2016 AQA IGCSE Mathematics (9260) 模擬試題連答案詳解

The roots of a quadratic equation are the values of \(x\) that make the equation equal to zero.

For roots 3 and -5, the corresponding factors of the quadratic are \((x - 3)\) and \((x + 5)\).

Therefore, the equation is \((x - 3)(x + 5) = 0\).

B1 for \((x - 3)(x + 5) = 0\)

Using the product rule of counting, the total number of combinations is calculated by multiplying the number of choices for each component:

\(4 \times 6 \times 5 = 120\).

B1 for 120

Let \(y = 5x\).

Rearranging to make \(x\) the subject:

\(x = \frac{y}{5}\)

Therefore, the inverse function is:

\(f^{-1}(x) = \frac{x}{5}\).

B1 for \(\frac{x}{5}\)

Subtracting the components of vector \(\mathbf{b}\) from vector \(\mathbf{a}\):

\(\mathbf{a} - \mathbf{b} = \begin{pmatrix} 6 - (-1) \\ -3 - 4 \end{pmatrix} = \begin{pmatrix} 7 \\ -7 \end{pmatrix}\)

B1 for \(\begin{pmatrix} 7 \\ -7 \end{pmatrix}\)

The area of a triangle with sides \(a\) and \(b\) and included angle \(C\) is given by:

\(\text{Area} = \frac{1}{2} a b \sin C\)

Substituting our values:

\(\text{Area} = \frac{1}{2} \times 6 \times 12 \times \sin y = 36 \sin y\)

B1 for \(36 \sin y\)

We are given that \(y^2 = 5\), which implies \(y = \sqrt{5}\).

To find \(y^3\):

\(y^3 = y^2 \times y = 5 \times \sqrt{5} = 5\sqrt{5}\).

B1 for \(5\sqrt{5}\)

Squaring the fraction yields:

\(y^2 = \left(\frac{3\sqrt{x}}{4}\right)^2 = \frac{3^2 \times (\sqrt{x})^2}{4^2} = \frac{9x}{16}\).

B1 for \(\frac{9x}{16}\)

To find the prime factors of 540, we can use a factor tree or division method:

\(540 = 2 \times 270\)
\(270 = 2 \times 135\)
\(135 = 3 \times 45\)
\(45 = 3 \times 15\)
\(15 = 3 \times 5\)

Therefore, \(540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^3 \times 5\).

M1: Correct prime factor tree or division showing at least two correct prime factors (e.g. 2 and 270, or 5 and 108).
A1: \(2^2 \times 3^3 \times 5\) or \(2 \times 2 \times 3 \times 3 \times 3 \times 5\).

First, expand the brackets:
\((2x - 3)(x + 5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15\)

Now, subtract \(x^2\):
\(2x^2 + 7x - 15 - x^2 = x^2 + 7x - 15\)

M1: Expanded expression of 4 terms with at least 3 correct: \(2x^2 + 10x - 3x - 15\) or simplified to \(2x^2 + 7x - 15\).
A1: \(x^2 + 7x - 15\).

Subtract 5 from both sides:
\(-3x \le 12\)

Divide both sides by \(-3\) and reverse the inequality sign:
\(x \ge -4\)

M1: \(-3x \le 12\) or \(3x \ge -12\) or an equation/inequality with one error in sign or arithmetic, e.g. \(x \le -4\).
A1: \(x \ge -4\).

The volume of a cylinder is given by the formula:
\(V = \pi r^2 h\)

Substitute the given values:
\(V = \pi \times 3^2 \times 8\)
\(V = \pi \times 9 \times 8\)
\(V = 72\pi\text{ cm}^3\)

M1: \(\text{Volume} = \pi \times 3^2 \times 8\)
A1: \(72\pi\)

First, rearrange the equation of L into gradient-intercept form (\(y = mx + c\)):
\(2y = 5x + 8\)
\(y = 2.5x + 4\)

The gradient of L is \(2.5\) (or \(\frac{5}{2}\)).

The gradient of the perpendicular line is \(-\frac{1}{m}\):
\(-\frac{1}{2.5} = -0.4\) (or \(-\frac{2}{5}\)).

M1: Rearranges equation to find the gradient of L is \(2.5\) or \(\frac{5}{2}\).
A1: \(-0.4\) or \(-\frac{2}{5}\) (or equivalent).

The total number of counters is \(5 + 3 = 8\).
The probability of choosing a red counter is \(\frac{5}{8}\).
The probability of choosing a blue counter is \(\frac{3}{8}\).

Since the counter is replaced, the events are independent.
\(P(\text{both same color}) = P(RR) + P(BB)\)
\(P(RR) = \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}\)
\(P(BB) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{64}\)

\(P(\text{both same color}) = \frac{25}{64} + \frac{9}{64} = \frac{34}{64} = \frac{17}{32}\) (or \(0.53125\))

M1: \(\left(\frac{5}{8}\right)^2\) or \(\left(\frac{3}{8}\right)^2\) seen, or a correct sum of two probability products.
A1: \(\frac{17}{32}\) or \(0.53125\) (or equivalent).

Write the relationship as an equation:
\(y = kx^2\)

Substitute the given values to find \(k\):
\(36 = k \times 3^2\)
\(36 = 9k \implies k = 4\)

So, the equation is:
\(y = 4x^2\)

When \(x = 5\):
\(y = 4 \times 5^2\)
\(y = 4 \times 25 = 100\)

M1: Formulates \(y = kx^2\) and correctly finds \(k = 4\), or writes \(\frac{y}{5^2} = \frac{36}{3^2}\).
A1: 100.

Let \(x = 0.2\dot{7} = 0.2777...\)
Multiply by 10:
\(10x = 2.7777...\)

Multiply by 100:
\(100x = 27.7777...\)

Subtract the first equation from the second:
\(100x - 10x = 27.7777... - 2.7777...\)
\(90x = 25\)
\(x = \frac{25}{90} = \frac{5}{18}\)

M1: Sets up two equations that allow subtraction to eliminate the recurring part, e.g. \(10x = 2.77...\) and \(100x = 27.77...\) leading to \(90x = 25\), or equivalent.
A1: \(\frac{5}{18}\) (must be in simplest form).

First, differentiate \( y \) with respect to \( x \) to find the gradient function: \( \frac{dy}{dx} = 8x - 3 \). Next, substitute \( x = 2 \) into the gradient function: \( \text{Gradient} = 8(2) - 3 = 16 - 3 = 13 \).

M1 for \( 8x - 3 \) (at least one term differentiated correctly)
A1 for 13

Divide the coefficients: \( 1.5 \div 6 = 0.25 \). Subtract the powers of 10: \( 10^8 \div 10^{-4} = 10^{8 - (-4)} = 10^{12} \). Combine them: \( 0.25 \times 10^{12} \). Convert to standard form: \( 2.5 \times 10^{11} \).

M1 for \( 0.25 \times 10^{12} \) or \( 2.5 \times 10^k \) where \( k \) is an integer
A1 for \( 2.5 \times 10^{11} \)

Multiply the matrix by the coordinate vector of \( P \): \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 0(3) + (-1)(-4) \\ 1(3) + 0(-4) \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \). Therefore, the coordinates of \( P' \) are \( (4, 3) \).

M1 for correct matrix multiplication setup or showing one coordinate is correct
A1 for \( (4, 3) \)

Divide the coefficients: \( \frac{12}{3} = 4 \). Simplify the \( x \) terms: \( \frac{x^5}{x^2} = x^{5-2} = x^3 \). Simplify the \( y \) terms: \( \frac{y^2}{y^6} = y^{2-6} = y^{-4} = \frac{1}{y^4} \). Combine the parts to get: \( \frac{4x^3}{y^4} \).

M1 for at least two of the three components correct (coefficient 4, \( x^3 \), or \( y^{-4} \))
A1 for \( \frac{4x^3}{y^4} \) or \( 4x^3 y^{-4} \)

At the end of 2 years, the value of the investment is:
\[ 5000 \times 1.03^2 = 5000 \times 1.0609 = \$5304.50 \]

At the end of 4 years (after 2 more years at 4.5% interest), the value of the investment is:
\[ 5304.50 \times 1.045^2 = 5304.50 \times 1.092025 = \$5792.65... \]

Rounding to the nearest dollar gives $5793.

M1: For calculating the value after 2 years: \( 5000 \times 1.03^2 \) or \( 5304.50 \)
M1: For multiplying their 2-year value by \( 1.045^2 \)
A1: For \( 5792.65... \)
A1ft: For rounding their final answer to the nearest integer: \( 5793 \)

First, find a common denominator for the two fractions, which is \((2x - 1)(x + 2)\).

Express each fraction with the common denominator:
\[ \frac{4(x + 2)}{(2x - 1)(x + 2)} - \frac{3(2x - 1)}{(2x - 1)(x + 2)} \]

Combine the numerators:
\[ \frac{4(x + 2) - 3(2x - 1)}{(2x - 1)(x + 2)} \]

Expand the terms in the numerator:
\[ 4x + 8 - 6x + 3 = 11 - 2x \]

This gives the final simplified fraction:
\[ \frac{11 - 2x}{(2x - 1)(x + 2)} \]

B1: For identifying \((2x - 1)(x + 2)\) or \(2x^2 + 3x - 2\) as the common denominator
M1: For \(4(x + 2) - 3(2x - 1)\) as the combined numerator
M1: For expanding the numerator to \(4x + 8 - 6x + 3\)
A1: For the correct simplified fraction: \( \frac{11 - 2x}{(2x - 1)(x + 2)} \) or \( \frac{11 - 2x}{2x^2 + 3x - 2} \)

First, calculate the area of the triangular cross-section:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 15 = 60\text{ cm}^2 \]

Next, calculate the volume of the prism:
\[ \text{Volume} = \text{Area of cross-section} \times \text{length} = 60 \times 25 = 1500\text{ cm}^3 \]

Now, calculate the mass of the prism in grams:
\[ \text{Mass (g)} = \text{Volume} \times \text{density} = 1500 \times 8.4 = 12600\text{ g} \]

Convert the mass into kilograms:
\[ \text{Mass (kg)} = \frac{12600}{1000} = 12.6\text{ kg} \]

M1: For calculating the cross-sectional area: \( \frac{1}{2} \times 8 \times 15 = 60 \)
M1: For calculating the volume: \( \text{their } 60 \times 25 = 1500 \)
M1: For calculating the mass in grams: \( \text{their } 1500 \times 8.4 = 12600 \)
A1: For \( 12.6 \text{ (kg)} \)

We can solve the quadratic equation by completing the square:
\[ x^2 - 10x + 18 = 0 \]
\[ (x - 5)^2 - 25 + 18 = 0 \]
\[ (x - 5)^2 - 7 = 0 \]
\[ (x - 5)^2 = 7 \]
\[ x - 5 = \pm \sqrt{7} \]
\[ x = 5 \pm \sqrt{7} \]

Alternatively, using the quadratic formula:
\[ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(18)}}{2} \]
\[ x = \frac{10 \pm \sqrt{100 - 72}}{2} = \frac{10 \pm \sqrt{28}}{2} \]
Since \( \sqrt{28} = 2\sqrt{7} \):
\[ x = \frac{10 \pm 2\sqrt{7}}{2} = 5 \pm \sqrt{7} \]

M1: For completing the square to get \( (x - 5)^2 \pm \dots \) or substituting correctly into the quadratic formula: \( \frac{10 \pm \sqrt{(-10)^2 - 4 \times 1 \times 18}}{2} \)
M1dep: For simplifying to \( (x - 5)^2 = 7 \) or \( \frac{10 \pm \sqrt{28}}{2} \)
A1: For simplifying the radical to \( 5 \pm \frac{\sqrt{28}}{2} \) or \( 5 \pm \sqrt{7} \) (condone one arithmetic slip in previous stages)
A1: For the final answer in the correct form: \( 5 \pm \sqrt{7} \)

The two pens are of different colours if we select either (Blue then Red) or (Red then Blue).

Probability of selecting Blue then Red:
\[ P(\text{Blue, Red}) = \frac{7}{10} \times \frac{3}{9} = \frac{21}{90} \]

Probability of selecting Red then Blue:
\[ P(\text{Red, Blue}) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90} \]

Total probability of different colours:
\[ P(\text{Different}) = \frac{21}{90} + \frac{21}{90} = \frac{42}{90} = \frac{7}{15} \]

Alternatively, using the complement of selecting the same colours:
\[ P(\text{Same}) = P(\text{Blue, Blue}) + P(\text{Red, Red}) = \left(\frac{7}{10} \times \frac{6}{9}\right) + \left(\frac{3}{10} \times \frac{2}{9}\right) = \frac{42}{90} + \frac{6}{90} = \frac{48}{90} \]
\[ P(\text{Different}) = 1 - P(\text{Same}) = 1 - \frac{48}{90} = \frac{42}{90} = \frac{7}{15} \]

M1: For writing down at least one correct product of probabilities with a descending denominator, e.g. \( \frac{7}{10} \times \frac{3}{9} \) or \( \frac{7}{10} \times \frac{6}{9} \)
M1: For identifying both products required for the chosen method: e.g. \( \frac{7}{10} \times \frac{3}{9} \) and \( \frac{3}{10} \times \frac{7}{9} \)
M1dep: For summing their two correct different-colour probabilities: \( \frac{21}{90} + \frac{21}{90} \) (or equivalent for the complement method)
A1: For \( \frac{7}{15} \) or \( \frac{42}{90} \) (or equivalent simplified fraction, decimal \( 0.467 \dots \) or percentage \( 46.7\% \))

First, find the total number of parts in the ratio:
\[ 3 + 5 = 8 \text{ parts} \]

Next, determine the weight of each type of bean in the 40 kg blend:
\[ \text{Weight of Premium} = \frac{3}{8} \times 40 = 15\text{ kg} \]
\[ \text{Weight of Standard} = \frac{5}{8} \times 40 = 25\text{ kg} \]

Now, calculate the cost for each component:
\[ \text{Cost of Premium} = 15\text{ kg} \times \$16.00 = \$240.00 \]
\[ \text{Cost of Standard} = 25\text{ kg} \times \$10.00 = \$250.00 \]

Finally, sum the costs to find the total cost of the blend:
\[ \text{Total Cost} = \$240.00 + \$250.00 = \$490.00 \]

M1: For dividing the total weight by the sum of parts in the ratio: \( 40 \div (3 + 5) = 5 \)
M1dep: For calculating the weights of both components: \( 3 \times 5 = 15\text{ kg} \) and \( 5 \times 5 = 25\text{ kg} \)
M1dep: For calculating the cost of both components: \( 15 \times 16 = 240 \) and \( 25 \times 10 = 250 \)
A1: For the final answer: \( 490 \) (or \( \$490.00 \))

First, find the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 6x^2 + b \]

Since \( (2, 10) \) is a stationary point, \( \frac{dy}{dx} = 0 \) when \( x = 2 \):
\[ 6(2)^2 + b = 0 \]
\[ 24 + b = 0 \implies b = -24 \]

Since \( (2, 10) \) lies on the curve, substitute \( x = 2 \) and \( y = 10 \) into the original equation:
\[ 10 = 2(2)^3 + b(2) + c \]
\[ 10 = 16 + 2b + c \]

Substitute \( b = -24 \) into the equation:
\[ 10 = 16 + 2(-24) + c \]
\[ 10 = 16 - 48 + c \]
\[ 10 = -32 + c \implies c = 42 \]

M1: Differentiates \( y \) to get \( 6x^2 + b \) and sets to \( 0 \) for \( x = 2 \)
A1: Finds \( b = -24 \)
M1: Substitutes \( x = 2 \), \( y = 10 \), and their \( b \) into \( y = 2x^3 + bx + c \)
A1: Finds \( c = 42 \)

Since AD is a diameter of the circle, the angle in a semicircle is a right angle:
\[ \angle ABD = 90^\circ \]

In triangle ABD, the angles sum to 180°:
\[ \angle BAD = 180^\circ - 90^\circ - 25^\circ = 65^\circ \]

ABCD is a cyclic quadrilateral because all four vertices lie on the circumference of the circle.
The opposite angles of a cyclic quadrilateral sum to 180°:
\[ \angle BCD + \angle BAD = 180^\circ \]
\[ \angle BCD = 180^\circ - 65^\circ = 115^\circ \]

B1: States that \( \angle ABD = 90^\circ \) (reason: angle in a semicircle is a right angle)
M1: Calculates \( \angle BAD = 180 - 90 - 25 = 65^\circ \)
M1: Identifies ABCD as a cyclic quadrilateral and states that opposite angles sum to 180° (\( \angle BCD = 180 - \angle BAD \))
A1: Finds \( \angle BCD = 115^\circ \)

To find the stationary point, we first differentiate the curve:
\(\frac{dy}{dx} = 6x^2 + 2ax + b\)

Since there is a stationary point at \((1, 2)\), the gradient at \(x = 1\) must be zero:
\(6(1)^2 + 2a(1) + b = 0 \implies 2a + b = -6\) (Equation 1)

Also, the point \((1, 2)\) lies on the curve, so we substitute \(x = 1\) and \(y = 2\) into the original equation:
\(2 = 2(1)^3 + a(1)^2 + b(1) + 5 \implies 2 = 2 + a + b + 5 \implies a + b = -5\) (Equation 2)

Now we solve the simultaneous equations:
Subtracting Equation 2 from Equation 1:
\((2a + b) - (a + b) = -6 - (-5) \implies a = -1\)

Substitute \(a = -1\) into Equation 2:
\(-1 + b = -5 \implies b = -4\)

M1: Differentiates \(y\) to find \(\frac{dy}{dx} = 6x^2 + 2ax + b\) (at least one term correct)
M1: Substitutes \(x = 1\) into \(\frac{dy}{dx}\) and sets to 0 to find \(2a + b = -6\)
M1: Substitutes \(x = 1\) and \(y = 2\) into the original equation to find \(a + b = -5\)
M1: Method to solve simultaneous equations to find \(a\) or \(b\)
A1: \(a = -1\) and \(b = -4\)

Let the three consecutive even integers be \(2n\), \(2n+2\), and \(2n+4\), where \(n\) is an integer.

The sum of their squares is:
\(S = (2n)^2 + (2n+2)^2 + (2n+4)^2\)

Expanding each term:
\(S = 4n^2 + (4n^2 + 8n + 4) + (4n^2 + 16n + 16)\)

Simplifying the expression:
\(S = 12n^2 + 24n + 20\)

We can rewrite this expression to show it is 8 more than a multiple of 12:
\(S = 12n^2 + 24n + 12 + 8\)

Factoring out 12 from the first three terms:
\(S = 12(n^2 + 2n + 1) + 8\)

Since \(n\) is an integer, \((n^2 + 2n + 1)\) must also be an integer. Therefore, \(12(n^2 + 2n + 1)\) is a multiple of 12. This proves that the sum of their squares is always 8 more than a multiple of 12.

B1: Sets up expressions for three consecutive even integers (e.g., \(2n\), \(2n+2\), \(2n+4\))
M1: Writes down the sum of their squares: \((2n)^2 + (2n+2)^2 + (2n+4)^2\)
M1: Expands the brackets correctly to get \(4n^2 + 4n^2 + 8n + 4 + 4n^2 + 16n + 16\) (allow one expansion error)
A1: Simplifies the expression to \(12n^2 + 24n + 20\)
A1: Correctly completes the proof by factoring out 12 to show \(12(n^2+2n+1) + 8\) and states that \(n^2+2n+1\) is an integer

First, we find \(\overrightarrow{OB}\):
\(\overrightarrow{OB} = \overrightarrow{OC} + \overrightarrow{CB} = \mathbf{c} + (2\mathbf{a} + \mathbf{c}) = 2\mathbf{a} + 2\mathbf{c}\)

Next, we find \(\overrightarrow{OQ}\) using the ratio \(OQ : QB = 2 : 1\), which means \(\overrightarrow{OQ} = \frac{2}{3}\overrightarrow{OB}\):
\(\overrightarrow{OQ} = \frac{2}{3}(2\mathbf{a} + 2\mathbf{c}) = \frac{4}{3}\mathbf{a} + \frac{4}{3}\mathbf{c}\)

Now we find \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + (2\mathbf{a} + 2\mathbf{c}) = \mathbf{a} + 2\mathbf{c}\)

Using the ratio \(AP : PB = 1 : 2\), we find \(\overrightarrow{AP} = \frac{1}{3}\overrightarrow{AB}\):
\(\overrightarrow{AP} = \frac{1}{3}(\mathbf{a} + 2\mathbf{c}) = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\)

Now we find \(\overrightarrow{OP}\):
\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \left(\frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\right) = \frac{4}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\)

Finally, we find \(\overrightarrow{PQ}\):
\(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\frac{4}{3}\mathbf{a} + \frac{4}{3}\mathbf{c}\right) - \left(\frac{4}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\right) = \frac{2}{3}\mathbf{c}\)

Since \(\overrightarrow{PQ} = \frac{2}{3}\mathbf{c}\) and \(\overrightarrow{OC} = \mathbf{c}\), \(\overrightarrow{PQ}\) is a scalar multiple of \(\overrightarrow{OC}\), which proves that \(\overrightarrow{PQ}\) is parallel to \(\overrightarrow{OC}\).

M1: Finds \(\overrightarrow{OB} = 2\mathbf{a} + 2\mathbf{c}\)
M1: Finds \(\overrightarrow{OQ} = \frac{4}{3}\mathbf{a} + \frac{4}{3}\mathbf{c}\)
M1: Finds \(\overrightarrow{AB} = \mathbf{a} + 2\mathbf{c}\) and \(\overrightarrow{AP} = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\)
M1: Finds \(\overrightarrow{OP} = \frac{4}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}\)
A1: Simplifies \(\overrightarrow{PQ} = \frac{2}{3}\mathbf{c}\) and correctly concludes they are parallel because one is a scalar multiple of the other

Let the height of the rectangular box be \(h\) cm.

The total surface area of the box, \(A\), consists of the top and bottom squares, and four rectangular sides:
\(A = 2x^2 + 4xh = 24\)

We can rearrange this equation to express \(h\) in terms of \(x\):
\(4xh = 24 - 2x^2 \implies h = \frac{24 - 2x^2}{4x} = \frac{12 - x^2}{2x}

The volume, \)V\), of the rectangular box is:
\(V = x^2 h\)

Substitute \(h\) into the volume formula:
\(V = x^2 \left(\frac{12 - x^2}{2x}\right) = x \left(\frac{12 - x^2}{2}\right) = 6x - \frac{1}{2}x^3\)

To find the maximum volume, we differentiate \(V\) with respect to \(x\):
\(\frac{dV}{dx} = 6 - \frac{3}{2}x^2\)

Set \(\frac{dV}{dx} = 0\) to find the stationary point:
\(6 - \frac{3}{2}x^2 = 0 \implies \frac{3}{2}x^2 = 6 \implies x^2 = 4 \implies x = 2\) (since length \(x > 0\))

Substitute \(x = 2\) back into the volume formula to find the maximum volume:
\(V = 6(2) - \frac{1}{2}(2)^3 = 12 - 4 = 8\) cm\(^3\).

Thus, the maximum volume is proved to be \(8\) cm\(^3\).

M1: Sets up the equation for the surface area: \(2x^2 + 4xh = 24\) and expresses \(h\) in terms of \(x\) as \(h = \frac{12-x^2}{2x}\)
M1: Substitutes \(h\) into the volume formula to show \(V = 6x - \frac{1}{2}x^3\)
M1: Differentiates \(V\) with respect to \(x\) to get \(\frac{dV}{dx} = 6 - \frac{3}{2}x^2\)
M1: Sets \(\frac{dV}{dx} = 0\) and solves for \(x = 2\)
A1: Substitutes \(x = 2\) into the volume formula to obtain \(V = 8\) cm\(^3\) and completes the proof

Add 1 to all parts of the inequality: \(-3 + 1 \le 2x < 7 + 1\) which simplifies to \(-2 \le 2x < 8\). Divide all parts by 2 to solve for \(x\): \(-1 \le x < 4\).

B1 for \(-1 \le x < 4\)

A reflection in the \(y\)-axis maps any coordinate \((x, y)\) to \((-x, y)\). Using unit basis vectors, the vector \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) is mapped to \(\begin{pmatrix} -1 \\ 0 \end{pmatrix}\) and the vector \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) remains \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\). Placing these columns together yields the transformation matrix \(\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\).

B1 for \(\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\)

By definition, \(fg(x) = f(g(x))\). Substituting the function \(g(x)\) inside \(f(x)\) gives \(f(x - 3) = (x - 3)^2\).

B1 for \((x-3)^2\)

Since events \(A\) and \(B\) are independent, \(P(A \cap B) = P(A) \times P(B)\). Calculating this product gives: \(0.6 \times 0.3 = 0.18\).

B1 for 0.18

The back bearing is found by adding \(180^\circ\) to the forward bearing because it is less than \(180^\circ\). Thus, \(115^\circ + 180^\circ = 295^\circ\).

B1 for \(295^\circ\) (accept 295)

Since \(y \propto x^2\), we can write the equation as \(y = k x^2\) for some constant \(k\). Substituting \(x = 3\) and \(y = 36\) gives: \(36 = k (3^2) \implies 36 = 9k \implies k = 4\). Therefore, the equation is \(y = 4x^2\).

B1 for \(y = 4x^2\)

Let the original price be \(P\).

\(0.88 \times P = 396\)

\(P = \frac{396}{0.88} = 450\)

The original price is $450.

M1: For \(\frac{396}{0.88}\) or \(396 \div 88 \times 100\) or equivalent calculation
A1: 450 (condone $450)

Expand the brackets:
\(7x - 4 \le 3x + 24\)

Subtract \(3x\) from both sides:
\(4x - 4 \le 24\)

Add 4 to both sides:
\(4x \le 28\)

Divide by 4:
\(x \le 7\)

M1: For expanding brackets and attempting to collect terms correctly, e.g. \(4x \le 28\) or \(4x \le 20\) (allowing one arithmetic slip)
A1: \(x \le 7\) (or \(7 \ge x\))

The total surface area of a solid hemisphere is the curved surface area plus the area of its circular base:

\(\text{Total Surface Area} = 2\pi r^2 + \pi r^2 = 3\pi r^2\)

Substitute \(r = 5\):

\(\text{Total Surface Area} = 3 \times \pi \times 5^2 = 75\pi \approx 235.619\text{ cm}^2\)

To 3 significant figures, this is 236.

M1: For \(3 \times \pi \times 5^2\) or \(2 \times \pi \times 5^2 + \pi \times 5^2\) or equivalent correct method
A1: 236 (accept any value in the range [235.5, 236])

First, find the derivative \(\frac{dy}{dx}\) to represent the gradient function:

\(\frac{dy}{dx} = 6x - 4\)

Substitute \(x = 3\) into the derivative:

\(\text{Gradient} = 6(3) - 4 = 18 - 4 = 14\).

M1: For differentiating with at least one term correct (i.e., \(6x\) or \(-4\))
A1: 14

Multiply the coefficients and the powers of 10 separately:

\((3.5 \times 8) \times (10^5 \times 10^{-2})\)

\(= 28 \times 10^{5-2}\)

\(= 28 \times 10^3\)

Convert into correct standard form:

\(= 2.8 \times 10^4\)

M1: For a correct calculation leading to \(28 \times 10^3\) or \(28000\)
A1: \(2.8 \times 10^4\)

Simplify the coefficients:
\(\frac{15}{3} = 5\)

Simplify the powers of \(x\):
\(\frac{x^5}{x^2} = x^{5-2} = x^3\)

Simplify the powers of \(y\):
\(\frac{y^2}{y^6} = y^{2-6} = y^{-4} = \frac{1}{y^4}\)

Combine the simplified terms:
\(\frac{5x^3}{y^4}\)

M1: For a product with at least two of the three components simplified correctly (e.g. \(5\), \(x^3\), or \(y^{-4}\))
A1: \(\frac{5x^3}{y^4}\) or \(5x^3y^{-4}\)

First, calculate \(2\mathbf{a}\):

\(2\mathbf{a} = 2 \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}\)

Next, calculate \(3\mathbf{b}\):

\(3\mathbf{b} = 3 \begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 12 \end{pmatrix}\)

Now add the two vectors:

\(2\mathbf{a} + 3\mathbf{b} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} + \begin{pmatrix} -3 \\ 12 \end{pmatrix} = \begin{pmatrix} 6 + (-3) \\ -4 + 12 \end{pmatrix} = \begin{pmatrix} 3 \\ 8 \end{pmatrix}\)

M1: For correctly evaluating \(\begin{pmatrix} 6 \\ -4 \end{pmatrix}\) or \(\begin{pmatrix} -3 \\ 12 \end{pmatrix}\), or for a final column vector with one correct coordinate
A1: \(\begin{pmatrix} 3 \\ 8 \end{pmatrix}\)

The total number of counters is \(4 + 6 = 10\).

Since replacement occurs, the probabilities stay the same for the second draw.

There are two ways to get different colours: Green then Yellow (GY) or Yellow then Green (YG).

\(P(\text{GY}) = \frac{4}{10} \times \frac{6}{10} = \frac{24}{100} = 0.24\)

\(P(\text{YG}) = \frac{6}{10} \times \frac{4}{10} = \frac{24}{100} = 0.24\)

Total probability of different colours:

\(P(\text{different}) = 0.24 + 0.24 = 0.48\)

M1: For calculating the probability of a single valid outcome, e.g. \(\frac{4}{10} \times \frac{6}{10}\) (or \(0.24\)), or for showing the structure of adding both combinations oe
A1: 0.48 or \(\frac{12}{25}\) or equivalent correct fraction/percentage

Let the hypotenuse be \(h\).

Using trigonometry:
\(\cos(43^\circ) = \dfrac{9.6}{h}\)

Rearranging to solve for \(h\):
\(h = \dfrac{9.6}{\cos(43^\circ)}\)

\(h \approx 13.126\text{ cm}\)

To 3 significant figures, \(h = 13.1\text{ cm}\).

M1 for writing a correct trigonometric ratio, eg \(\cos(43^\circ) = \dfrac{9.6}{h}\) or \(\dfrac{9.6}{\cos(43^\circ)}\)

A1 for \(13.1\) (accept answers in the range \([13.1, 13.13]\))

First, multiply both sides by the common denominator \((x-2)(x+1)\) to get \(4(x+1) - 3(x-2) = (x-2)(x+1)\). Expanding the brackets gives \(4x + 4 - 3x + 6 = x^2 - x - 2\). Simplifying leads to \(x + 10 = x^2 - x - 2\). Rearranging into standard quadratic form gives \(x^2 - 2x - 12 = 0\). Applying the quadratic formula: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-12)}}{2} = \frac{2 \pm \sqrt{52}}{2} = 1 \pm \sqrt{13}\). Thus, \(x \approx 4.61\) and \(x \approx -2.61\).

M1 for attempts to multiply through by \((x-2)(x+1)\) or common denominator. M1 for reducing to the quadratic form \(x^2 - 2x - 12 = 0\) (allow 1 sign error). M1 for substituting correctly into the quadratic formula. A1 for 4.61 and -2.61 (both correct to 3 s.f.).

Let the total mass of Alloy Z be 70 kg. Since Alloy X and Alloy Y are mixed in the ratio \(4 : 3\), we have 40 kg of Alloy X and 30 kg of Alloy Y. The mass of Metal B in Alloy X is \(\frac{3}{8} \times 40 = 15\) kg. The mass of Metal B in Alloy Y is \(\frac{2}{3} \times 30 = 20\) kg. The total mass of Metal B in Alloy Z is \(15 + 20 = 35\) kg. The percentage of Metal B in Alloy Z is \(\frac{35}{70} \times 100 = 50\%\).

M1 for finding the fraction of Metal B in Alloy X (\(\frac{3}{8}\)) and Alloy Y (\(\frac{2}{3}\)). M1 for using a consistent total mass (e.g. 70) or working with algebraic fractions of Z (e.g. \(\frac{4}{7}\) and \(\frac{3}{7}\)). M1 for calculating the total mass or fraction of Metal B: \(15 + 20 = 35\) kg (or \(\frac{1}{2}\) of the total). A1 for 50%.

The volume of a hemisphere of radius \(r\) is \(V_{\text{hemisphere}} = \frac{2}{3}\pi r^3\). The volume of a cone is \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\). Since \(h = 3r\), \(V_{\text{cone}} = \frac{1}{3}\pi r^2 (3r) = \pi r^3\). The total volume is \(V_{\text{total}} = \frac{2}{3}\pi r^3 + \pi r^3 = \frac{5}{3}\pi r^3\). Setting this equal to the given volume: \(\frac{5}{3}\pi r^3 = 240\pi\). Dividing both sides by \(\pi\) gives \(\frac{5}{3}r^3 = 240\), which simplifies to \(r^3 = 144\). Taking the cube root, \(r = \sqrt[3]{144} \approx 5.24\) cm.

M1 for writing correct expressions for either the volume of the hemisphere (\(\frac{2}{3}\pi r^3\)) or the cone (\(\pi r^3\)). M1 for setting up the equation for the total volume: \(\frac{5}{3}\pi r^3 = 240\pi\). M1 for rearranging to find \(r^3 = 144\). A1 for 5.24 (accept [5.24, 5.242]).

First, differentiate \(y\) to find the gradient function: \(\frac{dy}{dx} = 6x^2 - 18x - 24\). Set the gradient to zero at stationary points: \(6x^2 - 18x - 24 = 0\), which simplifies to \(x^2 - 3x - 4 = 0\). Factorising gives \((x - 4)(x + 1) = 0\), yielding stationary points at \(x = 4\) and \(x = -1\). Substituting back into the original curve equation: for \(x = 4\), \(y = 2(4)^3 - 9(4)^2 - 24(4) + 7 = 128 - 144 - 96 + 7 = -105\); for \(x = -1\), \(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7 = -2 - 9 + 24 + 7 = 20\).

M1 for correct differentiation of at least two terms to get \(\frac{dy}{dx} = 6x^2 - 18x - 24\). M1 for setting \(\frac{dy}{dx} = 0\) and solving to find \(x = 4\) and \(x = -1\). M1 for substituting at least one of their x-values back into the original curve equation to find a y-value. A1 for both coordinates (4, -105) and (-1, 20) correct.

In parallelogram \(ABCD\), \(\vec{BC} = \vec{AD} = \mathbf{q}\). Since \(M\) is the midpoint of \(BC\), \(\vec{MC} = \frac{1}{2}\mathbf{q}\) and \(\vec{BM} = \frac{1}{2}\mathbf{q}\). Also, \(\vec{CD} = -\mathbf{p}\). Since \(DN : NC = 3 : 1\), \(\vec{CN} = \frac{1}{4}\vec{CD} = -\frac{1}{4}\mathbf{p}\). Thus, \(\vec{MN} = \vec{MC} + \vec{CN} = \frac{1}{2}\mathbf{q} - \frac{1}{4}\mathbf{p}\). Then, \(\vec{ME} = 3\vec{MN} = \frac{3}{2}\mathbf{q} - \frac{3}{4}\mathbf{p}\). Finally, \(\vec{AE} = \vec{AB} + \vec{BM} + \vec{ME} = \mathbf{p} + \frac{1}{2}\mathbf{q} + \frac{3}{2}\mathbf{q} - \frac{3}{4}\mathbf{p} = \frac{1}{4}\mathbf{p} + 2\mathbf{q}\).

M1 for finding \(\vec{MC} = \frac{1}{2}\mathbf{q}\) and \(\vec{CN} = -\frac{1}{4}\mathbf{p}\) (or equivalent vector path). M1 for finding \(\vec{MN} = \frac{1}{2}\mathbf{q} - \frac{1}{4}\mathbf{p}\). M1 for finding \(\vec{ME} = \frac{3}{2}\mathbf{q} - \frac{3}{4}\mathbf{p}\). A1 for \(\frac{1}{4}\mathbf{p} + 2\mathbf{q}\) or equivalent simplified vector.

Let \(P\) be the initial investment. At the end of year 1, the value is \(P \times 1.15\). At the end of year 2, the value is \(P \times 1.15 \times 0.92 = 1.058P\). At the end of year 3, the value is \(1.058P \times 1.05 = 1.1109P\). We are given this final value is \(\$11,109\), so \(1.1109P = 11109\). Solving for \(P\) gives \(P = \frac{11109}{1.1109} = 10000\).

M1 for establishing the multipliers 1.15 and 0.92. M1 for establishing the multiplier 1.05. M1 for setting up the equation \(1.1109P = 11109\) (or \(P \times 1.15 \times 0.92 \times 1.05 = 11109\)). A1 for 10000.

The formula for the total surface area of a cylinder is \(A = 2\pi r^2 + 2\pi r h\).
Substituting \(r = 2x\) and \(h = x + 3\):
\(2\pi(2x)^2 + 2\pi(2x)(x + 3) = 160\pi\)

Divide both sides of the equation by \(2\pi\):
\((2x)^2 + 2x(x + 3) = 80\)
\(4x^2 + 2x^2 + 6x = 80\)
\(6x^2 + 6x - 80 = 0\)

Divide by 2:
\(3x^2 + 3x - 40 = 0\)

Using the quadratic formula where \(a = 3\), \(b = 3\), and \(c = -40\):
\(x = \frac{-3 \pm \sqrt{3^2 - 4(3)(-40)}}{2(3)}
\)x = \frac{-3 \pm \sqrt{9 + 480}}{6}
\(x = \frac{-3 \pm \sqrt{489}}{6}

Since \)x\) must be positive:
\(x = \frac{-3 + 22.113}{6} \approx 3.1856\text{ cm}

To 3 significant figures, \)x = 3.19\).

M1: Equating the surface area formula to \(160\pi\), i.e., \(2\pi(2x)^2 + 2\pi(2x)(x + 3) = 160\pi\)
M1: Simplifying to a three-term quadratic equation, e.g., \(6x^2 + 6x - 80 = 0\) or \(3x^2 + 3x - 40 = 0\)
M1: Applying a correct method to solve their three-term quadratic equation, e.g., the quadratic formula \(x = \frac{-3 \pm \sqrt{3^2 - 4(3)(-40)}}{2(3)}\)
A1: \(3.19\) (accept answers in the range \([3.18, 3.19]\))

The total number of counters in the box is \(n + 4\).
The probability of selecting a blue counter first is \(\frac{n}{n + 4}\).
Since there is no replacement, the probability of selecting a blue counter second is \(\frac{n - 1}{n + 3}\).

The probability of selecting two blue counters is:
\(\frac{n}{n + 4} \times \frac{n - 1}{n + 3} = \frac{1}{3}\)
\(\frac{n^2 - n}{n^2 + 7n + 12} = \frac{1}{3}\)

Multiply both sides by \(3(n^2 + 7n + 12)\):
\(3(n^2 - n) = n^2 + 7n + 12\)
\(3n^2 - 3n = n^2 + 7n + 12\)

Rearrange to form a quadratic equation:
\(2n^2 - 10n - 12 = 0\)

Divide by 2:
\(n^2 - 5n - 6 = 0\)

Factorise:
\((n - 6)(n + 1) = 0\)

This gives \(n = 6\) or \(n = -1\).
Since the number of counters must be positive, \(n = 6\).

M1: Sets up the initial probability product, e.g., \(\frac{n}{n + 4} \times \frac{n - 1}{n + 3} = \frac{1}{3}\)
M1: Multiplies out to form a quadratic equation, e.g., \(3(n^2 - n) = n^2 + 7n + 12\)
M1: Simplifies to a standard quadratic form, e.g., \(n^2 - 5n - 6 = 0\) and attempts to solve by factoring or quadratic formula
A1: \(6\) (reject \(n = -1\))

The total area of the rectangular lawn is: \( (2x + 1)(x + 2) = 2x^2 + 5x + 2 \) The area of the square patio is: \( (x - 1)^2 = x^2 - 2x + 1 \) The remaining area is the difference: \( (2x^2 + 5x + 2) - (x^2 - 2x + 1) = x^2 + 7x + 1 \) We are given that the remaining area is \( 19\text{ m}^2 \), so: \( x^2 + 7x + 1 = 19 \) \( x^2 + 7x - 18 = 0 \) Factorising the quadratic equation: \( (x + 9)(x - 2) = 0 \) This gives \( x = -9 \) or \( x = 2 \). Since a length must be positive and the side of the patio is \( x - 1 \), we must have \( x > 1 \). Therefore, \( x = 2 \).

- **M1**: Sets up a correct expression for either the total area of the lawn, \( (2x + 1)(x + 2) \), or the area of the patio, \( (x - 1)^2 \).
- **M1**: Subtracts the patio area from the lawn area to form the quadratic expression \( x^2 + 7x + 1 \).
- **M1**: Equates to 19 and forms a solvable quadratic equation, e.g., \( x^2 + 7x - 18 = 0 \) or factorises to \( (x + 9)(x - 2) = 0 \).
- **A1**: Identifies \( x = 2 \) as the only valid solution (rejecting \( x = -9 \)).

Let the number of full-time workers be \( 5x \) and the number of part-time workers be \( 3x \). For the full-time workers: Male workers = \( 40\% \) of \( 5x = 0.40 \times 5x = 2x \), Female workers = \( 60\% \) of \( 5x = 0.60 \times 5x = 3x \). For the part-time workers: Female workers = \( 60\% \) of \( 3x = 0.60 \times 3x = 1.8x \), Male workers = \( 40\% \) of \( 3x = 0.40 \times 3x = 1.2x \). Now, calculate the total number of male and female workers: Total male workers = \( 2x + 1.2x = 3.2x \), Total female workers = \( 3x + 1.8x = 4.8x \). The difference between total female and total male workers is: \( 4.8x - 3.2x = 1.6x \). We are given that this difference is 80: \( 1.6x = 80 \implies x = 50 \). The total number of workers is: \( 5x + 3x = 8x \). Substituting \( x = 50 \): \( 8 \times 50 = 400 \).

- **M1**: Uses the ratio to define full-time as \( 5x \) and part-time as \( 3x \), or works with percentages of parts (e.g., \( 2 \) parts male FT, \( 3 \) parts female FT).
- **M1**: Calculates the expressions for total male workers (\( 3.2x \)) and total female workers (\( 4.8x \)).
- **M1**: Sets up the equation for the difference, e.g., \( 1.6x = 80 \), and solves for \( x \).
- **A1**: Correct total of 400 workers.

First, find the derivative \( \frac{dy}{dx} \) to determine the gradient function: \( \frac{dy}{dx} = 6x^2 - 18x - 24 \) At stationary points, the gradient is zero, so set \( \frac{dy}{dx} = 0 \): \( 6x^2 - 18x - 24 = 0 \implies x^2 - 3x - 4 = 0 \) Factorise the quadratic: \( (x - 4)(x + 1) = 0 \) This gives the \( x \)-coordinates of the stationary points: \( x = 4 \) and \( x = -1 \). Now, substitute these \( x \)-values back into the original curve equation to find the corresponding \( y \)-coordinates: For \( x = 4 \): \( y = 2(4)^3 - 9(4)^2 - 24(4) + 7 = 128 - 144 - 96 + 7 = -105 \). For \( x = -1 \): \( y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7 = -2 - 9 + 24 + 7 = 20 \). Thus, the coordinates of the stationary points are \( (4, -105) \) and \( (-1, 20) \).

- **M1**: Differentiates to find \( \frac{dy}{dx} = 6x^2 - 18x - 24 \) (at least one term correct).
- **M1**: Sets their derivative equal to 0 and solves the quadratic to find \( x = 4 \) and \( x = -1 \).
- **M1**: Substitutes at least one of their \( x \)-values back into the original equation to find the \( y \)-coordinate.
- **A1**: Both correct coordinate pairs: \( (4, -105) \) and \( (-1, 20) \) (accept in any order).

Let the volume of the cylinder be \(V = \pi r^2 h = 128\pi\).

Rearranging for \(h\):
\(h = \frac{128}{r^2}\)

The total surface area \(S\) of a solid cylinder is:
\(S = 2\pi r^2 + 2\pi r h\)

Substitute the expression for \(h\) into the surface area formula:
\(S = 2\pi r^2 + 2\pi r \left(\frac{128}{r^2}\right) = 2\pi r^2 + \frac{256\pi}{r}\)

To find the stationary points, differentiate \(S\) with respect to \(r\):
\(\frac{dS}{dr} = 4\pi r - \frac{256\pi}{r^2}\)

Set \(\frac{dS}{dr} = 0\):
\(4\pi r - \frac{256\pi}{r^2} = 0 \implies 4\pi r = \frac{256\pi}{r^2} \implies r^3 = 64 \implies r = 4\)

Substitute \(r = 4\) back into the expression for \(S\) to find the minimum surface area:
\(S_{\text{min}} = 2\pi (4)^2 + \frac{256\pi}{4} = 32\pi + 64\pi = 96\pi\text{ cm}^2\)

To prove that this is a minimum, find the second derivative:
\(\frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{r^3}\)

At \(r = 4\):
\(\frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{64} = 12\pi\)

Since \(\frac{d^2S}{dr^2} > 0\), the value of \(S = 96\pi\) is a minimum.

**M1** For setting up the volume equation \(\pi r^2 h = 128\pi\) and finding \(h = \frac{128}{r^2}\) (or equivalent expression).
**M1** For substituting \(h\) into the total surface area formula \(S = 2\pi r^2 + 2\pi r h\) to get \(S = 2\pi r^2 + \frac{256\pi}{r}\).
**M1** For differentiating \(S\) correctly to get \(\frac{dS}{dr} = 4\pi r - \frac{256\pi}{r^2}\).
**A1** For setting \(\frac{dS}{dr} = 0\) and solving to find \(r = 4\).
**A1** For calculating the minimum surface area as \(96\pi\) (or approx. \(301.6\)).
**B1** For finding \(\frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{r^3}\), evaluating it at \(r = 4\) as \(12\pi > 0\) to correctly justify it is a minimum.

First, express \(\vec{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\):
Since \(P\) is the midpoint of \(AB\):
\(\vec{OP} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)

Since \(X\) lies on \(OP\), we can write:
\(\vec{OX} = k\vec{OP} = \frac{k}{2}\mathbf{a} + \frac{k}{2}\mathbf{b}\) --- [Equation 1]

Now find an expression for \(\vec{OX}\) via point \(Q\):
Since \(OQ : QB = 2 : 1\), \(\vec{OQ} = \frac{2}{3}\mathbf{b}\).

Thus, \(\vec{AQ} = \vec{OQ} - \vec{OA} = \frac{2}{3}\mathbf{b} - \mathbf{a}\).

Since \(X\) lies on \(AQ\), we can write:
\(\vec{OX} = \vec{OA} + m\vec{AQ} = \mathbf{a} + m\left(\frac{2}{3}\mathbf{b} - \mathbf{a}\right) = (1 - m)\mathbf{a} + \frac{2m}{3}\mathbf{b}\) --- [Equation 2]

Equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from Equation 1 and Equation 2:
From \(\mathbf{a}\): \(\frac{k}{2} = 1 - m \implies m = 1 - \frac{k}{2}\)
From \(\mathbf{b}\): \(\frac{k}{2} = \frac{2m}{3}\)

Substitute the expression for \(m\) into the second equation:
\(\frac{k}{2} = \frac{2}{3}\left(1 - \frac{k}{2}\right)
\implies \frac{k}{2} = \frac{2}{3} - \frac{k}{3}
\implies \frac{k}{2} + \frac{k}{3} = \frac{2}{3}
\implies \frac{5k}{6} = \frac{2}{3}
\implies k = \frac{4}{5}\)

Therefore, \(\vec{OX} = \frac{4}{5}\vec{OP}\).

**M1** For finding \(\vec{OP} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).
**M1** For setting up \(\vec{OX} = k\vec{OP} = \frac{k}{2}\mathbf{a} + \frac{k}{2}\mathbf{b}\).
**M1** For finding \(\vec{AQ} = \frac{2}{3}\mathbf{b} - \mathbf{a}\) and writing \(\vec{OX} = (1 - m)\mathbf{a} + \frac{2m}{3}\mathbf{b}\).
**M1** For equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to obtain \(\frac{k}{2} = 1 - m\) and \(\frac{k}{2} = \frac{2m}{3}\).
**A1** For solving the simultaneous equations to find \(k = \frac{4}{5}\) or \(m = \frac{3}{5}\).
**A1** For completing the proof with clear logical steps showing \(\vec{OX} = \frac{4}{5}\vec{OP}\).