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For the vertex:

The minimum value of the modulus term \(|3x - 6|\) is 0, which occurs when \(3x - 6 = 0\), giving \(x = 2\).

Substituting \(x = 2\) into the equation gives \(y = 0 - 2 = -2\).

So, the coordinates of the vertex are \((2, -2)\).

To solve the inequality \(|3x - 6| - 2 \le x\):

First, isolate the modulus term on one side:
\(|3x - 6| \le x + 2\)

Since the modulus is non-negative, this inequality only has solutions for \(x + 2 \ge 0 \implies x \ge -2\).

We can solve this by squaring both sides:

\((3x - 6)^2 \le (x + 2)^2\)

\(9x^2 - 36x + 36 \le x^2 + 4x + 4\)

\(8x^2 - 40x + 32 \le 0\)

Dividing the entire inequality by 8:

\(x^2 - 5x + 4 \le 0\)

Factorising the quadratic expression:

\((x - 1)(x - 4) \le 0\)

The critical values are \(x = 1\) and \(x = 4\).

For the quadratic to be less than or equal to zero, \(x\) must lie between these critical values.

Thus, the solution set is \(1 \le x \le 4\).

Vertex:
- B1: For identifying the correct x-coordinate of the vertex: \(x = 2\).
- B0.5: For identifying the correct y-coordinate of the vertex: \(y = -2\). (Accept the coordinate pair \((2, -2)\) written together for 1.5 marks).

Inequality:
- M1: For attempting to solve the inequality by either squaring both sides or by setting up and solving the two linear boundary equations \(3x - 6 = x + 2\) and \(-(3x - 6) = x + 2\).
- A1: For obtaining a correct quadratic form \(8x^2 - 40x + 32 = 0\) (or simplified to \(x^2 - 5x + 4 = 0\)), or for finding both boundary values \(x = 1\) and \(x = 4\).
- M1: For applying a correct method to find the inner interval of a 'less than or equal to' quadratic inequality (e.g., using a sketch, sign table, or test points).
- A1: For the correct final range \(1 \le x \le 4\) (accept interval notation \([1, 4]\)).

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

The straight line passes through the points \((1, 5)\) and \((4, 11)\).

First, calculate the gradient, \(m\):

\(m = \frac{11 - 5}{4 - 1} = \frac{6}{3} = 2\)

Using the straight-line equation \(Y - Y_1 = m(X - X_1)\) with point \((1, 5)\):

\(Y - 5 = 2(X - 1)\)

\(Y = 2X + 3\)

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

\(\ln y = 2x^2 + 3\)

(b) To express \(y\) in terms of \(x\), take the exponential of both sides of the equation from part (a):

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

Using the laws of indices, write the expression in the required form:

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

Comparing this with \(y = A e^{bx^2}\):

\(A = e^3\) (or approximately \(20.1\) to 3 significant figures)

\(b = 2\)

(a)
- M1: For a valid method to find the gradient of the straight line, showing \(m = 2\).
- A0.5: For writing the correct linear equation relating \(\ln y\) and \(x^2\): \(\ln y = 2x^2 + 3\).

(b)
- M1: For converting the logarithmic equation to exponential form: \(y = e^{2x^2 + 3}\).
- M1: For applying index laws to split the exponential term: \(y = e^3 \cdot e^{2x^2}\).
- A1: For identifying the correct value of the constant \(A = e^3\) (accept the decimal equivalent \(20.1\) rounded to 3 significant figures).
- A1: For identifying the correct value of the constant \(b = 2\).

From the first equation:
\(4^x \times 2^{2y} = 4096\)

Express each term using base 2:
\((2^2)^x \times 2^{2y} = 2^{12}\)
\(2^{2x + 2y} = 2^{12}\)

Equating exponents:
\(2x + 2y = 12 \implies x + y = 6 \implies y = 6 - x\)

From the second equation:
\(\log_2(x+1) + \log_2(y-1) = 3\)

Using the product law of logarithms:
\(\log_2[(x+1)(y-1)] = 3\)

Converting to exponential form:
\((x+1)(y-1) = 2^3 = 8\)

Substitute \(y = 6 - x\) into this equation:
\((x+1)((6-x)-1) = 8\)
\((x+1)(5-x) = 8\)
\(5x - x^2 + 5 - x = 8\)
\(-x^2 + 4x + 5 = 8\)
\(x^2 - 4x + 3 = 0\)

Factorising the quadratic equation:
\((x-1)(x-3) = 0\)

This gives:
\(x = 1 \quad \text{or} \quad x = 3\)

Using \(y = 6 - x\):
- If \(x = 1\), then \(y = 5\).
- If \(x = 3\), then \(y = 3\).

Both solutions are valid because they yield positive arguments for the logarithmic terms (\(x+1 > 0\) and \(y-1 > 0\)).

Thus, the solutions are \(x = 1, y = 5\) and \(x = 3, y = 3\).

M1: Expresses \(4^x \times 2^{2y} = 4096\) in base 2, e.g., \(2^{2x+2y} = 2^{12}\).
A1: Obtains the correct linear equation \(x+y=6\) (or equivalent).
M1: Applies the product law of logarithms to obtain \((x+1)(y-1) = 8\).
M1: Substitutes \(y = 6-x\) (or equivalent) into their equation to form a quadratic equation.
A1: Obtains the correct quadratic equation \(x^2 - 4x + 3 = 0\) and solves to find \(x = 1\) and \(x = 3\).
A1: Correctly pairs both sets of solutions: \(x = 1, y = 5\) and \(x = 3, y = 3\).

(a) Substituting the given coordinates into \(y = a b^x\):

For \((1, 6)\):
\(6 = a b^1 \implies 6 = ab \quad \text{--- (Equation 1)}\)

For \((3, 54)\):
\(54 = a b^3 \quad \text{--- (Equation 2)}\)

Dividing Equation 2 by Equation 1:
\(\frac{54}{6} = \frac{ab^3}{ab} \implies 9 = b^2\)

Since \(b > 0\) for exponential models, we have:
\(b = 3\)

Substitute \(b = 3\) back into Equation 1:
\(6 = a(3) \implies a = 2\)

So, \(a = 2\) and \(b = 3\).

(b) We are given the equation:
\(\log_3 y - \log_3(x+1) = x - 1\)

Substitute \(y = 2 \times 3^x\) into this equation:
\(\log_3(2 \times 3^x) - \log_3(x+1) = x - 1\)

Using the product law of logarithms:
\(\log_3 2 + \log_3(3^x) - \log_3(x+1) = x - 1\)

Since \(\log_3(3^x) = x\), this simplifies to:
\(\log_3 2 + x - \log_3(x+1) = x - 1\)

Subtracting \(x\) from both sides:
\(\log_3 2 - \log_3(x+1) = -1\)

Using the quotient law of logarithms:
\(\log_3\left(\frac{2}{x+1}\right) = -1\)

Converting to exponential form:
\(\frac{2}{x+1} = 3^{-1}\)
\(\frac{2}{x+1} = \frac{1}{3}\)

Cross-multiplying:
\(6 = x + 1 \implies x = 5\)

Part (a) [3 marks]:
M1: Sets up two equations using the coordinates: \(6 = ab\) and \(54 = ab^3\).
M1: Divides the equations to eliminate \(a\) and solve for \(b\).
A1: Correctly identifies \(b = 3\) and \(a = 2\).

Part (b) [4 marks]:
M1: Substitutes \(y = 2 \times 3^x\) into the logarithmic equation.
M1: Applies logarithm laws to expand \(\log_3(2 \times 3^x)\) to \(\log_3 2 + x\).
M1: Simplifies to obtain \(\log_3\left(\frac{2}{x+1}\right) = -1\) (or equivalent).
A1: Obtains the correct final answer \(x = 5\).

We use the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to rewrite the equation in terms of \(\tan \theta\). \(3(1 + \tan^2 \theta) + 5 \tan \theta = 5\) \(3 + 3 \tan^2 \theta + 5 \tan \theta = 5\) \(3 \tan^2 \theta + 5 \tan \theta - 2 = 0\). We factorise this quadratic equation: \((3 \tan \theta - 1)(\tan \theta + 2) = 0\). This gives two possible cases: Case 1: \(\tan \theta = \frac{1}{3}\). The basic angle is \(\tan^{-1}\left(\frac{1}{3}\right) \approx 18.43^\circ\). Since tangent is positive, \(\theta\) lies in the first and third quadrants: \(\theta \approx 18.4^\circ\) and \(\theta \approx 180^\circ + 18.43^\circ = 198.4^\circ\). Case 2: \(\tan \theta = -2\). The basic angle is \(\tan^{-1}(2) \approx 63.43^\circ\). Since tangent is negative, \(\theta\) lies in the second and fourth quadrants: \(\theta \approx 180^\circ - 63.43^\circ = 116.6^\circ\) and \(\theta \approx 360^\circ - 63.43^\circ = 296.6^\circ\). Thus, the solutions are \(\theta = 18.4^\circ, 116.6^\circ, 198.4^\circ, 296.6^\circ\) (each rounded to 1 decimal place).

M1: Substitution of \(\sec^2 \theta = 1 + \tan^2 \theta\) into the given equation. M1: Correctly factorises or solves the quadratic equation to obtain \(\tan \theta = \frac{1}{3}\) and \(\tan \theta = -2\). A1: Obtains at least two correct angles (e.g., \(18.4^\circ\) and \(116.6^\circ\)). A1.33: Obtains all four correct solutions: \(18.4^\circ, 116.6^\circ, 198.4^\circ, 296.6^\circ\) (rounded correctly to 1 decimal place). Deduct 0.5 marks for any extra values within the range.

For part (i): Combine the two fractions under a common denominator: \(\frac{\cos \theta(1 + \sin \theta) - \cos \theta(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} = \frac{\cos \theta + \cos \theta \sin \theta - \cos \theta + \cos \theta \sin \theta}{1 - \sin^2 \theta}\) \(= \frac{2 \cos \theta \sin \theta}{\cos^2 \theta} = 2 \left(\frac{\sin \theta}{\cos \theta}\right) = 2 \tan \theta\). Thus, \(k = 2\). For part (ii): Using the result from part (i), the equation becomes: \(2 \tan \theta = 3 \implies \tan \theta = 1.5\). The basic angle is \(\tan^{-1}(1.5) \approx 0.9828\) radians. Since tangent is positive, \(\theta\) lies in the first and third quadrants: \(\theta_1 \approx 0.983\) radians and \(\theta_2 = \pi + 0.9828 \approx 4.12\) radians.

M1 (Part i): Combines fractions with a common denominator and simplifies numerator. A1 (Part i): Correctly obtains \(2 \tan \theta\) (or identifies \(k = 2\)). M1 (Part ii): Sets up and attempts to solve \(2 \tan \theta = 3\) to find the basic angle in radians. A1.33 (Part ii): Obtains both correct solutions \(\theta \approx 0.983\) and \(\theta \approx 4.12\) (correct to 3 significant figures).

First, divide both sides by 2: \(\cos\left(2x - \frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}\). Let \(\phi = 2x - \frac{\pi}{3}\). Given the range \(0 \le x \le \pi\), we find the range for \(\phi\): \(0 \le 2x \le 2\pi \implies -\frac{\pi}{3} \le 2x - \frac{\pi}{3} \le \frac{5\pi}{3}\). The basic angle for \(\cos \phi = -\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\). Since cosine is negative, \(\phi\) must lie in the second and third quadrants: Quadrant 2: \(\phi = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\). Quadrant 3: \(\phi = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\). Both values lie within the interval \([-\frac{\pi}{3}, \frac{5\pi}{3}]\). Now we solve for \(x\): For \(2x - \frac{\pi}{3} = \frac{5\pi}{6} \implies 2x = \frac{5\pi}{6} + \frac{2\pi}{6} = \frac{7\pi}{6} \implies x = \frac{7\pi}{12}\). For \(2x - \frac{\pi}{3} = \frac{7\pi}{6} \implies 2x = \frac{7\pi}{6} + \frac{2\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \implies x = \frac{3\pi}{4}\). Thus, the exact solutions are \(x = \frac{7\pi}{12}\) and \(x = \frac{3\pi}{4}\).

M1: Isolates the trigonometric term to get \(\cos\left(2x - \frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}\). M1: Identifies the correct quadrant angles for the substituted variable \(2x - \frac{\pi}{3} = \frac{5\pi}{6}\) and \(\frac{7\pi}{6}\). A1: Obtains one correct value of \(x\) (either \(\frac{7\pi}{12}\) or \(\frac{3\pi}{4}\)). A1.33: Obtains both correct exact answers \(x = \frac{7\pi}{12}\) and \(x = \frac{3\pi}{4}\) with no extra incorrect values within the interval.

(a) Since the terms are in geometric progression, the common ratio \(r\) is constant:
\(\frac{k}{k+4} = \frac{2k-15}{k}\)

Cross-multiplying:
\(k^2 = (k+4)(2k-15)\)
\(k^2 = 2k^2 - 15k + 8k - 60\)
\(k^2 - 7k - 60 = 0\)

Factorising the quadratic equation:
\((k-12)(k+5) = 0\)

This gives \(k = 12\) or \(k = -5\).

If \(k = -5\), the terms are \(-1\), \(-5\), \(-25\), which gives a common ratio of \(5\). A geometric progression with \(|r| \ge 1\) does not have a sum to infinity, so we reject \(k = -5\).
Thus, \(k = 12\).

The first term is \(a = 12 + 4 = 16\) and the second term is \(12\).
The common ratio is:
\(r = \frac{12}{16} = 0.75\) (or \(\frac{3}{4}\)).

(b) The first term is \(a = 16\) and the common ratio is \(r = 0.75\).

The sum to infinity is:
\(S_\infty = \frac{a}{1-r} = \frac{16}{1 - 0.75} = 64\)

The sum of the first 6 terms is:
\(S_6 = \frac{a(1-r^6)}{1-r} = \frac{16(1 - 0.75^6)}{1 - 0.75} = 64(1 - 0.1779785...) = 52.609375\)

The difference between \(S_\infty\) and \(S_6\) is:
\(S_\infty - S_6 = 64 - 52.609375 = 11.390625\)

Rounding to 3 significant figures gives \(11.4\) (or exactly \(\frac{729}{64}\)).

(a)
- M1: For setting up the ratio equation \(\frac{k}{k+4} = \frac{2k-15}{k}\)
- A1: For expanding and simplifying to the quadratic equation \(k^2 - 7k - 60 = 0\)
- A1: For solving to find \(k = 12\) and explaining why \(k = -5\) is rejected (e.g., stating \(|r| < 1\) for sum to infinity to exist)
- B1: For finding the correct common ratio \(r = 0.75\) or \(\frac{3}{4}\)

(b)
- B1: For finding \(a = 16\) and calculating \(S_\infty = 64\)
- M1: For attempting to use the sum formula to find \(S_6\) or calculating the difference directly using \(a r^6 / (1-r)\)
- A1: For obtaining the correct difference of \(11.4\) (or \(\frac{729}{64}\) or \(11.390625\))

The volume \(V\) of a cone is given by \(V = \frac{1}{3}\pi r^2 h\). Since we are given \(h = 2r\), we can express the radius as \(r = \frac{h}{2}\). Substituting this into the volume formula gives: \(V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}\). Differentiating \(V\) with respect to \(h\) gives: \(\frac{dV}{dh} = \frac{3\pi h^2}{12} = \frac{\pi h^2}{4}\). Using the chain rule, we have: \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}\). Substituting the known rate of volume increase \(\frac{dV}{dt} = 12\) gives: \(12 = \frac{\pi h^2}{4} \times \frac{dh}{dt}\). At the instant when \(h = 6\text{ cm}\): \(12 = \frac{\pi (6)^2}{4} \times \frac{dh}{dt}\), which simplifies to \(12 = 9\pi \times \frac{dh}{dt}\). Solving for the rate of change of depth: \(\frac{dh}{dt} = \frac{12}{9\pi} = \frac{4}{3\pi}\text{ cm s}^{-1}\).

M1: Expresses the volume of the cone in terms of \(h\) only using \(r = \frac{h}{2}\). A1: Obtains the correct volume formula \(V = \frac{\pi h^3}{12}\). M1: Differentiates \(V\) with respect to \(h\) to find \(\frac{dV}{dh}\). A1: Obtains \(\frac{dV}{dh} = \frac{\pi h^2}{4}\). M1: Uses the chain rule connecting \(\frac{dV}{dt}\), \(\frac{dV}{dh}\), and \(\frac{dh}{dt}\). M1: Substitutes \(h = 6\) and \(\frac{dV}{dt} = 12\) into their expression. A1.75: Obtains the exact value of \(\frac{4}{3\pi}\) (or equivalent exact expression, or 0.424 to 3 significant figures).

(i) To find the points of intersection, equate the two equations: \(3\sqrt{x} = x + 2\). Squaring both sides yields \(9x = (x + 2)^2 = x^2 + 4x + 4\). Rearranging gives the quadratic equation \(x^2 - 5x + 4 = 0\), which factors to \((x - 1)(x - 4) = 0\). Thus, the \(x\)-coordinates of the intersection points are \(x = 1\) and \(x = 4\). Corresponding \(y\)-coordinates are \(y = 1 + 2 = 3\) and \(y = 4 + 2 = 6\). Thus, the coordinates are \(A(1, 3)\) and \(B(4, 6)\). (ii) The area of the enclosed region is given by the definite integral: \(\text{Area} = \int_{1}^{4} (3\sqrt{x} - (x + 2)) \, dx = \left[ 2x^{3/2} - \frac{1}{2}x^2 - 2x \right]_1^4\). Evaluating at the upper limit \(x = 4\): \(2(4)^{3/2} - \frac{1}{2}(4)^2 - 2(4) = 16 - 8 - 8 = 0\). Evaluating at the lower limit \(x = 1\): \(2(1)^{3/2} - \frac{1}{2}(1)^2 - 2(1) = 2 - 0.5 - 2 = -0.5\). The area is therefore \(0 - (-0.5) = 0.5\).

M1: Equates equations, squares both sides, and attempts to solve the resulting quadratic equation. A1: Obtains \(x = 1\) and \(x = 4\), with correct coordinates \(A(1, 3)\) and \(B(4, 6)\). M1: Sets up the correct definite integral of \((3\sqrt{x} - x - 2)\) from \(1\) to \(4\). M1: Integrates \(3x^{1/2}\) correctly to obtain \(2x^{3/2}\). A1: Integrates the remaining terms correctly to obtain \(-\frac{1}{2}x^2 - 2x\). M1: Substitutes the limits \(4\) and \(1\) into their integrated expression. A1.75: Obtains the correct final exact area of \(0.5\) (or \(\frac{1}{2}\)).

(i) Using the product rule to differentiate \(y = x^2 e^{-3x}\): \(\frac{dy}{dx} = 2x e^{-3x} + x^2 (-3 e^{-3x}) = x e^{-3x} (2 - 3x)\). For stationary points, set \(\frac{dy}{dx} = 0\). Since \(e^{-3x} \neq 0\), we must have \(x(2 - 3x) = 0\), which gives \(x = 0\) and \(x = \frac{2}{3}\). When \(x = 0\), \(y = 0\), giving the point \((0, 0)\). When \(x = \frac{2}{3}\), \(y = \left(\frac{2}{3}\right)^2 e^{-3(2/3)} = \frac{4}{9}e^{-2} = \frac{4}{9e^2}\), giving the point \(\left(\frac{2}{3}, \frac{4}{9e^2}\right)\). (ii) To determine the nature, find the second derivative of \(y\): \(\frac{d^2y}{dx^2} = \frac{d}{dx} [e^{-3x}(2x - 3x^2)] = -3e^{-3x}(2x - 3x^2) + e^{-3x}(2 - 6x) = e^{-3x}(9x^2 - 12x + 2)\). At \(x = 0\), \(\frac{d^2y}{dx^2} = e^{0}(2) = 2 > 0\), which is positive, so \((0, 0)\) is a local minimum. At \(x = \frac{2}{3}\), \(\frac{d^2y}{dx^2} = e^{-2}\left(9\left(\frac{4}{9}\right) - 12\left(\frac{2}{3}\right) + 2\right) = -2e^{-2} < 0\), which is negative, so \(\left(\frac{2}{3}, \frac{4}{9e^2}\right)\) is a local maximum.

M1: Applies the product rule to differentiate the function. A1: Obtains the correct derivative \(\frac{dy}{dx} = xe^{-3x}(2-3x)\). M1: Sets \(\frac{dy}{dx} = 0\) and solves for \(x\). A1: Finds both stationary coordinates \((0, 0)\) and \(\left(\frac{2}{3}, \frac{4}{9e^2}\right)\) exactly. M1: Finds the second derivative \(\frac{d^2y}{dx^2}\). A1: Correctly simplifies the second derivative to \(e^{-3x}(9x^2 - 12x + 2)\). A1.75: Correctly evaluates the sign of the second derivative at both points and concludes that \((0, 0)\) is a minimum and \(\left(\frac{2}{3}, \frac{4}{9e^2}\right)\) is a maximum.

(i) Displacement \(s\) is the integral of velocity with respect to time: \(s = \int (8 \cos(2t) - 4) \, dt = 4 \sin(2t) - 4t + C\). When \(t = 0\), the particle is at the point \(O\), so \(s = 0\). This gives \(0 = 4 \sin(0) - 4(0) + C \implies C = 0\). The displacement expression is \(s = 4 \sin(2t) - 4t\). For \(t = \frac{\pi}{4}\): \(s = 4 \sin\left(\frac{\pi}{2}\right) - 4\left(\frac{\pi}{4}\right) = 4(1) - \pi = 4 - \pi\text{ m}\). (ii) Acceleration \(a\) is the derivative of velocity with respect to time: \(a = \frac{dv}{dt} = \frac{d}{dt}(8 \cos(2t) - 4) = -16 \sin(2t)\). For \(t = \frac{\pi}{3}\): \(a = -16 \sin\left(\frac{2\pi}{3}\right) = -16 \left(\frac{\sqrt{3}}{2}\right) = -8\sqrt{3}\text{ m s}^{-2}\).

M1: Attempts to integrate \(v\) with respect to \(t\). A1: Obtains the correct integrated expression \(4\sin(2t) - 4t\). M1: Uses the initial condition \(s = 0\) at \(t = 0\) to show \(C = 0\). A1: Evaluates the displacement at \(t = \frac{\pi}{4}\) to obtain \(4 - \pi\). M1: Attempts to differentiate \(v\) to obtain the acceleration function. A1: Obtains the correct acceleration function \(a = -16\sin(2t)\). A1.75: Substitutes \(t = \frac{\pi}{3}\) to find the exact acceleration value \(-8\sqrt{3}\).

Equating the equations of the line and the curve:
\[x^2 + kx + 5 = 2x + k\]

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

For no points of intersection, the discriminant must be less than zero:
\[(k-2)^2 - 4(1)(5-k) < 0\]

Expanding and simplifying:
\[k^2 - 4k + 4 - 20 + 4k < 0\]
\[k^2 - 16 < 0\]

Solving this inequality gives:
\[-4 < k < 4\]

M1: Equating the line and curve equations and rearranging into quadratic form \(x^2 + (k-2)x + (5-k) = 0\) (allow one sign error).

M1: Correct use of discriminant \(b^2 - 4ac < 0\).

A1: Correct simplified quadratic inequality \(k^2 - 16 < 0\).

A1: Correct final range \(-4 < k < 4\) (or equivalent interval notation).

Express all terms as powers of the same base. For the first equation:
\[2^x \div (2^2)^y = 2^{-3} \Rightarrow 2^{x-2y} = 2^{-3}\]
This simplifies to:
\[x - 2y = -3 \quad \text{(Equation 1)}\]

For the second equation:
\[(3^2)^x \times 3^y = 3^4 \Rightarrow 3^{2x+y} = 3^4\]
This simplifies to:
\[2x + y = 4 \quad \text{(Equation 2)}\]

From Equation 2, express \(y\) in terms of \(x\):
\[y = 4 - 2x\]

Substitute this into Equation 1:
\[x - 2(4 - 2x) = -3\]
\[x - 8 + 4x = -3\]
\[5x = 5 \Rightarrow x = 1\]

Substitute \(x = 1\) back into the expression for \(y\):
\[y = 4 - 2(1) = 2\]

Therefore, the solution is \(x = 1\) and \(y = 2\).

M1: Correctly converting the first equation to index form \(x - 2y = -3\).

M1: Correctly converting the second equation to index form \(2x + y = 4\).

M1: Attempting to solve the two linear equations simultaneously.

A1: Obtaining correct values \(x = 1\) and \(y = 2\).

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

Using the Remainder Theorem, since dividing by \((x+1)\) leaves a remainder of \(-15\), we have \(p(-1) = -15\):
\[2(-1)^3 + a(-1)^2 + b(-1) - 6 = -15\]
\[-2 + a - b - 6 = -15\]
\[a - b = -7 \quad \text{(Equation 2)}\]

Add Equation 1 and Equation 2 to eliminate \(b\):
\[(2a + b) + (a - b) = -5 + (-7)\]
\[3a = -12 \Rightarrow a = -4\]

Substitute \(a = -4\) into Equation 2:
\[-4 - b = -7 \Rightarrow b = 3\]

So, \(a = -4\) and \(b = 3\).

M1: Substituting \(x = 2\) into \(p(x)\) and setting the expression to \(0\) to obtain a linear equation in \(a\) and \(b\).

M1: Substituting \(x = -1\) into \(p(x)\) and setting the expression to \(-15\) to obtain a second linear equation in \(a\) and \(b\).

M1: Solving the resulting simultaneous equations to find the value of \(a\) or \(b\).

A1: Finding the correct values \(a = -4\) and \(b = 3\).

(a) Rewrite the line equation as \(y = 2x + 5\).
Substitute this into the circle equation:
\(x^2 + (2x + 5)^2 - 2x - 4(2x + 5) - 20 = 0\)
\(x^2 + 4x^2 + 20x + 25 - 2x - 8x - 20 - 20 = 0\)
\(5x^2 + 10x - 15 = 0\)
Dividing by 5 gives:
\(x^2 + 2x - 3 = 0\)
\((x + 3)(x - 1) = 0\)
This yields \(x = -3\) and \(x = 1\).

For \(x = -3\):
\(y = 2(-3) + 5 = -1\)
For \(x = 1\):
\(y = 2(1) + 5 = 7\)

Thus, the coordinates of the points of intersection are \(A(-3, -1)\) and \(B(1, 7)\).

(b) The midpoint of the chord \(AB\) is:
\(M = \left(\frac{-3 + 1}{2}, \frac{-1 + 7}{2}\right) = (-1, 3)\)

The gradient of the line \(L\) is \(2\). The gradient of the perpendicular bisector is:
\(m = -\frac{1}{2}\)

The equation of the perpendicular bisector is:
\(y - 3 = -\frac{1}{2}(x - (-1))\)
\(y - 3 = -\frac{1}{2}x - \frac{1}{2}\)
\(y = -\frac{1}{2}x + \frac{5}{2}\) (or \(x + 2y - 5 = 0\)).

(c) The equation of the circle can be written as:
\((x-1)^2 - 1 + (y-2)^2 - 4 - 20 = 0 \implies (x-1)^2 + (y-2)^2 = 25\)
The center of the circle is \((1, 2)\).

Substitute the center \((1, 2)\) into the equation of the perpendicular bisector \(x + 2y - 5 = 0\):
\(1 + 2(2) - 5 = 1 + 4 - 5 = 0\)

Since the equation is satisfied, the perpendicular bisector of \(AB\) passes through the center of the circle.

(a)
M1: For substituting \(y = 2x + 5\) into the circle equation.
A1: For obtaining a correct simplified quadratic equation, e.g., \(5x^2 + 10x - 15 = 0\) or \(x^2 + 2x - 3 = 0\).
M1: For attempting to solve their 3-term quadratic equation.
A1: For both correct \(x\)-values (\(x = -3\) and \(x = 1\)).
A1: For both correct pairs of coordinates, \((-3, -1)\) and \((1, 7)\).

(b)
M1: For finding the midpoint of \(AB\) as \((-1, 3)\).
M1: For using the perpendicular gradient rule to get gradient \(-\frac{1}{2}\).
A1: For a correct equation of the perpendicular bisector, e.g., \(y = -\frac{1}{2}x + \frac{5}{2}\) or \(x + 2y - 5 = 0\).

(c)
B1: For identifying the center of the circle as \((1, 2)\).
B1: For substituting the center into the equation of the perpendicular bisector and showing that it lies on the line.

(a)(i) The word ALGORITHM has 9 distinct letters. The number of 5-letter codes with no restrictions is given by \(^{9}P_{5} = 9 \times 8 \times 7 \times 6 \times 5 = 15120\). (a)(ii) The letters consist of 3 vowels (A, O, I) and 6 consonants (L, G, R, T, H, M). The 5-letter code must end with a consonant, so there are 6 options for the last position. The letter 'O' must be in the code, and since it is not a consonant, it must be placed in one of the first 4 positions (4 options). The remaining 3 positions are filled by choosing and arranging 3 letters from the remaining 7 available letters, which can be done in \(^{7}P_{3} = 7 \times 6 \times 5 = 210\) ways. Thus, the total number of codes is \(6 \times 4 \times 210 = 5040\). (b)(i) The total number of people is \(5 + 7 = 12\). The number of ways to choose 6 people with no restrictions is \(^{12}C_{6} = 924\). (b)(ii) For a committee with at least 4 women, we consider three cases: Case 1: 4 women and 2 men: \(^{7}C_{4} \times ^{5}C_{2} = 35 \times 10 = 350\). Case 2: 5 women and 1 man: \(^{7}C_{5} \times ^{5}C_{1} = 21 \times 5 = 105\). Case 3: 6 women and 0 men: \(^{7}C_{6} \times ^{5}C_{0} = 7 \times 1 = 7\). Total number of committees = \(350 + 105 + 7 = 462\).

(a)(i) B1: For 15120 (or \(^{9}P_{5}\)). (a)(ii) M1: For identifying 6 options for the last position and 4 options for placing 'O' (e.g., seeing \(6 \times 4\)). M1: For multiplying by \(^{7}P_{3}\) (or 210) to fill the remaining positions. A1: For 5040. (b)(i) B1: For 924 (or \(^{12}C_{6}\)). (b)(ii) M1: For attempting to calculate at least two correct cases of the form \(^{7}C_{r} \times ^{5}C_{6-r}\) for \(r \ge 4\). M1: For adding three correct cases: \(^{7}C_{4}\times^{5}C_{2} + ^{7}C_{5}\times^{5}C_{1} + ^{7}C_{6}\times^{5}C_{0}\) (or equivalent subtraction from total). A1: For 462.

(i) Expressing \(\mathrm{f}(x)\) by completing the square: \(\mathrm{f}(x) = 2(x-3)^2 - 7\). For the inverse function to exist, \(\mathrm{f}\) must be a one-to-one function. The vertex of the quadratic curve is at \((3, -7)\). Since the domain is defined for \(x \le k\), the largest value of \(k\) is the x-coordinate of the vertex, which is \(3\). (ii) Let \(y = 2(x-3)^2 - 7\). Rearranging to make \(x\) the subject: \(y + 7 = 2(x-3)^2\) which gives \(\frac{y+7}{2} = (x-3)^2\). Taking the square root, since \(x \le 3\), we must choose the negative root: \(x - 3 = -\sqrt{\frac{y+7}{2}}\) hence \(x = 3 - \sqrt{\frac{y+7}{2}}\). Replacing \(y\) with \(x\), we obtain \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\). (iii) The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\). Since the vertex is at \((3, -7)\) and the curve opens upwards, the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \ge -7\). Thus, the domain of \(\mathrm{f}^{-1}\) is \(x \ge -7\). (iv) First find \(\mathrm{f}(2)\): \(\mathrm{f}(2) = 2(2)^2 - 12(2) + 11 = 8 - 24 + 11 = -5\). Then \(\mathrm{ff}(2) = \mathrm{f}(-5) = 2(-5)^2 - 12(-5) + 11 = 50 + 60 + 11 = 121\).

(i) M1: For attempting to find the x-coordinate of the vertex (e.g., by completing the square or using \(x = -b/(2a)\)). A1: For \(k = 3\). (ii) M1: For attempting to make \(x\) the subject of the quadratic expression. M1: For correctly identifying the negative square root due to the domain \(x \le 3\). A1: For \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\) (or equivalent). (iii) B1: For \(x \ge -7\) (accept interval notation \([-7, \infty)\)). (iv) B1: For 121.

(a) Start with the left-hand side (LHS) and combine the fractions using a common denominator:
\[\text{LHS} = \frac{\sin \theta(1 + \cos \theta) - \sin \theta(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}\]

Expand the numerator and simplify the denominator:
\[\text{LHS} = \frac{\sin \theta + \sin \theta \cos \theta - \sin \theta + \sin \theta \cos \theta}{1 - \cos^2 \theta}\]
\[\text{LHS} = \frac{2\sin \theta \cos \theta}{\sin^2 \theta}\]

Cancel \(\sin \theta\) from the numerator and denominator:
\[\text{LHS} = \frac{2\cos \theta}{\sin \theta} = 2\cot \theta = \text{RHS}\]

(b) Using the identity from part (a) with \(\theta = 2x\), the equation
\[\frac{\sin 2x}{1 - \cos 2x} - \frac{\sin 2x}{1 + \cos 2x} = 3\tan 2x\]

can be written as:
\[2\cot 2x = 3\tan 2x\]

Using the identity \(\cot 2x = \frac{1}{\tan 2x}\), we get:
\[\frac{2}{\tan 2x} = 3\tan 2x\]
\[3\tan^2 2x = 2\]
\[\tan^2 2x = \frac{2}{3}\]
\[\tan 2x = \pm\sqrt{\frac{2}{3}} \approx \pm 0.8165\]

For \(0^\circ \le x \le 180^\circ\), the range for \(2x\) is \(0^\circ \le 2x \le 360^\circ\).

The reference angle \(\alpha\) is:
\[\alpha = \tan^{-1}\left(\sqrt{\frac{2}{3}}\right) \approx 39.23^\circ\]

For \(\tan 2x = \sqrt{\frac{2}{3}}\):
\[2x = 39.23^\circ \implies x \approx 19.6^\circ\]
\[2x = 180^\circ + 39.23^\circ = 219.23^\circ \implies x \approx 109.6^\circ\]

For \(\tan 2x = -\sqrt{\frac{2}{3}}\):
\[2x = 180^\circ - 39.23^\circ = 140.77^\circ \implies x \approx 70.4^\circ\]
\[2x = 360^\circ - 39.23^\circ = 320.77^\circ \implies x \approx 160.4^\circ\]

Thus, the solutions are \(x = 19.6^\circ, 70.4^\circ, 109.6^\circ, 160.4^\circ\).

**Part (a)**
* **M1**: For combining fractions into a single fraction with a correct common denominator, e.g., \(\frac{\sin \theta(1 + \cos \theta) - \sin \theta(1 - \cos \theta)}{1 - \cos^2 \theta}\).
* **M1**: For correct algebraic simplification of the numerator to \(2\sin \theta \cos \theta\) and utilizing the identity \(1 - \cos^2 \theta = \sin^2 \theta\).
* **A1**: For completing the proof to obtain \(2\cot \theta\) convincingly with no errors seen.

**Part (b)**
* **M1**: For using the identity from part (a) to rewrite the equation as \(2\cot 2x = 3\tan 2x\).
* **M1**: For converting the equation into the form \(\tan^2 2x = k\) (where \(k = \frac{2}{3}\)).
* **A1**: For finding the correct reference angle or any two correct values of \(2x\) (e.g., \(2x \approx 39.2^\circ\) and \(2x \approx 219.2^\circ\)).
* **A1**: For any two correct solutions for \(x\) rounded to 1 decimal place (e.g., \(19.6^\circ\) and \(109.6^\circ\)).
* **A1**: For obtaining all four correct solutions \(19.6^\circ, 70.4^\circ, 109.6^\circ, 160.4^\circ\) and no extra solutions in the range.

First, find the \(y\)-coordinate of the curve at \(x = 4\):
\(y = \frac{3(4) - 2}{\sqrt{2(4) + 1}} = \frac{10}{\sqrt{9}} = \frac{10}{3}\).

Now, differentiate \(y\) with respect to \(x\) using the quotient rule:
Let \(u = 3x - 2 \implies \frac{du}{dx} = 3\)
Let \(v = (2x + 1)^{1/2} \implies \frac{dv}{dx} = \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 = (2x + 1)^{-1/2}\)

\(\frac{dy}{dx} = \frac{u'v - uv'}{v^2}\)
\(\frac{dy}{dx} = \frac{3(2x + 1)^{1/2} - (3x - 2)(2x + 1)^{-1/2}}{2x + 1}\)

Multiply the numerator and denominator by \((2x + 1)^{1/2}\):
\(\frac{dy}{dx} = \frac{3(2x + 1) - (3x - 2)}{(2x + 1)^{3/2}} = \frac{6x + 3 - 3x + 2}{(2x + 1)^{3/2}} = \frac{3x + 5}{(2x + 1)^{3/2}}\)

Find the gradient of the tangent at \(x = 4\):
\(m = \frac{3(4) + 5}{(2(4) + 1)^{3/2}} = \frac{17}{9^{3/2}} = \frac{17}{27}\)

The equation of the tangent is:
\(y - y_1 = m(x - x_1)\)
\(y - \frac{10}{3} = \frac{17}{27}(x - 4)\)

Multiply through by 27 to clear the fractions:
\(27y - 90 = 17(x - 4)\)
\(27y - 90 = 17x - 68\)
\(17x - 27y + 22 = 0\)

M1: Attempt to find the \(y\)-coordinate when \(x = 4\).
M1: Attempt to differentiate using the quotient rule or product rule.
A1: Correct simplified derivative, \(\frac{dy}{dx} = \frac{3x + 5}{(2x + 1)^{3/2}}\).
M1: Substitute \(x = 4\) into their derivative to find the gradient of the tangent.
A1: Correct equation of the tangent in the required form, \(17x - 27y + 22 = 0\) (or any integer multiple).

Use the double-angle trigonometric identity \(\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}\) to rewrite the integrand:
\(8\cos^2(2x) = 8 \left( \frac{1 + \cos(4x)}{2} \right) = 4 + 4\cos(4x)\).

Now, integrate each term with respect to \(x\):
\(\int (4 + 4\cos(4x)) \, dx = 4x + \sin(4x) + c\).

Evaluate the definite integral from \(0\) to \(\frac{\pi}{8}\):
\(\left[ 4x + \sin(4x) \right]_{0}^{\frac{\pi}{8}} = \left( 4\left(\frac{\pi}{8}\right) + \sin\left(4 \cdot \frac{\pi}{8}\right) \right) - \left( 4(0) + \sin(0) \right)\)
\(= \left( \frac{\pi}{2} + \sin\left(\frac{\pi}{2}\right) \right) - (0 + 0)\)
\(= \frac{\pi}{2} + 1\).

M1: Use the correct identity \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\) or equivalent to rewrite the integrand.
A1: Obtain the correct rewritten expression \(4 + 4\cos(4x)\).
M1: Integrate to obtain the form \(4x + k\sin(4x)\), where \(k\) is a constant.
A1: Obtain the correct integrated term \(4x + \sin(4x)\).
A1: Substitute the limits correctly to obtain the exact value \(\frac{\pi}{2} + 1\).

To find the stationary point, first determine the derivative \(\frac{dy}{dx}\) using the product rule:
\(\frac{dy}{dx} = (1)e^{-x} + (x - 2)(-e^{-x})\)
\(\frac{dy}{dx} = e^{-x}(1 - (x - 2))\)
\(\frac{dy}{dx} = (3 - x)e^{-x}\)

Set \(\frac{dy}{dx} = 0\) to find the stationary points:
\((3 - x)e^{-x} = 0\)

Since \(e^{-x} \neq 0\) for all real \(x\):
\(3 - x = 0 \implies x = 3\)

Find the corresponding \(y\)-coordinate:
\(y = (3 - 2)e^{-3} = e^{-3}\)

So, the coordinates of the stationary point are \((3, e^{-3})\).

To determine its nature, find the second derivative \(\frac{d^2y}{dx^2}\) using the product rule:
\(\frac{d^2y}{dx^2} = (-1)e^{-x} + (3 - x)(-e^{-x})\)
\(\frac{d^2y}{dx^2} = (x - 4)e^{-x}\)

Substitute \(x = 3\) into the second derivative:
\(\frac{d^2y}{dx^2} = (3 - 4)e^{-3} = -e^{-3}\)

Since \(-e^{-3} < 0\), the stationary point is a local maximum.

M1: Use the product rule to differentiate \(y = (x - 2)e^{-x}\).
A1: Obtain the correct derivative \(\frac{dy}{dx} = (3 - x)e^{-x}\).
M1: Set \(\frac{dy}{dx} = 0\) to find \(x = 3\) and evaluate the exact \(y\)-coordinate as \(e^{-3}\).
M1: Find the second derivative (or use a first-derivative sign test) and substitute their \(x\) value.
A1: Correctly identify the stationary point as a maximum with a valid mathematical justification.

(i) Since \(OP : PA = 2 : 1\), we have \(\vec{OP} = \frac{2}{3}\mathbf{a}\).
Since \(OQ : QB = 1 : 3\), we have \(\vec{OQ} = \frac{1}{4}\mathbf{b}\).
Therefore:
\(\vec{AQ} = \vec{AO} + \vec{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b} = \frac{1}{4}\mathbf{b} - \mathbf{a}\)
\(\vec{BP} = \vec{BO} + \vec{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\)

(ii) Using \(\vec{AX} = \lambda \vec{AQ}\):
\(\vec{OX} = \vec{OA} + \vec{AX} = \mathbf{a} + \lambda\left(-\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = (1-\lambda)\mathbf{a} + \frac{1}{4}\lambda\mathbf{b}\)

Using \(\vec{PX} = \mu \vec{PB}\):
Since \(\vec{PB} = -\vec{BP} = \mathbf{b} - \frac{2}{3}\mathbf{a}\):
\(\vec{OX} = \vec{OP} + \vec{PX} = \frac{2}{3}\mathbf{a} + \mu\left(\mathbf{b} - \frac{2}{3}\mathbf{a}\right) = \frac{2}{3}(1-\mu)\mathbf{a} + \mu\mathbf{b}\)

(iii) Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
From the \(\mathbf{b}\) components: \(\frac{1}{4}\lambda = \mu \implies \lambda = 4\mu\)
From the \(\mathbf{a}\) components: \(1-\lambda = \frac{2}{3}(1-\mu)\)
Substituting \(\lambda = 4\mu\) into the second equation:
\(1 - 4\mu = \frac{2}{3} - \frac{2}{3}\mu\)
Multiply by 3:
\(3 - 12\mu = 2 - 2\mu\)
\(1 = 10\mu \implies \mu = \frac{1}{10}\)
Substituting back to find \(\lambda\):
\(\lambda = 4\left(\frac{1}{10}\right) = \frac{2}{5}\)

(iv) We have \(\vec{PX} = \mu \vec{PB} = \frac{1}{10}\vec{PB}\).
This means that \(X\) lies on the line segment \(PB\) such that the distance \(PX\) is \(\frac{1}{10}\) of the total length \(PB\).
Therefore, \(XB = PB - PX = \frac{9}{10}PB\).
Thus, the ratio \(PX : XB = 1 : 9\).

(i)
B1: For \(\vec{AQ} = \frac{1}{4}\mathbf{b} - \mathbf{a}\) or equivalent.
B1: For \(\vec{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\) or equivalent.

(ii)
M1: For a valid vector path to express \(\vec{OX}\) in terms of \(\lambda\) or \(\mu\).
A1: For \(\vec{OX} = (1-\lambda)\mathbf{a} + \frac{1}{4}\lambda\mathbf{b}\) or equivalent.
A1: For \(\vec{OX} = \frac{2}{3}(1-\mu)\mathbf{a} + \mu\mathbf{b}\) or equivalent.

(iii)
M1: For equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form two simultaneous equations.
M1: For solving the simultaneous equations to find one variable.
A1: For obtaining both \(\lambda = \frac{2}{5}\) (or \(0.4\)) and \(\mu = \frac{1}{10}\) (or \(0.1\)).

(iv)
B1: For the ratio \(1 : 9\).