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To find the points of intersection, we equate the equations of the line and the curve:

\[kx - 3 = x^2 - 5x + 1\]

Rearranging into a standard quadratic form:

\[x^2 - (5 + k)x + 4 = 0\]

For the line and curve not to intersect, the quadratic equation must have no real roots. Therefore, its discriminant must be less than zero (\(b^2 - 4ac < 0\)):

\[[-(5+k)]^2 - 4(1)(4) < 0\]

\[(5+k)^2 - 16 < 0\]

\[(5+k)^2 < 16\]

Taking the square root on both sides:

\[-4 < 5 + k < 4\]

Subtracting 5 from all parts of the inequality:

\[-9 < k < -1\]

M1: Equating the curve and line equations and rearranging to obtain a 3-term quadratic equation in \(x\).
A1: Correct quadratic equation, e.g. \(x^2 - (5+k)x + 4 = 0\) (or equivalent).
M1: Applying the discriminant condition \(b^2 - 4ac < 0\).
A1.5: Correctly solving the inequality to obtain the range \(-9 < k < -1\).

(a) Since \(x > 0.5\), we have \(2x - 1 > 0\).
As a result, \(\frac{4}{2x - 1} > 0\).
Therefore, \(\mathrm{f}(x) = 3 - \frac{4}{2x - 1} < 3\).
So the range of \(\mathrm{f}\) is \(\mathrm{f}(x) < 3\).

(b) To find the inverse function, we set \(y = \mathrm{f}(x)\) and make \(x\) the subject:

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

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

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

\[2x = \frac{4}{3 - y} + 1\]

\[2x = \frac{4 + (3 - y)}{3 - y}\]

\[2x = \frac{7 - y}{3 - y}\]

\[x = \frac{7 - y}{2(3 - y)}\]

\[x = \frac{7 - y}{6 - 2y}\]

Thus, the inverse function is:

\[\mathrm{f}^{-1}(x) = \frac{7 - x}{6 - 2x}\]

The domain of \(\mathrm{f}^{-1}\) is the same as the range of \(\mathrm{f}\), which is \(x < 3\).

B1: Correct range \(\mathrm{f}(x) < 3\) (or \(y < 3\)) for part (a).
M1: Correct attempt to make \(x\) the subject of \(y = 3 - \frac{4}{2x-1}\).
A1: Correct intermediate step, e.g. \(2x - 1 = \frac{4}{3-y}\).
A1.5: Correct inverse function expression \(\mathrm{f}^{-1}(x) = \frac{7-x}{6-2x}\) (or equivalent) and correct domain \(x < 3\).

First, we find the coordinates of the midpoint of \(AB\), which lies on \(L_2\):

\[M = \left(\frac{2 + 8}{2}, \frac{5 + (-3)}{2}\right) = (5, 1)\]

Next, we calculate the gradient of the line \(L_1\) passing through \(A\) and \(B\):

\[m_{AB} = \frac{-3 - 5}{8 - 2} = \frac{-8}{6} = -\frac{4}{3}\]

Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_{\perp}\) satisfies \(m_{AB} \times m_{\perp} = -1\):

\[m_{\perp} = -\frac{1}{-4/3} = \frac{3}{4}\]

Now, we find the equation of \(L_2\) using the midpoint \((5, 1)\) and gradient \(\frac{3}{4}\):

\[y - 1 = \frac{3}{4}(x - 5)\]

Multiply both sides by 4 to eliminate the fraction:

\[4(y - 1) = 3(x - 5)\]

\[4y - 4 = 3x - 15\]

Rearranging to the requested form \(ay = bx + c\):

\[4y = 3x - 11\]

B1: Correct midpoint of \(AB\) at \((5, 1)\).
M1: Correct method to find the gradient of \(AB\) and obtain the perpendicular gradient.
A1: Correct perpendicular gradient \(m_{\perp} = \frac{3}{4}\).
A1.5: Correct equation in the specified integer form \(4y = 3x - 11\) (or any non-zero integer multiple).

We can write two equations based on the formulas for perimeter and area of a sector:

1. Perimeter \(P = 2r + r\theta = 30\)
2. Area \(A = \frac{1}{2}r^2\theta = 50\)

From the first equation, we can express \(r\theta\) as:

\[r\theta = 30 - 2r\]

Substitute this expression into the area equation \(\frac{1}{2}r(r\theta) = 50\):

\[\frac{1}{2}r(30 - 2r) = 50\]

Multiply both sides by 2:

\[r(30 - 2r) = 100\]

\[30r - 2r^2 = 100\]

Rearranging into standard quadratic form:

\[2r^2 - 30r + 100 = 0\]

Divide the equation by 2:

\[r^2 - 15r + 50 = 0\]

Factorize the quadratic equation:

\[(r - 5)(r - 10) = 0\]

This gives two possible values for the radius:

\[r = 5 \quad \text{or} \quad r = 10\]

Now, we find the corresponding values of \(\theta\) using \(\theta = \frac{30 - 2r}{r}\):

- If \(r = 5\):
\[\theta = \frac{30 - 2(5)}{5} = \frac{20}{5} = 4\text{ radians}\]

- If \(r = 10\):
\[\theta = \frac{30 - 2(10)}{10} = \frac{10}{10} = 1\text{ radian}\]

M1: Write down the correct equations for the perimeter and area of a sector.
A1: Eliminate \(\theta\) to obtain a correct quadratic equation in \(r\), e.g., \(r^2 - 15r + 50 = 0\).
A1: Solve the quadratic equation to find the two values \(r = 5\) and \(r = 10\).
A1.5: Calculate both correct corresponding values of \(\theta\): \(\theta = 4\) when \(r = 5\), and \(\theta = 1\) when \(r = 10\).

Using the trigonometric identity \(\cos^2 x = 1 - \sin^2 x\), we substitute this into the equation:

\[3(1 - \sin^2 x) - 5\sin x - 1 = 0\]

\[3 - 3\sin^2 x - 5\sin x - 1 = 0\]

\[-3\sin^2 x - 5\sin x + 2 = 0\]

Multiply by \(-1\) to get a positive leading coefficient:

\[3\sin^2 x + 5\sin x - 2 = 0\]

Factorize the quadratic expression:

\[(3\sin x - 1)(\sin x + 2) = 0\]

This yields two possible cases:

1. \(3\sin x - 1 = 0 \implies \sin x = \frac{1}{3}\)
2. \(\sin x + 2 = 0 \implies \sin x = -2\)

Since the range of the sine function is \([-1, 1]\), \(\sin x = -2\) has no solutions.

For \(\sin x = \frac{1}{3}\) in the interval \(0^\circ \le x \le 360^\circ\):

- The principal value is \(x = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ \approx 19.5^\circ\)
- The second solution is in the second quadrant: \(x = 180^\circ - 19.47^\circ = 160.53^\circ \approx 160.5^\circ\)

M1: Substitute \(\cos^2 x = 1 - \sin^2 x\) into the given equation.
A1: Formulate the correct quadratic equation \(3\sin^2 x + 5\sin x - 2 = 0\).
M1: Factorize or solve the quadratic equation to find \(\sin x = \frac{1}{3}\).
A1.5: Obtain the two solutions \(19.5^\circ\) and \(160.5^\circ\) (rounded to 1 decimal place), and reject \(\sin x = -2\).

Let \(a\) be the first term and \(d\) be the common difference of the arithmetic progression.

Using the formula for the \(n\)-th term, \(u_n = a + (n-1)d\):

\[u_3 = a + 2d = 11 \quad \text{--- (Equation 1)}\]

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

\[S_{10} = \frac{10}{2}[2a + 9d] = 210\]

\[5(2a + 9d) = 210\]

Divide both sides by 5:

\[2a + 9d = 42 \quad \text{--- (Equation 2)}\]

To solve the system of simultaneous equations, multiply Equation 1 by 2:

\[2a + 4d = 22 \quad \text{--- (Equation 3)}\]

Subtract Equation 3 from Equation 2:

\[(2a + 9d) - (2a + 4d) = 42 - 22\]

\[5d = 20 \implies d = 4\]

Substitute \(d = 4\) back into Equation 1:

\[a + 2(4) = 11 \implies a + 8 = 11 \implies a = 3\]

Thus, the first term is \(a = 3\) and the common difference is \(d = 4\).

M1: Write down the correct formula for the third term, \(a + 2d = 11\).
M1: Write down the correct equation for the sum of 10 terms, e.g., \(5(2a + 9d) = 210\).
A1: Eliminate one variable to obtain a single linear equation in terms of \(a\) or \(d\).
A1.5: Obtain both correct values: first term \(a = 3\) and common difference \(d = 4\).

To find the stationary points, we first find the first derivative of \(y\) with respect to \(x\) using the chain rule:

\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3(2x - 3)^2 \times 2 - 24 = 6(2x - 3)^2 - 24\]

At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):

\[6(2x - 3)^2 - 24 = 0\]

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

Taking the square root on both sides:

\[2x - 3 = 2 \quad \text{or} \quad 2x - 3 = -2\]

This gives two values of \(x\):
- If \(2x - 3 = 2 \implies 2x = 5 \implies x = 2.5\)
- If \(2x - 3 = -2 \implies 2x = 1 \implies x = 0.5\)

Now, we find the corresponding \(y\)-coordinates of these points:
- For \(x = 2.5\):
\[y = (2(2.5) - 3)^3 - 24(2.5) = (5 - 3)^3 - 60 = 2^3 - 60 = -52\]
So, one stationary point is \((2.5, -52)\).

- For \(x = 0.5\):
\[y = (2(0.5) - 3)^3 - 24(0.5) = (1 - 3)^3 - 12 = (-2)^3 - 12 = -20\]
So, the other stationary point is \((0.5, -20)\).

To determine the nature of these points, we calculate the second derivative:

\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(2x - 3) \times 2 = 24(2x - 3)\]

Now evaluate the second derivative at each of our \(x\)-values:
- At \(x = 2.5\):
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 24(2(2.5) - 3) = 24(2) = 48 > 0\]
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the point \((2.5, -52)\) is a **minimum** point.

- At \(x = 0.5\):
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 24(2(0.5) - 3) = 24(-2) = -48 < 0\]
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0\), the point \((0.5, -20)\) is a **maximum** point.

M1: Correct derivative of the function using the chain rule to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6(2x - 3)^2 - 24\).
A1: Set the derivative to 0 and find both correct values, \(x = 0.5\) and \(x = 2.5\).
A1: Calculate both correct \(y\)-coordinates, \(y = -20\) and \(y = -52\).
M1: Evaluate the second derivative (or first derivative sign test) at both \(x\)-coordinates.
A0.5: Correctly state that \((0.5, -20)\) is a maximum and \((2.5, -52)\) is a minimum.

(a) We substitute \( x = 2y - 6 \) from the line equation into the curve equation:
\( y^2 = 2(2y - 6) + 9 \)
\( y^2 = 4y - 12 + 9 \)
\( y^2 - 4y + 3 = 0 \)

Solving this quadratic equation by factoring:
\( (y-1)(y-3) = 0 \)
This gives \( y = 1 \) or \( y = 3 \).

Substitute these back to find \( x \):
For \( y = 1 \): \( x = 2(1) - 6 = -4 \)
For \( y = 3 \): \( x = 2(3) - 6 = 0 \)

Thus, the points of intersection are \( A(-4, 1) \) and \( B(0, 3) \).

(b) To find the perpendicular bisector of \( AB \):
First, find the midpoint \( M \) of \( AB \):
\( M = \left( \frac{-4 + 0}{2}, \frac{1 + 3}{2} \right) = (-2, 2) \)

Next, find the gradient of the line segment \( AB \):
\( m_{AB} = \frac{3 - 1}{0 - (-4)} = \frac{2}{4} = \frac{1}{2} \)

The gradient of the perpendicular bisector is the negative reciprocal:
\( m_{\perp} = -\frac{1}{1/2} = -2 \)

Now, use the point-slope form with midpoint \( M(-2, 2) \):
\( y - 2 = -2(x - (-2)) \)
\( y - 2 = -2x - 4 \)
\( y = -2x - 2 \)

Part (a):
M1: Substitute \( x = 2y - 6 \) or \( y = \frac{1}{2}x + 3 \) into the curve equation to obtain a quadratic in one variable.
A1: Solve the quadratic to find correct \( y \)-values (\( y = 1 \) and \( y = 3 \)) or \( x \)-values (\( x = -4 \) and \( x = 0 \)).
A1: Correct coordinates for both \( A \) and \( B \).

Part (b):
M1: Find the midpoint of \( AB \).
A1: Midpoint \( M(-2, 2) \).
M1: Find the gradient of \( AB \) and deduce the perpendicular gradient.
A1: Perpendicular gradient \( m = -2 \).
A1: State correct final equation in the required form: \( y = -2x - 2 \).

(a) Write the equation of the curve as:
\( y = 4x + 9(x-1)^{-1} \)

Differentiate with respect to \( x \):
\( \frac{dy}{dx} = 4 - 9(x-1)^{-2} = 4 - \frac{9}{(x-1)^2} \)

At the stationary point, \( \frac{dy}{dx} = 0 \):
\( 4 - \frac{9}{(x-1)^2} = 0 \Rightarrow (x-1)^2 = \frac{9}{4} \)

Since \( x > 1 \), we take the positive square root:
\( x - 1 = \frac{3}{2} \Rightarrow x = 2.5 \)

Substitute \( x = 2.5 \) into the original equation to find \( y \):
\( y = 4(2.5) + \frac{9}{2.5 - 1} = 10 + \frac{9}{1.5} = 10 + 6 = 16 \)

So the stationary point is \( (2.5, 16) \).

(b) Differentiate \( \frac{dy}{dx} \) again to find the second derivative:
\( \frac{d^2y}{dx^2} = 18(x-1)^{-3} = \frac{18}{(x-1)^3} \)

At \( x = 2.5 \):
\( \frac{d^2y}{dx^2} = \frac{18}{(1.5)^3} = \frac{18}{3.375} = \frac{16}{3} \)

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

(c) Using the chain rule for rates of change:
\( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

At \( x = 4 \):
\( \frac{dy}{dx} = 4 - \frac{9}{(4-1)^2} = 4 - \frac{9}{9} = 3 \)

Given \( \frac{dx}{dt} = 0.5 \):
\( \frac{dy}{dt} = 3 \times 0.5 = 1.5 \) units per second.

Part (a):
M1: Attempt to differentiate \( y \) (at least one term correct).
A1: Correct derivative \( \frac{dy}{dx} = 4 - \frac{9}{(x-1)^2} \).
M1: Set \( \frac{dy}{dx} = 0 \) and solve for \( x \) (rejecting negative root since \( x > 1 \)).
A1: Find correct coordinates \( (2.5, 16) \) (or equivalent fractions).

Part (b):
M1: Differentiate \( \frac{dy}{dx} \) to obtain \( \frac{d^2y}{dx^2} = \frac{k}{(x-1)^3} \).
A1: Evaluate at \( x = 2.5 \) and state that \( \frac{d^2y}{dx^2} > 0 \), hence it is a minimum.

Part (c):
M1: Apply the chain rule \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \).
A1: Calculate \( \frac{dy}{dx} = 3 \) at \( x = 4 \), leading to the correct rate of \( 1.5 \) units per second.

(a) Write the curve equation as \( y = (2x + 1)^{1/2} \).
Using the chain rule to differentiate:
\( \frac{dy}{dx} = \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x + 1}} \)

At the point \( P(4, 3) \):
\( \frac{dy}{dx} = \frac{1}{\sqrt{2(4) + 1}} = \frac{1}{3} \)

The gradient of the normal is the negative reciprocal:
\( m = -3 \)

The equation of the normal is:
\( y - 3 = -3(x - 4) \)
\( y = -3x + 15 \)

(b) Let's identify the boundaries of the region \( R \):
The curve \( y = \sqrt{2x + 1} \) intersects the x-axis when \( y = 0 \Rightarrow 2x + 1 = 0 \Rightarrow x = -0.5 \).
The normal \( y = -3x + 15 \) intersects the x-axis when \( y = 0 \Rightarrow -3x + 15 = 0 \Rightarrow x = 5 \).
The boundary changes at the intersection point \( P(4, 3) \).
Thus, the volume of revolution is split into two parts:
1) From \( x = -0.5 \) to \( x = 4 \), the boundary is the curve \( y = \sqrt{2x + 1} \).
2) From \( x = 4 \) to \( x = 5 \), the boundary is the normal line \( y = -3x + 15 \).

The formula for the volume of revolution about the x-axis is \( V = \pi \int y^2 \, dx \).

First part:
\( V_1 = \pi \int_{-0.5}^{4} (\sqrt{2x + 1})^2 \, dx = \pi \int_{-0.5}^{4} (2x + 1) \, dx \)
\( V_1 = \pi \left[ x^2 + x \right]_{-0.5}^{4} \)
\( V_1 = \pi \left[ (16 + 4) - (0.25 - 0.5) \right] = \pi [ 20 - (-0.25) ] = 20.25\pi \)

Second part:
\( V_2 = \pi \int_{4}^{5} (-3x + 15)^2 \, dx \)
Using substitution or direct integration:
\( V_2 = \pi \left[ \frac{(-3x + 15)^3}{3 \times (-3)} \right]_{4}^{5} = \pi \left[ -\frac{1}{9}(15 - 3x)^3 \right]_{4}^{5} \)
At \( x = 5 \): \( -\frac{1}{9}(0)^3 = 0 \)
At \( x = 4 \): \( -\frac{1}{9}(3)^3 = -3 \)
\( V_2 = \pi [ 0 - (-3) ] = 3\pi \)

Total volume:
\( V = V_1 + V_2 = 20.25\pi + 3\pi = 23.25\pi = \frac{93}{4}\pi \).

Part (a):
M1: Differentiate the curve using chain rule.
A1: Correct derivative \( \frac{dy}{dx} = \frac{1}{\sqrt{2x+1}} \).
M1: Find the gradient of the normal and write the equation of the normal.
A1: Correct equation of the normal: \( y = -3x + 15 \) (or equivalent).

Part (b):
M1: Identify the integration limits: \( x = -0.5 \) for the curve and \( x = 5 \) for the normal.
M1: Express total volume as the sum of two integrals (or one integral and one cone volume).
A1: Integrate \( (2x+1) \) correctly to get \( x^2 + x \), and evaluate to get \( 20.25\pi \).
A1: Integrate \( (-3x+15)^2 \) correctly (or use cone formula) to get \( 3\pi \).
A1: Add the two volumes to obtain the correct final answer: \( 23.25\pi \) or \( \frac{93}{4}\pi \).

(a) Complete the square for both \( x \) and \( y \):
\( (x - 2)^2 - 4 + (y - 3)^2 - 9 + 8 = 0 \)
\( (x - 2)^2 + (y - 3)^2 = 5 \)

Thus, the center of the circle is \( (2, 3) \) and the radius is \( \sqrt{5} \).

(b) To find the intersection with \( y = x \), substitute \( y = x \) into the circle's equation:
\( x^2 + x^2 - 4x - 6x + 8 = 0 \)
\( 2x^2 - 10x + 8 = 0 \)
\( x^2 - 5x + 4 = 0 \)

Factorizing this quadratic equation:
\( (x - 1)(x - 4) = 0 \)
This gives \( x = 1 \) or \( x = 4 \).

Since \( y = x \), the points of intersection are \( (1, 1) \) and \( (4, 4) \).

(c) Substitute \( x = 13 - 2y \) into the circle's equation \( (x-2)^2 + (y-3)^2 = 5 \):
\( (13 - 2y - 2)^2 + (y - 3)^2 = 5 \)
\( (11 - 2y)^2 + (y - 3)^2 = 5 \)
\( (121 - 44y + 4y^2) + (y^2 - 6y + 9) = 5 \)
\( 5y^2 - 50y + 130 = 5 \)
\( 5y^2 - 50y + 125 = 0 \)

Divide the entire equation by 5:
\( y^2 - 10y + 25 = 0 \)
\( (y - 5)^2 = 0 \)

Since this equation has a single repeated root \( y = 5 \), the line \( x + 2y = 13 \) is indeed a tangent to the circle.

Substitute \( y = 5 \) back to find \( x \):
\( x = 13 - 2(5) = 3 \)

Thus, the point of contact is \( (3, 5) \).

Part (a):
M1: Attempt to complete the square for \( x \) and \( y \).
A1: Center \( (2, 3) \) correct.
A1: Radius \( \sqrt{5} \) correct.

Part (b):
M1: Substitute \( y = x \) into the circle equation to obtain a 3-term quadratic in \( x \).
A1: Solve the quadratic to find \( x = 1, 4 \).
A1: State the coordinates \( (1, 1) \) and \( (4, 4) \).

Part (c):
M1: Substitute \( x = 13 - 2y \) into the circle equation and simplify to a quadratic equation.
A1: Show that there is a repeated root (discriminant = 0 or factored to perfect square \( (y-5)^2 = 0 \)), confirming it is a tangent.
A1: Find the correct coordinates of the point of contact: \( (3, 5) \).

(a) A stationary point occurs where \( \frac{dy}{dx} = 0 \). Since the stationary point is at \( (1, -2) \):
\( k\sqrt{1} - \frac{4}{\sqrt{1}} = 0 \)
\( k - 4 = 0 \Rightarrow k = 4 \)

To find the equation of the curve, integrate the gradient function with \( k = 4 \):
\( \frac{dy}{dx} = 4x^{1/2} - 4x^{-1/2} \)
\( y = \int (4x^{1/2} - 4x^{-1/2}) \, dx \)
\( y = 4 \cdot \frac{x^{3/2}}{3/2} - 4 \cdot \frac{x^{1/2}}{1/2} + C \)
\( y = \frac{8}{3}x^{3/2} - 8x^{1/2} + C \)

Since the curve passes through the point \( (1, -2) \):
\( -2 = \frac{8}{3}(1)^{3/2} - 8(1)^{1/2} + C \)
\( -2 = \frac{8}{3} - 8 + C \)
\( -2 = -\frac{16}{3} + C \)
\( C = \frac{10}{3} \)

Thus, the equation of the curve is:
\( y = \frac{8}{3}x^{3/2} - 8x^{1/2} + \frac{10}{3} \)

(b) To find where the gradient of the curve is 6:
\( 4\sqrt{x} - \frac{4}{\sqrt{x}} = 6 \)

Let \( u = \sqrt{x} \) (where \( u > 0 \)):
\( 4u - \frac{4}{u} = 6 \)
\( 4u^2 - 4 = 6u \)
\( 4u^2 - 6u - 4 = 0 \)
\( 2u^2 - 3u - 2 = 0 \)

Factorize the quadratic:
\( (2u + 1)(u - 2) = 0 \)

This gives \( u = -0.5 \) or \( u = 2 \).
Since \( u = \sqrt{x} \geq 0 \), we must have \( u = 2 \).

Thus:
\( \sqrt{x} = 2 \Rightarrow x = 4 \).

Part (a):
M1: Substitute \( x = 1 \) and set \( \frac{dy}{dx} = 0 \) to show \( k = 4 \).
M1: Integrate \( 4x^{1/2} - 4x^{-1/2} \) with at least one fractional index term integrated correctly.
A1: Correct integrated expression \( y = \frac{8}{3}x^{3/2} - 8x^{1/2} + C \).
M1: Substitute coordinates of \( (1, -2) \) to find the integration constant \( C \).
A1: State the final curve equation: \( y = \frac{8}{3}x^{3/2} - 8x^{1/2} + \frac{10}{3} \).

Part (b):
M1: Set \( \frac{dy}{dx} = 6 \).
M1: Form a quadratic equation in terms of \( \sqrt{x} \) (or make equivalent substitution).
A1: Solve the quadratic to find \( \sqrt{x} = 2 \) (and reject negative solution).
A1: Correctly find the x-coordinate: \( x = 4 \).

First, we rewrite the term \(2\ln(x)\) as \
\ln(x^2)\) using logarithmic properties. This gives:
\(\ln(2x + 3) - \ln(x^2) = \ln(3)\)

Next, combine the logarithms on the left-hand side:
\(\ln\left(\frac{2x + 3}{x^2}\right) = \ln(3)\)

Take the exponential of both sides to remove the logarithms:
\(\frac{2x + 3}{x^2} = 3\)

Multiply by \(x^2\) to form a quadratic equation:
\(2x + 3 = 3x^2 \implies 3x^2 - 2x - 3 = 0\)

Apply the quadratic formula to solve for \(x\):
\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-3)}}{2(3)} = \frac{2 \pm \sqrt{40}}{6} = \frac{1 \pm \sqrt{10}}{3}\)

This yields two possible values:
\(x \approx 1.387\) or \(x \approx -0.721\)

Since the term \(\ln(x)\) is only defined for \(x > 0\), we must reject the negative root \(x \approx -0.721\).

Therefore, the only valid solution is:
\(x = 1.39\) (correct to 3 significant figures).

M1: For expressing \(2\ln(x)\) as \(\ln(x^2)\) and combining the LHS into a single logarithm.
M1: For removing logarithms to obtain a polynomial equation in \(x\).
A1: For obtaining the correct quadratic equation \(3x^2 - 2x - 3 = 0\).
M1: For solving their 3-term quadratic equation to find two roots.
B1: For justifying the rejection of the negative root (e.g., stating that \(x > 0\) for \(\ln(x)\) to be defined).
A1.5: For the correct final answer \(1.39\) (rounded to 3 s.f.).

First, we use the double-angle trigonometric identity to express \(\cos^2(x)\) in an integrable form:
\(\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\)

Substitute this back into the integrand:
\(\int_{0}^{\frac{\pi}{6}} \left( \sin(2x) + \frac{1}{2} + \frac{1}{2}\cos(2x) \right) \text{d}x\)

Now, integrate term by term:
\(\int \sin(2x) \text{d}x = -\frac{1}{2}\cos(2x)\)
\(\int \frac{1}{2} \text{d}x = \frac{1}{2}x\)
\(\int \frac{1}{2}\cos(2x) \text{d}x = \frac{1}{4}\sin(2x)\)

This gives the antiderivative:
\(F(x) = -\frac{1}{2}\cos(2x) + \frac{1}{2}x + \frac{1}{4}\sin(2x)\)

Substitute the upper limit \(x = \frac{\pi}{6}\):
\(F\left(\frac{\pi}{6}\right) = -\frac{1}{2}\cos\left(\frac{\pi}{3}\right) + \frac{1}{2}\left(\frac{\pi}{6}\right) + \frac{1}{4}\sin\left(\frac{\pi}{3}\right)\)
\(F\left(\frac{\pi}{6}\right) = -\frac{1}{2}\left(\frac{1}{2}\right) + \frac{\pi}{12} + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) = -\frac{1}{4} + \frac{\pi}{12} + \frac{\sqrt{3}}{8}\)

Substitute the lower limit \(x = 0\):
\(F(0) = -\frac{1}{2}\cos(0) + 0 + \frac{1}{4}\sin(0) = -\frac{1}{2}(1) = -\frac{1}{2}\)

Subtract the value at the lower limit from the value at the upper limit:
\(F\left(\frac{\pi}{6}\right) - F(0) = \left( -\frac{1}{4} + \frac{\pi}{12} + \frac{\sqrt{3}}{8} \right) - \left( -\frac{1}{2} \right)\)
\(= \frac{1}{4} + \frac{\pi}{12} + \frac{\sqrt{3}}{8}\)

This can also be written under a common denominator as \(\frac{6 + 2\pi + 3\sqrt{3}}{24}\).

M1: For using the correct double-angle identity \(\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\).
A1: For obtaining the correct simplified integrand \(\sin(2x) + \frac{1}{2} + \frac{1}{2}\cos(2x)\).
B1: For integrating \(\sin(2x)\) to obtain \(-\frac{1}{2}\cos(2x)\).
B1: For integrating \(\frac{1}{2}\cos(2x)\) to obtain \(\frac{1}{4}\sin(2x)\).
M1: For substituting the limits \(\frac{\pi}{6}\) and \(0\) into their integrated expression.
A1.5: For obtaining the exact value \(\frac{1}{4} + \frac{\pi}{12} + \frac{\sqrt{3}}{8}\) or equivalent.

First, take the natural logarithm of both sides of the equation \(y = A e^{k x^2}\):
\(\ln(y) = \ln\left(A e^{k x^2}\right)\)
\(\ln(y) = \ln(A) + \ln\left(e^{k x^2}\right)\)
\(\ln(y) = \ln(A) + k x^2\)

Letting \(Y = \ln(y)\) and \(X = x^2\), this equation can be written as \(Y = k X + \ln(A)\), which is the equation of a straight line with gradient \(k\) and vertical intercept \(\ln(A)\).

We are given two points on the line, which correspond to coordinates \((X, Y)\):
\((X_1, Y_1) = (2, 5.8)\)
\((X_2, Y_2) = (6, 11.4)\)

Calculate the gradient \(k\):
\(k = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{11.4 - 5.8}{6 - 2} = \frac{5.6}{4} = 1.4\)

To find the vertical intercept \(\ln(A)\), substitute \(k = 1.4\) and the point \((2, 5.8)\) into the linear equation:
\(5.8 = 1.4(2) + \ln(A)\)
\(5.8 = 2.8 + \ln(A)\)
\(\ln(A) = 3.0\)

Now solve for \(A\):
\(A = e^{3.0} \approx 20.0855\)

Rounding to 3 significant figures, we get:
\(A = 20.1\)

Thus, the values of the constants are \(k = 1.4\) and \(A = 20.1\).

M1: For taking natural logarithms of both sides to obtain \(\ln(y) = \ln(A) + k x^2\).
M1: For recognizing that \(k\) represents the gradient of the graph of \(\ln(y)\) against \(x^2\).
A1.5: For calculating \(k = 1.4\) correctly.
M1: For substituting one of the coordinates and their value of \(k\) to find \(\ln(A)\).
A2: For calculating \(A = e^3 \approx 20.1\) (1 mark for \(A = e^3\), 1 mark for rounding to 3 s.f.).

To integrate \(\frac{2x + 3}{2x + 1}\), we first write the numerator in terms of the denominator using algebraic division or inspection:
\(\frac{2x + 3}{2x + 1} = \frac{(2x + 1) + 2}{2x + 1} = \frac{2x + 1}{2x + 1} + \frac{2}{2x + 1} = 1 + \frac{2}{2x + 1}\)

Now, integrate each term with respect to \(x\):
\(\int \left( 1 + \frac{2}{2x + 1} \right) \text{d}x = x + \ln|2x + 1|\)

Evaluate the definite integral using the limits from \(1\) to \(3\):
\(\left[ x + \ln(2x + 1) \right]_1^3 = (3 + \ln(2(3) + 1)) - (1 + \ln(2(1) + 1))\)
\(= (3 + \ln(7)) - (1 + \ln(3))\)

Subtract the terms:
\(= (3 - 1) + (\ln(7) - \ln(3))\)
\(= 2 + \ln\left(\frac{7}{3}\right)\)

This is in the form \(a + \ln(b)\) where \(a = 2\) and \(b = \frac{7}{3}\).

M1.5: For rewriting the algebraic fraction using division or inspection.
A1: For obtaining the correct rewritten expression \(1 + \frac{2}{2x + 1}\).
M1.5: For integrating to obtain \(x + \ln(2x + 1)\) (applying the reverse chain rule correctly).
M1: For substituting limits \(3\) and \(1\) into their integrated expression.
M1: For applying logarithm laws to combine the terms into the form \(a + \ln(b)\).
A0.5: For the correct final exact answer \(2 + \ln\left(\frac{7}{3}\right)\).

(i) We let \(7\cos\theta + 24\sin\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha\).
Comparing coefficients:
\(R\cos\alpha = 7\) and \(R\sin\alpha = 24\).
Squaring and adding these equations:
\(R^2 = 7^2 + 24^2 = 49 + 576 = 625 \implies R = 25\).
Dividing the equations:
\(\tan\alpha = \frac{24}{7} \implies \alpha = \tan^{-1}\left(\frac{24}{7}\right) \approx 1.287\) radians (correct to 3 decimal places).
So, \(7\cos\theta + 24\sin\theta = 25\cos(\theta - 1.287)\).

(ii)(a) The expression is \(25\cos(\theta - 1.287) + 15\).
The maximum value of the cosine function is 1, so the maximum value of the expression is \(25(1) + 15 = 40\).
This maximum occurs when \(\cos(\theta - 1.287) = 1 \implies \theta - 1.287 = 0 \implies \theta = 1.287 \approx 1.29\) radians.

(ii)(b) We solve \(25\cos(\theta - 1.287) = 12.5\):
\(\cos(\theta - 1.287) = 0.5\).
Taking the inverse cosine:
\(\theta - 1.287 = \pm \frac{\pi}{3} \approx \pm 1.0472\).

Case 1:
\(\theta - 1.287 = 1.0472 \implies \theta = 2.3342 \approx 2.33\) radians.

Case 2:
\(\theta - 1.287 = -1.0472 \implies \theta = 0.2402 \approx 0.240\) radians.

Both values are in the range \(0 < \theta < 2\pi\).

(i)
M1: For attempting to find \(R = \sqrt{7^2 + 24^2}\).
A1: For obtaining \(R = 25\).
A1: For obtaining \(\alpha \approx 1.287\).

(ii)(a)
B1: For stating maximum value is 40.
B1: For finding \(\theta \approx 1.29\) (or 1.287).

(ii)(b)
M1: For setting \(25\cos(\theta - 1.287) = 12.5\) and finding at least one correct angle for \(\theta - 1.287\).
A1: For \(\theta \approx 0.240\) (or 0.24).
A1: For \(\theta \approx 2.33\).

(i) We use the product rule to differentiate \(y = e^{-x} \sin x\):
\(\frac{dy}{dx} = \frac{d}{dx}(e^{-x}) \cdot \sin x + e^{-x} \cdot \frac{d}{dx}(\sin x)\)
\(\frac{dy}{dx} = -e^{-x}\sin x + e^{-x}\cos x = e^{-x}(\cos x - \sin x)\).
At a stationary point, \(\frac{dy}{dx} = 0 \implies e^{-x}(\cos x - \sin x) = 0\).
Since \(e^{-x} \neq 0\) for all real \(x\), we must have:
\(\cos x - \sin x = 0 \implies \sin x = \cos x\).
Dividing by \(\cos x\) (valid since \(\cos x \neq 0\) when \(\sin x = \cos x\)), we get \(\tan x = 1\).

(ii) Within the interval \(0 \le x \le \pi\), the equation \(\tan x = 1\) has the unique solution:
\(x = \frac{\pi}{4}\).
Substituting \(x = \frac{\pi}{4}\) back into the curve's equation:
\(y = e^{-\pi/4} \sin\left(\frac{\pi}{4}\right) = e^{-\pi/4} \left(\frac{\sqrt{2}}{2}\right)\).
Thus, the exact coordinates of the stationary point are \(\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}e^{-\pi/4}\right)\).

(iii) We find the second derivative using the product rule on \(\frac{dy}{dx} = e^{-x}(\cos x - \sin x)\):
\(\frac{d^2y}{dx^2} = -e^{-x}(\cos x - \sin x) + e^{-x}(-\sin x - \cos x)\)
\(\frac{d^2y}{dx^2} = e^{-x}(-\cos x + \sin x - \sin x - \cos x) = -2e^{-x}\cos x\).
Substituting \(x = \frac{\pi}{4}\):
\(\frac{d^2y}{dx^2} = -2e^{-\pi/4}\cos\left(\frac{\pi}{4}\right) = -2e^{-\pi/4}\left(\frac{\sqrt{2}}{2}\right) = -\sqrt{2}e^{-\pi/4}\).
Since \(e^{-\pi/4} > 0\), we have \(-\sqrt{2}e^{-\pi/4} < 0\).
Since the second derivative is negative, the stationary point is a local maximum.

(i)
M1: For attempting product rule to find \(\frac{dy}{dx}\).
A1: For correct derivative \(\frac{dy}{dx} = e^{-x}(\cos x - \sin x)\).
A1: For equating to zero and clearly showing the step leading to \(\tan x = 1\).

(ii)
B1: For finding \(x = \frac{\pi}{4}\).
B1: For obtaining the exact \(y\)-coordinate \(\frac{\sqrt{2}}{2}e^{-\pi/4}\) (or \(\frac{1}{\sqrt{2}}e^{-\pi/4}\)).

(iii)
M1: For differentiating \(\frac{dy}{dx}\) using product rule.
A1: For obtaining the correct simplified form \(\frac{d^2y}{dx^2} = -2e^{-x}\cos x\).
A1: For evaluating at \(x = \frac{\pi}{4}\) to show it is negative (\(-\sqrt{2}e^{-\pi/4}\)) and concluding it is a maximum.

(i) Differentiating \(x\) and \(y\) with respect to \(\theta\):
\(\frac{dx}{d\theta} = 2 - 2\cos 2\theta\).
Using the identity \(\cos 2\theta = 1 - 2\sin^2\theta\):
\(\frac{dx}{d\theta} = 2 - 2(1 - 2\sin^2\theta) = 4\sin^2\theta\).

Next, differentiating \(y\):
\(\frac{dy}{d\theta} = -4\sin\theta\).

Using the chain rule to find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-4\sin\theta}{4\sin^2\theta} = -\frac{1}{\sin\theta} = -\csc\theta\).

(ii) At \(\theta = \frac{\pi}{3}\):
\(x = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\).
\(y = 4\cos\left(\frac{\pi}{3}\right) = 4(0.5) = 2\).

The gradient of the tangent is:
\(m = -\csc\left(\frac{\pi}{3}\right) = -\frac{1}{\sin(\pi/3)} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\).

The equation of the tangent is:
\(y - y_1 = m(x - x_1)\)
\(y - 2 = -\frac{2\sqrt{3}}{3}\left(x - \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)\right)\)
\(y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + 1\)
\(y + \frac{2\sqrt{3}}{3}x = 3 + \frac{4\sqrt{3}\pi}{9}\).

(i)
B1: For finding \(\frac{dy}{d\theta} = -4\sin\theta\).
M1: For attempting to find \(\frac{dx}{d\theta}\) and applying the double-angle identity for \(\cos 2\theta\).
A1: For obtaining \(\frac{dx}{d\theta} = 4\sin^2\theta\).
A1: For correctly completing the division to show \(\frac{dy}{dx} = -\csc\theta\).

(ii)
B1: For finding exact coordinates \(x = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\) and \(y = 2\).
B1: For finding exact gradient \(m = -\frac{2\sqrt{3}}{3}\).
M1: For substituting their coordinates and gradient into the line equation \(y - y_1 = m(x - x_1)\).
A1: For obtaining \(y + \frac{2\sqrt{3}}{3}x = 3 + \frac{4\sqrt{3}\pi}{9}\) or equivalent exact form.