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We use the product rule \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\).
Let \(u = (3x-5)^4\) and \(v = 2x+1\).
Then \(\frac{du}{dx} = 4(3x-5)^3 \times 3 = 12(3x-5)^3\) and \(\frac{dv}{dx} = 2\).
So, \(\frac{dy}{dx} = (3x-5)^4(2) + (2x+1)(12(3x-5)^3)\).
Substituting \(x = 2\):
\(\frac{dy}{dx} = (3(2)-5)^4(2) + (2(2)+1)(12(3(2)-5)^3)\)
\(\frac{dy}{dx} = (1)^4(2) + (5)(12(1)^3) = 2 + 60 = 62\).

M1: For applying the product rule correctly, including chain rule on \((3x-5)^4\).
A1: For the correct derivative expression \(2(3x-5)^4 + 12(2x+1)(3x-5)^3\).
A0.5: For substituting \(x = 2\) to get 62.

Using the laws of logarithms:
\(\log_3\left(\frac{x + 4}{x - 2}\right) = 2\).
Converting to exponential form:
\(\frac{x + 4}{x - 2} = 3^2\)
\(\frac{x + 4}{x - 2} = 9\)
\(x + 4 = 9(x - 2)\)
\(x + 4 = 9x - 18\)
\(8x = 22\)
\(x = \frac{22}{8} = \frac{11}{4}\).

M1: For applying the subtraction law of logarithms to combine terms.
M1: For rewriting the logarithmic equation in exponential form \(3^2 = 9\).
A0.5: For solving the linear equation to get \(x = \frac{11}{4}\) (or 2.75).

Since the terms are in geometric progression, the common ratio \(r\) is constant:
\(r = \frac{k}{k+2} = \frac{k-1}{k}\).
Cross-multiplying gives:
\(k^2 = (k-1)(k+2)\)
\(k^2 = k^2 + k - 2\)
\(k = 2\).
Using \(k = 2\), the first term is \(a = 2 + 2 = 4\), and the common ratio is \(r = \frac{2}{4} = 0.5\).
The sum to infinity is:
\(S_\infty = \frac{a}{1-r} = \frac{4}{1 - 0.5} = 8\).

M1: For setting up the ratio equation and solving for \(k = 2\).
M1: For finding the first term \(a = 4\) and the common ratio \(r = 0.5\).
A0.5: For correctly calculating \(S_\infty = 8\).

Since the vectors are parallel, their corresponding components are proportional:
\(\frac{3}{p-1} = \frac{p}{4}\)
Cross-multiplying:
\(12 = p(p-1)\)
\(p^2 - p - 12 = 0\)
Factoring the quadratic equation:
\((p-4)(p+3) = 0\)
This gives \(p = 4\) or \(p = -3\).
Since \(p\) must be a positive constant, we reject \(p = -3\). Therefore, \(p = 4\).

M1: For equating the ratio of components to establish a quadratic equation in \(p\).
M1: For solving the quadratic equation to find the roots \(4\) and \(-3\).
A0.5: For selecting \(p = 4\) based on the condition that \(p\) is positive.

For the 4-digit number to be greater than 7000, the first digit must be 7, 8, or 9. This gives 3 choices for the first digit.
Since no digits are repeated, the remaining 3 positions must be filled from the remaining 5 digits.
The number of ways to arrange 3 digits from 5 is \(5 \times 4 \times 3 = 60\).
Therefore, the total number of valid 4-digit numbers is \(3 \times 60 = 180\).

M1: For identifying that the first digit has 3 possible choices (7, 8, or 9).
M1: For calculating the arrangements for the remaining 3 positions from the 5 remaining digits (e.g., \(P(5, 3) = 60\)).
A0.5: For multiplying the choices to get 180.

First, we find the value of \(\mathrm{f}(0)\):
\(\mathrm{f}(0) = \mathrm{e}^{2(0)} - 3 = \mathrm{e}^0 - 3 = 1 - 3 = -2\).
Next, we substitute this value into \(\mathrm{g}(x)\):
\(\mathrm{g}\mathrm{f}(0) = \mathrm{g}(-2) = \ln(-2 + 4) = \ln(2)\).

M1: For correctly evaluating \(\mathrm{f}(0) = -2\).
M1: For substituting their value of \(\mathrm{f}(0)\) into \(\mathrm{g}(x)\).
A0.5: For the correct exact value of \(\ln(2)\) (or equivalent log form).

To find the equation of the curve, we integrate \(\frac{dy}{dx}\) with respect to \(x\):

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

Using the integration rule \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\):

\(y = 3\left(\frac{x^{3/2}}{3/2}\right) - 4\left(\frac{x^{1/2}}{1/2}\right) + C\)
\(y = 2x^{3/2} - 8x^{1/2} + C\)

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

\(5 = 2(4)^{3/2} - 8(4)^{1/2} + C\)
\(5 = 2(8) - 8(2) + C\)
\(5 = 16 - 16 + C \implies C = 5\)

Thus, the equation of the curve is:

\(y = 2x^{3/2} - 8x^{1/2} + 5\)

- **M1**: Attempt to integrate, with at least one fractional power increased by 1.
- **A1**: Correct term \(2x^{3/2}\).
- **A1**: Correct term \(-8x^{1/2}\).
- **M1**: Substitute \(x=4\) and \(y=5\) into an integrated expression containing a constant of integration \(C\) to solve for \(C\).
- **A0.5**: Correct final equation \(y = 2x^{3/2} - 8x^{1/2} + 5\) (or equivalent, such as \(y = 2x\sqrt{x} - 8\sqrt{x} + 5\)).

Since the points \(A\), \(B\), and \(C\) lie on a straight line, the vectors \(\vec{AB}\) and \(\vec{BC}\) are parallel (collinear).

First, find \(\vec{AB}\):
\(\vec{AB} = \vec{OB} - \vec{OA} = (4\mathbf{i} + \mathbf{j}) - (\mathbf{i} + 3\mathbf{j}) = 3\mathbf{i} - 2\mathbf{j}\)

Next, find \(\vec{BC}\):
\(\vec{BC} = \vec{OC} - \vec{OB} = (p\mathbf{i} - 3\mathbf{j}) - (4\mathbf{i} + \mathbf{j}) = (p - 4)\mathbf{i} - 4\mathbf{j}\)

Since \(\vec{BC}\) is parallel to \(\vec{AB}\), there exists a scalar \(k\) such that:
\(\vec{BC} = k\vec{AB}\)
\((p - 4)\mathbf{i} - 4\mathbf{j} = k(3\mathbf{i} - 2\mathbf{j})\)

Comparing the \(\mathbf{j}\) components:
\(-4 = -2k \implies k = 2\)

Comparing the \(\mathbf{i}\) components:
\(p - 4 = 3k \implies p - 4 = 3(2) \implies p - 4 = 6 \implies p = 10\)

- **M1**: Method for finding vector \(\vec{AB}\) or \(\vec{AC}\) or \(\vec{BC}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\).
- **A1**: Correct vector \(\vec{AB} = 3\mathbf{i} - 2\mathbf{j}\) (or equivalent direction vector).
- **M1**: Setting up a scalar relationship for collinearity, e.g., \(\vec{BC} = k\vec{AB}\) or \(\frac{p-4}{3} = \frac{-4}{-2}\).
- **A1**: Finding the correct scale factor \(k = 2\) (or equivalent ratio equation).
- **A0.5**: Correct final value of \(p = 10\).

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

The second term is given by:
\(ar = 12 \implies a = \frac{12}{r}\) --- (Equation 1)

The sum to infinity is given by:
\(S_{\infty} = \frac{a}{1-r} = 64 \implies a = 64(1 - r)\) --- (Equation 2)

Equating the two expressions for \(a\):
\(\frac{12}{r} = 64(1 - r)\)
\(12 = 64r(1 - r)\)
\(12 = 64r - 64r^2\)
\(64r^2 - 64r + 12 = 0\)

Divide the entire equation by 4:
\(16r^2 - 16r + 3 = 0\)

Factorize the quadratic equation:
\((4r - 3)(4r - 1) = 0\)

This gives two possible values for \(r\):
\(r = \frac{3}{4}\) or \(r = \frac{1}{4}\)

Since we are given that \(r > 0.5\), we choose:
\(r = \frac{3}{4}\)

Now substitute \(r = \frac{3}{4}\) into Equation 1 to find \(a\):
\(a = \frac{12}{3/4} = 12 \times \frac{4}{3} = 16\)

Thus, the first term \(a = 16\) and the common ratio \(r = \frac{3}{4}\) (or \(0.75\)).

- **M1**: Write down the correct formulas for the second term (\(ar = 12\)) and sum to infinity (\(\frac{a}{1-r} = 64\)).
- **M1**: Eliminate \(a\) to form a quadratic equation in \(r\) (e.g., \(16r^2 - 16r + 3 = 0\)).
- **A1**: Solve the quadratic equation correctly to find \(r = \frac{1}{4}\) and \(r = \frac{3}{4}\).
- **A1**: Select the correct value \(r = \frac{3}{4}\) using the condition \(r > 0.5\) and substitute to find \(a = 16\).
- **A0.5**: Correct final values: \(a = 16\) and \(r = \frac{3}{4}\).

**(i) Find an expression for \(\mathrm{f}^{-1}(x)\):**
Let \(y = \mathrm{f}(x)\):
\(y = 3 + \sqrt{2x - 5}\)

Rearrange to make \(x\) the subject:
\(y - 3 = \sqrt{2x - 5}\)

Square both sides:
\((y - 3)^2 = 2x - 5\)

Add 5 to both sides:
\(2x = (y - 3)^2 + 5\)

Divide by 2:
\(x = \frac{1}{2}(y - 3)^2 + \frac{5}{2}\)

Therefore, the inverse function is:
\(\mathrm{f}^{-1}(x) = \frac{1}{2}(x - 3)^2 + \frac{5}{2}\) (or \(\frac{(x-3)^2 + 5}{2}\))

**(ii) Find the domain of \(\mathrm{f}^{-1}\):**
The domain of the inverse function \(\mathrm{f}^{-1}\) is the range of the original function \(\mathrm{f\)}.
Since the domain of \(\mathrm{f\)} is \(x \ge 2.5\), the minimum value of \(\sqrt{2x - 5}\) is \(0\) (which occurs when \(x = 2.5\)).
Therefore, \(\mathrm{f}(x) \ge 3 + 0 \implies \mathrm{f}(x) \ge 3\).
So, the range of \(\mathrm{f}\) is \([3, \infty)\), which means the domain of \(\mathrm{f}^{-1}\) is \(x \ge 3\).

- **M1**: Equate \(\mathrm{f}(x)\) to \(y\), subtract \(3\) from both sides, and square to eliminate the square root.
- **A1**: Correctly obtain \((y-3)^2 = 2x - 5\).
- **A1**: Make \(x\) the subject and write the inverse function as \(\mathrm{f}^{-1}(x) = \frac{1}{2}(x - 3)^2 + \frac{5}{2}\) or equivalent form.
- **M1**: Identify that the domain of \(\mathrm{f}^{-1}\) is equivalent to the range of \(\mathrm{f}\).
- **A0.5**: Correctly state the domain as \(x \ge 3\) (or \([3, \infty)\)).

First, rearrange the equation to group the logarithmic terms together on one side:
\(\log_2(x + 1) + \log_2(x - 1) - \log_2(x - 2) = 3\)

Apply the laws of logarithms:
\(\log_b(A) + \log_b(B) = \log_b(AB)\) and \(\log_b(A) - \log_b(C) = \log_b\left(\frac{A}{C}\right)\)

This gives:
\(\log_2\left( \frac{(x + 1)(x - 1)}{x - 2} \right) = 3\)

Convert the logarithmic equation to exponential form:
\(\frac{(x + 1)(x - 1)}{x - 2} = 2^3\)
\(\frac{x^2 - 1}{x - 2} = 8\)

Multiply both sides by \(x - 2\):
\(x^2 - 1 = 8(x - 2)\)
\(x^2 - 1 = 8x - 16\)

Rearrange into a quadratic equation form:
\(x^2 - 8x + 15 = 0\)

Factorize the quadratic equation:
\((x - 3)(x - 5) = 0\)

So, the potential solutions are:
\(x = 3\) or \(x = 5\)

We must check if these values lie in the domain of the original logarithmic terms. The logarithmic arguments require:
\(x + 1 > 0 \implies x > -1\)
\(x - 2 > 0 \implies x > 2\)
\(x - 1 > 0 \implies x > 1\)

Thus, the overall domain requirement is \(x > 2\).
Since both \(x = 3\) and \(x = 5\) are greater than \(2\), both are valid solutions.

- **M1**: Rearrange the equation and apply logarithmic addition/subtraction laws to combine terms into a single logarithm.
- **A1**: Correctly convert from log form to exponential form: \(\frac{x^2 - 1}{x - 2} = 8\).
- **M1**: Expand and rearrange to form a standard quadratic equation \(x^2 - 8x + 15 = 0\).
- **A1**: Solve the quadratic equation to get \(x = 3\) and \(x = 5\).
- **A0.5**: Verify that both solutions satisfy the domain restriction \(x > 2\).

To find the area under the curve, we calculate the definite integral:

\(\int_{0}^{3} 12(2x+3)^{-2} \mathrm{d}x\)

We integrate using the reverse chain rule:

\(\int 12(2x+3)^{-2} \mathrm{d}x = \frac{12}{-1 \times 2} (2x+3)^{-1} = -6(2x+3)^{-1} = -\frac{6}{2x+3}\)

Now, we evaluate this integral from \(x = 0\) to \(x = 3\):

\(\left[ -\frac{6}{2x+3} \right]_{0}^{3} = \left( -\frac{6}{2(3)+3} \right) - \left( -\frac{6}{2(0)+3} \right)\)

\(= \left( -\frac{6}{9} \right) - \left( -\frac{6}{3} \right)\)

\(= -\frac{2}{3} - (-2)\)

\(= -\frac{2}{3} + 2 = \frac{4}{3}\)

Thus, the exact area of the region is \(\frac{4}{3}\) (or \(1\frac{1}{3}\)).

M1: For an attempt to integrate of the form \(k(2x+3)^{-1}\)
A1: For the correct integrated term \(-6(2x+3)^{-1}\)
M1: For the correct application of limits \(0\) and \(3\) into their integrated expression
A1: For evaluating both limits correctly to obtain \(-\frac{2}{3}\) and \(-2\)
A1: For the final exact value of \(\frac{4}{3}\) or \(1\frac{1}{3}\)

We rewrite the equation as \(y = (x-2)(2x+5)^{\frac{1}{2}}\).

Using the product rule for differentiation, \(\frac{\mathrm{d}}{\mathrm{d}x}[uv] = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\):
Let \(u = x-2\) and \(v = (2x+5)^{\frac{1}{2}}\).
Then \( \frac{\mathrm{d}u}{\mathrm{d}x} = 1 \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(2x+5)^{-\frac{1}{2}} \times 2 = \frac{1}{\sqrt{2x+5}} \).

Now, substitute into the product rule formula:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = (x-2)\left( \frac{1}{\sqrt{2x+5}} \right) + \sqrt{2x+5}\)

To find the stationary point, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(\frac{x-2}{\sqrt{2x+5}} + \sqrt{2x+5} = 0\)

Multiply through by \(\sqrt{2x+5}\):
\(x - 2 + (2x + 5) = 0\)
\(3x + 3 = 0 \implies x = -1\)

Now find the corresponding \(y\)-coordinate by substituting \(x = -1\) into the original curve equation:
\(y = (-1 - 2)\sqrt{2(-1) + 5}\)
\(y = -3\sqrt{3}\)

Thus, the exact coordinates of the stationary point are \((-1, -3\sqrt{3})\).

M1: For applying the product rule to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\), with at least one term correct
A1: For a fully correct derivative \(\frac{x-2}{\sqrt{2x+5}} + \sqrt{2x+5}\) (or equivalent)
M1: For setting their derivative to 0 and attempting to clear the fraction
A1: For finding \(x = -1\)
M1: For substituting their \(x = -1\) into the original equation to find \(y\)
A1: For the correct exact coordinates \((-1, -3\sqrt{3})\)

Let the first term be \(a\) and the common ratio be \(r\), where \(a > 0\) and \(0 < r < 1\).

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

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

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

Since the terms of the progression are positive, \(r\) must be positive:
\(r = \sqrt{\frac{4}{9}} = \frac{2}{3}\)

Now substitute \(r = \frac{2}{3}\) back into the expression for \(a\):
\(a = 27\left(1 - \frac{2}{3}\right) = 27\left(\frac{1}{3}\right) = 9\)

Thus, the first term is \(a = 9\) and the common ratio is \(r = \frac{2}{3}\).

B1: For setting up the sum of the first two terms: \(a(1+r) = 15\)
B1: For setting up the sum to infinity: \(\frac{a}{1-r} = 27\)
M1: For eliminating one variable to obtain a single equation in terms of \(r\) (or \(a\))
A1: For obtaining \(r^2 = \frac{4}{9}\) (or equivalent simplified quadratic)
A1: For solving to find \(r = \frac{2}{3}\) (explaining or showing rejection of negative root)
A1: For solving to find \(a = 9\)

(a) We first complete the square for \(\mathrm{f}(x)\):
\(\mathrm{f}(x) = 2(x^2 - 6x) + 13 = 2\left[(x-3)^2 - 9\right] + 13 = 2(x-3)^2 - 18 + 13 = 2(x-3)^2 - 5\)

The vertex of the quadratic curve is at \((3, -5)\). For the function to have an inverse, it must be a one-to-one function. This requires the domain \(x \ge k\) to be restricted to one side of the vertex. Thus, the smallest value of \(k\) is the \(x\)-coordinate of the vertex, which is \(k = 3\).

(b) Set \(y = 2(x-3)^2 - 5\) for \(x \ge 3\). Rearrange to make \(x\) the subject:
\(y + 5 = 2(x-3)^2\)
\(\frac{y+5}{2} = (x-3)^2\)

Since \(x \ge 3\), we take the positive square root:
\(x - 3 = \sqrt{\frac{y+5}{2}}\)
\(x = 3 + \sqrt{\frac{y+5}{2}}\)

Thus, the inverse function is:
\(\mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\)

The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\). Since the minimum value of \(\mathrm{f}(x)\) is \(-5\) at \(x=3\), the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \ge -5\). Therefore, the domain of \(\mathrm{f}^{-1}(x)\) is \(x \ge -5\).

M1: For attempting to complete the square on \(2x^2 - 12x + 13\)
A1: For achieving \(2(x-3)^2 - 5\)
B1: For stating \(k = 3\) (must follow from their vertex)
M1: For attempting to make \(x\) the subject of \(y = 2(x-3)^2 - 5\)
A1: For the correct expression \(\mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\)
A1: For stating the domain of the inverse function is \(x \ge -5\)

We apply the laws of logarithms to combine the terms on the left side:

1) Apply the power law, \(k\log_b a = \log_b(a^k)\):
\(2\log_2(x-2) = \log_2\left((x-2)^2\right)\)

2) Substitute this back into the equation:
\( \log_2\left((x-2)^2\right) - \log_2(x+1) = 2 \)

3) Apply the quotient law, \(\log_b a - \log_b c = \log_b\left(\frac{a}{c}\right)\):
\(\log_2\left( \frac{(x-2)^2}{x+1} \right) = 2\)

4) Rewrite the logarithmic equation in exponential form:
\(\frac{(x-2)^2}{x+1} = 2^2\)
\(\frac{x^2 - 4x + 4}{x+1} = 4\)

5) Solve the resulting quadratic equation:
\(x^2 - 4x + 4 = 4(x+1)\)
\(x^2 - 4x + 4 = 4x + 4\)
\(x^2 - 8x = 0\)
\(x(x - 8) = 0\)

This gives the potential solutions \(x = 0\) and \(x = 8\).

However, the original logarithmic term \(\log_2(x-2)\) requires \(x - 2 > 0 \implies x > 2\). Therefore, \(x = 0\) is an extraneous solution and must be rejected.

Thus, the only valid solution is \(x = 8\).

M1: For applying the power law to obtain \(\log_2(x-2)^2\)
M1: For applying the division law to combine logarithmic terms into a single quotient
M1: For converting from logarithmic form to exponential form correctly: \(\dots = 4\)
A1: For expanding and simplifying to a correct quadratic equation, e.g. \(x^2 - 8x = 0\)
A1: For solving the quadratic to find \(x = 0\) and \(x = 8\)
A1: For rejecting \(x = 0\) with explicit reference to the domain \(x > 2\) to leave only \(x = 8\)

First, find the vectors \(\vec{AB}\) and \(\vec{BC}\):

\(\vec{AB} = \mathbf{b} - \mathbf{a} = (\mu\mathbf{i} + 8\mathbf{j}) - (3\mathbf{i} + 5\mathbf{j}) = (\mu - 3)\mathbf{i} + 3\mathbf{j}\)

\(\vec{BC} = \mathbf{c} - \mathbf{b} = (11\mathbf{i} + 17\mathbf{j}) - (\mu\mathbf{i} + 8\mathbf{j}) = (11 - \mu)\mathbf{i} + 9\mathbf{j}\)

Since the point \(B\) lies on the line segment \(AC\) and divides it in the ratio \(1:3\), the vector \(\vec{BC}\) must be in the same direction as \(\vec{AB}\) and exactly 3 times its length. Therefore, we can write:

\(\vec{BC} = 3\vec{AB}\)

Substitute the components of the vectors:

\((11 - \mu)\mathbf{i} + 9\mathbf{j} = 3\left[ (\mu - 3)\mathbf{i} + 3\mathbf{j} \right]\)

Equating the \(\mathbf{j}\) components:
\(9 = 3(3)\) (This is consistent and confirms our scale factor)

Equating the \(\mathbf{i}\) components:
\(11 - \mu = 3(\mu - 3)\)
\(11 - \mu = 3\mu - 9\)
\(20 = 4\mu\)
\(\mu = 5\)

Thus, the value of \(\mu\) is 5.

M1: For finding an expression for \(\vec{AB}\) in terms of \(\mu\)
A1: For \(\vec{AB} = (\mu - 3)\mathbf{i} + 3\mathbf{j}\)
M1: For finding an expression for \(\vec{BC}\) (or \(\vec{AC}\)) in terms of \(\mu\)
A1: For \(\vec{BC} = (11 - \mu)\mathbf{i} + 9\mathbf{j}\) (or \(\vec{AC} = 8\mathbf{i} + 12\mathbf{j}\))
M1: For setting up a correct scale vector equation, e.g., \(\vec{BC} = 3\vec{AB}\) or \(\vec{AC} = 4\vec{AB}\)
A1: For solving to find \(\mu = 5\)

Using the quotient rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3(2x + 1) - 2(3x - 2)}{(2x + 1)^2}\). Simplifying the numerator gives \(\frac{6x + 3 - 6x + 4}{(2x + 1)^2} = \frac{7}{(2x + 1)^2}\). Substituting \(x = 1\) gives \(\frac{7}{(2(1) + 1)^2} = \frac{7}{9}\).

M1: For a correct attempt to apply the quotient rule.
A1: For the correct gradient of \(\frac{7}{9}\).

First rewrite the integrand: \(\int_{1}^{4} 3x^{-1/2} \, \mathrm{d}x = \left[ 6x^{1/2} \right]_{1}^{4}\). Evaluating this at the limits gives \(6\sqrt{4} - 6\sqrt{1} = 12 - 6 = 6\).

M1: For integrating to the form \(k x^{1/2}\) where \(k \ne 0\).
A1: For the correct answer 6.

First, find the magnitude of the vector: \(\sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\). Divide the vector by its magnitude to obtain the unit vector: \(\frac{1}{13}(5\mathbf{i} - 12\mathbf{j}) = \frac{5}{13}\mathbf{i} - \frac{12}{13}\mathbf{j}\).

M1: For finding the magnitude of the vector is 13.
A1: For the correct unit vector in terms of \(\mathbf{i}\) and \(\mathbf{j}\) (or equivalent column vector format).

Using the arithmetic series sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n = 15\), \(a = 7\), and \(d = 4\): \(S_{15} = \frac{15}{2}[2(7) + (15 - 1)4] = \frac{15}{2}[14 + 14(4)] = \frac{15}{2}[70] = 525\).

M1: For correct substitution into the arithmetic sum formula.
A1: For 525.

Factor out the coefficient of \(x^2\) from the first two terms: \(\mathrm{f}(x) = 3(x^2 - 4x) + 7\). Complete the square inside the bracket: \(3[(x - 2)^2 - 4] + 7 = 3(x - 2)^2 - 12 + 7 = 3(x - 2)^2 - 5\).

M1: For an attempt to complete the square by factoring out 3 or writing \(3(x - b)^2 + c\).
A1: For \(3(x - 2)^2 - 5\).

Convert the logarithmic equation to exponential form: \(2x + 5 = 3^2\), which simplifies to \(2x + 5 = 9\). Subtracting 5 from both sides gives \(2x = 4\), so \(x = 2\).

M1: For rewriting the equation correctly in exponential form, i.e., \(2x + 5 = 3^2\).
A1: For \(x = 2\).

The word 'PRISM' contains 5 distinct letters. Since the order of letters matters and no repetition is allowed, we find the number of permutations of 4 letters from 5: \(^{5}\mathrm{P}_{4} = 5 \times 4 \times 3 \times 2 = 120\).

M1: For calculating \(^{5}\mathrm{P}_{4}\) or writing \(5 \times 4 \times 3 \times 2\).
A1: For 120.

The basic angle is \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ\). Since the sine value is negative, the solutions lie in the third and fourth quadrants. Third quadrant: \(\theta = 180^\circ + 60^\circ = 240^\circ\). Fourth quadrant: \(\theta = 360^\circ - 60^\circ = 300^\circ\).

M1: For finding the basic angle of \(60^\circ\) or identifying one correct quadrant solution.
A1: For both \(240^\circ\) and \(300^\circ\).

Using the laws of logarithms:

\(\log_a \left(\frac{125}{8}\right) = \frac{3}{2}\)

Convert the logarithmic equation to its exponential form:

\(a^{\frac{3}{2}} = \frac{125}{8}\)

Raise both sides to the power of \(\frac{2}{3}\):

\(a = \left(\frac{125}{8}\right)^{\frac{2}{3}}\)

\(a = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25\)

M1: for applying the subtraction law of logarithms to obtain \(\log_a \left(\frac{125}{8}\right) = \frac{3}{2}\) or writing the equivalent exponential form \(a^{\frac{3}{2}} = \frac{125}{8}\).
A1: for \(a = 6.25\) (or \(\frac{25}{4}\)).

Using the laws of logarithms:

\(\log_a \left(\frac{125}{8}\right) = \frac{3}{2}\)

Convert the logarithmic equation to its exponential form:

\(a^{\frac{3}{2}} = \frac{125}{8}\)

Raise both sides to the power of \(\frac{2}{3}\):

\(a = \left(\frac{125}{8}\right)^{\frac{2}{3}}\)

\(a = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25\)

M1: for applying the subtraction law of logarithms to obtain \(\log_a \left(\frac{125}{8}\right) = \frac{3}{2}\) or writing the equivalent exponential form \(a^{\frac{3}{2}} = \frac{125}{8}\).
A1: for \(a = 6.25\) (or \(\frac{25}{4}\)).

First, find the derivative of \( y = x^2 \ln x \) using the product rule:
\(\frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = x(2\ln x + 1)\).
At the stationary point, \(\frac{dy}{dx} = 0\):
Since \( x > 0 \), we have:
\(2\ln x + 1 = 0 \implies \ln x = -\frac{1}{2} \implies x = e^{-1/2}\).
Substitute \( x = e^{-1/2} \) into the original equation to find \( y \):
\(y = (e^{-1/2})^2 \ln(e^{-1/2}) = e^{-1} \left(-\frac{1}{2}\right) = -\frac{1}{2e}\).
To determine the nature of the stationary point, find the second derivative:
\(\frac{d^2y}{dx^2} = \frac{d}{dx} [x(2\ln x + 1)] = 1(2\ln x + 1) + x\left(\frac{2}{x}\right) = 2\ln x + 3\).
At \( x = e^{-1/2} \):
\(\frac{d^2y}{dx^2} = 2\left(-\frac{1}{2}\right) + 3 = 2\).
Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

M1: Attempt to differentiate \( y = x^2 \ln x \) using the product rule.
A1: Obtain \(\frac{dy}{dx} = 2x \ln x + x\) (or equivalent) and set to 0 to find \( x = e^{-1/2} \) (or \( \frac{1}{\sqrt{e}} \)).
A1: Find the exact \( y \)-coordinate as \( -\frac{1}{2e} \).
B1: Find \(\frac{d^2y}{dx^2}\) (or use first derivative test) and conclude it is a minimum.

Find the vector \(\vec{AB}\):
\(\vec{AB} = \mathbf{b} - \mathbf{a} = (k\mathbf{i} + 5\mathbf{j}) - (2\mathbf{i} - 3\mathbf{j}) = (k-2)\mathbf{i} + 8\mathbf{j}\).
Since \(\vec{AB}\) is parallel to \(3\mathbf{i} + 4\mathbf{j}\), we can write:
\(\vec{AB} = \lambda (3\mathbf{i} + 4\mathbf{j})\) for some scalar \(\lambda\).
Equating the \(\mathbf{j}\) components:
\(8 = 4\lambda \implies \lambda = 2\).
Equating the \(\mathbf{i}\) components:
\(k-2 = 3\lambda = 3(2) = 6\).
Solving for \(k\):
\(k = 8\).

M1: Obtain an expression for \(\vec{AB}\) in terms of \(k\).
A1: Correctly find \(\vec{AB} = (k-2)\mathbf{i} + 8\mathbf{j}\).
M1: Set up a ratio or scalar multiple equation relating \(\vec{AB}\) and \(3\mathbf{i} + 4\mathbf{j}\).
A1: Correctly solve to find \(k = 8\).

The formula for the sum to infinity is:
\(S_{\infty} = \frac{a}{1-r} = 27 \implies a = 27(1-r)\).
The formula for the sum of the first three terms is:
\(S_3 = a + ar + ar^2 = a(1 + r + r^2) = 26\).
Substitute \( a = 27(1-r) \) into the second equation:
\(27(1-r)(1+r+r^2) = 26\).
Since \((1-r)(1+r+r^2) = 1-r^3\), we have:
\(27(1-r^3) = 26 \implies 1-r^3 = \frac{26}{27} \implies r^3 = \frac{1}{27}\).
Since the common ratio is positive:
\(r = \frac{1}{3}\).
Substitute \(r = \frac{1}{3}\) back to find \(a\):
\(a = 27\left(1 - \frac{1}{3}\right) = 27 \times \frac{2}{3} = 18\).

M1: Write down the equations for \(S_{\infty}\) and \(S_3\) in terms of \(a\) and \(r\).
M1: Substitute \(a = 27(1-r)\) into the \(S_3\) formula and attempt to simplify.
A1: Correctly solve for \(r\) to obtain \(r = \frac{1}{3}\).
A1: Correctly find \(a = 18\).

First, find the inverse function \( f^{-1}(x) \):
Let \( y = f(x) = \frac{3x-1}{x+2} \).
y(x+2) = 3x-1 \implies yx + 2y = 3x - 1.
Rearranging to make \( x \) the subject:
2y + 1 = 3x - yx = x(3-y) \implies x = \frac{2y+1}{3-y}.
So, \( f^{-1}(x) = \frac{2x+1}{3-x} \) for \( x \neq 3 \).
Now, substitute \( g(x) = 2x+1 \) into \( f^{-1}(x) \):
\(f^{-1}g(x) = f^{-1}(2x+1) = \frac{2(2x+1) + 1}{3 - (2x+1)} = \frac{4x+2+1}{3-2x-1} = \frac{4x+3}{2-2x}\).
This function is undefined when the denominator is equal to zero:
2 - 2x = 0 \implies x = 1.

M1: Attempt to find \( f^{-1}(x) \) by setting \( y = \frac{3x-1}{x+2} \) and making \( x \) the subject.
A1: Correct expression for \( f^{-1}(x) = \frac{2x+1}{3-x} \).
M1: Find \( f^{-1}g(x) \) by replacing \( x \) with \( 2x+1 \) in their \( f^{-1}(x) \).
A1: Correctly simplified expression \( \frac{4x+3}{2(1-x)} \) (or equivalent) and state \( x = 1 \) is undefined.

Using the laws of logarithms:
\(2\log_4 x = \log_4(x^2)\).
The equation becomes:
\(\log_4(x^2) - \log_4(x+3) = 1 \implies \log_4\left(\frac{x^2}{x+3}\right) = 1\).
Convert from logarithmic to exponential form:
\(\frac{x^2}{x+3} = 4^1 = 4\).
Multiply both sides by \( (x+3) \):
\(x^2 = 4(x+3) \implies x^2 = 4x + 12 \implies x^2 - 4x - 12 = 0\).
Factorise the quadratic equation:
\((x-6)(x+2) = 0\).
This gives \( x = 6 \) or \( x = -2 \).
Since the argument of a logarithm must be positive, \( x > 0 \).
Thus, we must reject \( x = -2 \).
The only valid solution is \( x = 6 \).

M1: Apply log laws to write the LHS as a single logarithm, e.g., \(\log_4\left(\frac{x^2}{x+3}\right)\).
M1: Convert logarithmic equation to exponential form to obtain \(\frac{x^2}{x+3} = 4\).
A1: Solve the resulting quadratic equation to obtain \( x = 6 \) and \( x = -2 \).
A1: Reject \( x = -2 \) with a valid reason or state only \( x = 6 \) as the final answer.

(a) There are 10 people in total (6 men + 4 women) and we need to choose 5:
Number of ways = \(\binom{10}{5} = \frac{10!}{5!5!} = 252\).

(b) For the committee to contain more women than men, the possible selections of (women, men) are:
- Case 1: 3 women and 2 men
Number of ways = \(\binom{4}{3} \times \binom{6}{2} = 4 \times 15 = 60\).
- Case 2: 4 women and 1 man
Number of ways = \(\binom{4}{4} \times \binom{6}{1} = 1 \times 6 = 6\).
- Case 3: 5 women and 0 men is impossible because there are only 4 women.
Total number of ways = \(60 + 6 = 66\).

B1: Correctly find 252 for part (a).
M1: Identify the two valid cases (3W, 2M) and (4W, 1M) for part (b).
M1: Calculate the combinations for each case and sum them.
A1: Correctly obtain 66 for part (b).

First, find the center of the circle by completing the square:
\(x^2 - 6x + y^2 + 4y = 12 \implies (x-3)^2 - 9 + (y+2)^2 - 4 = 12\).
\((x-3)^2 + (y+2)^2 = 25\).
So the center of the circle is \( C(3, -2) \).
The gradient of the radius \( CP \) is:
\(m_{\text{radius}} = \frac{2 - (-2)}{6 - 3} = \frac{4}{3}\).
Since the tangent is perpendicular to the radius, the gradient of the tangent is:
\(m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{3}{4}\).
The equation of the tangent at \( P(6, 2) \) is:
\(y - 2 = -\frac{3}{4}(x - 6)\).
Multiply by 4:
\(4(y - 2) = -3(x - 6) \implies 4y - 8 = -3x + 18 \implies 3x + 4y = 26\).

M1: Complete the square to find the coordinates of the center \( C(3, -2) \).
M1: Find the gradient of the radius \( CP \).
A1: State the perpendicular gradient as \(-\frac{3}{4}\).
A1: Obtain the correct equation of the tangent in any correct form (e.g., \( 3x + 4y = 26 \)).

Using the double angle identity \(\sin(2\theta) = 2\sin\theta\cos\theta\):
\(3(2\sin\theta\cos\theta) = 2\cos\theta \implies 6\sin\theta\cos\theta - 2\cos\theta = 0\).
Factorise out \(2\cos\theta\):
\(2\cos\theta(3\sin\theta - 1) = 0\).
This gives two cases:
Case 1: \(\cos\theta = 0\)
In the range \(0^\circ \le \theta \le 180^\circ\):
\(\theta = 90^\circ\).
Case 2: \(3\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{3}\)
The basic angle is \(\alpha = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ\).
Since \(\sin\theta\) is positive, \(\theta\) lies in the first or second quadrant:
\(\theta = 19.5^\circ\) (to 1 d.p.)
\(\theta = 180^\circ - 19.47^\circ = 160.5^\circ\) (to 1 d.p.).
Thus, the solutions are \(\theta = 19.5^\circ, 90^\circ, 160.5^\circ\).

M1: Apply the double angle identity \(\sin(2\theta) = 2\sin\theta\cos\theta\).
M1: Factorise the equation to obtain \(\cos\theta = 0\) and \(\sin\theta = \frac{1}{3}\).
A1: Correctly find \(\theta = 90^\circ\).
A1: Correctly find \(\theta = 19.5^\circ\) and \(160.5^\circ\) (accept answers rounded to 1 decimal place; reject extra solutions outside the range).

Using the addition law of logarithms: \(\log_3 ((x + 2)(x - 4)) = 3\). Convert the logarithmic equation into exponential form: \((x + 2)(x - 4) = 3^3\), which simplifies to \(x^2 - 2x - 8 = 27\). Rearranging to form a quadratic equation: \(x^2 - 2x - 35 = 0\). Factoring the quadratic expression gives \((x - 7)(x + 5) = 0\), which yields the possible solutions \(x = 7\) or \(x = -5\). Since the arguments of logarithms must be strictly positive, we require both \(x + 2 > 0\) and \(x - 4 > 0\). Testing the values: for \(x = 7\), both terms are positive; for \(x = -5\), the terms are negative and thus undefined. Therefore, we reject \(x = -5\) and the only valid solution is \(x = 7\).

M1: For applying the product law of logarithms to obtain \(\log_3 ((x+2)(x-4)) = 3\). M1: For converting to exponential form to obtain \((x+2)(x-4) = 27\) or \(x^2 - 2x - 8 = 27\). A1: For forming the correct quadratic equation \(x^2 - 2x - 35 = 0\) and finding the roots \(x = 7\) and \(x = -5\). A1: For identifying \(x = 7\) as the only valid solution and rejecting \(x = -5\).

At a stationary point, \(\frac{dy}{dx} = 0\). Differentiating the curve equation, we get \(\frac{dy}{dx} = -4\sin(2x)\). Setting this to 0: \(-4\sin(2x) = 0 \implies \sin(2x) = 0\). In the interval \(0 < x < \pi\), we have \(0 < 2x < 2\pi\), so \(2x = \pi \implies x = \frac{\pi}{2}\). Substituting this back to find \(y\): \(y = 3 + 2\cos(\pi) = 3 - 2 = 1\). Thus, the stationary point is \(\left(\frac{\pi}{2}, 1\right)\). For the area, we integrate \(y\) from \(0\) to \(\frac{\pi}{4}\): Area \(= \int_{0}^{\pi/4} (3 + 2\cos(2x)) \, dx = \left[ 3x + \sin(2x) \right]_{0}^{\pi/4} = \left( 3\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{2}\right) \right) - (0 + \sin(0)) = \frac{3\pi}{4} + 1\).

M1: Attempts to differentiate to obtain \(k\sin(2x)\)
A1: Correct derivative \(-4\sin(2x)\)
A1: Correct stationary point \(\left(\frac{\pi}{2}, 1\right)\)
M1: Integrates the given function with correct trig integration form
A1: Correct antiderivative \(3x + \sin(2x)\)
A1: Correct substitution of limits to obtain \(\frac{3\pi}{4} + 1\)

From the first equation, using log laws: \(\ln(xy) = \ln 10 \implies xy = 10\). From the second equation, write both sides with base 3: \(3^x \cdot (3^2)^y = 3^9 \implies 3^{x+2y} = 3^9 \implies x + 2y = 9\). Express \(x\) in terms of \(y\): \(x = 9 - 2y\). Substitute into \(xy = 10\): \((9-2y)y = 10 \implies 9y - 2y^2 = 10 \implies 2y^2 - 9y + 10 = 0\). Factor the quadratic: \((2y-5)(y-2) = 0 \implies y = 2.5\) or \(y = 2\). If \(y = 2.5\), then \(x = 9 - 2(2.5) = 4\). If \(y = 2\), then \(x = 9 - 2(2) = 5\). Both pairs yield positive values, so the logarithmic terms remain defined.

M1.5: Applies logarithm properties to obtain \(xy = 10\)
M1.5: Simplifies index equation to linear equation \(x + 2y = 9\)
M1.5: Eliminates a variable and forms a three-term quadratic equation
M1: Correctly solves their quadratic equation for either \(x\) or \(y\)
A1: Obtains both correct coordinate pairs: \((4, 2.5)\) and \((5, 2)\)

Let the arithmetic progression (AP) have first term \(a\) and common difference \(d\). The given terms of the AP are \(T_1 = a\), \(T_3 = a+2d\), and \(T_{11} = a+10d\). These correspond to terms of a geometric progression, so: \(\frac{a+2d}{a} = \frac{a+10d}{a+2d} \implies (a+2d)^2 = a(a+10d) \implies a^2+4ad+4d^2 = a^2+10ad \implies 4d^2 = 6ad\). Given \(d \neq 0\), dividing by \(2d\) yields \(2d = 3a \implies d = 1.5a\). The sum of the first 5 terms of the AP is given by: \(S_5 = \frac{5}{2}(2a+4d) = 80 \implies 5(a+2d) = 80 \implies a+2d = 16\). Substituting \(d = 1.5a\) into this equation: \(a+2(1.5a) = 16 \implies 4a = 16 \implies a = 4\). Thus, \(d = 1.5(4) = 6\). The common ratio of the GP is \(r = \frac{a+2d}{a} = \frac{4+12}{4} = 4\).

M1: Expresses GP relation using AP terms: \((a+2d)^2 = a(a+10d)\)
M1: Simplifies equation to find a relationship between \(a\) and \(d\), e.g., \(2d = 3a\)
M1: Uses the AP sum formula to form \(5(a+2d) = 80\)
A1.5: Solves the equations simultaneously to find \(a = 4\) and \(d = 6\)
M1: Attempts to find the common ratio using their values of \(a\) and \(d\)
A1: Correctly evaluates the common ratio as 4

(i) Since \(Q\) is on \(AB\) such that \(AQ:QB = 3:2\), we have \(\vec{OQ} = \vec{OA} + \frac{3}{5}\vec{AB} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\). (ii) Since \(OP:PA = 2:1\), we have \(\vec{OP} = \frac{2}{3}\mathbf{a}\). Now, express \(\vec{OX}\) in two ways: First, \(\vec{OX} = \mu\vec{OQ} = \frac{2\mu}{5}\mathbf{a} + \frac{3\mu}{5}\mathbf{b}\). Second, \(\vec{OX} = \vec{OB} + \vec{BX} = \mathbf{b} + \lambda\vec{BP} = \mathbf{b} + \lambda(\vec{OP} - \vec{OB}) = \mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2\lambda}{3}\mathbf{a} + (1 - \lambda)\mathbf{b}\). Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): From \(\mathbf{a}\): \(\frac{2\mu}{5} = \frac{2\lambda}{3} \implies \lambda = \frac{3}{5}\mu\). From \(\mathbf{b}\): \(\frac{3\mu}{5} = 1 - \lambda\). Substitute \(\lambda\): \(\frac{3\mu}{5} = 1 - \frac{3}{5}\mu \implies \frac{6\mu}{5} = 1 \implies \mu = \frac{5}{6}\). Then \(\lambda = \frac{3}{5}\left(\frac{5}{6}\right) = \frac{1}{2}\).

M1: Expresses \(\vec{OQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\)
A1: Correctly simplifies to \(\vec{OQ} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\)
M1: Writes an expression for \(\vec{OX}\) using \(\mu\)
M1: Writes an expression for \(\vec{OX}\) using \(\lambda\)
M1.5: Equates coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form a system of simultaneous equations
A1: Correctly solves to find \(\mu = \frac{5}{6}\) and \(\lambda = \frac{1}{2}\)