2025 Cambridge IGCSE Mathematics - Additional (0606) 模拟试题及答案详解

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

Since the remainder when \(\mathrm{p}(x)\) is divided by \(x + 1\) is \(-12\), by the Remainder Theorem we have:
\(\mathrm{p}(-1) = -12\)
\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12\)

-2 + a - b - 6 = -12\)

a - b = -4\) --- (Equation 2)

Adding Equation 1 and Equation 2:
\((2a + b) + (a - b) = -5 + (-4)\)
\(3a = -9 \implies a = -3\)

Substituting \(a = -3\) into Equation 2:
\(-3 - b = -4 \implies b = 1\)

M1: Attempts to use the factor theorem with \(p(2) = 0\) to obtain a linear equation in \(a\) and \(b\).
A1: Correctly obtains \(2a + b = -5\) (or equivalent).
M1: Attempts to use the remainder theorem with \(p(-1) = -12\) to obtain a second linear equation in \(a\) and \(b\), and solves the simultaneous equations.
A1: Correctly finds \(a = -3\) and \(b = 1\).

Using the laws of logarithms:
\(2\log_3 x = \log_3 (x^2)\)

So the equation becomes:
\(\log_3 (x^2) - \log_3 (x + 6) = 1\)
\(\log_3 \left(\frac{x^2}{x+6}\right) = 1\)

Convert from logarithmic to exponential form:
\(\frac{x^2}{x+6} = 3^1\)

\(x^2 = 3(x + 6)\)

\(x^2 - 3x - 18 = 0\)

Factorise the quadratic equation:
\((x - 6)(x + 3) = 0\)

This gives \(x = 6\) or \(x = -3\).

Since the domain of \(\log_3 x\) requires \(x > 0\), the only valid solution is \(x = 6\).

M1: Applies the power law of logarithms to write \(2\log_3 x\) as \(\log_3 (x^2)\).
M1: Applies the quotient law of logarithms to combine terms into a single logarithm.
M1: Eliminates the logarithm to form a quadratic equation and attempts to solve it.
A1: Correctly identifies \(x = 6\) as the only valid solution (rejecting \(x = -3\)).

Expand the inequality:
\(2x^2 + 6x - x - 3 > 15\)

\(2x^2 + 5x - 3 > 15\)

Rearrange to form a quadratic inequality:
\(2x^2 + 5x - 18 > 0\)

Factorise the quadratic expression:
\((2x + 9)(x - 2) > 0\)

The critical values where the expression equals zero are \(x = -\frac{9}{2} = -4.5\) and \(x = 2\).

Since we require the expression to be greater than zero, the solution set is:
\(x < -4.5\) or \(x > 2\).

M1: Expands and rearranges the inequality into the form \(2x^2 + 5x - 18 > 0\).
M1: Factorises the quadratic expression to find critical values of \(-4.5\) and \(2\).
A1: Correctly identifies the range of values as \(x < -4.5\) or \(x > 2\) (or equivalent interval notation).

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

We are given:
\(u_3 = ar^2 = 18\) --- (1)

\(u_6 = ar^5 = -\frac{16}{3}\) --- (2)

Divide (2) by (1):
\(\frac{ar^5}{ar^2} = \frac{-16/3}{18}\)

\(r^3 = -\frac{16}{54} = -\frac{8}{27}\)

Taking the cube root of both sides:
\(r = -\frac{2}{3}\)

Substitute \(r = -\frac{2}{3}\) back into (1) to find \(a\):
\(a\left(-\frac{2}{3}\right)^2 = 18\)

\(a\left(\frac{4}{9}\right) = 18 \implies a = 18 \times \frac{9}{4} = \frac{81}{2}\)

The sum to infinity \(S_{\infty}\) is given by:
\(S_{\infty} = \frac{a}{1 - r} = \frac{81/2}{1 - (-2/3)} = \frac{81/2}{5/3} = \frac{81}{2} \times \frac{3}{5} = \frac{243}{10} = 24.3\)

M1: Sets up the equations for \(u_3\) and \(u_6\) and divides them to obtain an expression for \(r^3\).
A1: Correctly finds \(r = -\frac{2}{3}\).
M1: Substitutes \(r\) back to find \(a = \frac{81}{2}\) and uses the sum to infinity formula.
A1: Correctly calculates \(S_{\infty} = 24.3\) (or \(\frac{243}{10}\)).

Let \(y = 3 + \ln(x - 1)\).

We want to rearrange the equation to make \(x\) the subject:
\(y - 3 = \ln(x - 1)\)

Take the exponential of both sides:
\(\mathrm{e}^{y-3} = x - 1\)

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

Replace \(y\) with \(x\) to find the inverse function:
\(\mathrm{f}^{-1}(x) = 1 + \mathrm{e}^{x-3}\)

M1: Sets \(y = 3 + \ln(x - 1)\) and attempts to make \(\ln(x-1)\) the subject.
M1: Converts the logarithmic equation into exponential form correctly.
A1: Obtains \(\mathrm{f}^{-1}(x) = 1 + \mathrm{e}^{x-3}\) (or equivalent).

Using the product rule to differentiate \(y = (2x - 3)(4x + 1)^{1/2}\):

Let \(u = 2x - 3 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2\)

Let \(v = (4x + 1)^{1/2} \implies \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(4x + 1)^{-1/2} \times 4 = 2(4x + 1)^{-1/2}\)

Applying the product rule \(\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\):

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

Substitute \(x = 2\) into the derivative:

\(4x + 1 = 4(2) + 1 = 9\)

\(2x - 3 = 2(2) - 3 = 1\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = (1) \left(2(9)^{-1/2}\right) + 2(9)^{1/2}\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = (1) \left(\frac{2}{3}\right) + 2(3) = \frac{2}{3} + 6 = \frac{20}{3}\)

M1: Recognises the need for product rule and differentiates \(2x-3\) and/or \(\sqrt{4x+1}\) correctly.
A1: Correct unsimplified derivative expression.
M1: Substitute \(x = 2\) into their derivative.
A1: Correctly evaluates the gradient as \(\frac{20}{3}\) (or equivalent exact form).

Use the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(2(1 - \sin^2 \theta) + 3\sin \theta = 3\)

\(2 - 2\sin^2 \theta + 3\sin \theta = 3\)

Rearranging gives a quadratic in \(\sin \theta\):
\(2\sin^2 \theta - 3\sin \theta + 1 = 0\)

Factorise the quadratic:
\((2\sin \theta - 1)(\sin \theta - 1) = 0\)

This gives:
\(\sin \theta = \frac{1}{2}\) or \(\sin \theta = 1\)

For \(\sin \theta = \frac{1}{2}\):
\(\theta = 30^\circ, 150^\circ\)

For \(\sin \theta = 1\):
\(\theta = 90^\circ\)

All values are within the range \(0^\circ \le \theta \le 360^\circ\).

So, \(\theta = 30^\circ, 90^\circ, 150^\circ\).

M1: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to form an equation in terms of \(\sin \theta\) only.
M1: Solves the quadratic equation to obtain two values for \(\sin \theta\).
A1: Identifies \(\sin \theta = \frac{1}{2}\) and \(\sin \theta = 1\).
A1: Correctly finds all three angles: \(30^\circ\), \(90^\circ\), and \(150^\circ\).

Equate the two expressions for \(y\):
\(2x^2 - 3x + 4 = 2x + 7\)

Rearrange the terms to set the quadratic to zero:
\(2x^2 - 5x - 3 = 0\)

Factorise the quadratic expression:
\((2x + 1)(x - 3) = 0\)

This gives the \(x\)-coordinates of the intersection points:
\(x = -\frac{1}{2} = -0.5\) or \(x = 3\)

Substitute each \(x\)-value back into the linear equation \(y = 2x + 7\) to find the corresponding \(y\)-values:

For \(x = -0.5\):
\(y = 2(-0.5) + 7 = -1 + 7 = 6\)

For \(x = 3\):
\(y = 2(3) + 7 = 6 + 7 = 13\)

So the solutions are \((-0.5, 6)\) and \((3, 13)\).

M1: Equates the two equations to eliminate \(y\) and form a quadratic in \(x\).
A1: Correctly simplifies to \(2x^2 - 5x - 3 = 0\) (or equivalent).
M1: Solves the quadratic to find two values of \(x\) and attempts to find corresponding \(y\) values.
A1: Correctly obtains both coordinate pairs: \((-0.5, 6)\) and \((3, 13)\).

Using the Factor Theorem, since \(x - 2\) is a factor of \(\mathrm{p}(x)\), we have: \(\mathrm{p}(2) = 0\) which gives \(3(2)^3 + a(2)^2 - 12(2) + b = 0 \implies 24 + 4a - 24 + b = 0 \implies 4a + b = 0\) (Equation 1). Using the Remainder Theorem, since dividing \(\mathrm{p}(x)\) by \(x + 1\) gives a remainder of \(-9\), we have: \(\mathrm{p}(-1) = -9\) which gives \(3(-1)^3 + a(-1)^2 - 12(-1) + b = -9 \implies -3 + a + 12 + b = -9 \implies a + b + 9 = -9 \implies a + b = -18\) (Equation 2). Subtracting Equation 2 from Equation 1 gives: \(3a = 18 \implies a = 6\). Substituting \(a = 6\) into Equation 1 gives: \(4(6) + b = 0 \implies b = -24\).

B1: for \(4a + b = 0\) (or equivalent) derived from \(\mathrm{p}(2) = 0\). B1: for \(a + b = -18\) (or equivalent) derived from \(\mathrm{p}(-1) = -9\). M1: for a valid method to solve their two simultaneous linear equations. A1: for \(a = 6\) and \(b = -24\).

Using the laws of logarithms, we can combine the terms on the left-hand side: \(\log_4((y + 2)(y - 4)) = 2\). Convert the logarithmic equation into exponential form: \((y + 2)(y - 4) = 4^2 \implies y^2 - 2y - 8 = 16\). Rearrange into a quadratic equation: \(y^2 - 2y - 24 = 0\). Factorising gives: \((y - 6)(y + 4) = 0\), which yields potential solutions \(y = 6\) or \(y = -4\). Checking these solutions in the original equation: for \(y = 6\), both logarithm arguments are positive (valid); for \(y = -4\), the arguments are negative (invalid). Thus, the only valid solution is \(y = 6\).

M1: for correctly combining logarithms to obtain \(\log_4((y + 2)(y - 4)) = 2\). M1: for converting to exponential form: \((y + 2)(y - 4) = 16\). A1: for solving the quadratic equation to obtain \(y = 6\) and \(y = -4\). A1: for identifying \(y = 6\) as the only valid solution and rejecting \(y = -4\).

First, expand the expression: \(2x^2 + 6x - x - 3 > 4 \implies 2x^2 + 5x - 3 > 4\). Rearrange to get zero on the right: \(2x^2 + 5x - 7 > 0\). Factorising the quadratic gives: \((2x + 7)(x - 1) > 0\). The critical values are \(x = -3.5\) and \(x = 1\). Since the inequality is \(> 0\), the set of values is \(x < -3.5\) or \(x > 1\).

M1: for expanding and rearranging to obtain \(2x^2 + 5x - 7 > 0\). M1: for finding the critical values \(x = -3.5\) and \(x = 1\). A1: for the correct final set of values: \(x < -3.5\) or \(x > 1\).

Using the quotient rule with \(u = \ln(3x - 1)\) and \(v = x\), we have \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{3}{3x - 1}\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 1\). The quotient rule gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x \cdot \frac{3}{3x - 1} - \ln(3x - 1) \cdot 1}{x^2}\). Substituting \(x = 1\) into the derivative gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 \cdot \frac{3}{3(1) - 1} - \ln(3(1) - 1)}{1^2} = \frac{3}{2} - \ln(2)\).

B1: for finding the derivative of the numerator: \(\frac{3}{3x - 1}\). M1: for a correct application of the quotient rule. M1: for substituting \(x = 1\) into their derivative. A1: for the correct exact value: \(1.5 - \ln(2)\) or equivalent exact forms.

Using the Factor Theorem, since \(p(x)\) is divisible by \(x - 2\), we have \(p(2) = 0\):
\(3(2)^3 + a(2)^2 + b(2) - 12 = 0\)
\(24 + 4a + 2b - 12 = 0\)
\(4a + 2b = -12\)
\(2a + b = -6\) --- (Equation 1)

Using the Remainder Theorem, since \(p(x)\) divided by \(x + 1\) has a remainder of \(-18\), we have \(p(-1) = -18\):
\(3(-1)^3 + a(-1)^2 + b(-1) - 12 = -18\)
\(-3 + a - b - 12 = -18\)
\(a - b = -3\) --- (Equation 2)

Add Equation 1 and Equation 2:
\((2a + b) + (a - b) = -6 + (-3)\)
\(3a = -9 \implies a = -3\)

Substitute \(a = -3\) into Equation 2:
\(-3 - b = -3 \implies b = 0\)

Thus, \(a = -3\) and \(b = 0\).

B1: for setting \(p(2) = 0\) and obtaining a correct simplified linear equation in \(a\) and \(b\), e.g. \(2a + b = -6\).
B1: for setting \(p(-1) = -18\) and obtaining a correct simplified linear equation in \(a\) and \(b\), e.g. \(a - b = -3\).
M1: for a valid method to solve their simultaneous equations to find at least one of the variables.
A1: for both \(a = -3\) and \(b = 0\) correctly found.

First, use the change of base formula to write \(\log_4(x - 1)\) in base 2:
\(\log_4(x - 1) = \frac{\log_2(x - 1)}{\log_2 4} = \frac{1}{2}\log_2(x - 1)\)

Substitute this back into the original equation:
\(\log_2(x + 3) - \frac{1}{2}\log_2(x - 1) = 2\)

Multiply the entire equation by 2:
\(2\log_2(x + 3) - \log_2(x - 1) = 4\)

Apply the power law and the quotient law of logarithms:
\(\log_2\left(\frac{(x + 3)^2}{x - 1}\right) = 4\)

Convert the logarithmic equation to an exponential equation:
\(\frac{(x + 3)^2}{x - 1} = 2^4 = 16\)

Expand and simplify to form a quadratic equation:
\(x^2 + 6x + 9 = 16(x - 1)\)
\(x^2 + 6x + 9 = 16x - 16\)
\(x^2 - 10x + 25 = 0\)

Factorise the quadratic:
\((x - 5)^2 = 0 \implies x = 5\)

Since \(x = 5\) satisfies the initial constraints \(x + 3 > 0\) and \(x - 1 > 0\), the solution is \(x = 5\).

B1: for using the change of base law correctly to show \(\log_4(x-1) = \frac{1}{2}\log_2(x-1)\).
M1: for combining terms into a single logarithm using correct laws, e.g., \(\log_2\frac{(x+3)^2}{x-1} = 4\).
M1: for removing the logarithm to form a quadratic equation, e.g., \((x+3)^2 = 16(x-1)\).
A1: for solving the quadratic correctly to obtain \(x=5\) only.

Using the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\), substitute this into the equation:
\(2(1 - \sin^2 \theta) + \sin \theta - 1 = 0\)
\(2 - 2\sin^2 \theta + \sin \theta - 1 = 0\)
\(-2\sin^2 \theta + \sin \theta + 1 = 0\)
\(2\sin^2 \theta - \sin \theta - 1 = 0\)

Factorise the quadratic expression in terms of \(\sin \theta\):
\((2\sin \theta + 1)(\sin \theta - 1) = 0\)

This gives two cases:
1) \(\sin \theta = 1\)
In the range \(0^\circ \le \theta \le 360^\circ\), the only solution is:
\(\theta = 90^\circ\)

2) \(\sin \theta = -\frac{1}{2}\)
The reference angle is \(30^\circ\). Since \(\sin \theta\) is negative, \(\theta\) lies in the third and fourth quadrants:
Third quadrant: \(\theta = 180^\circ + 30^\circ = 210^\circ\)
Fourth quadrant: \(\theta = 360^\circ - 30^\circ = 330^\circ\)

Therefore, the complete set of solutions is \(\theta = 90^\circ, 210^\circ, 330^\circ\).

B1: for substituting \(\cos^2\theta = 1-\sin^2\theta\) to form a three-term quadratic in \(\sin\theta\), e.g. \(2\sin^2\theta - \sin\theta - 1 = 0\).
M1: for factorising or solving their quadratic equation to find values of \(\sin\theta\).
A1: for the solution \(\theta = 90^\circ\).
A1: for both solutions \(\theta = 210^\circ\) and \(\theta = 330^\circ\) with no extra values in the range.

Since the terms are consecutive in a geometric progression, the ratio between consecutive terms is constant:
\(\frac{k}{k + 2} = \frac{k - 1.5}{k}\)

Cross-multiplying yields:
\(k^2 = (k + 2)(k - 1.5)\)
\(k^2 = k^2 - 1.5k + 2k - 3\)
\(k^2 = k^2 + 0.5k - 3\)

Subtract \(k^2\) from both sides:
\(0.5k - 3 = 0\)
\(0.5k = 3 \implies k = 6\)

Now, substitute \(k = 6\) back to find the common ratio \(r\):
\(r = \frac{k}{k + 2} = \frac{6}{6 + 2} = \frac{6}{8} = 0.75\) (or \(\frac{3}{4}\)).

M1: for equating the ratios to set up the equation \(\frac{k}{k+2} = \frac{k-1.5}{k}\) or equivalent.
M1: for expanding and solving the linear equation in \(k\).
A1: for \(k = 6\).
A1: for the common ratio \(r = 0.75\) (or \(\frac{3}{4}\)).

We can rewrite the equation of the curve as \(y = (3x - 1)(2x + 1)^{\frac{1}{2}}\). Apply the product rule and chain rule to find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{d}{dx}(3x - 1) \cdot (2x + 1)^{\frac{1}{2}} + (3x - 1) \cdot \frac{d}{dx}\left[(2x + 1)^{\frac{1}{2}}\right]\)

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

\(\frac{dy}{dx} = 3\sqrt{2x + 1} + \frac{3x - 1}{\sqrt{2x + 1}}\)

Now, substitute \(x = 4\) to find the gradient at this point:
\(\frac{dy}{dx}\Big|_{x=4} = 3\sqrt{2(4) + 1} + \frac{3(4) - 1}{\sqrt{2(4) + 1}}\)

\(\frac{dy}{dx}\Big|_{x=4} = 3\sqrt{9} + \frac{11}{\sqrt{9}}\)

\(\frac{dy}{dx}\Big|_{x=4} = 3(3) + \frac{11}{3} = 9 + \frac{11}{3} = \frac{38}{3}\) (or \(12\frac{2}{3}\)).

M1: for attempting to differentiate using the product rule.
A1: for the correct derivative expression, e.g., \(3(2x+1)^{1/2} + (3x-1)(2x+1)^{-1/2}\).
M1: for substituting \(x = 4\) into their derivative.
A1: for the correct final answer \(\frac{38}{3}\) (or exact equivalent).

First expand the left-hand side of the inequality:
\(2x^2 + 6x - x - 3 < x + 13\)
\(2x^2 + 5x - 3 < x + 13\)

Rearrange to form a single quadratic inequality on one side:
\(2x^2 + 4x - 16 < 0\)

Divide the entire inequality by 2:
\(x^2 + 2x - 8 < 0\)

Factorise the quadratic expression:
\((x + 4)(x - 2) < 0\)

The critical values are \(x = -4\) and \(x = 2\).

Since the inequality is less than zero, the solution is the region between the critical values:
\(-4 < x < 2\).

M1: for expanding and rearranging to form a three-term quadratic inequality, e.g., \(2x^2 + 4x - 16 < 0\).
M1: for factorising their quadratic expression to find the critical values.
A1: for finding the correct critical values \(x = -4\) and \(x = 2\).
A1: for the correct inequality range \(-4 < x < 2\) (or equivalent notation).

To find the inverse function, let \(y = \mathrm{f}(x)\):
\(y = \ln(3x - 2)\)

Convert from logarithmic to exponential form:
\(e^y = 3x - 2\)

Solve for \(x\):
\(3x = e^y + 2 \implies x = \frac{e^y + 2}{3}\)

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

The domain of \(\mathrm{f}^{-1}\) is the range of the original function \(\mathrm{f}\). Since the range of \(\mathrm{f}(x) = \ln(3x - 2)\) is all real values of \(y\), the domain of \(\mathrm{f}^{-1}\) is \(x \in \mathbb{R}\) (all real numbers).

M1: for writing \(y = \ln(3x - 2)\) and attempting to make \(x\) the subject by using exponentiation.
A1: for obtaining a correct step, e.g. \(3x = e^y + 2\).
A1: for \(f^{-1}(x) = \frac{e^x + 2}{3}\) (must be written in terms of \(x\)).
B1: for stating the domain is \(x \in \mathbb{R}\) (or equivalent, such as all real values of \(x\)).

The center of the circle is the midpoint of the diameter line segment \(AB\):
\(\text{Center} = \left(\frac{-1 + 7}{2}, \frac{5 + (-1)}{2}\right) = (3, 2)\)

The radius of the circle is the distance from the center \((3, 2)\) to one of the endpoints, say \(B(7, -1)\):
\(r^2 = (7 - 3)^2 + (-1 - 2)^2 = 4^2 + (-3)^2 = 16 + 9 = 25\)
Thus, the radius \(r = 5\).

Using the standard equation of a circle \((x - a)^2 + (y - b)^2 = r^2\):
\((x - 3)^2 + (y - 2)^2 = 25\)
(or expanded form: \(x^2 + y^2 - 6x - 4y - 12 = 0\)).

M1: for finding the midpoint of \(AB\) to find the center of the circle.
A1: for correct center at \((3, 2)\).
M1: for calculating \(r^2\) or the radius using a distance formula from their center to a point on the circle.
A1: for the correct equation of the circle: \((x - 3)^2 + (y - 2)^2 = 25\) (or exact expanded equivalent).

Let \(u = \log_3(2x-1)\).\
Using the change of base formula, we have \(\log_{2x-1} 3 = \frac{1}{\log_3(2x-1)} = \frac{1}{u}\).\
Substituting these into the original equation gives:\
\(u + \frac{3}{u} = 4\)\
Multiply the entire equation by \(u\):\
\(u^2 - 4u + 3 = 0\)\
\((u-1)(u-3) = 0\)\
So, \(u = 1\) or \(u = 3\).\
\
Case 1: \(\log_3(2x-1) = 1\)\
\(2x-1 = 3^1\)\
\(2x = 4\)\
\(x = 2\)\
\
Case 2: \(\log_3(2x-1) = 3\)\
\(2x-1 = 3^3\)\
\(2x-1 = 27\)\
\(2x = 28\)\
\(x = 14\)\
\
Both values of \(x\) satisfy the conditions \(2x-1 > 0\) and \(2x-1 \neq 1\).\
Therefore, the solutions are \(x = 2\) and \(x = 14\).

M1: Use the change of base rule to express the equation in terms of \(\log_3(2x-1)\) or equivalent.\
M1: Form and solve a 3-term quadratic equation in terms of \(u\) (where \(u = \log_3(2x-1)\)).\
A1: Find the correct values \(u = 1\) and \(u = 3\) (or solve directly for \(2x-1 = 3\) and \(2x-1 = 27\)).\
A1: Obtain both correct final answers \(x = 2\) and \(x = 14\).

Since \(x-2\) is a factor of \(\mathrm{p}(x)\), we have \(\mathrm{p}(2) = 0\):\
\(3(2)^3 + a(2)^2 + b(2) - 12 = 0\)\
\(24 + 4a + 2b - 12 = 0\)\
\(4a + 2b = -12 \implies 2a + b = -6\) --- (Equation 1)\
\
When \(\mathrm{p}(x)\) is divided by \(x+1\), the remainder is \(-18\), so \(\mathrm{p}(-1) = -18\):\
\(3(-1)^3 + a(-1)^2 + b(-1) - 12 = -18\)\
\(-3 + a - b - 12 = -18\)\
\(a - b = -3\) --- (Equation 2)\
\
Adding Equation 1 and Equation 2:\
\(3a = -9 \implies a = -3\)\
Substituting \(a = -3\) into Equation 2:\
\(-3 - b = -3 \implies b = 0\)\
\
Thus, \(\mathrm{p}(x) = 3x^3 - 3x^2 - 12 = 3(x^3 - x^2 - 4)\).\
Since \(x-2\) is a factor of \(\mathrm{p}(x)\), we can divide \(x^3 - x^2 - 4\) by \(x-2\) using algebraic long division or synthetic division:\
\(x^3 - x^2 - 4 = (x-2)(x^2 + x + 2)\)\
\
To find any other real roots, we set the quadratic factor to zero:\
\(x^2 + x + 2 = 0\)\
Calculating the discriminant of this quadratic equation:\
\(\Delta = b^2 - 4ac = 1^2 - 4(1)(2) = 1 - 8 = -7\)\
\
Since the discriminant is negative (\(-7 < 0\)), the quadratic factor has no real roots.\
Therefore, the equation \(\mathrm{p}(x) = 0\) has only one real root, which is \(x = 2\).

M1: Set up the equation \(2a + b = -6\) using the factor theorem \(\mathrm{p}(2) = 0\).\
M1: Set up the equation \(a - b = -3\) using the remainder theorem \(\mathrm{p}(-1) = -18\).\
A1: Solve the simultaneous equations to find \(a = -3\) and \(b = 0\).\
M1: Divide \(\mathrm{p}(x)\) by \(x-2\) to obtain the quadratic factor \(x^2 + x + 2\) (or \(3x^2 + 3x + 6\)).\
A1: Use the discriminant of the quadratic factor (showing \(1^2 - 4(1)(2) = -7 < 0\)) to correctly conclude that there are no other real roots.

To find $\text{f}'(x)$, integrate $\text{f}''(x)$:
$$\text{f}'(x) = \int 6(2x-3)^{-\frac{3}{2}} \, dx = \frac{6}{-\frac{1}{2} \times 2}(2x-3)^{-\frac{1}{2}} + c = -6(2x-3)^{-\frac{1}{2}} + c$$

Using the given gradient of $-4$ at $x = 2$:
$$-4 = -6(2(2)-3)^{-\frac{1}{2}} + c$$
$$-4 = -6(1) + c \implies c = 2$$

So, the first derivative is:
$$\text{f}'(x) = -6(2x-3)^{-\frac{1}{2}} + 2$$

To find $\text{f}(x)$, integrate $\text{f}'(x)$:
$$\text{f}(x) = \int \left( -6(2x-3)^{-\frac{1}{2}} + 2 \right) \, dx = -6\frac{(2x-3)^{\frac{1}{2}}}{\frac{1}{2} \times 2} + 2x + d = -6(2x-3)^{\frac{1}{2}} + 2x + d$$

Using the point $(2, 10)$ on the curve:
$$10 = -6(2(2)-3)^{\frac{1}{2}} + 2(2) + d$$
$$10 = -6(1) + 4 + d \implies 10 = -2 + d \implies d = 12$$

So, the equation of the curve is:
$$\text{f}(x) = -6(2x-3)^{\frac{1}{2}} + 2x + 12$$

At a stationary point, $\text{f}'(x) = 0$:
$$-6(2x-3)^{-\frac{1}{2}} + 2 = 0$$
$$2 = \frac{6}{\sqrt{2x-3}} \implies \sqrt{2x-3} = 3$$
$$2x-3 = 9 \implies 2x = 12 \implies x = 6$$

Find the $y$-coordinate when $x = 6$:
$$y = -6(2(6)-3)^{\frac{1}{2}} + 2(6) + 12 = -6(3) + 12 + 12 = 6$$

Thus, the coordinates of the stationary point are $(6, 6)$.

**M1** for attempting to integrate $\text{f}''(x)$ of the form $k(2x-3)^{-\frac{1}{2}}$
**A1** for correct integration: $\text{f}'(x) = -6(2x-3)^{-\frac{1}{2}} + c$
**M1** for substituting $x = 2$ and gradient $= -4$ into their $\text{f}'(x)$ to find $c$
**A1** for $c = 2$ leading to $\text{f}'(x) = -6(2x-3)^{-\frac{1}{2}} + 2$
**M1** for attempting to integrate their $\text{f}'(x)$ to find $\text{f}(x)$
**A1** for correct integration: $\text{f}(x) = -6(2x-3)^{\frac{1}{2}} + 2x + d$
**M1** for substituting $(2, 10)$ into their $\text{f}(x)$ to find $d$
**A1** for $d = 12$ leading to $\text{f}(x) = -6(2x-3)^{\frac{1}{2}} + 2x + 12$
**M1** for setting their $\text{f}'(x) = 0$ and attempting to solve for $x$
**A1** for $x = 6$
**A1** for $y = 6$, giving the stationary point coordinates as $(6, 6)$

For (a):
Let \( y = 3\mathrm{e}^{2x} - 4 \).
\( y + 4 = 3\mathrm{e}^{2x} \)
\( \mathrm{e}^{2x} = \frac{y+4}{3} \)
\( 2x = \ln\left(\frac{y+4}{3}\right) \)
\( x = \frac{1}{2}\ln\left(\frac{y+4}{3}\right) \)
So, \( \mathrm{f}^{-1}(x) = \frac{1}{2}\ln\left(\frac{x+4}{3}\right) \).

For (b):
The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \).
Since the range of \( \mathrm{f} \) is \( \mathrm{f}(x) > -4 \), the domain of \( \mathrm{f}^{-1} \) is \( x > -4 \).

M1: For attempting to make \( x \) the subject of \( y = 3\mathrm{e}^{2x} - 4 \)
M1: For correct use of logarithms to isolate \( x \)
A1: For \( \mathrm{f}^{-1}(x) = \frac{1}{2}\ln\left(\frac{x+4}{3}\right) \) or equivalent
B1: For \( x > -4 \)

Using the product rule:
\( \frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x} \)
where \( u = x^2 \) and \( v = \ln(2x-3) \).
\( \frac{\mathrm{d}u}{\mathrm{d}x} = 2x \)
\( \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{2}{2x-3} \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 \left(\frac{2}{2x-3}\right) + 2x \ln(2x-3) \)
Substitute \( x = 2 \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 2^2 \left(\frac{2}{4-3}\right) + 2(2) \ln(4-3) = 4(2) + 4\ln(1) = 8 + 0 = 8 \).

M1: For applying the product rule to differentiate \( x^2 \ln(2x-3) \)
A1: For correct derivative of \( \ln(2x-3) \), which is \( \frac{2}{2x-3} \)
A1: For a fully correct expression for \( \frac{\mathrm{d}y}{\mathrm{d}x} \)
A1: For substituting \( x = 2 \) to get 8

Substitute \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( 3(1 - \cos^2 \theta) - 5\cos\theta - 1 = 0 \)
\( 3 - 3\cos^2\theta - 5\cos\theta - 1 = 0 \)
\( 3\cos^2\theta + 5\cos\theta - 2 = 0 \)
Factorise the quadratic:
\( (3\cos\theta - 1)(\cos\theta + 2) = 0 \)
This gives \( \cos\theta = \frac{1}{3} \) or \( \cos\theta = -2 \) (which has no solution).
For \( \cos\theta = \frac{1}{3} \):
\( \theta = 70.5^\circ \) and \( \theta = 360^\circ - 70.5^\circ = 289.5^\circ \).

M1: For using \( \sin^2\theta = 1 - \cos^2\theta \) to obtain a quadratic in \( \cos\theta \)
A1: For obtaining \( 3\cos^2\theta + 5\cos\theta - 2 = 0 \)
M1: For solving their quadratic in \( \cos\theta \)
A1: For \( \theta = 70.5^\circ \) (accept 70.5 or 70.53)
A1: For \( \theta = 289.5^\circ \) (accept 289.5 or 289.47)

For a geometric progression, \( u_n = a r^{n-1} \).
\( u_3 = a r^2 = 12 \)
\( u_6 = a r^5 = 96 \)
Dividing the two equations:
\( \frac{a r^5}{a r^2} = \frac{96}{12} \implies r^3 = 8 \implies r = 2 \).
Substitute \( r = 2 \) into \( a r^2 = 12 \):
\( a (2^2) = 12 \implies 4a = 12 \implies a = 3 \).
The sum of the first \( n \) terms is \( S_n = \frac{a(r^n - 1)}{r - 1} \).
For \( n = 8 \):
\( S_8 = \frac{3(2^8 - 1)}{2 - 1} = 3(256 - 1) = 3 \times 255 = 765 \).

M1: For using \( u_n = a r^{n-1} \) to set up two equations
M1: For solving to find \( r \)
A1: For \( r = 2 \) and \( a = 3 \)
A1: For finding \( S_8 = 765 \)

Use the laws of logarithms:
\( \log_3(x-2)^2 - \log_3(x+4) = 1 \)
\( \log_3\left(\frac{(x-2)^2}{x+4}\right) = 1 \)
Convert from logarithmic to exponential form:
\( \frac{(x-2)^2}{x+4} = 3^1 \)
\( (x-2)^2 = 3(x+4) \)
\( x^2 - 4x + 4 = 3x + 12 \)
\( x^2 - 7x - 8 = 0 \)
Factorise:
\( (x - 8)(x + 1) = 0 \)
So \( x = 8 \) or \( x = -1 \).
Since the logarithm is only defined for positive arguments, we require \( x - 2 > 0 \implies x > 2 \).
Therefore, the only solution is \( x = 8 \).

M1: For using the power law of logarithms, \( 2\log_3(x-2) = \log_3(x-2)^2 \)
M1: For using the quotient rule of logarithms, \( \log_3\left(\frac{(x-2)^2}{x+4}\right) = 1 \)
M1: For converting to quadratic equation: \( x^2 - 7x - 8 = 0 \)
A1: For solving to find \( x = 8 \) and \( x = -1 \)
A1: For selecting \( x = 8 \) as the only valid solution

Find the vectors \( \overrightarrow{AC} \) and \( \overrightarrow{AB} \):
\( \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (8\mathbf{i} + 7\mathbf{j}) - (4\mathbf{i} + 3\mathbf{j}) = 4\mathbf{i} + 4\mathbf{j} \)
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (10\mathbf{i} + y\mathbf{j}) - (4\mathbf{i} + 3\mathbf{j}) = 6\mathbf{i} + (y - 3)\mathbf{j} \)
Since \( C \) lies on the line \( AB \, \), the vectors \( \overrightarrow{AC} \) and \( \overrightarrow{AB} \) are parallel.
Therefore, \( \overrightarrow{AB} = \mu \overrightarrow{AC} \) for some scalar \( \mu \).
Equating the components:
\( 6 = 4\mu \implies \mu = 1.5 \)
\( y - 3 = 4\mu \implies y - 3 = 4(1.5) = 6 \)
\( y = 9 \).

M1: For finding a correct expression for \( \overrightarrow{AC} \)
M1: For finding a correct expression for \( \overrightarrow{AB} \) in terms of \( y \)
M1: For setting up a parallel vector relation (e.g., ratio of components)
A1: For obtaining \( y = 9 \)

The committee must have at least 3 men. Since the total size of the committee is 5, the possible cases are:
Case 1: 3 men and 2 women
Number of ways = \( \binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120 \)

Case 2: 4 men and 1 woman
Number of ways = \( \binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60 \)

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

Total number of different committees = \( 120 + 60 + 6 = 186 \).

M1: For considering three cases: 3M/2W, 4M/1W, and 5M
M1: For correctly calculating combinations for at least two cases
A1: For showing all three case values (120, 60, 6) correctly
A1: For the final total of 186

The perimeter of the sector is given by \( P = 2r + r\theta = 36 \).
The area of the sector is given by \( A = \frac{1}{2}r^2\theta = 80 \).
From the perimeter equation:
\( r\theta = 36 - 2r \)
Substitute \( r\theta \) into the area equation:
\( \frac{1}{2}r(r\theta) = 80 \implies \frac{1}{2}r(36 - 2r) = 80 \)
\( r(18 - r) = 80 \)
\( 18r - r^2 = 80 \)
\( r^2 - 18r + 80 = 0 \)
Factorise the quadratic equation:
\( (r - 10)(r - 8) = 0 \)
So the possible values of \( r \) are \( r = 10 \) and \( r = 8 \).

M1: For writing down the equations \( 2r + r\theta = 36 \) and \( \frac{1}{2}r^2\theta = 80 \)
M1: For eliminating \( \theta \) to obtain an equation in terms of \( r \) only
A1: For the quadratic equation \( r^2 - 18r + 80 = 0 \) or equivalent
M1: For attempting to solve their quadratic equation
A1: For \( r = 8 \) and \( r = 10 \)

We can calculate the number of ways using two main stages: selection and arrangement. Stage 1: Selection. 1. Choose 3 letters from the 5 available letters: \(\binom{5}{3} = 10\) ways. 2. Choose 2 digits from the 4 available digits: \(\binom{4}{2} = 6\) ways. Total combinations of 3 letters and 2 digits = \(10 \times 6 = 60\). Stage 2: Arrangement. For each chosen group of 3 letters and 2 digits: Choose 1 of the 3 letters for the first position: 3 ways. Choose 1 of the 2 digits for the fifth (last) position: 2 ways. Arrange the remaining 2 letters and 1 digit in the middle 3 positions: \(3! = 6\) ways. Total arrangements per group = \(3 \times 2 \times 6 = 36\). Total passcodes = \(60 \times 36 = 2160\).

M1 for choosing 3 letters and 2 digits, i.e., \(\binom{5}{3} \times \binom{4}{2}\), or equivalent slot-allocation step. M1 for considering the arrangements with constraints, e.g., \(3 \times 2 \times 6\) or showing a product of appropriate choices. A1 for 2160.

Use the laws of logarithms to combine the terms on the left-hand side: \(\log_3((x - 4)(x + 2)) = 3\). Convert the logarithmic equation to an exponential equation: \((x - 4)(x + 2) = 3^3\), which simplifies to \(x^2 - 2x - 8 = 27\), then \(x^2 - 2x - 35 = 0\). Factorising the quadratic equation gives \((x - 7)(x + 5) = 0\). This gives the potential solutions: \(x = 7\) or \(x = -5\). We must check the validity of these solutions in the original equation. For \(\log_3(x - 4)\) to be defined, we must have \(x - 4 > 0\), which means \(x > 4\). Therefore, \(x = -5\) is rejected. The only valid solution is \(x = 7\).

M1 for combining logarithms: \(\log_3((x - 4)(x + 2))\). M1 for removing logarithms: \((x - 4)(x + 2) = 3^3\) or 27. M1 for solving the quadratic equation to find \(x = 7\) and \(x = -5\). A1 for \(x = 7\) only (rejecting \(x = -5\)).

The centre of the circle, \((a, b)\), is the midpoint of the diameter \(AB\): \(a = \frac{-3 + 5}{2} = 1\) and \(b = \frac{2 + (-4)}{2} = -1\). So the centre is \((1, -1)\). The radius \(r\) is half the length of the diameter \(AB\). First, find the length of \(AB\): \(AB = \sqrt{(5 - (-3))^2 + (-4 - 2)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\). Therefore, the radius is \(r = \frac{10}{2} = 5\). The equation of the circle is \((x - 1)^2 + (y - (-1))^2 = 5^2\), which simplifies to \((x - 1)^2 + (y + 1)^2 = 25\).

B1 for finding the centre of the circle as \((1, -1)\). M1 for finding the radius or radius squared (e.g., \(r^2 = 25\) or \(r = 5\)). A1 for \((x - 1)^2 + (y + 1)^2 = 25\).

First, calculate the displacement vector \(\vec{AB}\): \(\vec{AB} = \vec{r}_B - \vec{r}_A = \begin{pmatrix} 8 \\ -2 \end{pmatrix} - \begin{pmatrix} -4 \\ 7 \end{pmatrix} = \begin{pmatrix} 8 - (-4) \\ -2 - 7 \end{pmatrix} = \begin{pmatrix} 12 \\ -9 \end{pmatrix}\). Since velocity is displacement divided by time: \(\vec{v} = \frac{\vec{AB}}{t} = \frac{1}{3} \begin{pmatrix} 12 \\ -9 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}\).

M1 for finding the displacement vector \(\vec{AB} = \begin{pmatrix} 12 \\ -9 \end{pmatrix}\) or showing a correct subtraction step. A1 for \(\begin{pmatrix} 4 \\ -3 \end{pmatrix}\).

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to rewrite the equation in terms of \(\sin \theta\): \(2(1 - \sin^2 \theta) + 3 \sin \theta - 3 = 0\), which simplifies to \(2 - 2 \sin^2 \theta + 3 \sin \theta - 3 = 0\), and further to \(-2 \sin^2 \theta + 3 \sin \theta - 1 = 0\). Multiplying the entire equation by \(-1\) gives: \(2 \sin^2 \theta - 3 \sin \theta + 1 = 0\). Factorising the quadratic equation gives: \((2 \sin \theta - 1)(\sin \theta - 1) = 0\). This yields two possible cases: Case 1: \(2 \sin \theta - 1 = 0 \Rightarrow \sin \theta = \frac{1}{2}\), which gives \(\theta = 30^\circ\) and \(\theta = 180^\circ - 30^\circ = 150^\circ\). Case 2: \(\sin \theta - 1 = 0 \Rightarrow \sin \theta = 1\), which gives \(\theta = 90^\circ\). Thus, the solutions are \(\theta = 30^\circ, 90^\circ, 150^\circ\).

M1 for substituting \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic in \(\sin \theta\). M1 for factorising or solving their quadratic equation in \(\sin \theta\) to obtain \(\sin \theta = \frac{1}{2}\) and \(\sin \theta = 1\). A1 for finding two correct values from \(\{30^\circ, 90^\circ, 150^\circ\}\). A1 for finding all three correct values \(\theta = 30^\circ, 90^\circ, 150^\circ\) and no extras in the range.

First, find the derivative \(\frac{dy}{dx}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(3x - 1)\frac{d}{dx}(x^2) - x^2\frac{d}{dx}(3x - 1)}{(3x - 1)^2} = \frac{(3x - 1)(2x) - x^2(3)}{(3x - 1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x - 1)^2} = \frac{3x^2 - 2x}{(3x - 1)^2}\). For stationary points, set \(\frac{dy}{dx} = 0\): \(\frac{3x^2 - 2x}{(3x - 1)^2} = 0\), which leads to \(3x^2 - 2x = 0 \Rightarrow x(3x - 2) = 0\). This gives \(x = 0\) or \(x = \frac{2}{3}\). Since the domain is \(x > \frac{1}{3}\), we reject \(x = 0\). Therefore, \(x = \frac{2}{3}\). Substitute \(x = \frac{2}{3}\) back into the original curve equation to find the corresponding \(y\)-coordinate: \(y = \frac{\left(\frac{2}{3}\right)^2}{3\left(\frac{2}{3}\right) - 1} = \frac{\frac{4}{9}}{2 - 1} = \frac{4}{9}\). Thus, the coordinates of the stationary point are \(\left(\frac{2}{3}, \frac{4}{9}\right)\).

M1 for attempting to use the quotient rule to differentiate \(y\). A1 for obtaining a correct expression for \(\frac{dy}{dx}\), e.g., \(\frac{3x^2 - 2x}{(3x - 1)^2}\). M1 for setting \(\frac{dy}{dx} = 0\) and solving to find \(x = \frac{2}{3}\) (allowing \(x = 0\) to be found but rejected). A1 for the correct coordinates \(\left(\frac{2}{3}, \frac{4}{9}\right)\).

To find where the line and curve intersect, we set their equations equal: \(kx - 5 = x^2 - 4x + 4\). Rearrange into standard quadratic form: \(x^2 - (4 + k)x + 9 = 0\). For the line not to intersect the curve, this quadratic equation must have no real roots, meaning the discriminant \(\Delta\) must be negative: \(\Delta = B^2 - 4AC < 0\), which translates to \((-(4 + k))^2 - 4(1)(9) < 0\), then \((4 + k)^2 - 36 < 0\), and so \((4 + k)^2 < 36\). Taking the square root on both sides: \(-6 < 4 + k < 6\). Subtracting 4 from all parts: \(-10 < k < 2\).

M1 for equating the line and curve equations and forming a 3-term quadratic in \(x\). M1 for using the condition \(\Delta < 0\) on their quadratic. M1 for finding the critical values \(k = -10\) and \(k = 2\) by solving \((4 + k)^2 - 36 = 0\). A1 for the final range \(-10 < k < 2\) or equivalent.

Let the first term of the progression be \(a\) and the common ratio be \(r\). The \(n\)-th term of a geometric progression is given by \(u_n = a r^{n-1}\). We are given: \(u_3 = a r^2 = 12\) (Equation 1) and \(u_6 = a r^5 = \frac{32}{9}\) (Equation 2). Dividing Equation 2 by Equation 1: \(\frac{a r^5}{a r^2} = \frac{32/9}{12} \Rightarrow r^3 = \frac{32}{108} = \frac{8}{27}\). Taking the cube root gives \(r = \frac{2}{3}\). Substitute \(r = \frac{2}{3}\) back into Equation 1 to find \(a\): \(a \left(\frac{2}{3}\right)^2 = 12 \Rightarrow a \left(\frac{4}{9}\right) = 12 \Rightarrow a = 12 \times \frac{9}{4} = 27\). Since \(|r| < 1\), the sum to infinity exists and is calculated as: \(S_\infty = \frac{a}{1 - r} = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 81\).

M1 for setting up the equations \(a r^2 = 12\) and \(a r^5 = \frac{32}{9}\). M1 for dividing the equations to find \(r^3\) and solving to find \(r = \frac{2}{3}\). M1 for substituting \(r\) back to find \(a = 27\). A1 for \(S_\infty = 81\).

From the first equation, we get:
\[y = 5 - 2x\]
Substitute this expression into the second equation:
\[x^2 + 3x(5 - 2x) - (5 - 2x)^2 = 1\]
Expand the terms:
\[x^2 + 15x - 6x^2 - (25 - 20x + 4x^2) = 1\]
\[-5x^2 + 15x - 25 + 20x - 4x^2 = 1\]
\[-9x^2 + 35x - 25 = 1\]
\[9x^2 - 35x + 26 = 0\]
Factorise the quadratic equation:
\[(9x - 26)(x - 1) = 0\]
This gives:
\[x = 1 \quad \text{or} \quad x = \frac{26}{9}\]
Find the corresponding \(y\) values using \(y = 5 - 2x\):
- If \(x = 1\), \(y = 5 - 2(1) = 3\)
- If \(x = \frac{26}{9}\), \(y = 5 - 2\left(\frac{26}{9}\right) = \frac{45 - 52}{9} = -\frac{7}{9}\] Therefore, the solutions are \)x = 1, y = 3\) and \(x = \frac{26}{9}, y = -\frac{7}{9}\).

M1: For expressing one variable in terms of the other (e.g., \(y = 5 - 2x\)) and substituting it into the quadratic equation.
A1: For obtaining a correct simplified 3-term quadratic in one variable, e.g., \(9x^2 - 35x + 26 = 0\).
M1: For a valid method to solve their 3-term quadratic.
A1: For both correct values of \(x\) (or \(y\)), \(x = 1\) and \(x = \frac{26}{9}\).
A1: For the correct corresponding values of the other variable, \(y = 3\) and \(y = -\frac{7}{9}\), correctly paired.

Using the change of base formula, we can write \(\log_x 5 = \frac{1}{\log_5 x}\).
The equation becomes:
\[2 \log_5 x + \frac{3}{\log_5 x} = 7\]
Let \(u = \log_5 x\):
\[2u + \frac{3}{u} = 7\]
Multiply the entire equation by \(u\):
\[2u^2 + 3 = 7u\]
\[2u^2 - 7u + 3 = 0\]
Factorise the quadratic equation:
\[(2u - 1)(u - 3) = 0\]
This gives:
\[u = \frac{1}{2} \quad \text{or} \quad u = 3\]
Now substitute back \(u = \log_5 x\):
- For \(\log_5 x = \frac{1}{2} \implies x = 5^{1/2} = \sqrt{5}\)
- For \(\log_5 x = 3 \implies x = 5^3 = 125\)
Thus, the solutions are \(x = \sqrt{5}\) and \(x = 125\).

M1: For applying the change of base rule correctly to write \(\log_x 5 = \frac{1}{\log_5 x}\).
M1: For setting up a quadratic equation in terms of \(u\) (where \(u = \log_5 x\)), obtaining \(2u^2 - 7u + 3 = 0\).
A1: For solving the quadratic correctly to get \(u = \frac{1}{2}\) and \(u = 3\).
M1: For solving \(\log_5 x = u\) for at least one value of \(u\).
A1: For both correct final answers: \(x = \sqrt{5}\) and \(x = 125\).

(i) Total number of people is \(6 + 5 = 11\).
Number of ways to choose 5 from 11:
\[\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462\]

(ii) For more men than women, we have the following cases:
- 5 men and 0 women: \(\binom{6}{5} \times \binom{5}{0} = 6\)
- 4 men and 1 woman: \(\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75\)
- 3 men and 2 women: \(\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200\)
Total number of ways = \(6 + 75 + 200 = 281\).

(iii) Let the particular man be \(M_1\) and the particular woman be \(W_1\).
The number of committees containing *both* \(M_1\) and \(W_1\) requires choosing 3 more people from the remaining 9:
\[\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\]
Subtract this from the total unrestricted committees:
\[462 - 84 = 378\]

B1: For 462 (part i).
M1: For considering the three cases (5M 0W, 4M 1W, 3M 2W) and showing a sum of products of combinations (part ii).
A1: For 281 (part ii).
M1: For a correct method to handle the restriction, e.g., subtracting committees containing both from the total (462 - \(\binom{9}{3}\)) (part iii).
A1: For 378 (part iii).

(i) First find the vector \(\overrightarrow{AB}\):
\[\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (15\mathbf{i} - 7\mathbf{j}) - (3\mathbf{i} + \mathbf{j}) = 12\mathbf{i} - 8\mathbf{j}\]
Now find \(\overrightarrow{AC}\):
\[\overrightarrow{AC} = \frac{3}{4}\overrightarrow{AB} = \frac{3}{4}(12\mathbf{i} - 8\mathbf{j}) = 9\mathbf{i} - 6\mathbf{j}\]
Find the position vector of \(C\), which is \(\overrightarrow{OC}\):
\[\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = (3\mathbf{i} + \mathbf{j}) + (9\mathbf{i} - 6\mathbf{j}) = 12\mathbf{i} - 5\mathbf{j}\]

(ii) Find the magnitude of \(\overrightarrow{OC}\):
\[|\overrightarrow{OC}| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\]
Find the unit vector in the direction of \(\overrightarrow{OC}\):
\[\widehat{\overrightarrow{OC}} = \frac{1}{13}(12\mathbf{i} - 5\mathbf{j}) = \frac{12}{13}\mathbf{i} - \frac{5}{13}\mathbf{j}\]

M1: For a correct method to find \(\overrightarrow{AB}\), e.g., subtracting \(\mathbf{a}\) from \(\mathbf{b}\).
M1: For adding \(\overrightarrow{AC}\) to \(\mathbf{a}\) to find \(\mathbf{c}\).
A1: For the correct position vector of \(C\): \(12\mathbf{i} - 5\mathbf{j}\).
M1: For a correct method to find the magnitude of their position vector \(\mathbf{c}\) and dividing \(\mathbf{c}\) by this magnitude.
A1: For \(\frac{12}{13}\mathbf{i} - \frac{5}{13}\mathbf{j}\) or \(\frac{1}{13}(12\mathbf{i} - 5\mathbf{j})\).

(i) The total surface area of a solid cylinder is given by:
\[A = 2\pi r^2 + 2\pi rh = 54\pi\]
Divide by \(2\pi\):
\[r^2 + rh = 27 \implies rh = 27 - r^2 \implies h = \frac{27 - r^2}{r}\]
The volume of the cylinder is:
\[V = \pi r^2 h = \pi r^2 \left(\frac{27 - r^2}{r}\right) = \pi r(27 - r^2) = 27\pi r - \pi r^3\]

(ii) Differentiate \(V\) with respect to \(r\):
\[\frac{dV}{dr} = 27\pi - 3\pi r^2\]
When \(r = 2\):
\[\frac{dV}{dr} = 27\pi - 3\pi(2)^2 = 27\pi - 12\pi = 15\pi\]
Using the chain rule to find the rate of change of volume with respect to time:
\[\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\]
Given that \(\frac{dr}{dt} = 0.2\):
\[\frac{dV}{dt} = 15\pi \times 0.2 = 3\pi\]
Thus, the rate of change of volume is \(3\pi\) \(\text{cm}^3\text{/s}\).

M1: For writing down the formula for the total surface area and expressing \(h\) in terms of \(r\).
A1: For substituting \(h\) into the volume formula and simplifying to show the required expression convincingly.
M1: For differentiating their \(V\) with respect to \(r\) correctly to find \(\frac{dV}{dr}\).
M1: For using the chain rule \(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\) with \(\frac{dr}{dt} = 0.2\) and \(r = 2\).
A1: For \(3\pi\) (or approximately 9.42).

(i) The \(n\)-th term of a geometric progression is given by \(u_n = ar^{n-1}\).
Given:
\[u_3 = ar^2 = 12\]
\[u_6 = ar^5 = 96\]
Divide the second equation by the first:
\[\frac{ar^5}{ar^2} = \frac{96}{12} \implies r^3 = 8 \implies r = 2\]
Substitute \(r = 2\) back into the equation for \(u_3\):
\[a(2)^2 = 12 \implies 4a = 12 \implies a = 3\]
So, the common ratio is \(r = 2\) and the first term is \(a = 3\).

(ii) The sum of the first \(n\) terms of a geometric progression is given by:
\[S_n = \frac{a(r^n - 1)}{r - 1}\]
For \(n = 8\):
\[S_8 = \frac{3(2^8 - 1)}{2 - 1} = 3(256 - 1) = 3(255) = 765\]

M1: For setting up the equations \(ar^2 = 12\) and \(ar^5 = 96\).
M1: For solving to find \(r\).
A1: For both correct values: \(r = 2\) and \(a = 3\).
M1: For using the sum formula \(S_n = \frac{a(r^n - 1)}{r - 1}\) with their \(a\), \(r\) and \(n=8\).
A1: For 765.

Using the double-angle identity \(2 \sin A \cos A = \sin 2A\), let \(A = 2x\).
Then \(4 \sin 2x \cos 2x = 2(2 \sin 2x \cos 2x) = 2 \sin 4x\).
The equation becomes:
\[2 \sin 4x = \sqrt{3} \implies \sin 4x = \frac{\sqrt{3}}{2}\]
Since \(0 \le x \le \pi\), we have \(0 \le 4x \le 4\pi\).
The solutions for \(\sin \phi = \frac{\sqrt{3}}{2}\) in the interval \([0, 4\pi]\) are:
\[4x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}\]
Dividing each value by 4 gives:
\[x = \frac{\pi}{12}, \frac{\pi}{6}, \frac{7\pi}{12}, \frac{2\pi}{3}\]
These are all within the interval \([0, \pi]\).

M1: For using the double-angle identity to simplify the LHS to \(2 \sin 4x\).
A1: For obtaining the simplified equation \(\sin 4x = \frac{\sqrt{3}}{2}\).
M1: For finding at least two correct angles for \(4x\) within the appropriate range.
A1: For finding any two correct final values of \(x\).
A1: For all four correct values of \(x\): \(\frac{\pi}{12}, \frac{\pi}{6}, \frac{7\pi}{12}, \frac{2\pi}{3}\), and no others in the range.

The formula for the perimeter of a sector is:
\[P = 2r + r\theta = 30\]
The formula for the area of a sector is:
\[A = \frac{1}{2}r^2\theta = 50\]
From the perimeter equation, express \(\theta\) in terms of \(r\):
\[r\theta = 30 - 2r \implies \theta = \frac{30 - 2r}{r}\]
Substitute this expression into the area equation:
\[\frac{1}{2}r^2\left(\frac{30 - 2r}{r}\right) = 50\]
Simplify the equation:
\[\frac{1}{2}r(30 - 2r) = 50\]
\[15r - r^2 = 50\]
\[r^2 - 15r + 50 = 0\]
Factorise the quadratic equation:
\[(r - 5)(r - 10) = 0\]
This gives \(r = 5\) or \(r = 10\).

Find the corresponding values of \(\theta\) using \(\theta = \frac{30 - 2r}{r}\):
- If \(r = 5\), then \(\theta = \frac{30 - 10}{5} = 2\) radians.
- If \(r = 10\), then \(\theta = \frac{30 - 20}{10} = 1\) radian.
Both pairs \((r=5, \theta=2)\) and \((r=10, \theta=1)\) are valid since \(\theta > 0\).

M1: For using the sector perimeter and area formulas to set up two simultaneous equations.
M1: For substituting one equation into the other to eliminate \(\theta\).
A1: For obtaining a correct quadratic equation in one variable, e.g., \(r^2 - 15r + 50 = 0\).
A1: For solving to find both correct values of \(r\) (\(5\) and \(10\)).
A1: For finding the correct corresponding values of \(\theta\) (\(2\) and \(1\)).

To find the points of intersection of the two curves, we equate the equations:
\[ 10\mathrm{e}^{-x} = 11 - 3\mathrm{e}^x \]

Multiply through by \( \mathrm{e}^x \):
\[ 10 = 11\mathrm{e}^x - 3\mathrm{e}^{2x} \]

Rearrange into a quadratic equation in terms of \( \mathrm{e}^x \):
\[ 3\mathrm{e}^{2x} - 11\mathrm{e}^x + 10 = 0 \]

Let \( u = \mathrm{e}^x \):
\[ 3u^2 - 11u + 10 = 0 \]
\[ (3u - 5)(u - 2) = 0 \]

This gives:
\[ u = \frac{5}{3} \quad \text{or} \quad u = 2 \]

Therefore, the \( x \)-coordinates of the intersection points are:
\[ x = \ln\left(\frac{5}{3}\right) \quad \text{and} \quad x = \ln 2 \]

Since \( 11 - 3\mathrm{e}^x > 10\mathrm{e}^{-x} \) for \( \ln\left(\frac{5}{3}\right) < x < \ln 2 \), the area \( A \) of the enclosed region is given by:
\[ A = \int_{\ln(5/3)}^{\ln 2} \left( (11 - 3\mathrm{e}^x) - 10\mathrm{e}^{-x} \right) \mathrm{d}x \]

Integrate the expression:
\[ \int (11 - 3\mathrm{e}^x - 10\mathrm{e}^{-x}) \mathrm{d}x = \left[ 11x - 3\mathrm{e}^x + 10\mathrm{e}^{-x} \right]_{\ln(5/3)}^{\ln 2} \]

Substitute the upper limit \( x = \ln 2 \):
\[ 11\ln 2 - 3\mathrm{e}^{\ln 2} + 10\mathrm{e}^{-\ln 2} = 11\ln 2 - 3(2) + 10\left(\frac{1}{2}\right) = 11\ln 2 - 6 + 5 = 11\ln 2 - 1 \]

Substitute the lower limit \( x = \ln(5/3) \):
\[ 11\ln\left(\frac{5}{3}\right) - 3\mathrm{e}^{\ln(5/3)} + 10\mathrm{e}^{-\ln(5/3)} = 11\ln\left(\frac{5}{3}\right) - 3\left(\frac{5}{3}\right) + 10\left(\frac{3}{5}\right) = 11\ln\left(\frac{5}{3}\right) - 5 + 6 = 11\ln\left(\frac{5}{3}\right) + 1 \]

Subtract the lower limit evaluation from the upper limit evaluation:
\[ A = (11\ln 2 - 1) - \left(11\ln\left(\frac{5}{3}\right) + 1\right) \]
\[ A = 11\ln 2 - 11\ln\left(\frac{5}{3}\right) - 2 \]
\[ A = 11\ln\left( \frac{2}{5/3} \right) - 2 \]
\[ A = 11\ln\left(\frac{6}{5}\right) - 2 \]
\[ A = -2 + 11\ln(1.2) \]

So, \( a = -2 \) and \( b = 11 \).

**M1**: For equating \( 10\mathrm{e}^{-x} = 11 - 3\mathrm{e}^x \) and attempting to multiply by \( \mathrm{e}^x \)
**A1**: For obtaining the correct quadratic equation \( 3\mathrm{e}^{2x} - 11\mathrm{e}^x + 10 = 0 \)
**M1**: For solving their quadratic in \( \mathrm{e}^x \) to find two positive values for \( \mathrm{e}^x \)
**A1**: For obtaining both correct limits \( x = \ln 2 \) and \( x = \ln(5/3) \) (or equivalent exact values)
**M1**: For a correct plan to integrate \( \int (11 - 3\mathrm{e}^x - 10\mathrm{e}^{-x}) \mathrm{d}x \)
**A2**: For correct integration \( 11x - 3\mathrm{e}^x + 10\mathrm{e}^{-x} \) (Award **A1** if only one error made)
**M1**: For substituting their limits into their integrated expression
**M1**: For showing clear subtraction and application of logarithmic laws to combine terms
**A1**: For the correct final answer \( -2 + 11\ln(1.2) \) (or \( a = -2, b = 11 \))