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\(\frac{dy}{dx} = -2k e^{-2x}\). At \(x=0\), \(\frac{dy}{dx} = -2k e^0 = -2k\). Given that the gradient is \(-6\), we have \(-2k = -6\), which gives \(k = 3\).

M1 for differentiating \(k e^{-2x}\) to obtain \(a k e^{-2x}\) (where \(a \neq 1\)). A1.2 for setting \(-2k = -6\) and solving to get \(k = 3\).

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), the equation becomes \(2(1 - \sin^2 \theta) + \sin \theta - 1 = 0\), which simplifies to \(2\sin^2 \theta - \sin \theta - 1 = 0\). Factoring gives \((2\sin \theta + 1)(\sin \theta - 1) = 0\). Thus, \(\sin \theta = -\frac{1}{2}\) or \(\sin \theta = 1\). For \(0^\circ \le \theta \le 360^\circ\): \(\sin \theta = 1 \implies \theta = 90^\circ\); \(\sin \theta = -\frac{1}{2} \implies \theta = 210^\circ, 330^\circ\).

M1 for substituting \(\cos^2 \theta = 1 - \sin^2 \theta\) and forming a quadratic in \(\sin \theta\). A1.2 for finding all three correct angles: \(90^\circ\), \(210^\circ\), and \(330^\circ\).

Using the addition rule for logarithms, \(\log_3 ((x + 4)(x - 2)) = 3\). Converting to exponential form gives \((x + 4)(x - 2) = 3^3\), which simplifies to \(x^2 + 2x - 8 = 27\). This gives \(x^2 + 2x - 35 = 0\), which factors to \((x + 7)(x - 5) = 0\). Thus, \(x = -7\) or \(x = 5\). Since the arguments of logarithms must be positive, \(x - 2 > 0\) requires \(x > 2\). Therefore, the only valid solution is \(x = 5\).

M1 for combining the logarithms and writing the equation in exponential form \(x^2 + 2x - 8 = 27\). A1.2 for solving the quadratic equation to get \(x = 5\) and correctly rejecting \(x = -7\).

Equating the line and the curve: \(x^2 - 4x + 12 = 2x + k \implies x^2 - 6x + (12 - k) = 0\). For no intersection, the discriminant must be less than 0: \(b^2 - 4ac < 0 \implies (-6)^2 - 4(1)(12 - k) < 0\). This simplifies to \(36 - 48 + 4k < 0 \implies 4k < 12 \implies k < 3\).

M1 for equating equations and writing the discriminant condition \(36 - 4(12 - k) < 0\). A1.2 for the correct final inequality \(k < 3\).

First, express \(\mathbf{a} + 2\mathbf{b}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\): \(\mathbf{a} + 2\mathbf{b} = (3\mathbf{i} - 4\mathbf{j}) + 2(p\mathbf{i} + 6\mathbf{j}) = (3 + 2p)\mathbf{i} + 8\mathbf{j}\). Since this vector is parallel to \(\mathbf{i} + 2\mathbf{j}\), the ratio of their components must be equal: \(\frac{3 + 2p}{1} = \frac{8}{2}\). This gives \(3 + 2p = 4 \implies 2p = 1 \implies p = 0.5\).

M1 for writing \(\mathbf{a} + 2\mathbf{b}\) as \((3 + 2p)\mathbf{i} + 8\mathbf{j}\) and setting up a ratio or scaling equation. A1.2 for obtaining \(p = 0.5\) (or \(\frac{1}{2}\)).

Since the terms form a geometric progression, the common ratio is constant: \(\frac{x - 1}{x + 1} = \frac{x - 2}{x - 1}\). Cross-multiplying gives \((x - 1)^2 = (x + 1)(x - 2)\). Expanding both sides yields \(x^2 - 2x + 1 = x^2 - x - 2\). Simplifying this linear equation gives \(-x = -3\), which results in \(x = 3\).

M1 for setting up the ratio equation \((x - 1)^2 = (x + 1)(x - 2)\) and expanding it. A1.2 for solving correctly to find \(x = 3\).

The number of ways to choose 2 men from 5 is \(\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10\). The number of ways to choose 2 women from 4 is \(\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6\). Using the multiplication principle, the total number of different committees is \(10 \times 6 = 60\).

M1 for calculating \(\binom{5}{2}\) or \(\binom{4}{2}\) or multiplying two combinations. A1.2 for the correct answer 60.

First, evaluate the inner function \(\mathrm{g}(3)\): \(\mathrm{g}(3) = \frac{5}{3 - 1} = \frac{5}{2}\). Next, substitute this value into the outer function \(\mathrm{f}(x)\): \(\mathrm{fg}(3) = \mathrm{f}\left(\frac{5}{2}\right) = 2\left(\frac{5}{2}\right) + 3 = 5 + 3 = 8\).

M1 for finding \(\mathrm{g}(3) = 2.5\) or writing the expression \(\mathrm{fg}(x) = 2\left(\frac{5}{x-1}\right) + 3\). A1.2 for obtaining the correct value of 8.

To find where the line and curve do not intersect, we set their equations equal to each other and require that the resulting quadratic equation has no real roots (discriminant \(< 0\)).

Equating the two expressions:
\(2x^2 + 5x + 5 = kx - 3\)

Rearranging into standard quadratic form:
\(2x^2 + (5 - k)x + 8 = 0\)

For no points of intersection, the discriminant \(\Delta = b^2 - 4ac < 0\):
\((5 - k)^2 - 4(2)(8) < 0\)

\((5 - k)^2 - 64 < 0\)

\((5 - k)^2 < 64\)

Taking the square root on both sides:
\(-8 < 5 - k < 8\)

Subtracting 5 from all parts:
\(-13 < -k < 3\)

Multiplying by \(-1\) (and reversing the inequalities):
\(-3 < k < 13\)

M1: For equating the line and the curve and setting up a quadratic equation in terms of \(x\), then applying the condition \(b^2 - 4ac < 0\).
A1: For obtaining \((5 - k)^2 < 64\) or the critical values \(k = -3\) and \(k = 13\).
A0.2: For the correct final inequality range \(-3 < k < 13\).

Using the laws of logarithms, we can combine the subtraction on the left-hand side into a single logarithm of a quotient:
\(\log_3 \left(\frac{x+4}{x-2}\right) = 2\)

Rewrite the logarithmic equation in exponential form:
\(\frac{x+4}{x-2} = 3^2\)

\(\frac{x+4}{x-2} = 9\)

Multiply both sides by \(x-2\) (where \(x \neq 2\)):
\(x + 4 = 9(x - 2)\)

\(x + 4 = 9x - 18\)

Rearranging the terms to solve for \(x\):

\(8x = 22\)

\(x = \frac{22}{8} = \frac{11}{4} = 2.75\)

Check validity: Substituting \(x = 2.75\) into the original log terms gives argument values \(6.75\) and \(0.75\), which are both positive. Hence, the solution is valid.

M1: For applying the subtraction law of logarithms to write as a single quotient, and converting to index form: \\frac{x+4}{x-2} = 3^2\).
A1: For correctly expanding and obtaining a linear equation such as \(8x = 22\).
A0.2: For the correct final value of \(x = \frac{11}{4}\) (or \(2.75\)).

Use the trigonometric identity \(\sin^2\theta = 1 - \cos^2\theta\) to express the equation solely in terms of \(\cos\theta\):
\(2(1 - \cos^2\theta) - \cos\theta - 1 = 0\)

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

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

Multiply the entire equation by \(-1\) to make the leading coefficient positive:
\(2\cos^2\theta + \cos\theta - 1 = 0\)

Factorise the quadratic expression:
\((2\cos\theta - 1)(\cos\theta + 1) = 0\)

This yields two cases:
1) \(2\cos\theta - 1 = 0 \implies \cos\theta = \frac{1}{2}\)
In the interval \(0^\circ \le \theta \le 360^\circ\), the solutions are:
\(\theta = 60^\circ\) (Quadrant I) and \(\theta = 360^\circ - 60^\circ = 300^\circ\) (Quadrant IV).

2) \(\cos\theta + 1 = 0 \implies \cos\theta = -1\)
In the interval \(0^\circ \le \theta \le 360^\circ\), the solution is:
\(\theta = 180^\circ\).

Combining all solutions, we get \(\theta = 60^\circ, 180^\circ, 300^\circ\).

M1: For using the identity \(\sin^2\theta = 1 - \cos^2\theta\) to obtain a quadratic in terms of \(\cos\theta\) and factorising to find \(\cos\theta = \frac{1}{2}\) and \(\cos\theta = -1\).
A1: For obtaining any two of the correct angles \(60^\circ\), \(180^\circ\), or \(300^\circ\).
A0.2: For obtaining all three correct angles \(\theta = 60^\circ, 180^\circ, 300^\circ\) with no extra incorrect solutions in the range.

The inequality \(|2x^2 - 5x| \le 3\) is equivalent to the system of inequalities:
\(-3 \le 2x^2 - 5x \le 3\)

We solve these as two separate quadratic inequalities:

1. \(2x^2 - 5x \le 3\)
\(2x^2 - 5x - 3 \le 0\)
Factoring the quadratic:
\((2x + 1)(x - 3) \le 0\)
The critical values are \(x = -\frac{1}{2}\) and \(x = 3\).
Since we want the expression to be less than or equal to 0, the solution is:
\(-\frac{1}{2} \le x \le 3\)

2. \(2x^2 - 5x \ge -3\)
\(2x^2 - 5x + 3 \ge 0\)
Factoring the quadratic:
\((2x - 3)(x - 1) \ge 0\)
The critical values are \(x = 1\) and \(x = \frac{3}{2}\).
Since we want the expression to be greater than or equal to 0, the solution is:
\(x \le 1\) or \(x \ge \frac{3}{2}\)

Now, we find the intersection of the two solution sets:
We need \(-\frac{1}{2} \le x \le 3\) AND (\(x \le 1\) or \(x \ge \frac{3}{2}\)).
Combining these gives:
\(-\frac{1}{2} \le x \le 1\) and \(\frac{3}{2} \le x \le 3\)

M1: For translating the absolute value inequality into two quadratic inequalities: \(2x^2 - 5x \le 3\) and \(2x^2 - 5x \ge -3\).
M1: For factorising/solving \(2x^2 - 5x - 3 = 0\) to find critical values \(-\frac{1}{2}\) and \(3\).
A1: For obtaining the range \(-\frac{1}{2} \le x \le 3\).
M1: For factorising/solving \(2x^2 - 5x + 3 = 0\) to find critical values \(1\) and \(\frac{3}{2}\).
A1: For obtaining the range \(x \le 1\) or \(x \ge \frac{3}{2}\).
M1: For a clear attempt to find the intersection of the two sets (e.g., using a number line or algebraic comparison).
A1: For the correct final compound inequality: \(-\frac{1}{2} \le x \le 1\) and \(\frac{3}{2} \le x \le 3\) (accept equivalent interval notation).

Using the double-angle identity for sine, \(\sin 2\theta = 2\sin \theta \cos \theta\), we rewrite the equation:
\(\sqrt{3} (2\sin \theta \cos \theta) = 2\sin^2 \theta\)
\(2\sqrt{3} \sin \theta \cos \theta - 2\sin^2 \theta = 0\)

We factor out \(2\sin \theta\):
\(2\sin \theta (\sqrt{3} \cos \theta - \sin \theta) = 0\)

This gives two possible cases:

1. \(\sin \theta = 0\)
In the interval \(-\pi < \theta \le \pi\), the solutions are:
\(\theta = 0\) and \(\theta = \pi\).
(Note: \(-\pi\) is excluded as the interval is strictly greater than \(-\pi\).)

2. \(\sqrt{3} \cos \theta - \sin \theta = 0\)
\(\sin \theta = \sqrt{3} \cos \theta\)
Assuming \(\cos \theta \ne 0\), we divide both sides by \(\cos \theta\):
\(\tan \theta = \sqrt{3}\)
In the interval \(-\pi < \theta \le \pi\), the tangent function has a positive value in Quadrant I and Quadrant III.
The basic angle is \(\frac{\pi}{3}\).
In Quadrant I: \(\theta = \frac{\pi}{3}\)
In Quadrant III: \(\theta = \frac{\pi}{3} - \pi = -\frac{2\pi}{3}\).

Combining all the solutions in order:
\(\theta = -\frac{2\pi}{3}, 0, \frac{\pi}{3}, \pi\).

B1: For stating or using \(\sin 2\theta = 2\sin \theta \cos \theta\).
M1: For factorising the equation to obtain \(2\sin \theta (\sqrt{3} \cos \theta - \sin \theta) = 0\) or equivalent.
A1: For identifying the two equations to solve: \(\sin \theta = 0\) and \(\tan \theta = \sqrt{3}\).
A1: For finding \(\theta = 0\) and \(\theta = \pi\) from \(\sin \theta = 0\) (deduct 1 mark if \(-\pi\) is included).
M1: For solving \(\tan \theta = \sqrt{3}\) to find the principal angle \(\frac{\pi}{3}\).
A1: For finding \(\theta = \frac{\pi}{3}\) and \(\theta = -\frac{2\pi}{3}\).
A1: For listing all correct solutions and no others in the given range: \(\theta = -\frac{2\pi}{3}, 0, \frac{\pi}{3}, \pi\).

(a)
The first, third, and fourth terms of the arithmetic progression (AP) are:
\(T_1 = a\)
\(T_2 = a + 2d\)
\(T_3 = a + 3d\)

These terms form a geometric progression (GP), which means:
\(\frac{T_2}{T_1} = \frac{T_3}{T_2}\)
\(\frac{a+2d}{a} = \frac{a+3d}{a+2d}\)

Cross-multiplying:
\((a + 2d)^2 = a(a + 3d)\)
\(a^2 + 4ad + 4d^2 = a^2 + 3ad\)

Subtracting \(a^2 + 3ad\) from both sides:
\(ad + 4d^2 = 0\)
\(d(a + 4d) = 0\)

Since \(d \ne 0\) (given non-zero common difference):
\(a + 4d = 0 \Rightarrow a = -4d\) (as required).

Now, we substitute \(a = -4d\) back into the expression for the common ratio \(r\):
\(r = \frac{a+2d}{a} = \frac{-4d+2d}{-4d} = \frac{-2d}{-4d} = \frac{1}{2}\).

(b)
The first term of the GP is \(a\), and its common ratio is \(r = \frac{1}{2}\).
Since \(|r| < 1\), the sum to infinity \(S_\infty\) exists:
\(S_\infty = \frac{a}{1 - r}\)
\(16 = \frac{a}{1 - \frac{1}{2}}\)
\(16 = 2a \Rightarrow a = 8\)

Substitute \(a = 8\) into \(a = -4d\):
\(8 = -4d \Rightarrow d = -2\).

Part (a):
M1: For expressing the terms of the GP in terms of \(a\) and \(d\): \(a, a+2d, a+3d\).
M1: For setting up the GP relation: \((a+2d)^2 = a(a+3d)\) and expanding.
A1: For simplifying to obtain \(a = -4d\) clearly showing all steps.
A1: For substituting \(a = -4d\) to find the common ratio \(r = \frac{1}{2}\).

Part (b):
M1: For using the sum to infinity formula: \(\frac{a}{1 - 1/2} = 16\).
A1: For finding \(a = 8\).
A1: For finding \(d = -2\).

To find the stationary point, we differentiate \(y = \frac{\ln(2x)}{x^2}\) with respect to \(x\) using the quotient rule:
Let \(u = \ln(2x) \Rightarrow \frac{du}{dx} = \frac{2}{2x} = \frac{1}{x}\)
Let \(v = x^2 \Rightarrow \frac{dv}{dx} = 2x\)

Using the quotient rule \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):
\(\frac{dy}{dx} = \frac{x^2 \left(\frac{1}{x}\right) - \ln(2x)(2x)}{(x^2)^2}\)
\(\frac{dy}{dx} = \frac{x - 2x\ln(2x)}{x^4}\)
\(\frac{dy}{dx} = \frac{1 - 2\text{ln}(2x)}{x^3}\)

At a stationary point, \(\frac{dy}{dx} = 0\):
\(\frac{1 - 2\ln(2x)}{x^3} = 0\)
Since \(x > 0\), we have:
\(1 - 2\ln(2x) = 0\)
\(\ln(2x) = \frac{1}{2}\)
\(2x = e^{1/2} = \sqrt{e}\)
\(x = \frac{1}{2}\sqrt{e}\)

Now we find the \(y\)-coordinate:
\(y = \frac{\ln\left(2 \cdot \frac{1}{2}\sqrt{e}\right)}{\left(\frac{1}{2}\sqrt{e}\right)^2} = \frac{\ln(\sqrt{e})}{\frac{1}{4}e} = \frac{\frac{1}{2}}{\frac{1}{4}e} = \frac{2}{e}\)

So the stationary point is \(\left(\frac{1}{2}\sqrt{e}, \frac{2}{e}\right)\).

To determine the nature, we find the second derivative \(\frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2} = \frac{x^3 \frac{d}{dx}(1 - 2\ln(2x)) - (1 - 2\ln(2x)) \frac{d}{dx}(x^3)}{(x^3)^2}\)
\(\frac{d^2y}{dx^2} = \frac{x^3 \left(-\frac{2}{x}\right) - (1 - 2\ln(2x))(3x^2)}{x^6}\)
\(\frac{d^2y}{dx^2} = \frac{-2x^2 - 3x^2 + 6x^2\ln(2x)}{x^6}\)
\(\frac{d^2y}{dx^2} = \frac{-5 + 6\ln(2x)}{x^4}\)

Substituting \(\ln(2x) = \frac{1}{2}\):
\(\frac{d^2y}{dx^2} = \frac{-5 + 6\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\sqrt{e}\right)^4} = \frac{-5 + 3}{\frac{1}{16}e^2} = -\frac{32}{e^2}\)

Since \(-\frac{32}{e^2} < 0\), the stationary point is a local maximum.

M1: For a correct attempt to use the quotient rule (or product rule) to differentiate.
A1: For finding the derivative of \(\ln(2x)\) as \(\frac{1}{x}\).
A1: For obtaining the correct derivative \(\frac{dy}{dx} = \frac{1 - 2\ln(2x)}{x^3}\).
M1: For setting \(\frac{dy}{dx} = 0\) and solving for \(x\) to find \(x = \frac{1}{2}\sqrt{e}\).
A1: For finding the exact \(y\)-coordinate as \(y = \frac{2}{e}\).
M1: For finding \(\frac{d^2y}{dx^2}\) and substituting their \(x\) value (or checking the sign of \(\frac{dy}{dx}\) on either side).
A1: For showing the value is negative and concluding it is a local maximum.

(a)
The total number of people is \(6 + 5 = 11\).
We need to choose 5 representatives from 11 with no restrictions:
Number of ways = \(\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462\).

(b)
For there to be more men than women on a team of 5, the possible compositions of (men, women) are:
- Case 1: 5 men, 0 women
Number of ways = \(\binom{6}{5} \times \binom{5}{0} = 6 \times 1 = 6\).
- Case 2: 4 men, 1 woman
Number of ways = \(\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75\).
- Case 3: 3 men, 2 women
Number of ways = \(\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200\).

Total number of ways = \(6 + 75 + 200 = 281\).

(c)
Let the particular man be \(M_1\) and the particular woman be \(W_1\).
The total number of ways to choose the team without restrictions is 462.
We subtract the number of teams where both \(M_1\) and \(W_1\) are selected:
If both are on the team, we must choose the remaining 3 team members from the remaining 9 people (since \(M_1\) and \(W_1\) are fixed):
Number of restricted teams = \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\).

So, the number of ways where they are NOT both on the team = \(462 - 84 = 378\).

Part (a):
B1: For 462.

Part (b):
M1: For identifying the three valid cases: (5M, 0W), (4M, 1W), and (3M, 2W).
M1: For calculating the combinations for each case (e.g., finding at least two of 6, 75, or 200 correctly).
A1: For 281.

Part (c):
M1: For identifying the complementary method: (Total) - (Teams with both) OR summing the three cases: (Neither on team) + (Only M on team) + (Only W on team).
M1: For calculating \(\binom{9}{3} = 84\) or evaluating the three alternative cases: \(\binom{9}{5} + \binom{9}{4} + \binom{9}{4} = 126 + 126 + 126\).
A1: For 378.

First, we change the base of the second logarithm to 12.
Using the change-of-base formula:
\(\log_{144} \left(x - \frac{2}{3}\right) = \frac{\log_{12} \left(x - \frac{2}{3}\right)}{\log_{12} 144}\)
Since \(144 = 12^2\), we have \(\log_{12} 144 = 2\).
Therefore:
\(\log_{144} \left(x - \frac{2}{3}\right) = \frac{1}{2} \log_{12} \left(x - \frac{2}{3}\right)\).

Substitute this back into the original equation:
\(\log_{12} (x + 2) - \frac{1}{2} \log_{12} \left(x - \frac{2}{3}\right) = \frac{1}{2}\).

Multiply the entire equation by 2:
\(2\log_{12} (x + 2) - \log_{12} \left(x - \frac{2}{3}\right) = 1\).

Using the power law of logarithms, \(2\log_{12} (x + 2) = \log_{12} (x + 2)^2\):
\(\log_{12} (x + 2)^2 - \log_{12} \left(x - \frac{2}{3}\right) = 1\).

Using the division law of logarithms:
\(\log_{12} \left[ \frac{(x + 2)^2}{x - \frac{2}{3}} \right] = 1\).

Convert from logarithmic form to exponential form:
\(\frac{(x + 2)^2}{x - \frac{2}{3}} = 12^1 = 12\).

Cross-multiply:
\((x + 2)^2 = 12 \left(x - \frac{2}{3}\right)\)
\(x^2 + 4x + 4 = 12x - 8\)

Rearrange into a standard quadratic equation form:
\(x^2 - 8x + 12 = 0\).

Factor the quadratic equation:
\((x - 2)(x - 6) = 0\).

Thus, the potential solutions are:
\(x = 2\) or \(x = 6\).

Checking for validity in original logarithms:
- For \(x = 2\), the arguments are \(2 + 2 = 4 > 0\) and \(2 - 2/3 = 4/3 > 0\). (Valid)
- For \(x = 6\), the arguments are \(6 + 2 = 8 > 0\) and \(6 - 2/3 = 16/3 > 0\). (Valid)

Both values are valid, so \(x = 2\) or \(x = 6\).

M1: For applying change of base formula to express \(\log_{144} \left(x - \frac{2}{3}\right)\) in terms of base 12.
A1: For correctly finding \(\log_{144} \left(x - \frac{2}{3}\right) = \frac{1}{2} \log_{12} \left(x - \frac{2}{3}\right)\).
M1: For applying the power law to obtain \(\log_{12} (x + 2)^2\).
M1: For combining logarithms to obtain \(\log_{12} \left[ \frac{(x + 2)^2}{x - \frac{2}{3}} \right]\).
A1: For converting to exponential form to obtain \(\frac{(x + 2)^2}{x - \frac{2}{3}} = 12\).
M1: For expanding and simplifying to the quadratic equation \(x^2 - 8x + 12 = 0\).
A1: For obtaining both correct solutions \(x = 2\) and \(x = 6\).

(a)
The radius \(r\) of the circle is the distance from the center \(C(-1, 1)\) to the point \(P(3, 4)\):
\(r^2 = (3 - (-1))^2 + (4 - 1)^2 = 4^2 + 3^2 = 16 + 9 = 25\).
The standard equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\).
Thus, the equation of the circle is:
\((x + 1)^2 + (y - 1)^2 = 25\).

(b)
The gradient of the radius \(CP\) is:
\(m_{CP} = \frac{4 - 1}{3 - (-1)} = \frac{3}{4}\).
Since the tangent is perpendicular to the radius at the point of contact:
The gradient of the tangent, \(m_{\text{tangent}}\), is:
\(m_{\text{tangent}} = -\frac{1}{m_{CP}} = -\frac{4}{3}\).

The equation of the tangent passing through \(P(3, 4)\) is:
\(y - 4 = -\frac{4}{3}(x - 3)\)
\(3(y - 4) = -4(x - 3)\)
\(3y - 12 = -4x + 12\)
\(4x + 3y - 24 = 0\).

(c)
To find where the tangent intersects the \(y\)-axis (point \(Q\)), we set \(x = 0\) in the equation of the tangent:
\(4(0) + 3y - 24 = 0\)
\(3y = 24 \Rightarrow y = 8\).
So the coordinates of \(Q\) are \((0, 8)\).

The length of \(CQ\) is the distance between \(C(-1, 1)\) and \(Q(0, 8)\):
\(CQ = \sqrt{(0 - (-1))^2 + (8 - 1)^2} = \sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}\).

Part (a):
M1: For calculating \(r^2 = (3 - (-1))^2 + (4 - 1)^2\).
A1: For the correct equation of the circle: \((x + 1)^2 + (y - 1)^2 = 25\) (or equivalent expanded form).

Part (b):
M1: For calculating the gradient of \(CP\): \(m_{CP} = \frac{3}{4}\).
M1: For using the perpendicular gradient rule to get \(m_{\text{tangent}} = -\frac{4}{3}\) and attempting to find the equation of the line.
A1: For the correct equation in the specified form: \(4x + 3y - 24 = 0\) (or any integer multiple).

Part (c):
M1: For finding the coordinates of \(Q\) as \((0, 8)\).
A1: For calculating the distance \(CQ = \sqrt{50} = 5\sqrt{2}\).

(a)
For the equation \(x^2 + (k + 1)x + (2k - 1) = 0\) to have no real roots, the discriminant \(\Delta\) must be strictly less than 0.
\(\Delta = B^2 - 4AC < 0\)
Here, \(A = 1\), \(B = k + 1\), and \(C = 2k - 1\).

\(\Delta = (k + 1)^2 - 4(1)(2k - 1)\)
\(\Delta = k^2 + 2k + 1 - 8k + 4\)
\(\Delta = k^2 - 6k + 5\)

We want \(k^2 - 6k + 5 < 0\).
We factorise the quadratic expression:
\((k - 1)(k - 5) < 0\).

The critical values are \(k = 1\) and \(k = 5\).
For the product to be negative, \(k\) must lie between these two values:
\(1 < k < 5\).

(b)
Given \(k = 3\):
\(f(x) = x^2 + (3 + 1)x + (2(3) - 1)\)
\(f(x) = x^2 + 4x + 5\)

(i)
To express \(f(x)\) in the form \((x + a)^2 + b\), we complete the square:
\(f(x) = \left(x + \frac{4}{2}\right)^2 - \left(\frac{4}{2}\right)^2 + 5\)
\(f(x) = (x + 2)^2 - 4 + 5\)
\(f(x) = (x + 2)^2 + 1\).

(ii)
From the completed square form \(y = (x + 2)^2 + 1\):
The coordinates of the stationary point are \((-2, 1)\).
Since the coefficient of \(x^2\) is positive (\(1 > 0\)), the stationary point is a minimum.

Part (a):
M1: For attempting to find the discriminant of the quadratic equation.
A1: For obtaining \(\Delta = k^2 - 6k + 5\).
M1: For setting their discriminant \(< 0\) and attempting to factorise/find critical values.
A1: For the correct range: \(1 < k < 5\) (accept equivalent notation).

Part (b)(i):
M1: For substituting \(k = 3\) to get \(f(x) = x^2 + 4x + 5\) and attempting to complete the square.
A1: For the correct form: \((x + 2)^2 + 1\).

Part (b)(ii):
B1: For stating the coordinates of the stationary point as \((-2, 1)\) and identifying it as a minimum.

For two vectors to be perpendicular, their scalar (dot) product must be zero.

\(\mathbf{a} \cdot \mathbf{b} = 0\)

\(3(k+1) + (k-2)(-2) = 0\)

\(3k + 3 - 2k + 4 = 0\)

\(k + 7 = 0\)

\(k = -7\)

M1: For attempting to find the scalar product and equating to zero.
A1: For a correct simplified linear equation, e.g., \(k + 7 = 0\).
A0.5: For the correct answer \(k = -7\).

To find the range of values for which there is no intersection, we equate the two equations:

\(kx - 5 = x^2 - 3x - 1\)

Rearranging to standard quadratic form:

\(x^2 - (k+3)x + 4 = 0\)

Since the line and curve do not intersect, this equation has no real roots, so the discriminant \(\Delta < 0\):

\(b^2 - 4ac < 0\)

\((-(k+3))^2 - 4(1)(4) < 0\)

\((k+3)^2 - 16 < 0\)

\((k+3)^2 < 16\)

\(-4 < k+3 < 4\)

\(-7 < k < 1\)

M1: Equating equations and expressing as a single quadratic equation.
M1: Applying \(b^2 - 4ac < 0\).
A0.5: Finding the correct range \(-7 < k < 1\).

Let \(x = 2\theta - \frac{\pi}{6}\).

Since \(0 \le \theta \le \pi\), the interval for \(x\) is:

\(-\frac{\pi}{6} \le x \le \frac{11\pi}{6}\)

In this interval, the solutions to \(\sin x = \frac{\sqrt{3}}{2}\) are:

\(x = \frac{\pi}{3}\) and \(x = \frac{2\pi}{3}\)

For \(x = \frac{\pi}{3}\):

\(2\theta - \frac{\pi}{6} = \frac{\pi}{3} \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}\)

For \(x = \frac{2\pi}{3}\):

\(2\theta - \frac{\pi}{6} = \frac{2\pi}{3} \implies 2\theta = \frac{5\pi}{6} \implies \theta = \frac{5\pi}{12}\)

Both values lie in the given domain.

M1: For finding at least one principal value of \(2\theta - \frac{\pi}{6}\) (e.g. \(\frac{\pi}{3}\) or \(\frac{2\pi}{3}\)).
A1: For obtaining either \(\theta = \frac{\pi}{4}\) or \(\theta = \frac{5\pi}{12}\).
A0.5: For finding both correct values with no extra values in the range.

Using the properties of logarithms:

\(\log_3[(x-1)(x+5)] = 3\)

Convert from logarithmic to exponential form:

\((x-1)(x+5) = 3^3\)

\(x^2 + 4x - 5 = 27\)

\(x^2 + 4x - 32 = 0\)

Factorizing the quadratic equation:

\((x-4)(x+8) = 0\)

This gives \(x = 4\) or \(x = -8\).

However, we must check the constraints for the original logarithms:

\(x-1 > 0 \implies x > 1\)

\(x+5 > 0 \implies x > -5\)

Since \(x = -8\) does not satisfy these constraints, the only valid solution is \(x = 4\).

M1: Applying the addition rule of logarithms to obtain \(\log_3((x-1)(x+5)) = 3\).
M1: Converting to the quadratic equation \(x^2 + 4x - 32 = 0\) and attempting to solve.
A0.5: Rejecting the extraneous solution \(x = -8\) and stating \(x = 4\) as the only answer.

We must choose a committee of 4 people containing at least 2 women from 5 men and 6 women.

We consider the three possible scenarios:

Case 1: 2 women and 2 men
Number of ways = \(\binom{6}{2} \times \binom{5}{2} = 15 \times 10 = 150\)

Case 2: 3 women and 1 man
Number of ways = \(\binom{6}{3} \times \binom{5}{1} = 20 \times 5 = 100\)

Case 3: 4 women and 0 men
Number of ways = \(\binom{6}{4} \times \binom{5}{0} = 15 \times 1 = 15\)

Total number of ways = \(150 + 100 + 15 = 265\).

M1: For calculating the combinations for at least two of the correct cases.
A1: For correct calculations of all three cases (150, 100, and 15).
A0.5: For summing them to get the correct answer of 265.

Let \(Y = \ln y\) and \(X = x^2\).

The straight line equation in terms of \(Y\) and \(X\) is:

\(Y = mX + c\)

First, find the gradient \(m\):

\(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\)

Next, use the point \((2, 5)\) to find the vertical intercept \(c\):

\(5 = 3(2) + c \implies c = -1\)

So, the linear equation is:

\(Y = 3X - 1\)

Substitute back \(Y = \ln y\) and \(X = x^2\):

\(\ln y = 3x^2 - 1\)

Taking the exponential of both sides:

\(y = e^{3x^2 - 1}\)

M1: For finding the gradient \(m = 3\) and attempting to write down the linear equation \(Y = 3X + c\).
M1: For correctly identifying \(\ln y = 3x^2 - 1\).
A0.5: For expressing \(y\) explicitly as \(y = e^{3x^2 - 1}\).

The perimeter of a sector is given by:

\(P = 2r + r\theta = 20\)

From this, we can express \(\theta\) in terms of \(r\):

\(r\theta = 20 - 2r \implies \theta = \frac{20-2r}{r}\)

The area of a sector is given by:

\(A = \frac{1}{2}r^2\theta = 24\)

Substitute the expression for \(\theta\) into the area equation:

\(\frac{1}{2}r^2 \left(\frac{20-2r}{r}\right) = 24\)

\(\frac{1}{2}r(20 - 2r) = 24\)

\(10r - r^2 = 24\)

\(r^2 - 10r + 24 = 0\)

Factorizing the quadratic equation:

\((r-4)(r-6) = 0\)

This gives \(r = 4\) or \(r = 6\).

Both values yield positive angles (\(\theta = 3\) and \(\theta = \frac{4}{3}\) respectively), so both are valid.

M1: Writing down two correct equations for perimeter and area: \(2r + r\theta = 20\) and \(\frac{1}{2}r^2\theta = 24\).
M1: Eliminating \(\theta\) to form a quadratic equation in \(r\).
A0.5: Finding both correct values: \(r = 4\) and \(r = 6\).

First, find the y-coordinate of the point of contact when \(x = 1\):

\(y = (3(1) - 2)^4 = 1^4 = 1\)

So the point is \((1, 1)\).

Now, find the derivative \(\frac{dy}{dx}\) using the chain rule:

\(\frac{dy}{dx} = 4(3x - 2)^3 \times 3 = 12(3x - 2)^3\)

At \(x = 1\), the gradient of the tangent \(m\) is:

\(m = 12(3(1) - 2)^3 = 12(1)^3 = 12\)

Using the equation of a straight line:

\(y - y_1 = m(x - x_1)\)

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

\(y = 12x - 11\)

M1: For finding the correct derivative \(\frac{dy}{dx} = 12(3x-2)^3\).
M1: For evaluating both the y-coordinate \(y = 1\) and gradient \(m = 12\) at \(x = 1\).
A0.5: For stating the final equation of the tangent as \(y = 12x - 11\) (or equivalent form).

First, divide the equation by 4 to obtain \(\sin(2\theta - 10^\circ) = 0.75\). Let \(x = 2\theta - 10^\circ\). Given the domain \(0^\circ \le \theta \le 180^\circ\), the domain for \(x\) is \(-10^\circ \le x \le 350^\circ\). Finding the principal value gives \(x = \sin^{-1}(0.75) \approx 48.59^\circ\). The second angle within the domain is \(180^\circ - 48.59^\circ = 131.41^\circ\). Now solve for \(\theta\) for both cases: \(2\theta - 10^\circ = 48.59^\circ \Rightarrow 2\theta = 58.59^\circ \Rightarrow \theta \approx 29.3^\circ\), and \(2\theta - 10^\circ = 131.41^\circ \Rightarrow 2\theta = 141.41^\circ \Rightarrow \theta \approx 70.7^\circ\). Both values lie within the range \(0^\circ \le \theta \le 180^\circ\).

M1 for attempting to solve \(\sin(2\theta - 10^\circ) = 0.75\) and finding at least one correct value for \(2\theta - 10^\circ\). A1 for one correct angle \(\theta = 29.3^\circ\) (rounded to 1 decimal place). A0.5 for the second correct angle \(\theta = 70.7^\circ\) (rounded to 1 decimal place) with no extra values in range.

To find the gradient, differentiate \(y\) with respect to \(x\) using the product rule \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\), where \(u = 2x - 3\) and \(v = \ln(x+1)\). Differentiating both components gives \(\frac{du}{dx} = 2\) and \(\frac{dv}{dx} = \frac{1}{x+1}\). Applying the rule: \(\frac{dy}{dx} = (2x - 3)\left(\frac{1}{x+1}\right) + 2\ln(x+1)\). Substituting \(x = 0\) into the derivative: \(\frac{dy}{dx} = (2(0) - 3)\left(\frac{1}{0+1}\right) + 2\ln(1) = -3(1) + 2(0) = -3\).

M1 for a clear attempt to apply the product rule to differentiate the expression. A1 for the correct unsimplified derivative expression \(\frac{dy}{dx} = \frac{2x-3}{x+1} + 2\ln(x+1)\). A0.5 for correctly substituting \(x = 0\) to obtain the gradient of -3.

Let \(a\) be the first term and \(r\) be the common ratio. The \(n\)-th term of a geometric progression is \(u_n = ar^{n-1}\). From the given information, we have the simultaneous equations: \(ar^2 = 18\) and \(ar^5 = -486\). Divide the second equation by the first: \(\frac{ar^5}{ar^2} = \frac{-486}{18} \Rightarrow r^3 = -27\). Solving this gives \(r = -3\). Substitute \(r = -3\) back into the first equation: \(a(-3)^2 = 18 \Rightarrow 9a = 18 \Rightarrow a = 2\). Therefore, the common ratio is \(-3\) and the first term is 2.

M1 for setting up the equations \(ar^2 = 18\) and \(ar^5 = -486\) and attempting to divide them to eliminate \(a\). A1 for correctly finding the common ratio \(r = -3\). A0.5 for correctly finding the first term \(a = 2\).

Apply the product rule of logarithms to combine the terms on the left side: \(\log_3((x-2)(x+6)) = 2\). Express this in exponential form: \((x-2)(x+6) = 3^2 = 9\). Expand the expression: \(x^2 + 4x - 12 = 9\). Rearrange into a quadratic equation: \(x^2 + 4x - 21 = 0\). Factoring the quadratic gives: \((x+7)(x-3) = 0\), which yields solutions \(x = -7\) or \(x = 3\). Checking the original logarithmic terms, we require the arguments to be positive: \(x-2 > 0 \Rightarrow x > 2\). Since \(x = -7\) does not satisfy this condition, it must be rejected. Hence, the only valid solution is \(x = 3\).

M1 for using logarithmic laws to obtain \(\log_3((x-2)(x+6)) = 2\) and rewriting it in index form \((x-2)(x+6) = 9\). A1 for solving the quadratic equation to get \(x = 3\) and \(x = -7\). A0.5 for successfully identifying and rejecting the extraneous solution to state \(x = 3\) as the sole final answer.

First, find the points of intersection by setting the curve equal to the line: \(4x - x^2 = 3\). This simplifies to \(x^2 - 4x + 3 = 0\), which factorises to \((x-1)(x-3) = 0\). Thus, the limits of integration are \(x = 1\) and \(x = 3\). The area \(A\) is given by the integral of the upper curve minus the lower line: \(A = \int_{1}^{3} (4x - x^2 - 3) \, dx\). Integrating terms gives \(\left[ 2x^2 - \frac{x^3}{3} - 3x \right]_{1}^{3}\). Evaluating at the upper limit \(x = 3\) yields \(2(9) - \frac{27}{3} - 3(3) = 18 - 9 - 9 = 0\). Evaluating at the lower limit \(x = 1\) yields \(2(1) - \frac{1}{3} - 3 = 2 - \frac{1}{3} - 3 = -\frac{4}{3}\). The area is \(0 - (-\frac{4}{3}) = \frac{4}{3}\).

M1: Set curve equal to line and solve for limits of integration. A1: Correct limits \(x = 1\) and \(x = 3\). M1: Correct setup of integration \(\int_{1}^{3} (4x - x^2 - 3) \, dx\). M1: Integrate to get \(2x^2 - \frac{x^3}{3} - 3x\). A1.5: Substitute limits correctly to obtain \(\frac{4}{3}\) or \(1.33\).

Using the identity \(\sec^2 x = 1 + \tan^2 x\), rewrite the equation: \(2(1 + \tan^2 x) + \tan x - 3 = 0 \implies 2\tan^2 x + \tan x - 1 = 0\). Factoring the quadratic gives \((2\tan x - 1)(\tan x + 1) = 0\). Thus, \(\tan x = 0.5\) or \(\tan x = -1\). For \(\tan x = 0.5\) in \(0^\circ \le x \le 360^\circ\): \(x \approx 26.6^\circ\) or \(x \approx 180^\circ + 26.6^\circ = 206.6^\circ\). For \(\tan x = -1\) in \(0^\circ \le x \le 360^\circ\): \(x = 180^\circ - 45^\circ = 135^\circ\) or \(x = 360^\circ - 45^\circ = 315^\circ\). The solutions are \(x = 26.6^\circ, 135^\circ, 206.6^\circ, 315^\circ\).

M1: Substitute \(\sec^2 x = 1 + \tan^2 x\) into the equation. A1: Correct quadratic equation \(2\tan^2 x + \tan x - 1 = 0\). M1: Solve the quadratic to find \(\tan x = 0.5\) and \(\tan x = -1\). A1: Correct solutions for \(\tan x = -1\) (\(135^\circ\) and \(315^\circ\)). A1.5: Correct solutions for \(\tan x = 0.5\) (\(26.6^\circ\) and \(206.6^\circ\), accept answers rounded to 1 decimal place).

For part (a), using the formula for the \(n\)-th term of an arithmetic progression, \(u_n = a + (n-1)d\), we have \(u_3 = a + 2d = 11\) and \(u_{10} = a + 9d = 39\). Subtracting the first equation from the second gives \(7d = 28 \implies d = 4\). Substituting \(d = 4\) back into the first equation: \(a + 2(4) = 11 \implies a = 3\). For part (b), the sum of the first 20 terms is \(S_{20} = \frac{20}{2} [2a + 19d] = 10 [2(3) + 19(4)] = 10 [6 + 76] = 10(82) = 820\).

M1: Formulate simultaneous equations for \(a\) and \(d\). A1: Solve to find \(d = 4\). A1: Find \(a = 3\). M1: Use correct sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n = 20\). A1.5: Obtain \(S_{20} = 820\).

The perimeter of the sector is given by \(P = 2r + r\theta = 45\), which yields \(\theta = \frac{45 - 2r}{r}\). The area is \(A = \frac{1}{2}r^2\theta = 100\). Substitute \(\theta\) into the area equation: \(\frac{1}{2}r^2\left(\frac{45 - 2r}{r}\right) = 100 \implies \frac{1}{2}r(45 - 2r) = 100 \implies 45r - 2r^2 = 200 \implies 2r^2 - 45r + 200 = 0\). Factorise the quadratic equation: \((2r - 5)(r - 20) = 0\), yielding \(r = 2.5\) or \(r = 20\). If \(r = 2.5\), then \(\theta = \frac{45 - 2(2.5)}{2.5} = 16\). Since we are given \(\theta < 2\pi\), \(16\) is invalid. If \(r = 20\), then \(\theta = \frac{45 - 2(20)}{20} = 0.25\). This is valid since \(0.25 < 2\pi\). Thus, \(r = 20\) and \(\theta = 0.25\).

M1: Write expression for perimeter \(2r + r\theta = 45\) and area \(\frac{1}{2}r^2\theta = 100\). A1: Correctly substitute to show \(2r^2 - 45r + 200 = 0\). M1: Solve the quadratic to find \(r = 2.5\) and \(r = 20\). M1: Evaluate \(\theta\) for both values of \(r\) and apply restriction \(\theta < 2\pi\). A1.5: State final values \(r = 20\) and \(\theta = 0.25\).

Using the change of base rule, \(\log_4(x-1) = \frac{\log_2(x-1)}{\log_2(4)} = \frac{1}{2}\log_2(x-1)\). The equation becomes \(\log_2(x) + \frac{1}{2}\log_2(x-1) = 1\). Multiply the entire equation by 2: \(2\log_2(x) + \log_2(x-1) = 2 \implies \log_2(x^2) + \log_2(x-1) = 2 \implies \log_2(x^2(x-1)) = 2\). Converting to exponential form: \(x^2(x-1) = 2^2 = 4 \implies x^3 - x^2 - 4 = 0\). Testing small integer values, we find \(x = 2\) is a solution since \(2^3 - 2^2 - 4 = 8 - 4 - 4 = 0\). Factorising gives \((x-2)(x^2 + x + 2) = 0\). The quadratic term \(x^2 + x + 2 = 0\) has discriminant \(\Delta = 1^2 - 4(1)(2) = -7 < 0\), so there are no other real roots. Therefore, the only solution is \(x = 2\).

M1: Use change of base to rewrite \(\log_4(x-1)\) as \(\frac{1}{2}\log_2(x-1)\). M1: Combine log terms to get \(\log_2(x^2(x-1)) = 2\). A1: Obtain the cubic equation \(x^3 - x^2 - 4 = 0\). M1: Show that \(x = 2\) is a root and factorise to \((x-2)(x^2 + x + 2) = 0\). A1.5: Conclude \(x = 2\) as the only real solution, demonstrating that the quadratic factor has no real roots.

To find the points of intersection, set the line equal to the curve: \(2x^2 + 3x + 3 = kx - 5 \implies 2x^2 + (3 - k)x + 8 = 0\). For no intersection, the discriminant \(\Delta\) of this quadratic equation must be less than zero: \(\Delta = (3 - k)^2 - 4(2)(8) < 0 \implies (3 - k)^2 - 64 < 0\). This simplifies to \((3 - k)^2 < 64\), which means \(-8 < 3 - k < 8\). Subtracting 3 gives \(-11 < -k < 5\). Multiplying by \(-1\) and reversing the inequalities gives \(-5 < k < 11\).

M1: Set the equations equal to each other to form a single quadratic. A1: Correct quadratic in standard form \(2x^2 + (3 - k)x + 8 = 0\). M1: Set up the discriminant inequality \(\Delta < 0\). A1: Solve for critical values \(k = -5\) and \(k = 11\). A1.5: Correct final range \(-5 < k < 11\).

For part (a), choose 5 people from a total of 11: \(\binom{11}{5} = \frac{11!}{5!6!} = 462\). For part (b), more women than men means the combinations can be: 3 women and 2 men, 4 women and 1 man, or 5 women and 0 men. Number of ways for 3 women and 2 men: \(\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150\). Number of ways for 4 women and 1 man: \(\binom{5}{4} \times \binom{6}{1} = 5 \times 6 = 30\). Number of ways for 5 women and 0 men: \(\binom{5}{5} \times \binom{6}{0} = 1 \times 1 = 1\). Total ways = \(150 + 30 + 1 = 181\).

M1: Calculate total combinations without restriction: \(\binom{11}{5}\). A1: Correctly calculate 462. M1: Identify the three valid scenarios: (3W, 2M), (4W, 1M), (5W, 0M). M1: Calculate combinations for each scenario. A1.5: Sum to get correct total of 181.

For part (a), the position vector of \(P\) divides the line segment \(AB\) in the ratio \(1:2\). Therefore, \(\overrightarrow{OP} = \overrightarrow{OA} + \frac{1}{3}\overrightarrow{AB} = \overrightarrow{OA} + \frac{1}{3}(\overrightarrow{OB} - \overrightarrow{OA}) = \frac{2}{3}\overrightarrow{OA} + \frac{1}{3}\overrightarrow{OB}\). Substituting the vectors: \(\overrightarrow{OP} = \frac{2}{3}(2\mathbf{i} + 5\mathbf{j}) + \frac{1}{3}(8\mathbf{i} - \mathbf{j}) = \left(\frac{4}{3}\mathbf{i} + \frac{10}{3}\mathbf{j}\right) + \left(\frac{8}{3}\mathbf{i} - \frac{1}{3}\mathbf{j}\right) = 4\mathbf{i} + 3\mathbf{j}\). For part (b), first find the magnitude of \(\overrightarrow{OP}\): \(|\overrightarrow{OP}| = \sqrt{4^2 + 3^2} = 5\). The unit vector is \(\frac{\overrightarrow{OP}}{|\overrightarrow{OP}|} = \frac{1}{5}(4\mathbf{i} + 3\mathbf{j}) = 0.8\mathbf{i} + 0.6\mathbf{j}\).

M1: Use the ratio \(AP : PB = 1 : 2\) to set up vector addition for \(\overrightarrow{OP}\). A1.5: Obtain the correct position vector \(\overrightarrow{OP} = 4\mathbf{i} + 3\mathbf{j}\). M1: Calculate the magnitude of \(\overrightarrow{OP}\) as 5. A2: Compute the correct unit vector \(0.8\mathbf{i} + 0.6\mathbf{j}\) (or \(\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}\)).

**Part (a)**

Given the equation of the curve:
\[y = 16(2x - 1)^{-2}\]

We find the derivative \(\frac{dy}{dx}\) using the chain rule:
\[\frac{dy}{dx} = 16 \cdot (-2)(2x - 1)^{-3} \cdot 2 = -\frac{64}{(2x - 1)^3}\]

At the instant when \(x = 2.5\):
\[\frac{dy}{dx} = -\frac{64}{(2(2.5) - 1)^3} = -\frac{64}{4^3} = -1\]

Using the rate of change chain rule:
\[\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\]

Given that \(\frac{dx}{dt} = 0.3\):
\[\frac{dy}{dt} = -1 \cdot 0.3 = -0.3 \text{ units s}^{-1}\]

---

**Part (b)**

The coordinates of the vertices of triangle \(OAP\) are \(O(0,0)\), \(A(x,0)\), and \(P(x,y)\). This forms a right-angled triangle with base \(x\) and height \(y\).

The area \(S\) of the triangle is:
\[S = \frac{1}{2}xy\]

Differentiating with respect to \(t\) using the product rule:
\[\frac{dS}{dt} = \frac{1}{2} \left( x\frac{dy}{dt} + y\frac{dx}{dt} \right)\]

At the instant when \(x = 2.5\):
- \(y = \frac{16}{(2(2.5) - 1)^2} = 1\)
- \(\frac{dx}{dt} = 0.3\)
- \(\frac{dy}{dt} = -0.3\)

Substituting these values into the expression:
\[\frac{dS}{dt} = \frac{1}{2} \left( 2.5(-0.3) + 1(0.3) \right) = \frac{1}{2} (-0.75 + 0.3) = -0.225 \text{ units}^2\text{ s}^{-1}\]

**Part (a)**
* **M1**: For an attempt to differentiate \(y\) with respect to \(x\) using the chain rule to obtain an expression of the form \(k(2x-1)^{-3}\).
* **A1**: For obtaining the correct derivative \(\frac{dy}{dx} = -64(2x - 1)^{-3}\).
* **M1**: For applying \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\) using their value of \(\frac{dy}{dx}\) evaluated at \(x = 2.5\) and \(\frac{dx}{dt} = 0.3\).
* **A1**: For the correct value of \(-0.3\) (or equivalent).

**Part (b)**
* **B1**: For setting up the area expression \(S = \frac{1}{2}xy\) or substituting to get \(S = \frac{8x}{(2x-1)^2}\).
* **M1**: For applying the product rule or quotient rule to differentiate \(S\) and finding the rate of change of area.
* **A1**: For the correct rate of change of area of \(-0.225\) (or equivalent fraction \(-\frac{9}{40}\)).