and \(n = 48\).
- **A1**: For writing a correct expression for the sum, e.g., \(\frac{X(1.02^{48} - 1)}{0.02}\).
- **M1**: For equating their GP sum to their answer from part (a) (or \(12840\)) and attempting to solve for \(X\).
- **A1**: For obtaining a correct simplified equation for \(X\), e.g., \(79.35\dots X = 12840\) or \(1.587\dots X = 256.8\).
- **A1**: For obtaining \(X = 161.81\) (must be rounded to exactly 2 decimal places).

**(a) Find the coordinates of D:**
Since \(ABCD\) is a parallelogram, the diagonals \(AC\) and \(BD\) bisect each other at their midpoint \(M\).
Midpoint of \(AC\) is:
\(M = \left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5)\).
Let the coordinates of \(D\) be \((x, y)\). Since \(M\) is also the midpoint of \(BD\):
\(\frac{5+x}{2} = 2 \implies 5+x = 4 \implies x = -1\)
\(\frac{4+y}{2} = 5 \implies 4+y = 10 \implies y = 6\)
So, \(D = (-1, 6)\).

*Alternative vector method:*
\(\vec{BC} = \vec{AD}\)
\(\begin{pmatrix} 3-5 \\ 8-4 \end{pmatrix} = \begin{pmatrix} x_D-1 \\ y_D-2 \end{pmatrix} \implies \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} x_D-1 \\ y_D-2 \end{pmatrix}\)
\(x_D - 1 = -2 \implies x_D = -1\)
\(y_D - 2 = 4 \implies y_D = 6\)
So, \(D = (-1, 6)\).

**(b) Find the perpendicular bisector of AC:**
The midpoint of \(AC\) is \(M(2, 5)\).
The gradient of the line segment \(AC\) is:
\(m_{AC} = \frac{8-2}{3-1} = \frac{6}{2} = 3\).
The gradient of the perpendicular bisector is:
\(m_{\perp} = -\frac{1}{m_{AC}} = -\frac{1}{3}\).
Using the point-slope form with \(M(2, 5)\):
\(y - 5 = -\frac{1}{3}(x - 2)\)
\(3y - 15 = -(x - 2)\)
\(3y - 15 = -x + 2\)
\(x + 3y = 17\).

**(c) Relationship with B and shape identification:**
Substitute the coordinates of \(B(5, 4)\) into the equation of the perpendicular bisector:
\(LHS = 5 + 3(4) = 5 + 12 = 17 = RHS\).
Since the coordinates of \(B\) satisfy the equation, the perpendicular bisector of \(AC\) passes through \(B\).

Because \(B\) lies on the perpendicular bisector of \(AC\), the distance from \(B\) to \(A\) is equal to the distance from \(B\) to \(C\) (i.e., \(AB = BC\)).
A parallelogram with two adjacent sides of equal length is a **rhombus** (all four sides are equal: \(AB = BC = CD = DA\)).

**(a)**
* **M1**: For a valid method to find the midpoint of \(AC\) or setup a vector relationship (e.g., \(\vec{OD} = \vec{OA} + \vec{BC}\)).
* **M1**: For calculating the coordinates of \(D\) from their equation.
* **A1**: For obtaining \(D(-1, 6)\) correctly.

**(b)**
* **M1**: For finding the gradient of \(AC\) and attempting to find the perpendicular gradient \(-\frac{1}{m}\).
* **B1**: For stating or using the midpoint \(M(2, 5)\).
* **A1**: For the correct simplified equation \(x + 3y = 17\) (or any integer multiple thereof).

**(c)**
* **B1**: For substituting \(B(5, 4)\) into their equation from (b) and showing it is satisfied.
* **M1**: For explaining that \(B\) lying on the perpendicular bisector of \(AC\) implies \(AB = BC\) (or for explicitly calculating \(AB = \sqrt{20}\) and \(BC = \sqrt{20}\) and noting they are equal).
* **A1**: For concluding that the shape is a **rhombus** with a complete and correct justification.

\( \textbf{(a)} \)
To find the stationary points, we first find the first derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 5) = 6x^2 - 18x + 12 \)

At stationary points, \( \frac{dy}{dx} = 0 \):
\( 6x^2 - 18x + 12 = 0 \)
\( x^2 - 3x + 2 = 0 \)
\( (x - 1)(x - 2) = 0 \)

Thus, \( x = 1 \) or \( x = 2 \).

Now we find the corresponding \( y \)-coordinates:
- For \( x = 1 \):
\( y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10 \)
So, one stationary point is \( (1, 10) \).

- For \( x = 2 \):
\( y = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9 \)
So, the other stationary point is \( (2, 9) \).

\( \textbf{(b)} \)
To determine the nature of these points, we find the second derivative \( \frac{d^2y}{dx^2} \):
\( \frac{d^2y}{dx^2} = 12x - 18 \)

- At \( x = 1 \):
\( \frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0 \)
Since the second derivative is negative, \( (1, 10) \) is a local maximum.

- At \( x = 2 \):
\( \frac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0 \)
Since the second derivative is positive, \( (2, 9) \) is a local minimum.

\( \textbf{(c)} \)
We are given that \( \frac{dx}{dt} = 0.5 \) units per second.
We want to find \( \frac{dy}{dt} \) at \( x = 3 \).

Using the chain rule:
\( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

First, find \( \frac{dy}{dx} \) at \( x = 3 \):
\( \frac{dy}{dx} = 6(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12 \)

Now, substitute these values into the chain rule:
\( \frac{dy}{dt} = 12 \cdot 0.5 = 6 \) units per second.

\( \textbf{Part (a)} \)
* **M1**: Attempt to differentiate \( y = 2x^3 - 9x^2 + 12x + 5 \), with at least two terms differentiated correctly.
* **A1**: Correctly find \( \frac{dy}{dx} = 6x^2 - 18x + 12 \).
* **M1**: Set \( \frac{dy}{dx} = 0 \) and attempt to solve the resulting quadratic equation.
* **A1**: Correct \( x \)-values of \( x = 1 \) and \( x = 2 \).
* **A1**: Correctly calculate coordinates of both stationary points: \( (1, 10) \) and \( (2, 9) \).

\( \textbf{Part (b)} \)
* **M1**: Find the second derivative \( \frac{d^2y}{dx^2} \) and attempt to substitute at least one of their \( x \)-values into it (or use a sign change of gradient method).
* **A1**: Correctly identify \( (1, 10) \) as a maximum and \( (2, 9) \) as a minimum with valid mathematical justification.

\( \textbf{Part (c)} \)
* **M1**: Substitute \( x = 3 \) into their expression for \( \frac{dy}{dx} \).
* **A1**: Obtain \( \frac{dy}{dx} = 12 \).
* **M1**: State and apply the chain rule relation \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \) using \( \frac{dx}{dt} = 0.5 \).
* **A1**: Obtain the correct rate of change of \( 6 \) units per second.

\( \textbf{(a)} \)
To find the stationary points, we first find the first derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 5) = 6x^2 - 18x + 12 \)

At stationary points, \( \frac{dy}{dx} = 0 \):
\( 6x^2 - 18x + 12 = 0 \)
\( x^2 - 3x + 2 = 0 \)
\( (x - 1)(x - 2) = 0 \)

Thus, \( x = 1 \) or \( x = 2 \).

Now we find the corresponding \( y \)-coordinates:
- For \( x = 1 \):
\( y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10 \)
So, one stationary point is \( (1, 10) \).

- For \( x = 2 \):
\( y = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9 \)
So, the other stationary point is \( (2, 9) \).

\( \textbf{(b)} \)
To determine the nature of these points, we find the second derivative \( \frac{d^2y}{dx^2} \):
\( \frac{d^2y}{dx^2} = 12x - 18 \)

- At \( x = 1 \):
\( \frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0 \)
Since the second derivative is negative, \( (1, 10) \) is a local maximum.

- At \( x = 2 \):
\( \frac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0 \)
Since the second derivative is positive, \( (2, 9) \) is a local minimum.

\( \textbf{(c)} \)
We are given that \( \frac{dx}{dt} = 0.5 \) units per second.
We want to find \( \frac{dy}{dt} \) at \( x = 3 \).

Using the chain rule:
\( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

First, find \( \frac{dy}{dx} \) at \( x = 3 \):
\( \frac{dy}{dx} = 6(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12 \)

Now, substitute these values into the chain rule:
\( \frac{dy}{dt} = 12 \cdot 0.5 = 6 \) units per second.

\( \textbf{Part (a)} \)
* **M1**: Attempt to differentiate \( y = 2x^3 - 9x^2 + 12x + 5 \), with at least two terms differentiated correctly.
* **A1**: Correctly find \( \frac{dy}{dx} = 6x^2 - 18x + 12 \).
* **M1**: Set \( \frac{dy}{dx} = 0 \) and attempt to solve the resulting quadratic equation.
* **A1**: Correct \( x \)-values of \( x = 1 \) and \( x = 2 \).
* **A1**: Correctly calculate coordinates of both stationary points: \( (1, 10) \) and \( (2, 9) \).

\( \textbf{Part (b)} \)
* **M1**: Find the second derivative \( \frac{d^2y}{dx^2} \) and attempt to substitute at least one of their \( x \)-values into it (or use a sign change of gradient method).
* **A1**: Correctly identify \( (1, 10) \) as a maximum and \( (2, 9) \) as a minimum with valid mathematical justification.

\( \textbf{Part (c)} \)
* **M1**: Substitute \( x = 3 \) into their expression for \( \frac{dy}{dx} \).
* **A1**: Obtain \( \frac{dy}{dx} = 12 \).
* **M1**: State and apply the chain rule relation \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \) using \( \frac{dx}{dt} = 0.5 \).
* **A1**: Obtain the correct rate of change of \( 6 \) units per second.

(a) To transform the graph of \(y = \sin x\) into the graph of \(y = -2\sin\left(2\left(x - \frac{\pi}{12}\right)\right) + 4\), one valid sequence of transformations is:
1. Stretch horizontally with scale factor \(\frac{1}{2}\) parallel to the \(x\)-axis to obtain \(y = \sin(2x)\).
2. Translate by the vector \(\begin{pmatrix} \frac{\pi}{12} \\ 0 \end{pmatrix}\) (or shift \(\frac{\pi}{12}\) units to the right) to obtain \(y = \sin\left(2\left(x - \frac{\pi}{12}\right)\right) = \sin\left(2x - \frac{\pi}{6}\right)\).
3. Stretch vertically with scale factor 2 parallel to the \(y\)-axis and reflect in the \(x\)-axis (or stretch vertically by scale factor \(-2\) parallel to the \(y\)-axis) to obtain \(y = -2\sin\left(2x - \frac{\pi}{6}\right)\).
4. Translate by the vector \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\) (or shift 4 units upwards) to obtain \(y = 4 - 2\sin\left(2x - \frac{\pi}{6}\right)\).

Alternative sequence:
1. Translate by the vector \(\begin{pmatrix} \frac{\pi}{6} \\ 0 \end{pmatrix}\) to obtain \(y = \sin\left(x - \frac{\pi}{6}\right)\).
2. Stretch horizontally with scale factor \(\frac{1}{2}\) parallel to the \(x\)-axis to obtain \(y = \sin\left(2x - \frac{\pi}{6}\right)\).
3. Stretch vertically with scale factor 2 parallel to the \(y\)-axis and reflect in the \(x\)-axis.
4. Translate by the vector \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\).

(b) The function \(f(x)\) is continuous and strictly decreasing on the domain \(-\frac{\pi}{6} \le x \le \frac{\pi}{3}\).
At \(x = -\frac{\pi}{6}\), \(f\left(-\frac{\pi}{6}\right) = 4 - 2\sin\left(2\left(-\frac{\pi}{6}\right) - \frac{\pi}{6}\right) = 4 - 2\sin\left(-\frac{\pi}{2}\right) = 4 - 2(-1) = 6\).
At \(x = \frac{\pi}{3}\), \(f\left(\frac{\pi}{3}\right) = 4 - 2\sin\left(2\left(\frac{\pi}{3}\right) - \frac{\pi}{6}\right) = 4 - 2\sin\left(\frac{\pi}{2}\right) = 4 - 2(1) = 2\).
Therefore, the range of \(f\) is \(2 \le f(x) \le 6\).

(c) Set \(y = 4 - 2\sin\left(2x - \frac{\pi}{6}\right)\):
\(2\sin\left(2x - \frac{\pi}{6}\right) = 4 - y\)
\(\sin\left(2x - \frac{\pi}{6}\right) = \frac{4 - y}{2}\)
\(2x - \frac{\pi}{6} = \sin^{-1}\left(\frac{4 - y}{2}\right)\)
\(2x = \frac{\pi}{6} + \sin^{-1}\left(\frac{4 - y}{2}\right)\)
\(x = \frac{\pi}{12} + \frac{1}{2}\sin^{-1}\left(\frac{4 - y}{2}\right)\)

Replacing \(y\) with \(x\) gives:
\(f^{-1}(x) = \frac{\pi}{12} + \frac{1}{2}\sin^{-1}\left(\frac{4 - x}{2}\right)\) (or \(f^{-1}(x) = \frac{\pi}{12} + \frac{1}{2}\sin^{-1}\left(2 - \frac{x}{2}\right)\)).
The domain of \(f^{-1}\) is the range of \(f\), which is \(2 \le x \le 6\).

(d) We set \(f(g(x)) = 4\):
\(4 - 2\sin\left(2g(x) - \frac{\pi}{6}\right) = 4\)
\(\sin\left(2g(x) - \frac{\pi}{6}\right) = 0\)
Since the input \(g(x)\) must lie within the domain of \(f\) (i.e., \(-\frac{\pi}{6} \le g(x) \le \frac{\pi}{3}\)), we have:
\(-\frac{\pi}{3} \le 2g(x) \le \frac{2\pi}{3} \implies -\frac{\pi}{2} \le 2g(x) - \frac{\pi}{6} \le \frac{\pi}{2}\).
Within this interval, \(\sin\theta = 0\) has the unique solution \(\theta = 0\).
Therefore:
\(2g(x) - \frac{\pi}{6} = 0 \implies g(x) = \frac{\pi}{12}\).
Substituting \(g(x) = \frac{1}{x - 1}\):
\[\frac{1}{x - 1} = \frac{\pi}{12} \implies x - 1 = \frac{12}{\pi} \implies x = 1 + \frac{12}{\pi}\]
(Note that \(x = 1 + \frac{12}{\pi} \approx 4.82 > 1\), which is valid for the domain of \(g\)).

**(a)**
* **B1**: For correctly identifying a horizontal stretch with scale factor \(\frac{1}{2}\) parallel to the \(x\)-axis.
* **B1**: For a translation of \(\begin{pmatrix} \frac{\pi}{12} \\ 0 \end{pmatrix}\) or \(\begin{pmatrix} \frac{\pi}{6} \\ 0 \end{pmatrix}\) (must correspond correctly to the order of operations in the described sequence).
* **B1**: For a vertical stretch with scale factor 2 parallel to the \(y\)-axis combined with a reflection in the \(x\)-axis (or vertical stretch with scale factor \(-2\)).
* **B1**: For a translation of \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\).

**(b)**
* **M1**: For evaluating \(f(x)\) at the boundaries \(x = -\frac{\pi}{6}\) and \(x = \frac{\pi}{3}\) to find the maximum and minimum values.
* **A1**: For correct range: \(2 \le f(x) \le 6\) or \(2 \le y \le 6\) (accept interval notation \([2, 6]\); do not accept \(2 \le x \le 6\)).

**(c)**
* **M1**: For a valid attempt to make \(x\) the subject of \(y = 4 - 2\sin\left(2x - \frac{\pi}{6}\right)\) by isolating the sine term.
* **A1**: For obtaining \(2x - \frac{\pi}{6} = \sin^{-1}\left(\frac{4-y}{2}\right)\) (or equivalent).
* **A1**: For the correct expression \(f^{-1}(x) = \frac{\pi}{12} + \frac{1}{2}\sin^{-1}\left(\frac{4-x}{2}\right)\) in terms of \(x\).
* **B1**: For stating the correct domain \(2 \le x \le 6\) (accept \([2, 6]\)).

**(d)**
* **M1**: For setting \(4 - 2\sin\left(2g(x) - \frac{\pi}{6}\right) = 4\) and simplifying to \(\sin\left(2g(x) - \frac{\pi}{6}\right) = 0\).
* **A1**: For obtaining \(g(x) = \frac{\pi}{12}\).
* **A1**: For the correct exact answer \(x = 1 + \frac{12}{\pi}\) (or equivalent single fraction \(\frac{\pi + 12}{\pi}\)).

(a) We rewrite the equation of the circle \( C \) by completing the square for both \( x \) and \( y \) terms:
\(x^2 - 6x + y^2 - 8y + 5 = 0\)
\((x - 3)^2 - 9 + (y - 4)^2 - 16 + 5 = 0\)
\((x - 3)^2 + (y - 4)^2 = 20\)

Comparing this to the standard circle equation \((x - h)^2 + (y - k)^2 = r^2\):
- The center of the circle is \((3, 4)\).
- The radius of the circle is \(\sqrt{20} = 2\sqrt{5}\).

(b) **Method 1: Algebraic approach using the discriminant**
Substitute \( y = 2x + c \) into the equation of the circle:
\(x^2 + (2x + c)^2 - 6x - 8(2x + c) + 5 = 0\)
\(x^2 + 4x^2 + 4cx + c^2 - 6x - 16x - 8c + 5 = 0\)
\(5x^2 + (4c - 22)x + (c^2 - 8c + 5) = 0\)

For the line not to intersect the circle, this quadratic equation in \( x \) must have no real roots. Therefore, its discriminant must be strictly less than zero (\(\Delta < 0\)):
\(\Delta = b^2 - 4ac < 0\)
\((4c - 22)^2 - 4(5)(c^2 - 8c + 5) < 0\)
\(16c^2 - 176c + 484 - 20(c^2 - 8c + 5) < 0\)
\(16c^2 - 176c + 484 - 20c^2 + 160c - 100 < 0\)
\(-4c^2 - 16c + 384 < 0\)

Divide the entire inequality by \(-4\), remembering to reverse the inequality sign:
\(c^2 + 4c - 96 > 0\)

Factorise the quadratic expression:
\((c + 12)(c - 8) > 0\)

The critical values are \(c = -12\) and \(c = 8\).
Since we require the expression to be greater than zero, the set of values is:
\(c < -12\) or \(c > 8\)

**Method 2: Geometric approach using perpendicular distance**
The line \( L \) has equation \( 2x - y + c = 0 \).
For the line not to intersect the circle, the perpendicular distance \( d \) from the center of the circle \((3, 4)\) to the line must be strictly greater than the radius \(\sqrt{20}\).

Using the perpendicular distance formula \( d = \frac{|ax_1 + by_1 + d_0|}{\sqrt{a^2 + b^2}} \):
\(d = \frac{|2(3) - (4) + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|c + 2|}{\sqrt{5}}\)

Set \( d > \text{radius} \):
\(\frac{|c + 2|}{\sqrt{5}} > \sqrt{20}\)
\(|c + 2| > \sqrt{100}\)
\(|c + 2| > 10\)

This gives two inequalities:
\(c + 2 > 10 \implies c > 8\)
\(c + 2 < -10 \implies c < -12\)

Thus, the set of values is \(c < -12\) or \(c > 8\).

**(a)**
* **M1**: For attempting to complete the square for both \(x\) and \(y\).
* **A1**: For obtaining the correct center \((3, 4)\).
* **A1**: For obtaining the correct radius \(\sqrt{20}\) (or \(2\sqrt{5}\)).

**(b)**
* **M1**: For substituting \(y = 2x+c\) into the circle equation to obtain a quadratic equation in \(x\), or using the geometric distance formula.
* **A1**: For obtaining the correct 3-term quadratic in \(x\): \(5x^2 + (4c-22)x + (c^2-8c+5) = 0\), or for the correct distance expression \(\frac{|c+2|}{\sqrt{5}}\).
* **M1**: For setting discriminant \(\Delta < 0\), or setting distance \(d > \sqrt{20}\).
* **A1**: For obtaining the simplified inequality \(c^2 + 4c - 96 > 0\) (or equivalent, e.g., \(|c+2| > 10\)).
* **M1**: For finding the critical values \(c = -12\) and \(c = 8\) and choosing the outside regions.
* **A1**: For the correct final range of values: \(c < -12\) or \(c > 8\) (allow interval notation \(c \in (-\infty, -12) \cup (8, \infty)\); do not accept \(-12 > c > 8\)).

To find the exact value of the definite integral:

1. Integrate the function with respect to \(x\):
\(\int \frac{6}{3x+2} \, \mathrm{d}x = 6 \times \frac{1}{3} \ln(3x+2) = 2 \ln(3x+2)\)

2. Substitute the limits of integration, \(x = 4\) and \(x = 0\):
\(\left[ 2 \ln(3x+2) \right]_{0}^{4} = 2 \ln(3(4)+2) - 2 \ln(3(0)+2) = 2 \ln(14) - 2 \ln(2)\)

3. Simplify the expression using the laws of logarithms:
\(2 \ln(14) - 2 \ln(2) = 2 (\ln(14) - \ln(2)) = 2 \ln\left(\frac{14}{2}\right) = 2 \ln(7)\)

4. Write the answer in the form \(\ln k\):
\(2 \ln(7) = \ln(7^2) = \ln(49)\)

Thus, the exact value is \(\ln 49\).

M1: Obtain an integral of the form \(c \ln(3x+2)\), where \(c = 2\).
M1: Substitute the limits of integration correctly and apply logarithm laws to combine the terms into a single logarithm.
A1: Obtain the correct final answer of \(\ln 49\). (Do not accept decimal approximations; must be in the form \(\ln k\).)

To find the exact value of the definite integral:

1. Integrate the function with respect to \(x\):
\(\int \frac{6}{3x+2} \, \mathrm{d}x = 6 \times \frac{1}{3} \ln(3x+2) = 2 \ln(3x+2)\)

2. Substitute the limits of integration, \(x = 4\) and \(x = 0\):
\(\left[ 2 \ln(3x+2) \right]_{0}^{4} = 2 \ln(3(4)+2) - 2 \ln(3(0)+2) = 2 \ln(14) - 2 \ln(2)\)

3. Simplify the expression using the laws of logarithms:
\(2 \ln(14) - 2 \ln(2) = 2 (\ln(14) - \ln(2)) = 2 \ln\left(\frac{14}{2}\right) = 2 \ln(7)\)

4. Write the answer in the form \(\ln k\):
\(2 \ln(7) = \ln(7^2) = \ln(49)\)

Thus, the exact value is \(\ln 49\).

M1: Obtain an integral of the form \(c \ln(3x+2)\), where \(c = 2\).
M1: Substitute the limits of integration correctly and apply logarithm laws to combine the terms into a single logarithm.
A1: Obtain the correct final answer of \(\ln 49\). (Do not accept decimal approximations; must be in the form \(\ln k\).)

(i) The graph of \(y = |2x - 5|\) is a V-shape with its vertex at \((\frac{5}{2}, 0)\) and crossing the y-axis at \((0, 5)\). The graph of \(y = x + 1\) is a straight line with a gradient of 1, crossing the y-axis at \((0, 1)\) and the x-axis at \((-1, 0)\). (ii) To solve the inequality \(|2x - 5| < x + 1\), we first find the critical values by solving the equation \(|2x - 5| = x + 1\). Squaring both sides gives: \((2x - 5)^2 = (x + 1)^2\) which expands to \(4x^2 - 20x + 25 = x^2 + 2x + 1\). Simplifying this gives \(3x^2 - 22x + 24 = 0\), which factorises to \((3x - 4)(x - 6) = 0\), giving critical values \(x = \frac{4}{3}\) and \(x = 6\). Alternatively, we can solve the linear equations: Case 1: \(2x - 5 = x + 1\) which gives \(x = 6\); Case 2: \(-(2x - 5) = x + 1\) which gives \(3x = 4\), so \(x = \frac{4}{3}\). From the sketch, the V-shaped graph lies below the line between these two intersection points. Therefore, the solution to the inequality is \(\frac{4}{3} < x < 6\).

(i) B1: For a correctly shaped V-graph with vertex on the x-axis and a straight line with positive gradient. B1: For showing the correct intercepts: \((\frac{5}{2}, 0)\) and \((0, 5)\) for the modulus graph, and \((0, 1)\) and \((-1, 0)\) for the line. (ii) M1: For an attempt to solve the equation \(|2x - 5| = x + 1\) by squaring both sides to obtain a 3-term quadratic, or by solving the two linear equations \(2x - 5 = \pm(x + 1)\). A1: For obtaining the critical values \(x = \frac{4}{3}\) and \(x = 6\). A1: For the correct final range \(\frac{4}{3} < x < 6\) (or equivalent).

We are given the curve with equation:
\[y = \frac{e^{2x}}{2x - 3}\]

To find the stationary point, we differentiate \(y\) with respect to \(x\) using the quotient rule:
\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\]

Let \(u = e^{2x}\), which gives \(u' = 2e^{2x}\).
Let \(v = 2x - 3\), which gives \(v' = 2\).

Applying the quotient rule:
\[\frac{dy}{dx} = \frac{2e^{2x}(2x - 3) - 2e^{2x}}{(2x - 3)^2}\]

Simplify the numerator by factoring out \(2e^{2x}\):
\[\frac{dy}{dx} = \frac{2e^{2x}(2x - 3 - 1)}{(2x - 3)^2} = \frac{2e^{2x}(2x - 4)}{(2x - 3)^2}\]

At a stationary point, \(\frac{dy}{dx} = 0\):
\[\frac{2e^{2x}(2x - 4)}{(2x - 3)^2} = 0\]

Since \(2e^{2x} > 0\) for all real \(x\) and the denominator is non-zero for \(x > 1.5\), we solve:
\[2x - 4 = 0 \implies x = 2\]

Substituting \(x = 2\) back into the original equation to find the \(y\)-coordinate:
\[y = \frac{e^{2(2)}}{2(2) - 3} = \frac{e^4}{4 - 3} = e^4\]

Thus, the exact coordinates of the stationary point are \((2, e^4)\).

M1: For applying the quotient rule to differentiate \(y\), with correct structure \(\frac{u'v - uv'}{v^2}\).
A1: For obtaining a correct derivative, e.g., \(\frac{2e^{2x}(2x - 3) - 2e^{2x}}{(2x - 3)^2}\).
M1: For setting their derivative to 0 and attempting to solve for \(x\).
A1: For finding \(x = 2\).
A1: For finding the exact coordinates \((2, e^4)\).

(i) To find the intersection of the curve and the line, set their equations equal to each other:
\(\mathrm{e}^{2x} - 4 = 2x + 1\)

Rearrange to isolate the exponential term:
\(\mathrm{e}^{2x} = 2x + 5\)

Take the natural logarithm of both sides:
\(2x = \ln(2x + 5)\)

Divide by 2 to obtain the required equation:
\(x = \frac{1}{2}\ln(2x + 5)\)

(ii) Apply the iterative formula \(x_{n+1} = \frac{1}{2}\ln(2x_n + 5)\) starting with \(x_1 = 1\):
\(x_2 = \frac{1}{2}\ln(2(1) + 5) = \frac{1}{2}\ln 7 \approx 0.9730\)
\(x_3 = \frac{1}{2}\ln(2(0.9730) + 5) \approx 0.9691\)
\(x_4 = \frac{1}{2}\ln(2(0.9691) + 5) \approx 0.9685\)
\(x_5 = \frac{1}{2}\ln(2(0.9685) + 5) \approx 0.9684\)

Since successive approximations round to \(0.97\), the value of the \(x\)-coordinate correct to 2 decimal places is \(0.97\).

(iii) The curve \(y = \mathrm{e}^{2x} - 4\) crosses the \(x\)-axis where \(y = 0\):
\(\mathrm{e}^{2x} - 4 = 0 \implies 2x = \ln 4 \implies x = \ln 2\)

The region lies between \(x = 0\) and \(x = \ln 2\). Since \(\mathrm{e}^{2x} - 4 \le 0\) on this interval, the curve lies below the \(x\)-axis. The area is given by:
\(\text{Area} = \int_{0}^{\ln 2} (0 - (\mathrm{e}^{2x} - 4)) \, \mathrm{d}x\)
\(\text{Area} = \int_{0}^{\ln 2} (4 - \mathrm{e}^{2x}) \, \mathrm{d}x\)
\(\text{Area} = \left[ 4x - \frac{1}{2}\mathrm{e}^{2x} \right]_0^{\ln 2}\)
\(\text{Area} = \left( 4\ln 2 - \frac{1}{2}\mathrm{e}^{2\ln 2} \right) - \left( 0 - \frac{1}{2}\mathrm{e}^0 \right)\)
\(\text{Area} = \left( 4\ln 2 - \frac{1}{2}(4) \right) + \frac{1}{2}\)
\(\text{Area} = 4\ln 2 - 2 + \frac{1}{2} = 4\ln 2 - \frac{3}{2}\)

(i)
B1: For equating the equations and showing clearly that \(x = \frac{1}{2}\ln(2x+5)\).

(ii)
M1: For substituting \(x_1 = 1\) to calculate \(x_2\).
A1: For obtaining correct iterations \(0.9730, 0.9691, 0.9685, 0.9684\).
A1: For concluding \(0.97\) with sufficient iterations.

(iii)
M1: For setting \(y=0\) and correctly finding the upper limit as \(x = \ln 2\) (or equivalent).
M1: For integrating \(\mathrm{e}^{2x} - 4\) to obtain \(\frac{1}{2}\mathrm{e}^{2x} - 4x\) (or negative thereof).
A1: For substituting limits \(0\) and \ \ln 2\) correctly into their integrated function.
A1: For obtaining the exact area \(4\ln 2 - \frac{3}{2}\) (or \(\ln 16 - 1.5\)).

(i) Since \((x - 2)\) is a factor of \(p(x)\), we have \(p(2) = 0\) by the Factor Theorem:
\[2(2)^3 + a(2)^2 + b(2) - 6 = 0\]
\[16 + 4a + 2b - 6 = 0\]
\[4a + 2b = -10 \implies 2a + b = -5 \quad \text{--- (Equation 1)}\]

Since dividing \(p(x)\) by \((x + 1)\) gives a remainder of \(-12\), we have \(p(-1) = -12\) by the Remainder Theorem:
\[2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12\]
\[-2 + a - b - 6 = -12\]
\[a - b = -4 \quad \text{--- (Equation 2)}\]

Adding Equation 1 and Equation 2:
\[(2a + b) + (a - b) = -5 + (-4)\]
\[3a = -9 \implies a = -3\]

Substituting \(a = -3\) back into Equation 2:
\[-3 - b = -4 \implies b = 1\]

(ii) Substituting the values \(a = -3\) and \(b = 1\), the polynomial is:
\[p(x) = 2x^3 - 3x^2 + x - 6\]

Since \((x-2)\) is a factor, we can express \(p(x)\) as:
\[2x^3 - 3x^2 + x - 6 = (x-2)(2x^2 + kx + 3)\]

Comparing the coefficient of \(x^2\):
\[-4 + k = -3 \implies k = 1\]

Thus, the fully factorized form of \(p(x)\) is:
\[p(x) = (x-2)(2x^2 + x + 3)\]
*(Note: The quadratic factor \(2x^2 + x + 3\) has discriminant \(1^2 - 4(2)(3) = -23 < 0\), confirming it is irreducible.)*

(iii) Let \(x = \cot \phi\). The equation becomes:
\[2x^3 - 3x^2 + x - 6 = 0\]

Using the result from part (ii), this factorizes to:
\[(x - 2)(2x^2 + x + 3) = 0\]

Since the quadratic term has no real roots, the only real solution is:
\[x = 2 \implies \cot \phi = 2\]

Since \(\cot \phi = \frac{1}{\tan \phi}\), we have:
\[\tan \phi = 0.5\]

For the range \(0^\circ < \phi < 180^\circ\):
\[\phi = \tan^{-1}(0.5) \approx 26.565^\circ\]
Rounding to 1 decimal place gives:
\[\phi = 26.6^\circ\]

Part (i):
M1: Attempt to apply the factor theorem by substituting \(x = 2\) and setting \(p(2) = 0\).
M1: Attempt to apply the remainder theorem by substituting \(x = -1\) and setting \(p(-1) = -12\).
A1: Obtain any correct simplified linear equations, e.g., \(2a + b = -5\) and \(a - b = -4\).
A1: Correctly solve the simultaneous equations to find both \(a = -3\) and \(b = 1\).

Part (ii):
M1: Attempt polynomial division, synthetic division, or inspection to find the quadratic factor of \(p(x)\).
A1: Obtain the correct factorization \((x-2)(2x^2 + x + 3)\).

Part (iii):
M1: Formulate the substitution \(x = \cot \phi\) and recognize that \(\cot \phi = 2\) is the only real root.
M1: Write \(\tan \phi = 0.5\) and attempt to find \(\phi\) within the given interval.
A1: Obtain the final answer \(\phi = 26.6^\circ\). Deduct 1 mark if extra solutions are given within the range.

**(i)**
Substitute \(x = e\) and \(y = e\) into the left-hand side of the curve's equation:
\(\text{LHS} = e \ln(e) + e^2 - e^2\)
Since \(\ln(e) = 1\):
\(\text{LHS} = e(1) + 0 = e\)
Since \(\text{LHS} = \text{RHS} = e\), the point \(P(e, e)\) lies on the curve.

**(ii)**
Differentiate \(y \ln x + x^2 - y^2 = e\) implicitly with respect to \(x\):
Using the product rule on \(y \ln x\):
\(\frac{d}{dx}(y \ln x) = \frac{dy}{dx} \ln x + y \cdot \frac{1}{x}\)

Differentiating the remaining terms:
\(\frac{d}{dx}(x^2) = 2x\)
\(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\)
\(\frac{d}{dx}(e) = 0\)

Combining these terms gives:
\(\frac{dy}{dx} \ln x + \frac{y}{x} + 2x - 2y \frac{dy}{dx} = 0\)

Multiply the entire equation by \(x\) to clear the fraction:
\(x \frac{dy}{dx} \ln x + y + 2x^2 - 2xy \frac{dy}{dx} = 0\)

Group the \(\frac{dy}{dx}\) terms:
\(\frac{dy}{dx}(2xy - x \ln x) = 2x^2 + y\)

Divide to solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{2x^2 + y}{2xy - x \ln x}\) (as required)

Substitute \(x = e\) and \(y = e\) into the expression to find the gradient \(m\) at \(P\):
\(m = \frac{2e^2 + e}{2e^2 - e \ln e} = \frac{e(2e + 1)}{e(2e - 1)} = \frac{2e + 1}{2e - 1}\)

**(iii)**
The equation of the tangent at \(P(e, e)\) with gradient \(m = \frac{2e + 1}{2e - 1}\) is:
\(y - e = m(x - e)\)

Substitute the expression for \(m\):
\(y - e = \left(\frac{2e + 1}{2e - 1}\right)(x - e)\)
\(y = \left(\frac{2e + 1}{2e - 1}\right)x - e\left(\frac{2e + 1}{2e - 1}\right) + e\)

Simplify the constant term:
\(c = - e\left(\frac{2e + 1}{2e - 1}\right) + e = e\left(1 - \frac{2e + 1}{2e - 1}\right) = e\left(\frac{2e - 1 - (2e + 1)}{2e - 1}\right) = e\left(\frac{-2}{2e - 1}\right) = -\frac{2e}{2e - 1}\)

Thus, the equation of the tangent is:
\(y = \left(\frac{2e + 1}{2e - 1}\right)x - \frac{2e}{2e - 1}\)

**(i)**
* **M1**: Substute \(x = e\) and \(y = e\) into the equation of the curve.
* **A1**: Obtain \(e = e\) or equivalent convincing verification showing LHS = RHS.

**(ii)**
* **M1**: Apply product rule to differentiate \(y \ln x\), obtaining \(\frac{dy}{dx} \ln x + \frac{y}{x}\).
* **A1**: Correct differentiation of \(y \ln x\).
* **B1**: Correct differentiation of other terms (obtain \(2x\) and \(-2y \frac{dy}{dx}\)).
* **A1**: Correctly complete the algebraic steps to show the given expression for \(\frac{dy}{dx}\).
* **A1**: Substitute \((e, e)\) and obtain the exact gradient at \(P\) as \(\frac{2e + 1}{2e - 1}\).

**(iii)**
* **M1**: Use the straight line equation formula with their gradient from part (ii) and the point \((e, e)\).
* **A1**: Obtain the correct final equation \(y = \left(\frac{2e + 1}{2e - 1}\right)x - \frac{2e}{2e - 1}\) or equivalent.

**(a)**

We start with the given equation:
\[ 3\cos 2x + \sin(x - 60^{\circ}) = \sin(x + 60^{\circ}) \]

Using the compound angle formulae:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]

We expand both sine terms on the left-hand side and right-hand side:
\[ \sin(x + 60^{\circ}) = \sin x \cos 60^{\circ} + \cos x \sin 60^{\circ} \]
\[ \sin(x - 60^{\circ}) = \sin x \cos 60^{\circ} - \cos x \sin 60^{\circ} \]

Substituting these expansions into the equation gives:
\[ 3\cos 2x + (\sin x \cos 60^{\circ} - \cos x \sin 60^{\circ}) = \sin x \cos 60^{\circ} + \cos x \sin 60^{\circ} \]

Subtracting \(\sin x \cos 60^{\circ}\) from both sides and simplifying:
\[ 3\cos 2x - \cos x \sin 60^{\circ} = \cos x \sin 60^{\circ} \]
\[ 3\cos 2x = 2\cos x \sin 60^{\circ} \]

Using the exact value \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\):
\[ 3\cos 2x = 2\cos x \left(\frac{\sqrt{3}}{2}\right) \]
\[ 3\cos 2x = \sqrt{3}\cos x \]

Now, we apply the double angle identity for cosine to express the entire equation in terms of \(\cos x\):
\[ \cos 2x = 2\cos^2 x - 1 \]

Substituting this into the equation:
\[ 3(2\cos^2 x - 1) = \sqrt{3}\cos x \]
\[ 6\cos^2 x - 3 = \sqrt{3}\cos x \]
\[ 6\cos^2 x - \sqrt{3}\cos x - 3 = 0 \]

This completes the proof.

**(b)**

From part (a), we solve the quadratic equation:
\[ 6\cos^2 x - \sqrt{3}\cos x - 3 = 0 \]

Let \(u = \cos x\):
\[ 6u^2 - \sqrt{3}u - 3 = 0 \]

Applying the quadratic formula:
\[ u = \frac{-(-\sqrt{3}) \pm \sqrt{(-\sqrt{3})^2 - 4(6)(-3)}}{2(6)} \]
\[ u = \frac{\sqrt{3} \pm \sqrt{3 + 72}}{12} \]
\[ u = \frac{\sqrt{3} \pm \sqrt{75}}{12} \]

Since \(\sqrt{75} = 5\sqrt{3}\):
\[ u = \frac{\sqrt{3} \pm 5\sqrt{3}}{12} \]

This yields two possible values for \(\cos x\):
1. \(\cos x = \frac{6\sqrt{3}}{12} = \frac{\sqrt{3}}{2}\)
2. \(\cos x = \frac{-4\sqrt{3}}{12} = -\frac{\sqrt{3}}{3}\)

Now we find the values of \(x\) in the interval \(0^{\circ} \le x \le 360^{\circ}\) for each case:

**Case 1: \(\cos x = \frac{\sqrt{3}}{2}\)**
Since the cosine value is positive, \(x\) lies in the 1st and 4th quadrants.
\[ x = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^{\circ} \]
\[ x = 360^{\circ} - 30^{\circ} = 330^{\circ} \]

**Case 2: \(\cos x = -\frac{\sqrt{3}}{3}\)**
Since the cosine value is negative, \(x\) lies in the 2nd and 3rd quadrants.
First, find the reference angle \(\alpha\):
\[ \alpha = \cos^{-1}\left(\frac{\sqrt{3}}{3}\right) \approx 54.74^{\circ} \]

Now find \(x\):
\[ x = 180^{\circ} - 54.74^{\circ} \approx 125.3^{\circ} \]
\[ x = 180^{\circ} + 54.74^{\circ} \approx 234.7^{\circ} \]

Thus, the solutions in the interval are \(x = 30^{\circ}, 125.3^{\circ}, 234.7^{\circ}, 330^{\circ}\).

**(a)**
* **M1**: Attempt to expand \(\sin(x - 60^{\circ})\) and/or \(\sin(x + 60^{\circ})\) using compound angle formulae.
* **A1**: Obtain correct expansions for both terms (either in terms of \(\sin 60^{\circ}\) and \(\cos 60^{\circ}\) or with their exact values substituted).
* **M1**: Substitute the exact values of \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\) and \(\cos 60^{\circ} = \frac{1}{2}\) to simplify the equation to the form \(3\cos 2x = \sqrt{3}\cos x\).
* **M1**: Apply the double-angle identity \(\cos 2x = 2\cos^2 x - 1\) to obtain an equation in terms of \(\cos x\) only.
* **A1**: Complete the algebraic steps correctly to show \(6\cos^2 x - \sqrt{3}\cos x - 3 = 0\) with no errors seen.

**(b)**
* **M1**: Attempt to solve the quadratic equation in \(\cos x\) using the quadratic formula, factorization, or completing the square.
* **A1**: Obtain the two correct values \(\cos x = \frac{\sqrt{3}}{2}\) and \(\cos x = -\frac{\sqrt{3}}{3}\) (or equivalents like \(-0.577\)).
* **B1**: Obtain \(x = 30^{\circ}\).
* **B1**: Obtain \(x = 330^{\circ}\) (or FT from \(360^{\circ} - \text{their first positive solution}\)).
* **M1**: Correct method to find at least one angle in the range for \(\cos x = -\frac{\sqrt{3}}{3}\) (e.g., \(180^{\circ} - \alpha\) or \(180^{\circ} + \alpha\)).
* **A1**: Obtain both \(125.3^{\circ}\) and \(234.7^{\circ}\) (rounded to 1 decimal place, and no extra solutions in the range).