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Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), the equation becomes:

\(2 \sin \theta \left(\frac{\sin \theta}{\cos \theta}\right) = 3 \implies 2 \sin^2 \theta = 3 \cos \theta\).

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

\(2(1 - \cos^2 \theta) = 3 \cos \theta \implies 2 - 2 \cos^2 \theta = 3 \cos \theta\).

Rearranging into a quadratic in \(\cos \theta\):

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

Factorising this quadratic equation:

\((2 \cos \theta - 1)(\cos \theta + 2) = 0\).

This gives two possible solutions:

1) \(\cos \theta = \frac{1}{2}\)

2) \(\cos \theta = -2\) (which has no real solutions since \(-1 \le \cos \theta \le 1\)).

For \(\cos \theta = \frac{1}{2}\) in the interval \(0^\circ \le \theta \le 360^\circ\):

\(\theta = 60^\circ\) or \(\theta = 360^\circ - 60^\circ = 300^\circ\).

M1: Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and use \(\sin^2 \theta = 1 - \cos^2 \theta\) to form an equation in terms of \(\cos \theta\) only.
A1: Obtain the correct 3-term quadratic equation \(2 \cos^2 \theta + 3 \cos \theta - 2 = 0\).
M1: Solve the quadratic equation to get \(\cos \theta = \frac{1}{2}\) and find at least one correct value of \(\theta\).
A1: Obtain both \(\theta = 60^\circ\) and \(\theta = 300^\circ\) and no other solutions in the range.

Let \(u = x^{\frac{1}{2}}\). Since \(x > 0\), we have \(u > 0\). The equation can be rewritten as:

\(3u - \frac{2}{u} = 5\).

Multiplying both sides by \(u\) to clear the fraction:

\(3u^2 - 2 = 5u \implies 3u^2 - 5u - 2 = 0\).

Factorising the quadratic equation:

\((3u + 1)(u - 2) = 0\).

This yields two solutions for \(u\):

\(u = -\frac{1}{3}\) or \(u = 2\).

Since \(u = x^{\frac{1}{2}} \ge 0\), we discard \(u = -\frac{1}{3}\).

Thus, \(x^{\frac{1}{2}} = 2 \implies x = 2^2 = 4\).

M1: Substitute \(u = x^{1/2}\) (or multiply through by \(\sqrt{x}\)) to form a quadratic equation.
A1: Obtain the correct quadratic equation \(3u^2 - 5u - 2 = 0\) (or \(3x - 5\sqrt{x} - 2 = 0\)).
M1: Solve the quadratic equation to obtain \(u = 2\) or \(\sqrt{x} = 2\) (and state or imply the rejection of the negative root).
A1: Obtain the correct final answer \(x = 4\).

Multiply all terms by the common denominator \(2x(x-2)\):

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

Expand both sides:

\(2x^2 + 2(x^2 - 4x + 4) = 5x^2 - 10x\)

\(2x^2 + 2x^2 - 8x + 8 = 5x^2 - 10x\)

\(4x^2 - 8x + 8 = 5x^2 - 10x\).

Rearrange to form a quadratic equation:

\(5x^2 - 4x^2 - 10x + 8x - 8 = 0 \implies x^2 - 2x - 8 = 0\).

Factorise the quadratic:

\((x - 4)(x + 2) = 0\).

This gives the solutions:

\(x = 4\) or \(x = -2\).

*(Alternatively, let \(u = \frac{x}{x-2}\). Then \(u + \frac{1}{u} = \frac{5}{2} \implies 2u^2 - 5u + 2 = 0 \implies (2u-1)(u-2)=0 \implies u=2\) or \(u=\frac{1}{2}\). Solving these gives \(x=4\) and \(x=-2\).)*

M1: Attempt to eliminate fractions by multiplying by the common denominator \(2x(x-2)\) (or by using the substitution \(u = \frac{x}{x-2}\)).
A1: Obtain a correct quadratic equation, e.g., \(x^2 - 2x - 8 = 0\) or \(2u^2 - 5u + 2 = 0\).
M1: Correct method to solve their 3-term quadratic equation.
A1: Obtain both correct solutions \(x = 4\) and \(x = -2\).

(a) To find \(f(x)\), we integrate \(f'(x)\):
\[f(x) = \int (2x - 8x^{-2}) \, dx = x^2 - \frac{8x^{-1}}{-1} + C = x^2 + 8x^{-1} + C\]

Using the point \((1, 15)\) to find the constant of integration, \(C\):
\[15 = (1)^2 + \frac{8}{1} + C\]
\[15 = 9 + C \implies C = 6\]

Thus, the expression for \(f(x)\) is:
\[f(x) = x^2 + 8x^{-1} + 6 \quad \text{or} \quad f(x) = x^2 + \frac{8}{x} + 6\]

(b) The gradient of the normal is \(-\frac{1}{2}\), which means the gradient of the tangent, \(f'(x)\), is \(2\).
\[2x - \frac{8}{x^2} = 2\]
\[2x^3 - 8 = 2x^2\]
\[x^3 - x^2 - 4 = 0\]

By inspection or testing integer factors of \(4\), we find that \(x = 2\) is a root because:
\[2^3 - 2^2 - 4 = 8 - 4 - 4 = 0\]

Factorising \(x^3 - x^2 - 4 = 0\):
\[(x - 2)(x^2 + x + 2) = 0\]

Since the quadratic term \(x^2 + x + 2 = 0\) has a negative discriminant (\(1^2 - 4(1)(2) = -7 < 0\)), there are no other real roots. Therefore, \(x = 2\) is the only real solution.

Substituting \(x = 2\) into \(f(x)\) to find the \(y\)-coordinate:
\[y = 2^2 + \frac{8}{2} + 6 = 4 + 4 + 6 = 14\]

So, the coordinates of the point are \((2, 14)\).

(a)
- **M1**: Integrate \(f'(x)\) with at least one term correct.
- **A1**: Obtain \(x^2 + 8x^{-1}\) (allow unsimplified but must have resolved the negative sign).
- **M1**: Substitute \(x = 1, y = 15\) into their integrated expression including \(C\) and solve for \(C\).
- **A1**: State correct equation \(f(x) = x^2 + 8x^{-1} + 6\).

(b)
- **M1**: Form the equation \(f'(x) = 2\) and attempt to simplify to a cubic equation.
- **A1**: Show that \(x = 2\) is the only real root (must show factorization or explain why the remaining quadratic has no real roots).
- **A1**: Obtain \(y = 14\) and state the final coordinates \((2, 14)\).

(a) First, find the derivative \(\frac{dy}{dx}\) to determine the gradient of the tangent:
\[\frac{dy}{dx} = 2x + 16\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} = 2x - 8x^{-\frac{3}{2}}\]

Substitute \(x = 4\) to find the gradient of the tangent at point \(P\):
\[\frac{dy}{dx}\Big|_{x=4} = 2(4) - 8(4)^{-\frac{3}{2}} = 8 - 8\left(\frac{1}{8}\right) = 8 - 1 = 7\]

The gradient of the normal is the negative reciprocal of the tangent gradient:
\[m_{\text{normal}} = -\frac{1}{7}\]

The equation of the normal at \(P(4, 24)\) is:
\[y - 24 = -\frac{1}{7}(x - 4)\]
\[7(y - 24) = -(x - 4)\]
\[7y - 168 = -x + 4\]
\[x + 7y = 172\]

(b) The area is given by the definite integral of the curve from \(x = 1\) to \(x = 4\):
\[\text{Area} = \int_{1}^{4} \left(x^2 + 16x^{-\frac{1}{2}}\right) \, dx\]
\[\text{Area} = \left[ \frac{x^3}{3} + \frac{16x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{1}^{4} = \left[ \frac{x^3}{3} + 32x^{\frac{1}{2}} \right]_{1}^{4}\]

Substitute the limits of integration:
\[\text{Area} = \left( \frac{4^3}{3} + 32\sqrt{4} \right) - \left( \frac{1^3}{3} + 32\sqrt{1} \right)\]
\[\text{Area} = \left( \frac{64}{3} + 64 \right) - \left( \frac{1}{3} + 32 \right)\]
\[\text{Area} = \left( \frac{64}{3} - \frac{1}{3} \right) + (64 - 32)\]
\[\text{Area} = \frac{63}{3} + 32 = 21 + 32 = 53\]

(a)
- **M1**: Differentiate the equation of the curve to find \(\frac{dy}{dx}\) (at least one term correct).
- **A1**: Obtain \(\frac{dy}{dx} = 2x - 8x^{-\frac{3}{2}}\).
- **M1**: Evaluate \(\frac{dy}{dx}\) at \(x = 4\) and find the negative reciprocal to get the normal gradient.
- **A1**: Form the equation of the line and simplify to \(x + 7y = 172\).

(b)
- **M1**: Integrate the function \(x^2 + 16x^{-\frac{1}{2}}\) with at least one term correct.
- **M1**: Show correct substitution of limits \(4\) and \(1\) into their integrated expression.
- **A1**: Obtain the correct final numerical area of \(53\).

(a) A stationary point occurs where \(\frac{dy}{dx} = 0\):
\[3(4x - 1)^{\frac{1}{2}} - 6 = 0\]
\[3(4x - 1)^{\frac{1}{2}} = 6\]
\[(4x - 1)^{\frac{1}{2}} = 2\]
\[4x - 1 = 4\]
\[4x = 5 \implies x = 1.25 \quad \text{or} \quad x = \frac{5}{4}\]

To determine the nature of the stationary point, find the second derivative:
\[\frac{d^2y}{dx^2} = 3 \cdot \frac{1}{2}(4x - 1)^{-\frac{1}{2}} \cdot 4 = 6(4x - 1)^{-\frac{1}{2}}\]

Substitute \(x = 1.25\) into the second derivative:
\[\frac{d^2y}{dx^2}\Big|_{x=1.25} = 6(4(1.25) - 1)^{-\frac{1}{2}} = 6(4)^{-\frac{1}{2}} = \frac{6}{2} = 3\]

Since \(\frac{d^2y}{dx^2} = 3 > 0\), the stationary point is a **minimum**.

(b) To find the equation of the curve, integrate \(\frac{dy}{dx}\):
\[y = \int \left(3(4x - 1)^{\frac{1}{2}} - 6\right) \, dx\]
Using the reverse chain rule:
\[y = \frac{3(4x - 1)^{\frac{3}{2}}}{\frac{3}{2} \cdot 4} - 6x + C\]
\[y = \frac{3(4x - 1)^{\frac{3}{2}}}{6} - 6x + C = \frac{1}{2}(4x - 1)^{\frac{3}{2}} - 6x + C\]

Use the point \((2.5, 5.5)\) to determine the value of \(C\):
\[5.5 = \frac{1}{2}(4(2.5) - 1)^{\frac{3}{2}} - 6(2.5) + C\]
\[5.5 = \frac{1}{2}(9)^{\frac{3}{2}} - 15 + C\]
\[5.5 = \frac{1}{2}(27) - 15 + C\]
\[5.5 = 13.5 - 15 + C\]
\[5.5 = -1.5 + C \implies C = 7\]

Thus, the equation of the curve is:
\[y = \frac{1}{2}(4x - 1)^{\frac{3}{2}} - 6x + 7\]

(a)
- **M1**: Set \(\frac{dy}{dx} = 0\) and attempt to solve for \(x\).
- **A1**: Obtain \(x = 1.25\) (or \(\frac{5}{4}\)).
- **A1**: Obtain correct second derivative, evaluate it at their \(x\)-value, and state "minimum" based on a positive value.

(b)
- **M1**: Attempt integration of \(3(4x - 1)^{\frac{1}{2}}\) using reverse chain rule (division by both \(\frac{3}{2}\) and \(4\) must be attempted).
- **A1**: Obtain correct integrated terms: \(\frac{1}{2}(4x - 1)^{\frac{3}{2}} - 6x\).
- **M1**: Substitute \(x = 2.5, y = 5.5\) into their integrated expression including \(+C\) and solve for \(C\).
- **A1**: State correct final equation: \(y = \frac{1}{2}(4x - 1)^{\frac{3}{2}} - 6x + 7\).

Part (a): The gradient of the tangent line \(L\) is \(m_1 = 2\). Since the radius at the point of contact \(P\) is perpendicular to the tangent line, the gradient of the normal line \(AP\) is \(m_2 = -\frac{1}{2}\). The equation of the normal line through the center \(A(3, -2)\) is \(y - (-2) = -\frac{1}{2}(x - 3)\), which simplifies to \(y = -\frac{1}{2}x - \frac{1}{2}\). To find the coordinates of \(P\), we solve the intersection of the tangent \(L\) and the normal line: \(2x - 13 = -\frac{1}{2}x - \frac{1}{2} \implies 4x - 26 = -x - 1 \implies 5x = 25 \implies x = 5\). Substituting \(x = 5\) into the equation of \(L\) gives \(y = 2(5) - 13 = -3\). So, the point of contact is \(P(5, -3)\). The radius squared, \(r^2\), is the square of the distance \(AP\): \(r^2 = (5 - 3)^2 + (-3 - (-2))^2 = 2^2 + (-1)^2 = 5\). Thus, the equation of the circle \(C\) is \((x - 3)^2 + (y + 2)^2 = 5\). Part (b): Method 1: The line is \(y = 2x + k\), which is \(2x - y + k = 0\). The perpendicular distance from the center \(A(3, -2)\) to this line must equal the radius \(r = \sqrt{5}\). Using the distance formula: \(\frac{|2(3) - (-2) + k|}{\sqrt{2^2 + (-1)^2}} = \sqrt{5} \implies \frac{|8 + k|}{\sqrt{5}} = \sqrt{5} \implies |8 + k| = 5\). This gives \(8 + k = 5 \implies k = -3\), or \(8 + k = -5 \implies k = -13\). Since \(k \neq -13\), we have \(k = -3\). Method 2: The vector translation from \(P\) to the center \(A\) is \(\vec{PA} = \begin{pmatrix} 3-5 \\ -2-(-3) \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}\). The point of contact \(Q\) of the second parallel tangent is the diametrically opposite point of \(P\): \(\vec{AQ} = \vec{PA} = \begin{pmatrix} -2 \\ 1 \end{pmatrix} \implies Q = (3-2, -2+1) = (1, -1)\). Substituting \(Q(1, -1)\) into \(y = 2x + k\) gives \(-1 = 2(1) + k \implies k = -3\).

Part (a): M1 for finding normal gradient as \(-1/2\). M1 for finding normal equation \(y = -1/2x - 1/2\) and solving simultaneously with \(y = 2x - 13\). A1 for \(P(5, -3)\). M1 for finding radius squared \(r^2 = 5\). A1 for circle equation \((x - 3)^2 + (y + 2)^2 = 5\). Part (b): M1 for setting up perpendicular distance from \((3, -2)\) to \(2x - y + k = 0\) as \(\sqrt{5}\) (or finding opposite point \(Q(1, -1)\)). A1 for \(|8 + k| = 5\) (or \(Q(1, -1)\)). M1 for solving to get \(k = -3\) and \(k = -13\) (or substituting \(Q\) into \(y = 2x + k\)). A0.5 for rejecting \(k = -13\) and giving \(k = -3\) as final answer.

Part (a): Completing the square for both \(x\) and \(y\) in the circle's equation: \(x^2 - 2x + y^2 - 4y = 40 \implies (x - 1)^2 - 1 + (y - 2)^2 - 4 = 40 \implies (x - 1)^2 + (y - 2)^2 = 45\). The center of the circle is \((1, 2)\) and the radius is \(\sqrt{45} = 3\sqrt{5}\). Part (b): Method 1 (Algebraic): Substituting \(y = 2x + c\) into the circle equation: \(x^2 + (2x + c)^2 - 2x - 4(2x + c) - 40 = 0 \implies 5x^2 + (4c - 10)x + (c^2 - 4c - 40) = 0\). For two distinct points of intersection, the discriminant \(\Delta\) must be strictly greater than 0: \(\Delta = (4c - 10)^2 - 4(5)(c^2 - 4c - 40) > 0 \implies 16c^2 - 80c + 100 - 20c^2 + 80c + 800 > 0 \implies -4c^2 + 900 > 0 \implies c^2 < 225\). Thus, \(-15 < c < 15\). Method 2 (Geometric): For the line \(2x - y + c = 0\) to intersect the circle at two distinct points, the perpendicular distance from the center \((1, 2)\) to the line must be less than the radius \(\sqrt{45}\). Distance is given by \(\frac{|2(1) - 2 + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|c|}{\sqrt{5}}\). Setting this less than the radius: \(\frac{|c|}{\sqrt{5}} < \sqrt{45} \implies |c| < \sqrt{225} \implies |c| < 15 \implies -15 < c < 15\).

Part (a): M1 for completing the square for \(x\) and \(y\). A1 for center \((1, 2)\). A1 for radius \(\sqrt{45}\) or \(3\sqrt{5}\). Part (b): M1 for substituting \(y = 2x + c\) and expanding. A1 for correct quadratic equation in \(x\). M1 for expressing discriminant \(\Delta > 0\) (or using distance inequality). A1 for \(c^2 < 225\) (or \(|c| < 15\)). A1.5 for the final range \(-15 < c < 15\).

(a) To find the smallest value of \(k\) for which \(f\) has an inverse, we complete the square for \(f(x)\):
\[f(x) = 2(x^2 - 6x) + 13 = 2[(x - 3)^2 - 9] + 13 = 2(x - 3)^2 - 5\]
The vertex of this quadratic function is at \((3, -5)\). For the quadratic function to be one-to-one (and thus have an inverse), the domain must be restricted to one side of the vertex. Since the domain is defined for \(x \ge k\), the smallest value of \(k\) is the \(x\)-coordinate of the vertex, which is \(k = 3\).

(b) Using the completed square form with \(k = 3\):
\[y = 2(x - 3)^2 - 5\]
Since \(x \ge 3\), we have \(x - 3 \ge 0\). Rearranging to make \(x\) the subject:
\[y + 5 = 2(x - 3)^2 \implies \frac{y+5}{2} = (x - 3)^2\]
Taking the positive square root:
\[x - 3 = \sqrt{\frac{y+5}{2}} \implies x = 3 + \sqrt{\frac{y+5}{2}}\]
Thus, the inverse function is:
\[f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\]
The domain of \(f^{-1}\) is the range of \(f\). For \(x \ge 3\), the minimum value of \(f(x)\) is \(-5\), so the range of \(f\) is \(f(x) \ge -5\). Hence, the domain of \(f^{-1}\) is \(x \ge -5\).

(c) We solve \(fg(x) = 5\), which is \(f(g(x)) = 5\):
\[2(g(x) - 3)^2 - 5 = 5 \implies 2(g(x) - 3)^2 = 10 \implies (g(x) - 3)^2 = 5\]
Since the domain of \(f\) requires the input to be at least \(3\), we must have \(g(x) \ge 3\). Thus:
\[g(x) - 3 = \sqrt{5} \implies g(x) = 3 + \sqrt{5}\]
Now substitute the expression for \(g(x)\):
\[\frac{4}{x - 1} = 3 + \sqrt{5} \implies x - 1 = \frac{4}{3 + \sqrt{5}}\]
Rationalising the denominator:
\[x - 1 = \frac{4(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \frac{4(3 - \sqrt{5})}{9 - 5} = 3 - \sqrt{5}\]
Thus, \(x = 4 - \sqrt{5}\).

(d) The graph of \(y = f^{-1}(x)\) is the reflection of the graph of \(y = f(x)\) in the line \(y = x\), which is the line of symmetry.
To find the intersection of \(y = f(x)\) with this line of symmetry, we solve \(f(x) = x\) for \(x \ge 3\):
\[2x^2 - 12x + 13 = x \implies 2x^2 - 13x + 13 = 0\]
Using the quadratic formula:
\[x = \frac{13 \pm \sqrt{(-13)^2 - 4(2)(13)}}{2(2)} = \frac{13 \pm \sqrt{169 - 104}}{4} = \frac{13 \pm \sqrt{65}}{4}\]
Since \(\sqrt{65} \approx 8.06\), we have:
\[x = \frac{13 - \sqrt{65}}{4} \approx 1.23 \quad \text{(rejected since } x \ge 3\text{)}\]
\[x = \frac{13 + \sqrt{65}}{4} \approx 5.27 \quad \text{(accepted)}\]
Thus, the coordinates of the point of intersection are \(\left(\frac{13+\sqrt{65}}{4}, \frac{13+\sqrt{65}}{4}\right)\).

(a)
- M1: For attempting to complete the square or find the x-coordinate of the vertex of the quadratic using \(-\frac{b}{2a}\).
- A1: For obtaining \(k = 3\).

(b)
- M1: For attempting to rearrange \(y = 2(x-3)^2 - 5\) (or their completed square) to make \(x\) the subject.
- A1: For correctly reaching the square root step, i.e., \(x - 3 = \sqrt{\frac{y+5}{2}}\).
- A1: For the correct expression \(f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\) (must be in terms of \(x\)).
- B1: For stating the domain is \(x \ge -5\).

(c)
- M1: For substituting \(g(x)\) into the completed square form of \(f\) and setting equal to \(5\), i.e., \(2(g(x)-3)^2 - 5 = 5\).
- A1: For simplifying to find the correct value for \(g(x)\), which is \(g(x) = 3 + \sqrt{5}\) (the negative root must be rejected based on the domain of \(f\)).
- M1: For setting \(\frac{4}{x-1} = 3 + \sqrt{5}\) and attempting to solve for \(x\) by rationalising the denominator.
- A1: For obtaining the final exact value \(x = 4 - \sqrt{5}\).

(d)
- B1: For stating that the graphs are reflections of each other in the line \(y = x\) (or identifying \(y = x\) as the line of symmetry).
- B1: For solving \(2x^2 - 13x + 13 = 0\) to obtain the accepted coordinate \(x = \frac{13+\sqrt{65}}{4}\) and giving the correct coordinates \(\left(\frac{13+\sqrt{65}}{4}, \frac{13+\sqrt{65}}{4}\right)\).

(a) To express \(f(x)\) in the given form, we rewrite the numerator:
\[f(x) = \frac{2x + 3}{x - 2} = \frac{2(x - 2) + 7}{x - 2} = 2 + \frac{7}{x - 2}\]
Thus, \(A = 2\) and \(B = 7\).
Since \(x > 2\), the term \(x - 2 > 0\), which means \(\frac{7}{x - 2} > 0\). Therefore, \(f(x) > 2\).
So, the range of \(f\) is \(f(x) > 2\).

(b) To find the inverse function, we let \(y = \frac{2x+3}{x-2}\) and rearrange to make \(x\) the subject:
\[y(x - 2) = 2x + 3 \implies xy - 2y = 2x + 3 \implies xy - 2x = 2y + 3\]
\[x(y - 2) = 2y + 3 \implies x = \frac{2y + 3}{y - 2}\]
Thus:
\[f^{-1}(x) = \frac{2x + 3}{x - 2}\]
The domain of \(f^{-1}\) is the range of \(f\), which is \(x > 2\).
Since \(f^{-1}(x) = f(x)\), the function is a self-inverse function. This means that the reflection of the graph of \(y = f(x)\) in the line \(y = x\) results in the same graph. Therefore, the graph is symmetrical about the line \(y = x\).

(c) (i) The composite function \(fg(x)\) is defined if the output of \(g(x)\) lies within the domain of \(f\). The domain of \(f\) is \(x > 2\). Therefore, \(fg(x)\) is defined when \(g(x) > 2\).
Consequently, \(fg(x)\) is NOT defined when:
\[g(x) \le 2 \implies x^2 - 2x \le 2 \implies x^2 - 2x - 2 \le 0\]
To solve the quadratic inequality, we find the roots of \(x^2 - 2x - 2 = 0\):
\[x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}\]
Since the coefficient of \(x^2\) is positive, the quadratic is less than or equal to zero between these roots:
\[1 - \sqrt{3} \le x \le 1 + \sqrt{3}\]

(ii) To find \(fg(3)\), we first evaluate \(g(3)\):
\[g(3) = 3^2 - 2(3) = 9 - 6 = 3\]
Since \(3 > 2\), this is in the domain of \(f\). Now we substitute into \(f\):
\[fg(3) = f(3) = \frac{2(3) + 3}{3 - 2} = \frac{9}{1} = 9\]

(a)
- M1: For attempting to write \(f(x)\) in the form \(A + \frac{B}{x-2}\) via division or inspection.
- A1: For obtaining \(2 + \frac{7}{x-2}\).
- B1: For stating the correct range \(f(x) > 2\) (or \(y > 2\)).

(b)
- M1: For a correct algebraic method to make \(x\) the subject of \(y = \frac{2x+3}{x-2}\).
- A1: For the correct expression \(f^{-1}(x) = \frac{2x+3}{x-2}\) and stating domain as \(x > 2\).
- B1: For explaining that \(f(x) = f^{-1}(x)\) indicates the graph is its own reflection (self-inverse).
- B1: For identifying the line of symmetry as \(y = x\).

(c)(i)
- M1: For identifying that \(fg(x)\) is NOT defined when \(g(x) \le 2\).
- M1: For attempting to solve \(x^2 - 2x - 2 = 0\) using the quadratic formula.
- A1: For the correct interval \(1 - \sqrt{3} \le x \le 1 + \sqrt{3}\).

(c)(ii)
- M1: For evaluating \(g(3) = 3\) and substituting this into \(f\).
- A1: For obtaining the final answer of \(9\).

To solve the inequality \(|3x - 2| \ge 2|x + 4|\), we square both sides:

\[(3x - 2)^2 \ge 4(x + 4)^2\]

Expanding both sides:

\[9x^2 - 12x + 4 \ge 4(x^2 + 8x + 16)\]

\[9x^2 - 12x + 4 \ge 4x^2 + 32x + 64\]

Rearranging to form a quadratic inequality:

\[5x^2 - 44x - 60 \ge 0\]

Find the critical values by solving the corresponding quadratic equation:

\[(5x + 6)(x - 10) = 0\]

This gives critical values of \(x = -1.2\) and \(x = 10\).

Since we require the expression to be greater than or equal to zero, the solution is:

\[x \le -1.2 \text{ or } x \ge 10\]

M1: For squaring both sides and attempting expansion.
A1: For obtaining a correct three-term quadratic inequality, e.g., \(5x^2 - 44x - 60 \ge 0\).
M1: For solving their quadratic equation to find two critical values.
A1: For the correct final range: \(x \le -1.2\) or \(x \ge 10\) (or equivalent interval notation).

We differentiate \(y = \frac{\ln(x+1)}{x^2}\) using the quotient rule:

Let \(u = \ln(x+1) \implies \frac{du}{dx} = \frac{1}{x+1}\)

Let \(v = x^2 \implies \frac{dv}{dx} = 2x\)

Applying the quotient rule:

\[\frac{dy}{dx} = \frac{\frac{1}{x+1} \cdot x^2 - \ln(x+1) \cdot 2x}{x^4}\]

Now, substitute \(x = 1\) into the derivative:

\[\frac{dy}{dx}\Big|_{x=1} = \frac{\frac{1}{1+1} \cdot (1)^2 - \ln(2) \cdot 2(1)}{(1)^4}\]

\[\frac{dy}{dx}\Big|_{x=1} = \frac{\frac{1}{2} - 2\ln 2}{1} = \frac{1}{2} - 2\ln 2\]

Thus, \(a = \frac{1}{2}\) and \(b = -2\).

M1: For applying the quotient rule (or product rule) to differentiate \(y\).
A1: For obtaining a correct derivative expression, e.g., \(\frac{dy}{dx} = \frac{\frac{x^2}{x+1} - 2x\ln(x+1)}{x^4}\).
M1: For substituting \(x = 1\) into their derivative.
A1: For obtaining the correct exact value \(\frac{1}{2} - 2\ln 2\).

To find the stationary point, we differentiate \(y = \frac{e^{3x}}{2x + 1}\) using the quotient rule:

Let \(u = e^{3x} \implies \frac{du}{dx} = 3e^{3x}\)

Let \(v = 2x + 1 \implies \frac{dv}{dx} = 2\)

Using the quotient rule:

\[\frac{dy}{dx} = \frac{3e^{3x}(2x + 1) - 2e^{3x}}{(2x + 1)^2}\]

Simplifying the numerator:

\[\frac{dy}{dx} = \frac{e^{3x}[3(2x + 1) - 2]}{(2x + 1)^2} = \frac{e^{3x}(6x + 1)}{(2x + 1)^2}\]

For a stationary point, set \(\frac{dy}{dx} = 0\):

\[e^{3x}(6x + 1) = 0\]

Since \(e^{3x} \neq 0\) for all real \(x\), we solve:

\[6x + 1 = 0 \implies x = -\frac{1}{6}\]

Substitute \(x = -\frac{1}{6}\} back into the original equation to find the \)y\)-coordinate:

\[y = \frac{e^{3(-1/6)}}{2(-1/6) + 1} = \frac{e^{-1/2}}{-\frac{1}{3} + 1} = \frac{e^{-1/2}}{\frac{2}{3}} = \frac{3}{2\sqrt{e}}\]

Therefore, the exact coordinates of the stationary point are \(\left(-\frac{1}{6}, \frac{3}{2\sqrt{e}}\right)\).

M1: For attempting to differentiate using the quotient rule or product rule.
A1: For a correct derivative expression, e.g., \(\frac{dy}{dx} = \frac{e^{3x}(6x + 1)}{(2x + 1)^2}\).
M1: For setting their numerator to 0 and finding \(x = -\frac{1}{6}\).
A1: For finding the correct exact \(y\)-coordinate \(\frac{3}{2\sqrt{e}}\) (or \(\frac{3}{2}e^{-0.5}\)) and presenting the answer as coordinates.

(i) Let \( f(x) = e^{-x} - 2x^2 + 3 \).
\( f(1.2) = e^{-1.2} - 2(1.2)^2 + 3 \approx 0.3012 - 2.88 + 3 = 0.4212 > 0 \).
\( f(1.3) = e^{-1.3} - 2(1.3)^2 + 3 \approx 0.2725 - 3.38 + 3 = -0.1075 < 0 \).
Since \( f(x) \) is continuous and there is a change of sign between 1.2 and 1.3, the root \( \alpha \) must lie in the interval \( (1.2, 1.3) \).

(ii) Let \( x = \sqrt{1.5 + 0.5e^{-x}} \).
Squaring both sides: \( x^2 = 1.5 + 0.5e^{-x} \).
Multiplying by 2: \( 2x^2 = 3 + e^{-x} \).
Rearranging to match the original equation: \( e^{-x} = 2x^2 - 3 \).

(iii) Using \( x_{n+1} = \sqrt{1.5 + 0.5e^{-x_n}} \) with \( x_1 = 1.25 \):
\( x_2 = \sqrt{1.5 + 0.5e^{-1.25}} \approx 1.2819 \)
\( x_3 = \sqrt{1.5 + 0.5e^{-1.2819}} \approx 1.2801 \)
\( x_4 = \sqrt{1.5 + 0.5e^{-1.2801}} \approx 1.2802 \)
\( x_5 = \sqrt{1.5 + 0.5e^{-1.2802}} \approx 1.2802 \)
Since successive approximations converge to 1.2802, the root is \( \alpha = 1.28 \) correct to 2 decimal places.

(i) M1: For calculating the values of \( f(1.2) \) and \( f(1.3) \) with at least one correct calculation.
A1: For both values correct with a sign change noted and a clear conclusion.

(ii) M1: For squaring the iterative relation and attempting to rearrange it.
A1: For completing the steps correctly to obtain \( e^{-x} = 2x^2 - 3 \).

(iii) M1: For using the iterative formula at least twice to find successive values.
A1: For obtaining \( x_2 = 1.2819 \), \( x_3 = 1.2801 \), and \( x_4 = 1.2802 \) (or \( x_5 = 1.2802 \)).
A1: For the final answer of 1.28 with sufficient iterative evidence shown.

(i) Let \( f(x) = x^2 - \ln(x) - 4 \).
\( f(2.1) = 2.1^2 - \ln(2.1) - 4 \approx 4.41 - 0.7419 - 4 = -0.3319 < 0 \).
\( f(2.2) = 2.2^2 - \ln(2.2) - 4 \approx 4.84 - 0.7885 - 4 = 0.0515 > 0 \).
Since \( f(x) \) is continuous and there is a change of sign between 2.1 and 2.2, the root \( \alpha \) must lie in the interval \( (2.1, 2.2) \).

(ii) Let \( x = \sqrt{4 + \ln(x)} \).
Squaring both sides: \( x^2 = 4 + \ln(x) \).
Rearranging: \( x^2 - \ln(x) = 4 \), which is the equation of the intersection of the curve and the line.

(iii) Using \( x_{n+1} = \sqrt{4 + \ln(x_n)} \) with \( x_1 = 2.1 \):
\( x_2 = \sqrt{4 + \ln(2.1)} \approx 2.1776 \)
\( x_3 = \sqrt{4 + \ln(2.1776)} \approx 2.1859 \)
\( x_4 = \sqrt{4 + \ln(2.1859)} \approx 2.1868 \)
\( x_5 = \sqrt{4 + \ln(2.1868)} \approx 2.1869 \)
Since successive approximations converge to 2.1869, the root is \( \alpha = 2.19 \) correct to 2 decimal places.

(i) M1: For calculating the values of \( f(2.1) \) and \( f(2.2) \) with at least one correct calculation.
A1: For both values correct with a sign change noted and a clear conclusion.

(ii) M1: For squaring the iterative relation and attempting to rearrange it.
A1: For completing the steps correctly to obtain \( x^2 - \ln(x) = 4 \).

(iii) M1: For using the iterative formula at least twice to find successive values.
A1: For obtaining \( x_2 = 2.1776 \), \( x_3 = 2.1859 \), and \( x_4 = 2.1868 \) (or \( x_5 = 2.1869 \)).
A1: For the final answer of 2.19 with sufficient iterative evidence shown.

(a) Since \((x - 2)\) is a factor of \(P(x)\), by the Factor Theorem we have:
\(P(2) = 0 \Rightarrow 2(2)^3 + a(2)^2 + b(2) - 6 = 0\)
\(16 + 4a + 2b - 6 = 0 \Rightarrow 4a + 2b = -10 \Rightarrow 2a + b = -5\) (Equation 1)

By the Remainder Theorem, since dividing by \((x + 1)\) gives a remainder of \(-9\), we have:
\(P(-1) = -9 \Rightarrow 2(-1)^3 + a(-1)^2 + b(-1) - 6 = -9\)
\(-2 + a - b - 6 = -9 \Rightarrow a - b = -1\) (Equation 2)

Adding Equation 1 and Equation 2 gives:
\(3a = -6 \Rightarrow a = -2\)

Substituting \(a = -2\) into Equation 2:
\(-2 - b = -1 \Rightarrow b = -1\)

(b) Using \(a = -2\) and \(b = -1\), the polynomial is \(P(x) = 2x^3 - 2x^2 - x - 6\).
Since \((x - 2)\) is a factor, we can express \(P(x)\) as:
\(P(x) = (x - 2)(2x^2 + kx + 3)\)
Comparing coefficients of \(x^2\):
\(-4 + k = -2 \Rightarrow k = 2\)
So \(P(x) = (x - 2)(2x^2 + 2x + 3)\).
The quadratic factor \(2x^2 + 2x + 3\) has discriminant \(\Delta = 2^2 - 4(2)(3) = -20 < 0\), so it cannot be factorised further over the real numbers.

(c) The equation \(2e^{3z} - 2e^{2z} - e^z - 6 = 0\) can be rewritten by letting \(x = e^z\):
\(2x^3 - 2x^2 - x - 6 = 0 \Rightarrow P(x) = 0\)
From part (b), the only real solution to \(P(x) = 0\) is \(x = 2\).
Therefore, \(e^z = 2 \Rightarrow z = \ln 2\).

(d) We want to solve \(P(x) > (x - 2)(x^2 + 5)\).
Substituting the factorised form of \(P(x)\):
\((x - 2)(2x^2 + 2x + 3) > (x - 2)(x^2 + 5)\)
\((x - 2)(2x^2 + 2x + 3) - (x - 2)(x^2 + 5) > 0\)
\((x - 2)[(2x^2 + 2x + 3) - (x^2 + 5)] > 0\)
\((x - 2)(x^2 + 2x - 2) > 0\)

The roots of \(x^2 + 2x - 2 = 0\) are:
\(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-2)}}{2} = -1 \pm \sqrt{3}\)

The three critical values are \(-1 - \sqrt{3} \approx -2.73\), \(-1 + \sqrt{3} \approx 0.732\), and \(2\).
By testing intervals or sketching the cubic graph, we find the expression is positive when:
\(-1 - \sqrt{3} < x < -1 + \sqrt{3}\) or \(x > 2\).

(a)
- M1: Attempts to use the factor theorem with \(P(2) = 0\) and remainder theorem with \(P(-1) = -9\) to write two equations in terms of \(a\) and \(b\).
- A1: Obtains two correct equations, e.g., \(2a + b = -5\) and \(a - b = -1\).
- M1: Solves the simultaneous equations to find \(a\) and \(b\).
- A1: Obtains correct values \(a = -2\) and \(b = -1\).

(b)
- M1: Attempts to divide or use inspection to find the quadratic factor.
- A1: Obtains correct factorisation \((x - 2)(2x^2 + 2x + 3)\) and notes or shows that the quadratic has no real roots.

(c)
- M1: Recognises the substitution \(x = e^z\) and identifies that \(e^z = 2\).
- A1: Obtains exact answer \(z = \ln 2\).

(d)
- M1: Sets up the inequality \((x - 2)(x^2 + 2x - 2) > 0\) (or equivalent cubic inequality \(x^3 - 6x + 4 > 0\)).
- A1: Finds the roots of the quadratic factor, \(x = -1 \pm \sqrt{3}\).
- A1: Identifies the correct intervals \(-1 - \sqrt{3} < x < -1 + \sqrt{3}\) and \(x > 2\).

(a) First, find the derivative of \(x\) with respect to \(t\):
\(\frac{dx}{dt} = \frac{1}{t + 2}\)

Next, use the quotient rule to find the derivative of \(y\) with respect to \(t\):
\(\frac{dy}{dt} = \frac{\frac{d}{dt}(e^{2t}) \cdot (t + 1) - e^{2t} \cdot \frac{d}{dt}(t + 1)}{(t + 1)^2}\)
\(\frac{dy}{dt} = \frac{2e^{2t}(t + 1) - e^{2t}(1)}{(t + 1)^2}\)
\(\frac{dy}{dt} = \frac{e^{2t}(2t + 2 - 1)}{(t + 1)^2} = \frac{e^{2t}(2t + 1)}{(t + 1)^2}\)

Now use the chain rule to find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} =
\frac{\frac{e^{2t}(2t + 1)}{(t + 1)^2}}{\frac{1}{t + 2}} =
\frac{e^{2t}(2t + 1)(t + 2)}{(t + 1)^2}\) (as required).

(b) At a stationary point, \(\frac{dy}{dx} = 0\).
Since \(e^{2t} > 0\) for all real \(t\), this requires:
\((2t + 1)(t + 2) = 0\)
This gives \(t = -0.5\) or \(t = -2\).
Since the domain is \(t > -1\), the only valid value is \(t = -0.5\).

Substitute \(t = -0.5\) into the parametric equations:
\(x = \ln(-0.5 + 2) = \ln(1.5) = \ln\left(\frac{3}{2}\right)\)
\(y = \frac{e^{2(-0.5)}}{-0.5 + 1} = \frac{e^{-1}}{0.5} = \frac{2}{e}\)

Thus, the exact coordinates of the stationary point are \(\left(\ln\left(\frac{3}{2}\right), \frac{2}{e}\right)\).

(c) When \(t = 0\):
\(x = \ln(0 + 2) = \ln 2\)
\(y = \frac{e^0}{0 + 1} = 1\)

Calculate the gradient \(m\) of the tangent by substituting \(t = 0\) into \(\frac{dy}{dx}\):
\(m =
\frac{e^0(2(0) + 1)(0 + 2)}{(0 + 1)^2} =
\frac{1(1)(2)}{1} = 2\)

The equation of the tangent is given by:
\(y - y_1 = m(x - x_1)\)
\(y - 1 = 2(x - \ln 2)\)
\(y = 2x - 2\ln 2 + 1\)
Since \(2\ln 2 = \ln(2^2) = \ln 4\), we can write this as:
\(y = 2x + 1 - \ln 4\).

(a)
- B1: Correctly differentiates \(x\) to get \(\frac{dx}{dt} = \frac{1}{t+2}\).
- M1: Applies the quotient rule correctly to \(y\) with at least one term correctly differentiated.
- A1: Obtains correct simplified \(\frac{dy}{dt} = \frac{e^{2t}(2t + 1)}{(t + 1)^2}\).
- A1: Combines to show the given expression for \(\frac{dy}{dx}\) with clear intermediate steps.

(b)
- M1: Sets \(\frac{dy}{dx} = 0\) and solves for \(t\), choosing the correct value \(t = -0.5\) (and discarding \(t = -2\)).
- A1: Obtains exact \(x\)-coordinate \(\ln\left(\frac{3}{2}\right)\) (or \(\ln 1.5\)).
- A1: Obtains exact \(y\)-coordinate \(\frac{2}{e}\) (or \(2e^{-1}\)).

(c)
- B1: Finds correct coordinates at \(t = 0\), which are \(x = \ln 2\) and \(y = 1\).
- M1: Substitutes \(t = 0\) into the expression for \(\frac{dy}{dx}\) to find the gradient.
- A1: Obtains gradient \(m = 2\).
- A1: Formulates the equation of the tangent and simplifies to \(y = 2x + 1 - 2\ln 2\) (or \(y = 2x + 1 - \ln 4\)).