2026 Edexcel IAS-Level Mathematics (XMA01) 模擬試題連答案詳解

(a) Setting the equations equal:
\[ 2x^2 - kx + 8 = 3x + 6 \]
Subtracting \( 3x + 6 \) from both sides gives:
\[ 2x^2 - (k+3)x + 2 = 0 \]
which is the required equation.

(b) For \( L \) to be tangent to \( C \), the equation of intersection must have a repeated root, so the discriminant must be zero:
\[ b^2 - 4ac = 0 \implies (-(k+3))^2 - 4(2)(2) = 0 \]
\[ (k+3)^2 - 16 = 0 \implies k+3 = \pm 4 \]
Thus, \( k = 1 \) or \( k = -7 \).

(c) For the positive value \( k = 1 \), substitute back into the equation from part (a):
\[ 2x^2 - 4x + 2 = 0 \implies 2(x-1)^2 = 0 \implies x = 1 \]
Substitute \( x = 1 \) into \( y = 3x + 6 \):
\[ y = 3(1) + 6 = 9 \]
So the coordinates of the point of contact are \( (1, 9) \).

(a)
- M1: Sets the two equations equal to each other.
- A1: Rearranges correctly to the given form with no errors.

(b)
- M1: Identifies the need to use the discriminant \( b^2 - 4ac = 0 \).
- M1: Substitutes \( a = 2 \), \( b = -(k+3) \), and \( c = 2 \) into the discriminant.
- A1: Obtains a correct quadratic in \( k \) (e.g., \( k^2 + 6k - 7 = 0 \) or \( (k+3)^2 = 16 \)).
- A1: Correctly solves to find both \( k = 1 \) and \( k = -7 \).

(c)
- M1: Substitutes \( k = 1 \) into the equation from (a) and solves for \( x \).
- A1: Correct point of contact coordinates \( (1, 9) \).

(a) Rearranging \( 3x - 4y + 12 = 0 \) into slope-intercept form:
\[ 4y = 3x + 12 \implies y = \frac{3}{4}x + 3 \]
So the gradient of \( l_1 \) is \( m_1 = \frac{3}{4} \).
The perpendicular gradient is:
\[ m_2 = -\frac{1}{m_1} = -\frac{4}{3} \]
Using the point-slope formula with \( A(3, -1) \):
\[ y - (-1) = -\frac{4}{3}(x - 3) \]
\[ 3(y + 1) = -4(x - 3) \implies 3y + 3 = -4x + 12 \implies 4x + 3y - 9 = 0 \]

(b) Solve simultaneously:
1) \( 3x - 4y = -12 \)
2) \( 4x + 3y = 9 \)
Multiply (1) by 3 and (2) by 4:
\[ 9x - 12y = -36 \]
\[ 16x + 12y = 36 \]
Adding the equations:
\[ 25x = 0 \implies x = 0 \]
Substituting \( x = 0 \) into \( 4x + 3y = 9 \) yields:
\[ 3y = 9 \implies y = 3 \]
So \( B \) is \( (0, 3) \).

(c) The distance between \( A(3, -1) \) and \( B(0, 3) \) is:
\[ AB = \sqrt{(3 - 0)^2 + (-1 - 3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

(a)
- M1: Rearranges \( l_1 \) to find its gradient is \( \frac{3}{4} \).
- M1: Uses the perpendicular gradient rule \( m_2 = -1/m_1 \) and writes an equation using \( A(3, -1) \).
- A1: Obtains the correct integer form equation: \( 4x + 3y - 9 = 0 \).

(b)
- M1: Attempts to solve the simultaneous equations by elimination or substitution.
- A1: Correctly finds either \( x = 0 \) or \( y = 3 \).
- A1: Correct coordinates of \( B(0, 3) \).

(c)
- M1: Applies the distance formula correctly using coordinates of \( A \) and \( B \).
- A1: Obtains the exact length of \( 5 \).

(a) Using the identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \):
\[ 3\sin\theta\left(\frac{\sin\theta}{\cos\theta}\right) = 8 \implies 3\frac{\sin^2\theta}{\cos\theta} = 8 \implies 3\sin^2\theta = 8\cos\theta \]
Substitute \( \sin^2\theta = 1 - \cos^2\theta \):
\[ 3(1 - \cos^2\theta) = 8\cos\theta \implies 3 - 3\cos^2\theta = 8\cos\theta \]
Rearranging gives:
\[ 3\cos^2\theta + 8\cos\theta - 3 = 0 \]

(b) Let \( \theta = 2x - 10^\circ \). For \( 0 \le x < 180^\circ \), the range of \( \theta \) is \( -10^\circ \le \theta < 350^\circ \).
Solve the quadratic:
\[ (3\cos\theta - 1)(\cos\theta + 3) = 0 \]
Since \( \cos\theta = -3 \) has no real solutions, we solve:
\[ \cos\theta = \frac{1}{3} \]
\[ \theta \approx 70.5288^\circ \quad \text{or} \quad \theta \approx 360^\circ - 70.5288^\circ = 289.4712^\circ \]
Substitute back \( 2x - 10^\circ = \theta \):
1) \( 2x - 10^\circ = 70.5288^\circ \implies 2x = 80.5288^\circ \implies x \approx 40.3^\circ \)
2) \( 2x - 10^\circ = 289.4712^\circ \implies 2x = 299.4712^\circ \implies x \approx 149.7^\circ \)
Both solutions lie in the range \( 0 \le x < 180^\circ \).

(a)
- M1: Uses \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) correctly.
- M1: Employs the identity \( \sin^2\theta = 1 - \cos^2\theta \) to produce an equation in terms of \( \cos\theta \) only.
- A1: Correctly derives the given quadratic equation with no algebraic errors.

(b)
- M1: Solves the quadratic to find \( \cos\theta = \frac{1}{3} \).
- A1: Obtains at least one correct value of \( 2x - 10^\circ \), such as \( 70.5^\circ \) or \( 289.5^\circ \).
- M1: Sets up two equations to solve for \( x \).
- A1: Finds one correct value of \( x \) (either \( 40.3^\circ \) or \( 149.7^\circ \)).
- A1: Obtains both \( x \approx 40.3^\circ \) and \( x \approx 149.7^\circ \) and no others.

(a) Differentiate to find \( \frac{\mathrm{d}y}{\mathrm{d}x} \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 - 8x + 12 \]
Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) to locate stationary points:
\[ x^2 - 8x + 12 = 0 \implies (x-2)(x-6) = 0 \]
So \( x = 2 \) or \( x = 6 \).
Substituting \( x = 2 \) into the curve's equation:
\[ y = \frac{1}{3}(8) - 4(4) + 12(2) + 5 = \frac{8}{3} - 16 + 24 + 5 = \frac{47}{3} \]
Substituting \( x = 6 \):
\[ y = \frac{1}{3}(216) - 4(36) + 12(6) + 5 = 72 - 144 + 72 + 5 = 5 \]
The stationary points are \( \left(2, \frac{47}{3}\right) \) and \( (6, 5) \).

(b) Differentiate again to find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2x - 8 \]
At \( x = 2 \):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(2) - 8 = -4 < 0 \implies \text{local maximum at } \left(2, \frac{47}{3}\right) \]
At \( x = 6 \):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(6) - 8 = 4 > 0 \implies \text{local minimum at } (6, 5) \]

(a)
- M1: Attempts to differentiate with at least one non-zero power decreased by 1.
- A1: Correctly finds \( \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 - 8x + 12 \).
- M1: Sets their derivative equal to zero and solves the quadratic to find \( x \).
- A1: Finds one correct corresponding \( y \)-coordinate (either \( y = 47/3 \) or \( y = 5 \)).
- A1: Fully correct coordinates for both stationary points: \( \left(2, \frac{47}{3}\right) \) and \( (6, 5) \).

(b)
- M1: Correctly differentiates to find \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = Ax + B \).
- A1: Finds \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2x - 8 \).
- A1: Substitutes both \( x \) values, evaluates the sign of the second derivative, and correctly identifies \( \left(2, \frac{47}{3}\right) \) as a local maximum and \( (6, 5) \) as a local minimum.

(a) Set \( y = 0 \):
\[ 9x^{1/2} - x^{3/2} = 0 \implies x^{1/2}(9 - x) = 0 \]
This gives solutions \( x = 0 \) (the origin) and \( x = 9 \) (the point \( A(9, 0) \)).

(b) The area is given by the definite integral:
\[ R = \int_{0}^{9} (9x^{1/2} - x^{3/2}) \, \mathrm{d}x \]
Integrating term-by-term:
\[ \int (9x^{1/2} - x^{3/2}) \, \mathrm{d}x = 9\left(\frac{x^{3/2}}{3/2}\right) - \frac{x^{5/2}}{5/2} + C = 6x^{3/2} - \frac{2}{5}x^{5/2} + C \]
Substituting the limits 9 and 0:
\[ R = \left[ 6x^{3/2} - \frac{2}{5}x^{5/2} \right]_{0}^{9} = \left( 6(9^{3/2}) - \frac{2}{5}(9^{5/2}) \right) - (0) \]
\[ = 6(27) - \frac{2}{5}(243) = 162 - 97.2 = 64.8 \quad \text{or} \quad \frac{324}{5} \]

(a)
- M1: Sets \( y = 0 \) and attempts to factor out \( x^{1/2} \) or divide by it.
- A1: Obtains \( x = 0 \) and \( x = 9 \) correctly.

(b)
- M1: Sets up the integral \( \int (9x^{1/2} - x^{3/2}) \, \mathrm{d}x \) with correct limits \( 0 \) and \( 9 \).
- M1: Integrates at least one term correctly.
- A1: Fully correct integrated expression: \( 6x^{3/2} - 0.4x^{5/2} \).
- M1: Substitutes the limit \( x = 9 \) correctly into their integrated expression.
- A1: Obtains the correct numerical value \( 64.8 \) (or equivalent fraction \( 324/5 \)).

(a) The curve is a standard reciprocal graph translated 3 units vertically upwards.
- Asymptotes are \( x = 0 \) and \( y = 3 \).
- The curve does not cross the \( y \)-axis.
- For \( x \)-axis intersection, set \( y = 0 \):
\[ \frac{2}{x} + 3 = 0 \implies \frac{2}{x} = -3 \implies x = -\frac{2}{3} \]
So the intersection point is \( \left(-\frac{2}{3}, 0\right) \).
The branches lie in the region \( x > 0, y > 3 \) and \( x < 0, y < 3 \).

(b) Translating by \( \begin{pmatrix} -2 \\ 1 \end{pmatrix} \) means shifting 2 units left and 1 unit up:
\[ y = f(x+2) + 1 = \left( \frac{2}{x+2} + 3 \right) + 1 = \frac{2}{x+2} + 4 \]

(c) To find the crossing point with the \( y \)-axis, set \( x = 0 \):
\[ y = \frac{2}{0+2} + 4 = 1 + 4 = 5 \]

(a)
- B1: Correct reciprocal shape with two branches.
- B1: Correct asymptotes stated or drawn: \( x = 0 \) and \( y = 3 \).
- B1: Stating or marking the correct intercept \( \left(-\frac{2}{3}, 0\right) \).

(b)
- M1: Replaces \( x \) with \( x+2 \) or adds \( 1 \) to the function.
- A1: Obtains the correct equation \( y = \frac{2}{x+2} + 4 \) (or equivalent form like \( y = \frac{4x+10}{x+2} \)).

(c)
- M1: Substitutes \( x = 0 \) into their translated equation.
- A1: Correctly evaluates \( y = 5 \).

(a) Find the midpoint \( M \) of \( AB \):
\[ M = \left( \frac{-3 + 5}{2}, \frac{2 + 8}{2} \right) = (1, 5) \]
Find the gradient of \( AB \):
\[ m_{AB} = \frac{8 - 2}{5 - (-3)} = \frac{6}{8} = \frac{3}{4} \]
The gradient of the perpendicular bisector is:
\[ m = -\frac{4}{3} \]
Using the point-slope form through \( (1, 5) \):
\[ y - 5 = -\frac{4}{3}(x - 1) \implies y = -\frac{4}{3}x + \frac{4}{3} + 5 \implies y = -\frac{4}{3}x + \frac{19}{3} \]

(b) For intersection with the \( y \)-axis \( P \), set \( x = 0 \):
\[ P = \left( 0, \frac{19}{3} \right) \]
For intersection with the \( x \)-axis \( Q \), set \( y = 0 \):
\[ 0 = -\frac{4}{3}x + \frac{19}{3} \implies 4x = 19 \implies x = \frac{19}{4} \implies Q = \left( \frac{19}{4}, 0 \right) \]
The area of the right-angled triangle \( OPQ \) is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{19}{4} \times \frac{19}{3} = \frac{361}{24} \]

(a)
- M1: Finds the midpoint of \( AB \) correctly as \( (1, 5) \).
- M1: Calculates the gradient of \( AB \) as \( \frac{3}{4} \) and identifies the perpendicular gradient as \( -\frac{4}{3} \).
- M1: Uses their midpoint and perpendicular gradient to write a line equation.
- A1: Obtains \( y = -\frac{4}{3}x + \frac{19}{3} \) (accept equivalent fractions).

(b)
- M1: Identifies the intercepts of their line equation from (a).
- A1: Correct coordinates \( P\left(0, \frac{19}{3}\right) \) and \( Q\left(\frac{19}{4}, 0\right) \) (or equivalent lengths from the origin).
- M1: Evaluates the area using \( \frac{1}{2} \times OP \times OQ \).
- A1: Obtains the correct exact fraction \( \frac{361}{24} \).

(a) To find the gradient of the tangent at \( P(4, 12) \), substitute \( x = 4 \) into the derivative:
\[ m = 3\sqrt{4} - \frac{4}{\sqrt{4}} = 3(2) - \frac{4}{2} = 6 - 2 = 4 \]
Using the point-slope form with \( (4, 12) \):
\[ y - 12 = 4(x - 4) \implies y - 12 = 4x - 16 \implies y = 4x - 4 \]

(b) Integrate the gradient function to find \( y \):
\[ y = \int \left( 3x^{1/2} - 4x^{-1/2} \right) \, \mathrm{d}x \]
\[ y = 3\left( \frac{x^{3/2}}{3/2} \right) - 4\left( \frac{x^{1/2}}{1/2} \right) + K \]
\[ y = 2x^{3/2} - 8x^{1/2} + K \]
To find the constant \( K \), substitute the coordinates of \( P(4, 12) \):
\[ 12 = 2(4^{3/2}) - 8(4^{1/2}) + K \]
\[ 12 = 2(8) - 8(2) + K \]
\[ 12 = 16 - 16 + K \implies K = 12 \]
Thus, the equation of the curve is:
\[ y = 2x^{3/2} - 8x^{1/2} + 12 \]

(a)
- M1: Substitutes \( x = 4 \) into the given expression for \( \frac{\mathrm{d}y}{\mathrm{d}x} \).
- A1: Finds the gradient of the tangent is \( 4 \).
- A1: Writes down the correct tangent equation: \( y = 4x - 4 \).

(b)
- M1: Integrates the expression with at least one term integrated correctly (power increased by 1).
- A1: Obtains the correct integrated expression: \( 2x^{3/2} - 8x^{1/2} \).
- M1: Substitutes \( (4, 12) \) into their integrated expression to find the constant of integration.
- A1: Finds the constant of integration is \( 12 \).
- A1: State the final curve equation: \( y = 2x^{3/2} - 8x^{1/2} + 12 \).

**Part (a)**

First, express all terms of \(y\) in index form:
\[y = \frac{2}{3}x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 8x^{-\frac{1}{2}}\]

Differentiate with respect to \(x\):
\[\frac{dy}{dx} = \left(\frac{2}{3}\right)\left(\frac{3}{2}\right)x^{\frac{1}{2}} - (6)\left(\frac{1}{2}\right)x^{-\frac{1}{2}} + (8)\left(-\frac{1}{2}\right)x^{-\frac{3}{2}}\]

Simplifying each term yields:
\[\frac{dy}{dx} = x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}\]

**Part (b)**

At a stationary point, \(\frac{dy}{dx} = 0\):
\[x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}} = 0\]

Multiply the entire equation by \(x^{\frac{3}{2}}\) (since \(x > 0\)):
\[x^2 - 3x - 4 = 0\]

Factorising the quadratic equation:
\[(x - 4)(x + 1) = 0\]

Since \(x > 0\), we reject \(x = -1\). Therefore, the \(x\)-coordinate is \(x = 4\).

Substitute \(x = 4\) into the original equation to find the \(y\)-coordinate:
\[y = \frac{2}{3}(4)^{\frac{3}{2}} - 6(4)^{\frac{1}{2}} + \frac{8}{\sqrt{4}}\]
\[y = \frac{2}{3}(8) - 6(2) + \frac{8}{2}\]
\[y = \frac{16}{3} - 12 + 4 = -\frac{8}{3}\]

Thus, the coordinates of the stationary point are
\[\left(4, -\frac{8}{3}\right)\]

**Part (c)**

Find the second derivative by differentiating \(\frac{dy}{dx}\):
\[\frac{d^2y}{dx^2} = \frac{d}{dx}\left(x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}\right)\]
\[\frac{d^2y}{dx^2} = \frac{1}{2}x^{-\frac{1}{2}} + \frac{3}{2}x^{-\frac{3}{2}} + 6x^{-\frac{5}{2}}\]

Substitute \(x = 4\) into \(\frac{d^2y}{dx^2}\):
\[\frac{d^2y}{dx^2} = \frac{1}{2}(4)^{-\frac{1}{2}} + \frac{3}{2}(4)^{-\frac{3}{2}} + 6(4)^{-\frac{5}{2}}\]
\[\frac{d^2y}{dx^2} = \frac{1}{2}\left(\frac{1}{2}\right) + \frac{3}{2}\left(\frac{1}{8}\right) + 6\left(\frac{1}{32}\right)\]
\[\frac{d^2y}{dx^2} = \frac{1}{4} + \frac{3}{16} + \frac{6}{32} = \frac{8 + 6 + 6}{32} = \frac{20}{32} = \frac{5}{8}\]

Since \(\frac{d^2y}{dx^2} = \frac{5}{8} > 0\), the stationary point is a local minimum.

**Part (a)**
* **M1:** Attempts to differentiate at least one term of the form \(Ax^n \to Anx^{n-1}\).
* **A1:** Any two terms of their derivative correct.
* **A1:** Fully correct simplified derivative \(\frac{dy}{dx} = x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}\).

**Part (b)**
* **M1:** Sets their \(\frac{dy}{dx} = 0\) and multiplies through by \(x^{\frac{3}{2}}\), or equivalent valid algebraic manipulation, to obtain a three-term quadratic equation.
* **A1:** Correctly solves to find \(x = 4\) (rejecting or ignoring \(x = -1\)).
* **A1:** Evaluates \(y\) at their \(x = 4\) to obtain the coordinates \(\left(4, -\frac{8}{3}\right)\) or equivalent exact fractional values.

**Part (c)**
* **M1:** Differentiates their \(\frac{dy}{dx}\) to obtain \(\frac{d^2y}{dx^2}\) with at least two terms correct, and attempts to evaluate this at their \(x\) value.
* **A1:** Correct value of \(\frac{d^2y}{dx^2} = \frac{5}{8}\) (or successfully shows that \(\frac{d^2y}{dx^2} > 0\)) and concludes that the stationary point is a local minimum.

(a) Since \( (x-2) \) is a factor, \( \text{f}(2) = 0 \):
\( 2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5 \) (Equation 1).

Since dividing by \( (x+1) \) gives a remainder of \(-15\), \( \text{f}(-1) = -15 \):
\( 2(-1)^3 + a(-1)^2 + b(-1) - 6 = -15 \implies -2 + a - b - 6 = -15 \implies a - b = -7 \) (Equation 2).

Adding Equation 1 and Equation 2 gives:
\( 3a = -12 \implies a = -4 \).

Substituting \( a = -4 \) into Equation 2:
\( -4 - b = -7 \implies b = 3 \).

(b) Using the values of \( a \) and \( b \), we have \( \text{f}(x) = 2x^3 - 4x^2 + 3x - 6 \).
Factorise by grouping:
\( 2x^2(x - 2) + 3(x - 2) = (x - 2)(2x^2 + 3) \).
Since \( 2x^2 + 3 = 0 \) has no real roots, the polynomial is completely factorised.

(a) M1: Attempts to use the factor theorem with \( \text{f}(2) = 0 \) to obtain an equation in \( a \) and \( b \).
A1: Correct simplified equation, e.g., \( 2a + b = -5 \).
M1: Attempts to use the remainder theorem with \( \text{f}(-1) = -15 \) to obtain a second equation.
A1: Correct simplified equation, e.g., \( a - b = -7 \).
A1: Correct values \( a = -4 \) and \( b = 3 \).

(b) M1: Attempts to divide or group to find the quadratic factor.
A1: Correct completely factorised expression: \( (x-2)(2x^2 + 3) \).

(a) For \( x = 0.5 \):
\( y = \frac{4^{0.5}}{0.5+1} = \frac{2}{1.5} = 1.3333 \) (to 4 d.p.).

For \( x = 1.5 \):
\( y = \frac{4^{1.5}}{1.5+1} = \frac{8}{2.5} = 3.2000 \) (to 4 d.p.).

(b) The step width is \( h = 0.5 \).
Using the trapezium rule formula:
\( \int_{0}^{2} y \, \text{d}x \approx \frac{h}{2} [ y_0 + y_4 + 2(y_1 + y_2 + y_3) ] \)
\( \approx \frac{0.5}{2} [ 1 + 5.3333 + 2(1.3333 + 2 + 3.2000) ] \)
\( = 0.25 [ 6.3333 + 2(6.5333) ] \)
\( = 0.25 [ 6.3333 + 13.0666 ] = 0.25 [ 19.4 ] = 4.85 \) (to 3 s.f.).

(c) A more accurate estimate can be obtained by increasing the number of strips (or intervals), which decreases the width of each strip \( h \).

(a) B1: \( y = 1.3333 \) (awrt).
B1: \( y = 3.2000 \) (accept \( 3.2 \)).

(b) B1: Correct value of \( h = 0.5 \).
M1: Direct application of the trapezium rule formula with their values.
A1: Correct unsimplified expression inside the bracket, \( [ 1 + 5.3333 + 2(1.3333 + 2 + 3.2) ] \) or equivalent.
A1: \( 4.85 \) (must be to 3 s.f.).

(c) B1: Mentions increasing the number of strips/intervals/trapezia or decreasing the width of each strip.

Using the binomial theorem:
\( \left( 2 - \frac{x}{4} \right)^7 = 2^7 + \binom{7}{1} (2)^6 \left(-\frac{x}{4}\right) + \binom{7}{2} (2)^5 \left(-\frac{x}{4}\right)^2 + \binom{7}{3} (2)^4 \left(-\frac{x}{4}\right)^3 + \dots \)

Calculating each term:
Term 1: \( 128 \)
Term 2: \( 7 \times 64 \times \left(-\frac{x}{4}\right) = -112x \)
Term 3: \( 21 \times 32 \times \frac{x^2}{16} = 42x^2 \)
Term 4: \( 35 \times 16 \times \left(-\frac{x^3}{64}\right) = -8.75x^3 \) (or \( -\frac{35}{4}x^3 \))

So the expansion is:
\( 128 - 112x + 42x^2 - 8.75x^3 + \dots \)

For the second part, the \( x^3 \) term in the product \( (3 + 2x)(128 - 112x + 42x^2 - 8.75x^3 + \dots) \) comes from:
\( 3 \times (-8.75x^3) + 2x \times (42x^2) = -26.25x^3 + 84x^3 = 57.75x^3 \).

Therefore, the coefficient of \( x^3 \) is \( 57.75 \) (or \( \frac{231}{4} \)).

First Part:
B1: For 128.
M1: Standard structure of binomial term showing \( \binom{7}{r} a^{7-r} b^r \).
A1: For \( -112x \).
A1: For \( 42x^2 \).
A1: For \( -8.75x^3 \) (or equivalent fraction).

Second Part:
M1: Clearly identifies the two relevant terms that multiply to form the \( x^3 \) term, i.e., \( 3 \times \text{coefficient of } x^3 \) and \( 2 \times \text{coefficient of } x^2 \).
A1: Correct final coefficient of \( 57.75 \) (or equivalent fraction).

(a) Set the equations equal to find intersection points:
\( 9 - x^2 = 2x + 6 \implies x^2 + 2x - 3 = 0 \implies (x+3)(x-1) = 0 \).
This yields \( x = -3 \) and \( x = 1 \).
When \( x = -3 \), \( y = 2(-3) + 6 = 0 \).
When \( x = 1 \), \( y = 2(1) + 6 = 8 \).
The coordinates of the points of intersection are \( (-3, 0) \) and \( (1, 8) \).

(b) The area is given by the integral:
\( A = \int_{-3}^{1} [ (9 - x^2) - (2x + 6) ] \, \text{d}x = \int_{-3}^{1} (3 - 2x - x^2) \, \text{d}x \)
\( = \left[ 3x - x^2 - \frac{x^3}{3} \right]_{-3}^{1} \)
Evaluating at the upper limit \( x = 1 \):
\( 3(1) - 1^2 - \frac{1^3}{3} = 3 - 1 - \frac{1}{3} = \frac{5}{3} \).
Evaluating at the lower limit \( x = -3 \):
\( 3(-3) - (-3)^2 - \frac{(-3)^3}{3} = -9 - 9 - (-9) = -9 \).
Subtracting these:
\( A = \frac{5}{3} - (-9) = \frac{32}{3} \).

(a) M1: Equating the line and curve equations and attempting to solve the resulting quadratic.
A1: Correct \( x \) values \( x = -3, 1 \).
A1: Correct coordinates \( (-3, 0) \) and \( (1, 8) \).

(b) M1: Formulates a correct integral with their limits, integrating \( y_{\text{curve}} - y_{\text{line}} \).
M1: Attempts integration on each term, raising powers by 1.
A1: Correct integration: \( 3x - x^2 - \frac{x^3}{3} \) (ignore limits and constant for this mark).
M1: Substituting their limits \( -3 \) and \( 1 \) into their integrated expression.
A1: Correct exact area of \( \frac{32}{3} \) (or equivalent).

(a) Using the sum formulas:
\( S_{\infty} = \frac{a}{1-r} = 12 \implies a = 12(1-r) \).
\( S_3 = \frac{a(1-r^3)}{1-r} = 10.5 \).
Substitute \( a = 12(1-r) \) into the equation for \( S_3 \):
\( \frac{12(1-r)(1-r^3)}{1-r} = 10.5 \)
\( 12(1-r^3) = 10.5 \implies 1-r^3 = \frac{10.5}{12} = \frac{7}{8} \)
\( r^3 = 1 - \frac{7}{8} = \frac{1}{8} \) (as required).

(b) From part (a):
\( r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} = 0.5 \).
Using \( a = 12(1-r) \):
\( a = 12(1 - 0.5) = 6 \).

(c) The 5th term is given by:
\( u_5 = a r^4 = 6 \times (0.5)^4 = 6 \times \frac{1}{16} = \frac{3}{8} = 0.375 \).

(a) M1: Uses formula for sum to infinity to express \( a \) in terms of \( r \).
M1: Uses formula for sum of first 3 terms.
M1: Subsitutes for \( a \) (or \( 1-r \)) to obtain an equation in only \( r \).
A1*: Correct algebraic steps leading to \( r^3 = \frac{1}{8} \).

(b) B1: For \( r = 0.5 \) (or equivalent).
B1: For \( a = 6 \).

(c) M1: Uses the formula \( u_n = a r^{n-1} \) with \( n=5 \) and their values.
A1: \( 0.375 \) (or equivalent fraction).

(a) Using the factor theorem:
\( \text{g}(3) = 3(3)^3 - 5(3)^2 - 16(3) + 12 = 3(27) - 5(9) - 48 + 12 = 81 - 45 - 48 + 12 = 0 \).
Since \( \text{g}(3) = 0 \), \( (x-3) \) is a factor of \( \text{g}(x) \).

(b) Performing polynomial division of \( 3x^3 - 5x^2 - 16x + 12 \) by \( x-3 \):
\( 3x^3 - 5x^2 - 16x + 12 = (x-3)(3x^2 + 4x - 4) \).

(c) To solve the inequality, factorise the quadratic term:
\( 3x^2 + 4x - 4 = (3x-2)(x+2) \).
So, \( \text{g}(x) = (x-3)(3x-2)(x+2) \).
Roots of \( \text{g}(x) = 0 \) are \( x = -2 \), \( x = \frac{2}{3} \), and \( x = 3 \).
Considering the sign of the cubic curve:
\( \text{g}(x) \le 0 \) when \( x \le -2 \) or \( \frac{2}{3} \le x \le 3 \).

(a) B1: Evaluation of \( \text{g}(3) \) showing it is equal to 0 with a concluding statement.

(b) M1: Attempts algebraic division or matching of coefficients to find the quadratic factor.
A1: Correct quadratic term \( 3x^2 + 4x - 4 \).
A1: Expresses as \( (x-3)(3x^2 + 4x - 4) \).

(c) M1: Factorises the quadratic factor into two linear factors.
M1: Identifies the three roots \( -2, \frac{2}{3}, 3 \) and attempts to find intervals where the cubic is negative.
A1: Correct inequality solution: \( x \le -2 \) or \( \frac{2}{3} \le x \le 3 \) (accept interval notation).

(a) The 4th term is \( u_4 = a + 3d = 14 \) (Equation 1).
The sum of the first 10 terms is:
\( S_{10} = \frac{10}{2} [ 2a + (10-1)d ] = 5(2a + 9d) = 215 \implies 2a + 9d = 43 \) (Equation 2).

(b) Multiply Equation 1 by 2:
\( 2a + 6d = 28 \).
Subtract this from Equation 2:
\( 3d = 15 \implies d = 5 \).
Substitute \( d = 5 \) back into Equation 1:
\( a + 3(5) = 14 \implies a = -1 \).

(c) The sum of the first 25 terms is:
\( S_{25} = \frac{25}{2} [ 2a + 24d ] = 12.5 [ 2(-1) + 24(5) ] = 12.5 [ -2 + 120 ] = 12.5 [ 118 ] = 1475 \).

(a) B1: Correct equation for the 4th term, \( a + 3d = 14 \).
B1: Correct simplified equation for the sum of 10 terms, \( 2a + 9d = 43 \) (or unsimplified equal to 215).

(b) M1: Attempts to solve the simultaneous equations by eliminating one variable.
A1: Finds either \( d = 5 \) or \( a = -1 \).
A1: Finds both \( a = -1 \) and \( d = 5 \).

(c) M1: Applies the sum of arithmetic series formula with \( n = 25 \) and their values of \( a \) and \( d \).
A1: Correct answer: \( 1475 \).

(a) Set \( y = 0 \) to find the intersection with the \( x \)-axis:
\( 3\sqrt{x} - \frac{4}{\sqrt{x}} = 0 \implies 3x - 4 = 0 \implies x = \frac{4}{3} \).
So the \( x \)-coordinate of \( P \) is \( \frac{4}{3} \).

(b) The area is given by the definite integral:
\( A = \int_{4/3}^{4} \left( 3x^{1/2} - 4x^{-1/2} \right) \, \text{d}x \)
Integrate term-by-term:
\( \int \left( 3x^{1/2} - 4x^{-1/2} \right) \, \text{d}x = \left[ \frac{3x^{3/2}}{3/2} - \frac{4x^{1/2}}{1/2} \right] = \left[ 2x^{3/2} - 8x^{1/2} \right] \).
Evaluate at the upper limit \( x = 4 \):
\( 2(4)^{3/2} - 8(4)^{1/2} = 2(8) - 8(2) = 16 - 16 = 0 \).
Evaluate at the lower limit \( x = 4/3 \):
\( 2\left(\frac{4}{3}\right)^{3/2} - 8\left(\frac{4}{3}\right)^{1/2} = 2 \left( \frac{8}{3\sqrt{3}} \right) - 8 \left( \frac{2}{\sqrt{3}} \right) = \frac{16}{3\sqrt{3}} - \frac{48}{3\sqrt{3}} = -\frac{32}{3\sqrt{3}} \).
Subtract lower limit from upper limit:
\( A = 0 - \left(-\frac{32}{3\sqrt{3}}\right) = \frac{32}{3\sqrt{3}} = \frac{32\sqrt{3}}{9} \).

(a) M1: Equates curve to 0 and multiplies by \( \sqrt{x} \) to solve for \( x \).
A1: \( x = \frac{4}{3} \) (or equivalent).

(b) M1: Attempts to write the integrand in index form as \( 3x^{1/2} - 4x^{-1/2} \).
M1: Integrates term-by-term raising the powers by 1.
A1: Correct integrated expression: \( 2x^{3/2} - 8x^{1/2} \).
M1: Substitute limits \( 4 \) and their \( \frac{4}{3} \) into their integrated expression.
A1: Correct exact area: \( \frac{32\sqrt{3}}{9} \) (or \( \frac{32}{3\sqrt{3}} \)).

(a) The sum of the first two terms is given by \(S_2 = a + ar = a(1+r) = 15\). The sum to infinity is given by \(S_{\infty} = \frac{a}{1-r} = 27\), which implies \(a = 27(1-r)\). Substituting this expression for \(a\) into the first equation: \(27(1-r)(1+r) = 15 \Rightarrow 27(1-r^2) = 15\). Dividing both sides by 3 gives \(9(1-r^2) = 5 \Rightarrow 9 - 9r^2 = 5 \Rightarrow 9r^2 = 4\). (b) From \(9r^2 = 4\), we have \(r^2 = \frac{4}{9}\), which gives \(r = \pm \frac{2}{3}\). If \(r = \frac{2}{3}\), then \(a = 27\left(1-\frac{2}{3}\right) = 9\). If \(r = -\frac{2}{3}\), then \(a = 27\left(1-\left(-\frac{2}{3}\right)\right) = 45\). Thus, the two pairs of values are \(r = \frac{2}{3}, a = 9\) and \(r = -\frac{2}{3}, a = 45\). (c) For the first series where \(a = 9\) and \(r = \frac{2}{3}\): \(S_4 = \frac{9\left(1 - (2/3)^4\right)}{1 - 2/3} = \frac{9\left(1 - 16/81\right)}{1/3} = 27 \times \frac{65}{81} = \frac{65}{3}\). For the second series where \(a = 45\) and \(r = -\frac{2}{3}\): \(S_4 = \frac{45\left(1 - (-2/3)^4\right)}{1 - (-2/3)} = \frac{45\left(1 - 16/81\right)}{5/3} = 27 \times \frac{65}{81} = \frac{65}{3}\). Both series have a sum of \(\frac{65}{3}\) for their first 4 terms, which shows the sum is the same.

(a) M1: Attempts to write down equations for the sum of the first two terms and the sum to infinity, i.e., \(a(1+r) = 15\) and \(\frac{a}{1-r} = 27\). M1: Substitutes \(a = 27(1-r)\) into the first equation to form an equation in \(r\) only. A1*: Correctly simplifies to the given answer \(9r^2 = 4\) with no errors seen. (b) M1: Solves \(9r^2 = 4\) to find \(r = \pm\frac{2}{3}\) and attempts to find at least one value of \(a\). A1: Both pairs of values correct: \(r = \frac{2}{3}, a = 9\) and \(r = -\frac{2}{3}, a = 45\). (c) M1: Applies the sum formula \(S_n = \frac{a(1-r^n)}{1-r}\) or calculates the sum of the first four terms manually for at least one of the series. A1: Correctly evaluates the sum of the first four terms for both series, showing they both equal \(\frac{65}{3}\) (or \(21\frac{2}{3}\)). A1: Clear conclusion showing that both sums are equal.

(a) To find the points of intersection, set the equation of the curve equal to the equation of the line: \(10 + 3x - x^2 = 6 \Rightarrow x^2 - 3x - 4 = 0\). Factoring the quadratic: \((x - 4)(x + 1) = 0\). This gives the \(x\)-coordinates of the intersection points as \(x = -1\) and \(x = 4\). (b) The exact area of the region \(R\) is given by: \(\text{Area} = \int_{-1}^{4} \left( (10 + 3x - x^2) - 6 \right) \text{d}x = \int_{-1}^{4} (4 + 3x - x^2) \text{d}x\). Integrating term-by-term: \(\left[ 4x + \frac{3}{2}x^2 - \frac{1}{3}x^3 \right]_{-1}^{4}\). Evaluating at the upper limit \(x = 4\): \(4(4) + \frac{3}{2}(16) - \frac{1}{3}(64) = 16 + 24 - \frac{64}{3} = 40 - \frac{64}{3} = \frac{56}{3}\). Evaluating at the lower limit \(x = -1\): \(4(-1) + \frac{3}{2}(1) - \frac{1}{3}(-1) = -4 + \frac{3}{2} + \frac{1}{3} = -\frac{13}{6}\). Subtracting the lower limit value from the upper limit value: \(\text{Area} = \frac{56}{3} - \left(-\frac{13}{6}\right) = \frac{112}{6} + \frac{13}{6} = \frac{125}{6}\) (or \(20\frac{5}{6}\)).

(a) M1: Equates the curve and line equations to form a 3-term quadratic and attempts to solve. A1: Correct \(x\)-coordinates \(x = -1\) and \(x = 4\). (b) M1: Writes down a correct integral expression for the area with limits, either \(\int_{-1}^{4} (10 + 3x - x^2 - 6) \text{d}x\) or \(\int_{-1}^{4} (10 + 3x - x^2) \text{d}x - 6 \times (4 - (-1))\). M1: Integrates the expression with at least two terms correct. A1: Correct integration: \(4x + \frac{3}{2}x^2 - \frac{1}{3}x^3\). M1: Substitutes their limits \(4\) and \(-1\) into their integrated expression and subtracts the lower limit value from the upper limit value. A1: Obtains the correct exact area of \(\frac{125}{6}\) (or \(20\frac{5}{6}\)).

Resolving perpendicular to the slope for \(A\):
\(R = m_A g \cos \theta = 4 \times 9.8 \times 0.8 = 31.36\text{ N}\).

The maximum frictional force is:
\(F_{\text{max}} = \mu R = 0.5 \times 31.36 = 15.68\text{ N}\).

Resolving parallel to the slope, the component of the weight of \(A\) acting down the slope is:
\(m_A g \sin \theta = 4 \times 9.8 \times 0.6 = 23.52\text{ N}\).

The total resistance to moving up the slope is:
\(23.52 + 15.68 = 39.2\text{ N}\).

The weight of \(B\) is:
\(m_B g = 6 \times 9.8 = 58.8\text{ N}\).

Since \(58.8 > 39.2\text{ N}\), the system will move with \(B\) accelerating downwards and \(A\) accelerating up the slope.

(a) Equations of motion:
For \(B\): \(6g - T = 6a \implies 58.8 - T = 6a\)
For \(A\): \(T - 4g \sin \theta - F_{\text{max}} = 4a \implies T - 39.2 = 4a\)

Adding the two equations:
\(19.6 = 10a \implies a = 1.96\text{ m s}^{-2}\).

(b) Substituting \(a\) into the equation for \(A\):
\(T = 4(1.96) + 39.2 = 47.04\text{ N}\), which is \(47\text{ N}\) (to 2 s.f.).

(c) For the motion of \(B\):
\(u = 0\), \(a = 1.96\text{ m s}^{-2}\), \(s = 0.98\text{ m}\).
Using \(s = ut + \frac{1}{2}at^2\):
\(0.98 = 0.5 \times 1.96 \times t^2 \implies 0.98 = 0.98 t^2 \implies t^2 = 1 \implies t = 1.0\text{ s}\).

(a)
M1: For resolving forces perpendicular to the plane.
A1: For obtaining correct reaction force \(31.36\text{ N}\).
M1: For applying Newton's second law to both particles.
A1: For correct equations.
A1: For correct acceleration \(1.96\text{ m s}^{-2}\) (or \(2.0\text{ m s}^{-2}\)).

(b)
M1: For substituting acceleration to find tension.
A1: For correct tension \(47.04\text{ N}\) (accept \(47\text{ N}\) or \(47.0\text{ N}\)).

(c)
M1: For using SUVAT equation \(s = ut + 0.5at^2\).
A1: For correct substitution.
A1: For correct time \(1.0\text{ s}\) (or \(1\text{ s}\)).

(a) The velocity-time graph starts at \((0,0)\), rises linearly to \((20, V)\), remains constant at \(V\) until \(t = 20+T\), and then decreases linearly to \(0\) at \(t = 150\).

(b) The total time is \(2.5 \times 60 = 150\text{ s}\).
The duration of the three stages must sum to the total time:
\(20 + T + 30 = 150 \implies T = 100\text{ s}\).

(c) The total distance is \(2.4\text{ km} = 2400\text{ m}\).
This is represented by the area under the velocity-time graph (trapezium):
\(\text{Area} = \frac{1}{2}(T + \text{Total Time})V = \frac{1}{2}(100 + 150)V = 125 V\).
Setting this equal to the distance:
\(125 V = 2400 \implies V = 19.2\text{ m s}^{-1}\).

(d) The acceleration in the first stage is the gradient of the graph:
\(a = \frac{V - 0}{20} = \frac{19.2}{20} = 0.96\text{ m s}^{-2}\).

(a)
B1: For correct shape of graph (trapezium starting and ending at zero velocity).
B1: For labeling the axes and key times \(20\), \(20+T\), and \(150\) (or equivalent).

(b)
M1: For converting 2.5 minutes to 150 seconds.
A1: For showing \(T = 100\text{ s}\) with clear reasoning.

(c)
M1: For setting up an equation for the total area in terms of \(V\).
A1: For \(125V = 2400\).
A1: For \(V = 19.2\text{ m s}^{-1}\) (or \(19\text{ m s}^{-1}\)).

(d)
M1: For using gradient to find acceleration.
A1: For \(a = 0.96\text{ m s}^{-2}\).

Since \(\tan \theta = 0.75\), we have \(\sin \theta = 0.6\) and \(\cos \theta = 0.8\).
The weight of the particle is \(mg = 5 \times 9.8 = 49\text{ N}\).

(a) Resolving forces perpendicular to the inclined plane:
\(R = mg \cos \theta + H \sin \theta \implies R = 49(0.8) + H(0.6) = 39.2 + 0.6H\).

(b) Since the particle is on the point of slipping up the plane, the frictional force \(F\) acts down the plane and is at its maximum value:
\(F = F_{\text{max}} = \mu R = 0.4(39.2 + 0.6H) = 15.68 + 0.24H\).

Resolving parallel to the plane:
\(H \cos \theta - mg \sin \theta - F = 0\)
\(0.8H - 49(0.6) - (15.68 + 0.24H) = 0\)
\(0.8H - 29.4 - 15.68 - 0.24H = 0\)
\(0.56H = 45.08 \implies H = 80.5\text{ N}\).

(a)
M1: For resolving perpendicular to the plane.
A1: For including weight and horizontal force components.
A1: For obtaining \(R = 39.2 + 0.6H\) (or \(4.0g + 0.6H\)).

(b)
M1: For using \(F = \mu R\).
M1: For resolving parallel to the plane with correct direction of friction.
A1: For correct parallel equation in terms of \(H\).
M1: For solving the equation for \(H\).
A1: For \(H = 80.5\text{ N}\) (accept \(81\text{ N}\) or \(80.5\text{ N}\)).

Let the direction of motion of \(A\) before the collision be positive.
Before collision:
Velocity of \(A\) is \(3u\), velocity of \(B\) is \(-2u\).

After collision:
Velocity of \(A\) is \(-u\), velocity of \(B\) is \(4u\).

(a) Applying conservation of linear momentum:
\(3m(3u) + km(-2u) = 3m(-u) + km(4u)\)
\(9mu - 2kmu = -3mu + 4kmu\)
Dividing by \(mu\):
\(9 - 2k = -3 + 4k\)
\(12 = 6k \implies k = 2\).

(b) The impulse exerted on \(A\) by \(B\) is given by the change in momentum of \(A\):
\(I = m_A v_A - m_A u_A = 3m(-u) - 3m(3u) = -12mu\).
The magnitude of the impulse is \(12mu\).

(c) The assumption that the table is smooth means there is no friction between the particles and the table. Therefore, there are no external horizontal forces acting on the system, which justifies the use of conservation of linear momentum.

(a)
M1: For writing a momentum conservation equation with correct signs.
A1: For correct initial momentum term \(9mu - 2kmu\).
A1: For correct final momentum term \(-3mu + 4kmu\).
M1: For solving the linear equation for \(k\).
A1: For \(k = 2\).

(b)
M1: For calculating change of momentum of \(A\) (or \(B\)).
A1: For correct calculation.
A1: For giving the positive magnitude \(12mu\).

(c)
B1: For stating there is no friction.
B1: For explaining that this means momentum is conserved (no external horizontal forces).

(a) Using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\):
\(7\mathbf{i} - 5\mathbf{j} = (-\mathbf{i} + 3\mathbf{j}) + 4\mathbf{a}\)

\(4\mathbf{a} = 8\mathbf{i} - 8\mathbf{j} \implies \mathbf{a} = (2\mathbf{i} - 2\mathbf{j})\text{ m s}^{-2}\).

(b) Using Newton's second law:
\(\mathbf{F} = m\mathbf{a} = 0.5(2\mathbf{i} - 2\mathbf{j}) = (\mathbf{i} - \mathbf{j})\text{ N}\).

(c) Position vector at \(t = 3\):
\(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\)

\(\mathbf{r} = (2\mathbf{i} - 5\mathbf{j}) + 3(-\mathbf{i} + 3\mathbf{j}) + \frac{1}{2}(2\mathbf{i} - 2\mathbf{j})(3)^2\)

\(\mathbf{r} = (2\mathbf{i} - 5\mathbf{j}) + (-3\mathbf{i} + 9\mathbf{j}) + (9\mathbf{i} - 9\mathbf{j}) = (8\mathbf{i} - 5\mathbf{j})\text{ m}\).

(d) Velocity at \(t = 3\):
\(\mathbf{v} = \mathbf{u} + \mathbf{a}t = (-\mathbf{i} + 3\mathbf{j}) + 3(2\mathbf{i} - 2\mathbf{j}) = (5\mathbf{i} - 3\mathbf{j})\text{ m s}^{-1}\).

The speed is:
\(|\mathbf{v}| = \sqrt{5^2 + (-3)^2} = \sqrt{34} \approx 5.83\text{ m s}^{-1}\).

(a)
M1: For using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\).
A1: For \(\mathbf{a} = (2\mathbf{i} - 2\mathbf{j})\text{ m s}^{-2}\).

(b)
M1: For using \(\mathbf{F} = m\mathbf{a}\).
A1: For \(\mathbf{F} = (\mathbf{i} - \mathbf{j})\text{ N}\).

(c)
M1: For using integrated vector position formula.
A1: For correct substituting of values.
A1: For \(\mathbf{r} = (8\mathbf{i} - 5\mathbf{j})\text{ m}\).

(d)
M1: For finding velocity vector at \(t = 3\).
M1: For finding the magnitude of this velocity.
A1: For \(5.83\text{ m s}^{-1}\) (or \(\sqrt{34}\text{ m s}^{-1}\)).

(a) When the rod is on the point of tilting about \(C\), the reaction force at support \(D\) is zero: \(R_D = 0\).

Taking moments about \(C\):
The particle of mass \(6\text{ kg}\) is at \(A\), which is at distance \(1.5\text{ m}\) from \(C\).
The weight of the rod \(12g\) acts at the center of mass, which is at distance \(x - 1.5\text{ m}\) from \(C\).

Clockwise moments = Counter-clockwise moments:
\(12g \times (x - 1.5) = 6g \times 1.5\)
\(12(x - 1.5) = 9 \implies x - 1.5 = 0.75 \implies x = 2.25\text{ m}\).

(b) When the rod is on the point of tilting about \(D\), the reaction force at support \(C\) is zero: \(R_C = 0\).

The distance from \(A\) to \(D\) is \(AB - DB = 6 - 1 = 5\text{ m}\).
Since the center of mass is at \(2.25\text{ m}\) from \(A\), its distance from \(D\) is \(5 - 2.25 = 2.75\text{ m}\).

Taking moments about \(D\):
\(12g \times 2.75 = Mg \times 1\)
\(12 \times 2.75 = M \implies M = 33\text{ kg}\).

(a)
M1: For identifying that \(R_D = 0\) at tilting point.
M1: For taking moments about \(C\) with correct distances.
A1: For correct equation \(12g(x-1.5) = 6g(1.5)\).
M1: For solving for \(x\).
A1: For \(x = 2.25\text{ m}\).

(b)
M1: For identifying that \(R_C = 0\) at tilting point.
M1: For finding the correct distance from the center of mass to \(D\) (\(2.75\text{ m}\)).
M1: For taking moments about \(D\).
A1: For correct equation \(12g(2.75) = Mg(1)\).
A1: For \(M = 33\text{ kg}\).

(a) Resolving perpendicular to the plane:
\(R = mg \cos 20^\circ = 8 \times 9.8 \cos 20^\circ = 78.4 \cos 20^\circ \approx 73.67\text{ N}\), which is \(74\text{ N}\) (to 2 s.f.).

(b) The maximum frictional force is:
\(F_{\text{max}} = \mu R = 0.35 \times 73.67 = 25.79\text{ N}\), which is \(26\text{ N}\) (to 2 s.f.).

(c) The minimum value of \(P\) occurs when the box is on the point of sliding down the plane. In this case, the friction force acts up the plane.
Resolving parallel to the plane:
\(P + F_{\text{max}} = mg \sin 20^\circ\)
\(P = 8 \times 9.8 \sin 20^\circ - 25.79 = 26.81 - 25.79 = 1.02\text{ N}\), which is \(1.0\text{ N}\) (to 2 s.f.).

(d) The maximum value of \(P\) occurs when the box is on the point of sliding up the plane. In this case, the friction force acts down the plane.
Resolving parallel to the plane:
\(P = mg \sin 20^\circ + F_{\text{max}}\)
\(P = 26.81 + 25.79 = 52.6\text{ N}\), which is \(53\text{ N}\) (to 2 s.f.).

(a)
M1: For resolving forces perpendicular to the plane.
A1: For \(73.67\text{ N}\) or \(74\text{ N}\) (or \(73.7\text{ N}\)).

(b)
M1: For using \(F = \mu R\).
A1: For \(25.79\text{ N}\) or \(26\text{ N}\) (or \(25.8\text{ N}\)).

(c)
M1: For recognizing friction acts up the plane.
M1: For resolving parallel to the plane.
A1: For \(1.0\text{ N}\) (or \(1.02\text{ N}\)).

(d)
M1: For recognizing friction acts down the plane.
M1: For resolving parallel to the plane.
A1: For \(53\text{ N}\) (or \(52.6\text{ N}\)).

(a) \(\text{P}(A \cup B) = 1 - \text{P}(A' \cap B') = 1 - 0.25 = 0.75\)

(b) \(\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) \implies 0.75 = 0.4 + 0.5 - \text{P}(A \cap B) \implies \text{P}(A \cap B) = 0.15\)

(c) \(\text{P}(A) \times \text{P}(B) = 0.4 \times 0.5 = 0.2\). Since \(\text{P}(A \cap B) = 0.15 \neq 0.2\), the events are not independent.

(d) Since \(A\) and \(C\) are mutually exclusive, they do not overlap. The regions are:
- Only \(A\): \(\text{P}(A) - \text{P}(A \cap B) = 0.40 - 0.15 = 0.25\)
- Intersection \(A \cap B\): \(0.15\)
- Only \(B\): \(\text{P}(B) - \text{P}(A \cap B) - \text{P}(B \cap C) = 0.50 - 0.15 - 0.15 = 0.20\)
- Intersection \(B \cap C\): \(0.15\)
- Only \(C\): \(\text{P}(C) - \text{P}(B \cap C) = 0.25 - 0.15 = 0.10\)
- Outside region: \(1 - \text{P}(A \cup B \cup C) = 1 - 0.85 = 0.15\).

(a) M1 for using P(A u B) = 1 - P(A' n B'), A1 for 0.75 (2 marks)
(b) M1 for using the addition rule P(A u B) = P(A) + P(B) - P(A n B), A1 for 0.15 (2 marks)
(c) M1 for calculating P(A)P(B), A1 for concluding 'not independent' with correct comparison (2 marks)
(d) M1 for three circles (with A and C not overlapping), A1 for correctly placing 0.15 in both intersections, A1 for correct 'only' regions for A and C, A1 for correct 'only B' region, A1 for correct outside region 0.15 (5 marks)

(a) Since \(n = 15\):
- Median \(Q_2\) is the \(\frac{15+1}{2} = 8\)th value: \(Q_2 = 255\).
- \(Q_1\) is the \(\frac{15+1}{4} = 4\)th value: \(Q_1 = 252\).
- \(Q_3\) is the \(\frac{3(15+1)}{4} = 12\)th value: \(Q_3 = 262\).

(b) Interquartile range \(IQR = Q_3 - Q_1 = 262 - 252 = 10\).
- Lower limit: \(Q_1 - 1.5 \times IQR = 252 - 1.5 \times 10 = 237\).
- Upper limit: \(Q_3 + 1.5 \times IQR = 262 + 1.5 \times 10 = 277\).
- Since \(278 > 277\), it is an outlier. The next highest value is 267, which is within the range, so 278 is the only outlier.

(c) The box plot has:
- Box from 252 to 262 with median line at 255.
- Lower whisker extending to the minimum non-outlier value, which is 245.
- Upper whisker extending to the maximum non-outlier value, which is 267.
- Outlier at 278 plotted as an 'x' or '+'.

(d) Since \(Q_3 - Q_2 = 262 - 255 = 7\) and \(Q_2 - Q_1 = 255 - 252 = 3\), we have \(Q_3 - Q_2 > Q_2 - Q_1\), which indicates the distribution is positively skewed.

(a) B1 for Q2 = 255, B1 for Q1 = 252, B1 for Q3 = 262 (3 marks)
(b) M1 for IQR = 10, M1 for calculating 252 - 15 = 237 and 262 + 15 = 277, A1 for showing 278 > 277 and stating it is the only outlier (3 marks)
(c) B1 for correct box and median line, B1 for whiskers at 245 and 267, B1 for outlier at 278 (3 marks)
(d) M1 for comparing (Q3 - Q2) with (Q2 - Q1), A1 for positive skew with correct numerical justification (2 marks)

(a) \(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 385 - \frac{55^2}{10} = 385 - 302.5 = 82.5\)

\(S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 15300 - \frac{380^2}{10} = 15300 - 14440 = 860\)

\(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 2270 - \frac{55 \times 380}{10} = 2270 - 2090 = 180\)

(b) \(r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{180}{\sqrt{82.5 \times 860}} = \frac{180}{\sqrt{70950}} \approx 0.6757 \approx 0.676\) (to 3 s.f.)

(c) Since there is a moderate to strong positive linear correlation (\(r \approx 0.676\)), a linear regression model is suitable.

(d) \(b = \frac{S_{xy}}{S_{xx}} = \frac{180}{82.5} = \frac{24}{11} \approx 2.1818 \approx 2.18\)

\(a = \bar{y} - b\bar{x} = 38 - \left(\frac{24}{11}\right) \times 5.5 = 38 - 12 = 26.0\)

So the regression equation is \(y = 26.0 + 2.18x\) (with coefficients to 3 s.f.)

(e) For \(x = 8\), \(y = 26 + 2.1818 \times 8 = 26 + 17.4545 = 43.45\) thousand £, which is £43,500 (or 43.5 to 3 s.f.).

(a) M1 for clear method for S_xx, S_yy or S_xy, A1 for S_xx = 82.5 and S_yy = 860, A1 for S_xy = 180 (3 marks)
(b) M1 for correct formula for r, A1 for 0.676 (2 marks)
(c) B1 for stating suitable and referencing correlation (1 mark)
(d) M1 for b = S_xy / S_xx, A1 for b = 24/11 or 2.18, M1 for a = y_bar - b*x_bar, A1 for a = 26.0 (3 marks)
(e) M1 for substituting x = 8 into their regression line, A1 for 43.5 thousand (or 43.45) (2 marks)

(a) Since the sum of probabilities must equal 1:
\(\sum \text{P}(X=x) = k(4) + k(3) + k(2) + k(1) = 10k = 1 \implies k = 0.1\) (as required).

(b) \(\text{E}(X) = \sum x \text{P}(X=x) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 0.4 + 0.6 + 0.6 + 0.4 = 2.0\).

(c) First, find \(\text{E}(X^2)\):
\(\text{E}(X^2) = \sum x^2 \text{P}(X=x) = 1^2(0.4) + 2^2(0.3) + 3^2(0.2) + 4^2(0.1) = 0.4 + 1.2 + 1.8 + 1.6 = 5.0\).

Then, \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 5.0 - 2.0^2 = 1.0\).

(d) \(\text{Var}(3 - 2X) = (-2)^2 \text{Var}(X) = 4 \times 1.0 = 4\).

(a) M1 for setting sum of probabilities to 1, A1 for correctly expanding terms (4k + 3k + 2k + k = 10k), A1 for concluding k = 0.1 (3 marks)
(b) M1 for clear formula sum(x*P(X=x)), A1 for substitution of values, A1 for E(X) = 2 (3 marks)
(c) M1 for E(X^2) formula, A1 for E(X^2) = 5, A1 for Var(X) = 1 (3 marks)
(d) M1 for using Var(aX+b) = a^2*Var(X), A1 for 4 (2 marks)

(a) Standardising the given probabilities:
- For \(\text{P}(L < 1200) = 0.1587\), the z-score corresponding to a lower tail probability of 0.1587 is \(-1.00\) (since \(\Phi(1.00) = 0.8413\)).
So, \(\frac{1200 - \mu}{\sigma} = -1.00 \implies 1200 - \mu = -\sigma\).
- For \(\text{P}(L > 1500) = 0.0228\), this corresponds to an upper tail of 0.0228, so \(\text{P}(L < 1500) = 0.9772\). The z-score is \(2.00\).
So, \(\frac{1500 - \mu}{\sigma} = 2.00 \implies 1500 - \mu = 2\sigma\).

(b) Subtracting the first equation from the second:
\((1500 - \mu) - (1200 - \mu) = 2\sigma - (-\sigma) \implies 300 = 3\sigma \implies \sigma = 100\).
Substituting back:
\(1200 - \mu = -100 \implies \mu = 1300\).

(c) \(\text{P}(L > 1450) = \text{P}\left(Z > \frac{1450 - 1300}{100}\right) = \text{P}(Z > 1.5) = 1 - \Phi(1.50) = 1 - 0.9332 = 0.0668\).

(a) M1 for standardising with -1.00, A1 for (1200-mu)/sigma = -1.00, M1 for standardising with 2.00, A1 for (1500-mu)/sigma = 2.00 (4 marks)
(b) M1 for attempting to eliminate mu or sigma, A1 for sigma = 100, M1 for substituting back to find mu, A1 for mu = 1300 (4 marks)
(c) M1 for standardising 1450 with their mu and sigma, A1 for Z = 1.5, A1 for 0.0668 (3 marks)

(a) Cumulative frequencies: 6, 22, 42, 54, 60. Median is the 30th value, which lies in the \(16 \le t < 20\) class.
\(\text{Median} = 16 + \frac{30 - 22}{20} \times (20 - 16) = 16 + 1.6 = 17.6\) °C.

(b) Class midpoints \(x\) and coded values \(y = \frac{x - 18}{4}\):
- \(x=10 \implies y = -2\), frequency \(f=6\), \(fy = -12\)
- \(x=14 \implies y = -1\), frequency \(f=16\), \(fy = -16\)
- \(x=18 \implies y = 0\), frequency \(f=20\), \(fy = 0\)
- \(x=22 \implies y = 1\), frequency \(f=12\), \(fy = 12\)
- \(x=26 \implies y = 2\), frequency \(f=6\), \(fy = 12\)
\(\sum fy = -12 - 16 + 0 + 12 + 12 = -4\).
(i) Mean of \(y\): \(\bar{y} = \frac{-4}{60} = -\frac{1}{15} \approx -0.0667\) (to 3 s.f.).
(ii) Standard deviation of \(y\): \(\sigma_y = \sqrt{\frac{\sum fy^2}{N} - \bar{y}^2} = \sqrt{\frac{82}{60} - \left(-\frac{1}{15}\right)^2} = \sqrt{1.3667 - 0.0044} = \sqrt{1.3622} \approx 1.167 \approx 1.17\) (to 3 s.f.).

(c) Using the inverse of the coding:
- Mean of \(t\): \(\bar{t} = 4\bar{y} + 18 = 4\left(-\frac{1}{15}\right) + 18 = 17.73 \approx 17.7\) °C.
- Standard deviation of \(t\): \(\sigma_t = 4\sigma_y = 4 \times 1.1671 \approx 4.67\) °C.

(a) M1 for choosing class 16 - 20, M1 for linear interpolation formula, A1 for 17.6 (3 marks)
(b) (i) M1 for finding sum of fy, A1 for mean of y = -1/15 or -0.0667 (ii) M1 for standard deviation formula of y, A1 for 1.17 (5 marks total for b)
(c) M1 for mean of t = 4*mean_y + 18, A1 for mean = 17.7, A1 for s.d. = 4.67 (3 marks)

(a) First draw:
- \(\text{P}(R_1) = \frac{4}{10} = 0.4\)
- \(\text{P}(B_1) = \frac{6}{10} = 0.6\)

Second draw:
- If \(R_1\) occurs: we remove 1 red ball (3 remaining) and add 2 red balls. The box now contains \(3+2=5\) red balls and 6 blue balls (total 11). So \(\text{P}(R_2 \mid R_1) = \frac{5}{11}\) and \(\text{P}(B_2 \mid R_1) = \frac{6}{11}\).
- If \(B_1\) occurs: we remove 1 blue ball (5 remaining) and add 2 blue balls. The box now contains 4 red balls and \(5+2=7\) blue balls (total 11). So \(\text{P}(R_2 \mid B_1) = \frac{4}{11}\) and \(\text{P}(B_2 \mid B_1) = \frac{7}{11}\).

(b) \(\text{P}(R_2) = \text{P}(R_1 \cap R_2) + \text{P}(B_1 \cap R_2) = \left(\frac{4}{10} \times \frac{5}{11}\right) + \left(\frac{6}{10} \times \frac{4}{11}\right) = \frac{20}{110} + \frac{24}{110} = \frac{44}{110} = \frac{4}{10} = 0.4\).

(c) \(\text{P}(R_1 \mid R_2) = \frac{\text{P}(R_1 \cap R_2)}{\text{P}(R_2)} = \frac{20/110}{44/110} = \frac{20}{44} = \frac{5}{11} \approx 0.455\) (to 3 s.f.).

(a) M1 for first stage branches with 4/10 and 6/10, M1 for second stage from R1 with correct denominator 11, M1 for second stage from B1, A1 for fully correct tree diagram (4 marks)
(b) M1 for sum of two products, A1 for 20/110 and 24/110, A1 for 0.4 (3 marks)
(c) M1 for correct Bayes formula P(R1 n R2) / P(R2), A1 for numerator 20/110 or 2/11, A1 for denominator 44/110 or 0.4, A1 for 5/11 or 0.455 (4 marks)