2025 Edexcel IAS-Level Mathematics (XMA01) 模擬試題連答案詳解

(a) Using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\):

\(6\mathbf{i} - 3\mathbf{j} = -2\mathbf{i} + 5\mathbf{j} + 4\mathbf{a}\)

\(4\mathbf{a} = 8\mathbf{i} - 8\mathbf{j}\)

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

(b) Using \(\mathbf{F} = m\mathbf{a}\):

\(\mathbf{F} = 0.5(2\mathbf{i} - 2\mathbf{j}) = (\mathbf{i} - \mathbf{j})\text{ N}\)

(c) At \(t = 5\text{ s}\):

\(\mathbf{v} = \mathbf{u} + 5\mathbf{a} = (-2\mathbf{i} + 5\mathbf{j}) + 5(2\mathbf{i} - 2\mathbf{j}) = (8\mathbf{i} - 5\mathbf{j})\text{ m s}^{-1}\)

Speed \(= |\mathbf{v}| = \sqrt{8^2 + (-5)^2} = \sqrt{64 + 25} = \sqrt{89} \approx 9.43\text{ m s}^{-1}\) (to 3 s.f.)

(d) Displacement \(\mathbf{s}\) at \(t = 3\text{ s}\):

\(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\)

\(\mathbf{s} = 3(-2\mathbf{i} + 5\mathbf{j}) + \frac{1}{2}(9)(2\mathbf{i} - 2\mathbf{j}) = -6\mathbf{i} + 15\mathbf{j} + 9\mathbf{i} - 9\mathbf{j} = (3\mathbf{i} + 6\mathbf{j})\text{ m}\)

Distance \(= |\mathbf{s}| = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5} \approx 6.71\text{ m}\) (to 3 s.f.)

(a)
M1: Attempt to use \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) to find \(\mathbf{a}\).
A1: Correct vector \(\mathbf{a} = 2\mathbf{i} - 2\mathbf{j}\).

(b)
M1: Attempt to use \(\mathbf{F} = m\mathbf{a}\).
A1: Correct vector \(\mathbf{F} = \mathbf{i} - \mathbf{j}\).

(c)
M1: Substitute \(t=5\) into \(\mathbf{v} = \mathbf{u} + t\mathbf{a}\) and attempt to find the magnitude of the resulting vector.
A1: Correct speed of \(\sqrt{89}\text{ m s}^{-1}\) or \(9.43\text{ m s}^{-1}\).

(d)
M1: Attempt to use \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(t=3\) and find its magnitude.
A1: Correct distance of \(3\sqrt{5}\text{ m}\) or \(6.71\text{ m}\).

Since \(\tan \alpha = \frac{3}{4}\), we have \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\).

Resolving forces perpendicular to the plane:

\(R = mg \cos \alpha = 3 \times 9.8 \times 0.8 = 23.52\text{ N}\)

The maximum frictional force is:

\(F_{\max} = \mu R = 0.25 \times 23.52 = 5.88\text{ N}\)

The component of the weight down the plane is:

\(W_{\parallel} = mg \sin \alpha = 3 \times 9.8 \times 0.6 = 17.64\text{ N}\)

There are two limiting cases:

Case 1: The particle is on the point of slipping down the plane.
The frictional force acts up the plane.

\(P_{\min} + F_{\max} = mg \sin \alpha\)

\(P_{\min} + 5.88 = 17.64 \implies P_{\min} = 11.76\text{ N}\)

Case 2: The particle is on the point of slipping up the plane.
The frictional force acts down the plane.

\(P_{\max} = mg \sin \alpha + F_{\max}\)

\(P_{\max} = 17.64 + 5.88 = 23.52\text{ N}\)

Thus, for equilibrium, the range of possible values of \(P\) is:

\(11.76 \le P \le 23.52\)

Using 3 significant figures, this is \(11.8 \le P \le 23.5\) (or \(12 \le P \le 24\) to 2 s.f.).

M1: Resolving forces perpendicular to the plane to find \(R\) (using \(\cos \alpha = 0.8\)).
A1: Correct value of \(R = 23.52\text{ N}\).
B1: Calculating the maximum frictional force \(F_{\max} = 5.88\text{ N}\).
B1: Calculating the weight component down the slope \(17.64\text{ N}\).
M1: Setting up an equation for the lower limit where friction acts up the plane, and solving for \(P\).
A1: Correct minimum value \(P = 11.76\text{ N}\) (accept \(11.8\) or \(12\)).
M1: Setting up an equation for the upper limit where friction acts down the plane, and solving for \(P\).
A1: Correct maximum value \(P = 23.52\text{ N}\) (accept \(23.5\) or \(24\)) and expressing as a range.

(a)
The speed-time graph for \(P\) is a trapezium starting at \((0,0)\), rising to speed \(V\) at time \(t = 10\), remaining at speed \(V\) until time \(t = 10 + T\), and falling to the time axis at time \(t = 18 + T\).

The speed-time graph for \(Q\) is a horizontal line at \(v = 14\) from \(t = 0\) to \(t = 18 + T\).

(b)
For motorcycle \(Q\):
Using \(\text{distance} = \text{speed} \times \text{time}\):
\[420 = 14 \times t_{\text{total}}\]
\[t_{\text{total}} = \frac{420}{14} = 30\text{ s}\]

Since both vehicles take the same time to travel from \(A\) to \(B\):
\[10 + T + 8 = 30\]
\[18 + T = 30\]
\[T = 12\]

(c)
The total distance travelled by car \(P\) is the area under its speed-time graph:
\[\text{Distance} = \frac{1}{2}(a + b)h\]
where \(a = T = 12\), \(b = 30\), and \(h = V\).
\[420 = \frac{1}{2}(12 + 30)V\]
\[420 = 21V\]
\[V = 20\]

(d)
In the final stage of deceleration, the car decelerates from \(V = 20\text{ m/s}\) to rest in \(8\text{ s}\):
\[\text{Deceleration} = \frac{20 - 0}{8} = 2.5\text{ m/s}^2\]

(a)
- M1: For drawing a trapezium starting at the origin (for \(P\)) and a horizontal line (for \(Q\)) on the same axes.
- A1: For correctly shaped graphs with \(Q\) intersecting the sloping parts of \(P\).
- A1: For labelling key values: \(14\) and \(V\) on the vertical axis, and \(10\), \(10+T\), and \(18+T\) (or \(30\)) on the horizontal axis.

(b)
- M1: Using constant speed formula for \(Q\) to find total time: \(\frac{420}{14}\).
- A1: Finding total time \(t = 30\text{ s}\).
- A1: Equating total time for \(P\) to \(30\) to find \(T = 12\).

(c)
- M1: Formulating an equation for the total distance of \(P\) using the area of a trapezium or individual segments: \(\frac{1}{2}(10)V + T \cdot V + \frac{1}{2}(8)V = 420\).
- A1ft: Substituting their value of \(T\) to get a correct equation, e.g., \(21V = 420\).
- A1: For \(V = 20\).

(d)
- M1: For using \(v = u + at\) with their \(V\), \(v=0\), and \(t=8\) (or equivalent acceleration/deceleration calculation).
- A1: For deceleration \(= 2.5\text{ m/s}^2\) (or \(\frac{5}{2}\)).

Let \(a\) be the acceleration of the system and \(T\) be the tension in the string.

(a) Since the plane is inclined at angle \(\alpha\) where \(\tan \alpha = \frac{3}{4}\), we have:
\(\sin \alpha = \frac{3}{5} = 0.6\) and \(\cos \alpha = \frac{4}{5} = 0.8\).

For particle \(A\) on the rough incline, the normal reaction is:
\(R = 3mg \cos \alpha = 3mg(0.8) = 2.4mg\)

Since particle \(A\) moves up the plane, the friction force \(F\) acts down the plane and is at its limiting value:
\(F = \mu R = \frac{1}{3} \times 2.4mg = 0.8mg\)

The forces on \(A\) parallel to the plane are the tension \(T\) acting up the plane, the component of gravity \(3mg \sin \alpha\) acting down the plane, and the friction \(F\) acting down the plane.
The equation of motion for \(A\) is:
\(T - 3mg \sin \alpha - F = 3ma\)
\(T - 3mg(0.6) - 0.8mg = 3ma\)
\(T - 1.8mg - 0.8mg = 3ma\)
\(T - 2.6mg = 3ma\) [Equation 1]

For particle \(B\), which moves vertically downwards, the equation of motion is:
\(4mg - T = 4ma\) [Equation 2]

Adding Equation 1 and Equation 2 gives:
\(4mg - 2.6mg = 7ma\)
\(1.4mg = 7ma\)
\(a = 0.2g \text{ m/s}^2\) (or \(1.96 \text{ m/s}^2\))

(b) Substituting \(a = 0.2g\) into Equation 2:
\(4mg - T = 4m(0.2g)\)
\(T = 4mg - 0.8mg = 3.2mg\) (or \(\frac{16}{5}mg\))

(c) The force exerted by the string on the pulley is the resultant of two tensions of magnitude \(T\) acting at an angle of \(\phi = 90^\circ - \alpha\) to each other.

We know that \(\sin \alpha = 0.6\).
The magnitude of the resultant force \(R_p\) on the pulley is given by:
\(R_p = \sqrt{T^2 + T^2 + 2T^2 \cos(90^\circ - \alpha)}\)
\(R_p = \sqrt{2T^2 + 2T^2 \sin \alpha} = T\sqrt{2(1 + \sin \alpha)}\)
\(R_p = T\sqrt{2(1 + 0.6)} = T\sqrt{3.2}\)

Since \(\sqrt{3.2} = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5}\), we have:
\(R_p = 3.2mg \times \frac{4\sqrt{5}}{5} = 2.56\sqrt{5}mg\) (or \(\frac{64\sqrt{5}}{25}mg\))

(a)
M1: Attempting to resolve perpendicular to the plane for \(A\) to find \(R\).
A1: Correct expression for \(R = 2.4mg\).
M1: Use of \(F = \mu R\) and writing the equation of motion for \(A\) up the incline.
A1: Correct equation for \(A\): \(T - 2.6mg = 3ma\) (or equivalent).
M1: Writing the equation of motion for \(B\) and solving the simultaneous equations for \(a\).
A1: Correct acceleration \(a = 0.2g\) (accept \(1.96\text{ m/s}^2\)).

(b)
M1: Substituting their \(a\) into either equation of motion to find \(T\).
A1: Correct method leading to \(T\) in terms of \(m\) and \(g\).
A1: \(T = 3.2mg\) or \(\frac{16}{5}mg\) (accept \(31.36m\) if \(g=9.8\) is substituted).

(c)
M1: Understanding that the resultant force on the pulley is \(2T\cos(\phi/2)\) or \(\sqrt{2T^2(1+\sin\alpha)}\).
A1: Correct substitution of \(\sin\alpha = 0.6\) to obtain \(R_p = T\sqrt{3.2}\).
M1: Substituting their value of \(T\) into the resultant force formula.
A1: Correct exact final answer of \(2.56\sqrt{5}mg\) (or equivalent exact expression like \(\frac{64\sqrt{5}}{25}mg\); accept 3 s.f. decimal \(5.72mg\)).

Let the acceleration of the system be \(a = 0.8\text{ m/s}^2\).

(a) Considering the truck and car as a single combined system of mass \(1200 + 800 = 2000\text{ kg}\):
The total resistance to motion is \(400 + 200 = 600\text{ N}\).
Using Newton's second law:
\(D - 600 = 2000 \times 0.8\)
\(D - 600 = 1600\)
\(D = 2200\text{ N}\)

(b) Considering the motion of the car only:
The forces acting on the car are the tension \(T\) of the towbar in the direction of motion and the resistance of \(200\text{ N}\) opposing motion.
Using Newton's second law:
\(T - 200 = 800 \times 0.8\)
\(T - 200 = 640\)
\(T = 840\text{ N}\)

Alternatively, considering the truck only:
\(D - T - 400 = 1200 \times 0.8\)
\(2200 - T - 400 = 960\)
\(1800 - T = 960 \implies T = 840\text{ N}\)

(c) Once the towbar breaks, there is no tension force pulling the car forward. The only horizontal force acting on the car is the constant resistance of \(200\text{ N\) opposing motion.
Let \(a_C\) be the acceleration of the car:
\(-200 = 800 a_C \implies a_C = -0.25\text{ m/s}^2\)

The car's initial velocity is \(u = 15\text{ m/s}\) and its final velocity is \(v = 0\).
Using \(v^2 = u^2 + 2as\):
\(0^2 = 15^2 + 2(-0.25)s\)
\(0 = 225 - 0.5s\)
\(0.5s = 225 \implies s = 450\text{ m}\)

(d) After the towbar breaks, the tension acting on the truck is \(0\).
The truck has a driving force of \(D = 2200\text{ N}\) and a resistance of \(400\text{ N}\).
Let \(a_T\) be the new acceleration of the truck:
\(D - 400 = 1200 a_T\)
\(2200 - 400 = 1200 a_T\)
\(1800 = 1200 a_T\)
\(a_T = 1.5\text{ m/s}^2\)

(a)
M1: Attempting to write an equation of motion for the whole system with the correct number of terms.
A1: Correct equation, e.g., \(D - 600 = 2000 \times 0.8\).
A1: \(D = 2200\text{ N}\).

(b)
M1: Attempting to write an equation of motion for either the car or the truck.
A1: Correct equation for the car (\(T - 200 = 800 \times 0.8\)) or the truck (\(2200 - T - 400 = 1200 \times 0.8\)).
A1: \(T = 840\text{ N}\).

(c)
M1: Finding the deceleration of the car after the towbar breaks using \(F = ma\).
A1: Correct deceleration of \(0.25\text{ m/s}^2\) (or \(a = -0.25\)).
M1: Using a valid SUVAT equation (\(v^2 = u^2 + 2as\)) with \(v = 0\) and \(u = 15\) to find \(s\).
A1: Correct substitution of values into the SUVAT equation.
B1: Correct final distance of \(450\text{ m}\).

(d)
M1: Writing down the new equation of motion for the truck (with \(T = 0\)).
A1: Correct equation \(2200 - 400 = 1200 a_T\) (or equivalent).
A1: \(a_T = 1.5\text{ m/s}^2\).

**(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 \(C\) for the rod in equilibrium:
\(\text{Clockwise Moments} = \text{Anticlockwise Moments}\)

\(15g \times (x - 1.5) = 10g \times 1.5\)

Divide both sides by \(15g\):
\(x - 1.5 = 1\)
\(x = 2.5\)

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

The centre of mass \(G\) is at a distance of \(2.5\text{ m}\) from \(A\).
Since \(BD = 1\text{ m}\) and the total length is \(6\text{ m}\), the distance \(AD = 5\text{ m}\).
Therefore, the distance \(GD = AD - AG = 5 - 2.5 = 2.5\text{ m}\).

Taking moments about \(D\) for the rod in equilibrium:
\(\text{Clockwise Moments} = \text{Anticlockwise Moments}\)

\(Mg \times 1 = 15g \times 2.5\)

Divide both sides by \(g\):
\(M = 37.5\)

**(c)**
By modelling the beam as a rod, we assume that it remains straight and does not bend under the loads.

**(a)**
- **M1**: Sets the reaction force at \(D\) to zero (or takes moments about \(C\) without including \(R_D\)).
- **M1**: Formulates a moments equation about \(C\) containing the weight of the rod and the mass at \(A\).
- **A1**: Correct equation: \(15g(x - 1.5) = 10g \times 1.5\) (or equivalent).
- **A1**: Solves to find \(x = 2.5\).

**(b)**
- **M1**: Sets the reaction force at \(C\) to zero (or takes moments about \(D\) without including \(R_C\)).
- **B1**: Identifies or uses the correct distance \(GD = 2.5\text{ m}\).
- **M1**: Formulates a moments equation about \(D\) containing the weight of the rod and the mass \(M\) at \(B\).
- **A1**: Correct equation \(Mg(1) = 15g(2.5)\) and reaches \(M = 37.5\).

**(c)**
- **B1**: Mentions that the rod remains straight.
- **B1**: Mentions that it does not bend / forces act in a single plane.

(a) Since \(\tan \alpha = \frac{3}{4}\), we have \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\).

Resolving perpendicular to the plane:
\[ R = mg \cos \alpha = 6 \times 9.8 \times 0.8 = 47.04 \text{ N} \]

Resolving parallel to the plane (up the plane):
\[ T = mg \sin \alpha + F \]
\[ 50.96 = 6 \times 9.8 \times 0.6 + F \]
\[ 50.96 = 35.28 + F \]
\[ F = 15.68 \text{ N} \]

Using the law of friction \(F = \mu R\) for limiting equilibrium:
\[ 15.68 = \mu \times 47.04 \]
\[ \mu = \frac{15.68}{47.04} = \frac{1}{3} \approx 0.333 \]

(b) Let \(R'\) be the new normal reaction. Resolving perpendicular to the plane:
\[ R' = mg \cos \alpha + H \sin \alpha = 47.04 + 0.6H \]

Since the box is on the point of sliding down the plane, the friction force \(F'\) acts up the plane. Resolving parallel to the plane:
\[ H \cos \alpha + F' = mg \sin \alpha \]
\[ 0.8H + F' = 35.28 \implies F' = 35.28 - 0.8H \]

Since the box is in limiting equilibrium, \(F' = \mu R'\):
\[ 35.28 - 0.8H = \frac{1}{3}(47.04 + 0.6H) \]
\[ 3(35.28 - 0.8H) = 47.04 + 0.6H \]
\[ 105.84 - 2.4H = 47.04 + 0.6H \]
\[ 3H = 58.8 \]
\[ H = 19.6 \text{ N} \]

(a)
- M1: Resolves perpendicular to the plane to find an expression for \(R\).
- A1: Correct value or expression: \(R = 47.04\text{ N}\) (or \(4.8g\)).
- M1: Resolves parallel to the plane, including \(T\), component of weight, and friction \(F\).
- A1: Correct equation: \(50.96 = 35.28 + F\) (or \(5.2g = 3.6g + F\)).
- A1: Correct value for friction: \(F = 15.68\text{ N}\) (or \(1.6g\)).
- M1: Uses \(F = \mu R\) to solve for \(\mu\).
- A1: Correct coefficient of friction: \(\mu = \frac{1}{3}\) or \(0.33\) or \(0.333\).

(b)
- M1: Resolves perpendicular to the plane for the new scenario, including a component of \(H\).
- A1: Correct expression for the new normal reaction: \(R' = 47.04 + 0.6H\).
- M1: Resolves parallel to the plane (with friction acting up the plane) and sets up an equation in \(H\) using \(F' = \mu R'\).
- A1: Correct equation in \(H\), e.g., \(35.28 - 0.8H = \frac{1}{3}(47.04 + 0.6H)\) or equivalent.
- A1: Correct value for \(H\): \(19.6\) (accept \(20\) from 2 significant figures).

**Part (a)**

Rewrite \(f'(x)\) in index form:
\[f'(x) = 6x^{-1/2} - 3x^2 + 5\]

Integrate \(f'(x)\) with respect to \(x\) to find \(f(x)\):
\[f(x) = \int \left( 6x^{-1/2} - 3x^2 + 5 \right) dx\]
\[f(x) = \frac{6x^{1/2}}{1/2} - \frac{3x^3}{3} + 5x + C\]
\[f(x) = 12x^{1/2} - x^3 + 5x + C\]

Since \(C\) passes through \((4, -10)\), substitute \(x = 4\) and \(y = -10\) into the equation:
\[-10 = 12(4)^{1/2} - (4)^3 + 5(4) + C\]
\[-10 = 12(2) - 64 + 20 + C\]
\[-10 = 24 - 64 + 20 + C\]
\[-10 = -20 + C \implies C = 10\]

Therefore, the equation of the curve is:
\[f(x) = 12\sqrt{x} - x^3 + 5x + 10\]

**Part (b)**

To find the gradient of the tangent at \(x = 4\), substitute \(x = 4\) into \(f'(x)\):
\[f'(4) = \frac{6}{\sqrt{4}} - 3(4)^2 + 5 = 3 - 48 + 5 = -40\]

Using the gradient \(m = -40\) and point \((4, -10)\), find the equation of the tangent:
\[y - (-10) = -40(x - 4)\]
\[y + 10 = -40x + 160\]
\[y = -40x + 150\]

**Part (a)**

* **M1**: For an attempt to integrate: at least one \(x\) term has its power increased by 1.
* **A1**: Correct integration of any two terms (unsimplified forms allowed).
* **A1**: Fully correct integration, including the constant \(C\): \(12x^{1/2} - x^3 + 5x + C\).
* **M1**: Uses the boundary condition by substituting \(x = 4\) and \(y = -10\) to form an equation for \(C\).
* **A1**: Correct equation: \(f(x) = 12x^{1/2} - x^3 + 5x + 10\) (or exact equivalent simplified form).

**Part (b)**

* **M1**: Substitutes \(x = 4\) into \(f'(x)\) to find the gradient, and attempts the equation of the tangent using their gradient and the point \((4, -10)\).
* **A1**: Correct tangent equation: \(y = -40x + 150\).

**Part (a)**

Rewrite the derivative in index form:
\[f'(x) = 5x^{3/2} - 3x^{-2}\]

Integrate with respect to \(x\):
\[f(x) = \int \left( 5x^{3/2} - 3x^{-2} \right) dx\]
\[f(x) = \frac{5x^{5/2}}{5/2} - \frac{3x^{-1}}{-1} + C\]
\[f(x) = 2x^{5/2} + 3x^{-1} + C\]
\[f(x) = 2x^{5/2} + \frac{3}{x} + C\]

Substitute the point \((1, 10)\) to find the constant \(C\):
\[10 = 2(1)^{5/2} + \frac{3}{1} + C\]
\[10 = 2 + 3 + C\]
\[10 = 5 + C \implies C = 5\]

Thus, the equation of the curve is:
\[f(x) = 2x^{5/2} + \frac{3}{x} + 5\]

**Part (b)**

At \(x = 1\), the gradient of the tangent is given by:
\[f'(1) = 5(1)\sqrt{1} - \frac{3}{1^2} = 5 - 3 = 2\]

The gradient of the normal is the negative reciprocal of the tangent gradient:
\[m_{\text{normal}} = -\frac{1}{2}\]

Using the gradient \(-\frac{1}{2}\) and the point \((1, 10)\), the equation of the normal is:
\[y - 10 = -\frac{1}{2}(x - 1)\]
\[2(y - 10) = -(x - 1)\]
\[2y - 20 = -x + 1\]
\[x + 2y - 21 = 0\]

**Part (a)**

* **M1**: For expressing \(f'(x)\) as \(5x^{3/2} - 3x^{-2}\) and attempting to integrate (at least one power increased by 1).
* **A1**: At least one term integrated correctly (unsimplified forms accepted).
* **A1**: Fully correct integrated expression including the constant of integration, e.g., \(2x^{5/2} + 3x^{-1} + C\).
* **M1**: Substitutes \(x = 1\) and \(y = 10\) into their integrated expression to find \(C\).
* **A1**: \(f(x) = 2x^{5/2} + \frac{3}{x} + 5\) (or exact equivalent simplified expression).

**Part (b)**

* **M1**: Substitutes \(x = 1\) into \(f'(x)\) to get the tangent gradient \(2\), finds the normal gradient \(-\frac{1}{2}\), and attempts the equation of the normal using \((1, 10)\).
* **A1**: Correct equation in the specified form: \(x + 2y - 21 = 0\) (or any non-zero integer multiple, such as \(-x - 2y + 21 = 0\)).

**Part (a)**

First, find the gradient of the line segment \( AB \):
\[ m_{AB} = \frac{7 - (-1)}{6 - 2} = \frac{8}{4} = 2 \]

The gradient of the perpendicular bisector, \( m \), is the negative reciprocal of \( m_{AB} \):
\[ m = -\frac{1}{2} \]

Next, find the midpoint, \( M \), of \( AB \):
\[ M = \left( \frac{2 + 6}{2}, \frac{-1 + 7}{2} \right) = (4, 3) \]

The equation of the perpendicular bisector passing through \( M(4, 3) \) with gradient \( -\frac{1}{2} \) is:
\[ y - 3 = -\frac{1}{2}(x - 4) \]
\[ 2y - 6 = -x + 4 \]
\[ x + 2y - 10 = 0 \]

**Part (b)**

The point \( C \) has coordinates of the form \( (k, 2k) \). Since \( C \) lies on the perpendicular bisector:
\[ k + 2(2k) - 10 = 0 \]
\[ 5k = 10 \implies k = 2 \]
So, the coordinates of \( C \) are \( (2, 4) \).

The length of \( AB \) is:
\[ AB = \sqrt{(6-2)^2 + (7-(-1))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \]

Since \( C \) lies on the perpendicular bisector of \( AB \), the line segment \( MC \) is perpendicular to \( AB \) and represents the height of triangle \( ABC \).
The distance \( MC \) is:
\[ MC = \sqrt{(2-4)^2 + (4-3)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{5} \]

Therefore, the area of triangle \( ABC \) is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times MC \]
\[ \text{Area} = \frac{1}{2} \times 4\sqrt{5} \times \sqrt{5} = 2 \times 5 = 10 \]

**Part (a) [4 Marks]**
* **M1**: Attempts to find the gradient of the line segment \( AB \).
* **M1**: Finds the correct midpoint \( (4, 3) \) of \( AB \).
* **M1**: Uses the negative reciprocal gradient with their midpoint to form a linear equation.
* **A1**: Correct equation in the form \( ax + by + c = 0 \) (e.g., \( x + 2y - 10 = 0 \) or any non-zero integer multiple).

**Part (b) [2.5 Marks]**
* **M1**: Substitutes \( (k, 2k) \) into their line equation from part (a) to find the coordinates of \( C \).
* **M1**: Attempts to find the length of \( AB \) and the distance from \( C \) to the midpoint \( M \) (or uses another complete method such as the Shoelace formula with the three coordinates).
* **A0.5**: Correct exact area of \( 10 \).

**Part (a)**

The gradient of \( l_1 \) is \( 3 \).
Since \( l_2 \) is perpendicular to \( l_1 \), its gradient is:
\[ m_2 = -\frac{1}{3} \]

Using the point-slope formula with the point \( P(6, k) \):
\[ y - k = -\frac{1}{3}(x - 6) \]
\[ y = -\frac{1}{3}x + 2 + k \]
This can also be expressed as \( x + 3y - 3k - 6 = 0 \).

**Part (b)**

To find the coordinates of \( A \) (where \( l_2 \) crosses the \( x \)-axis), set \( y = 0 \):
\[ 0 = -\frac{1}{3}x + 2 + k \implies \frac{1}{3}x = k + 2 \implies x = 3k + 6 \]
So, \( A \) has coordinates \( (3k+6, 0) \).

To find the coordinates of \( B \) (where \( l_2 \) crosses the \( y \)-axis), set \( x = 0 \):
\[ y = k + 2 \]
So, \( B \) has coordinates \( (0, k+2) \).

Since \( k > 0 \), both \( 3k+6 \) and \( k+2 \) are positive.
Thus, the lengths of the sides \( OA \) and \( OB \) of the right-angled triangle \( OAB \) are \( 3k + 6 \) and \( k + 2 \) respectively.

Using the area of the triangle:
\[ \text{Area} = \frac{1}{2} \times OA \times OB \]
\[ 54 = \frac{1}{2} (3k + 6)(k + 2) \]
\[ 108 = 3(k + 2)(k + 2) \]
\[ 108 = 3(k + 2)^2 \]
\[ (k + 2)^2 = 36 \]

Taking the square root of both sides:
\[ k + 2 = \pm 6 \]
This yields:
\[ k + 2 = 6 \implies k = 4 \]
\[ k + 2 = -6 \implies k = -8 \]

Since \( k \) is specified as a positive constant, we reject \( k = -8 \).
Thus, \( k = 4 \).

**Part (a) [2 Marks]**
* **M1**: Identifies the perpendicular gradient as \( -\frac{1}{3} \) and attempts to write the line equation using \( P(6, k) \).
* **A1**: Correct equation in any form, e.g., \( y = -\frac{1}{3}x + 2 + k \) or \( x + 3y - 3k - 6 = 0 \).

**Part (b) [4.5 Marks]**
* **M1**: Attempts to find the coordinates of both \( A \) and \( B \) in terms of \( k \) by setting \( y = 0 \) and \( x = 0 \).
* **M1**: Sets up a correct expression for the area of triangle \( OAB \) in terms of \( k \) and equates it to 54.
* **A1**: Obtains a correct simplified quadratic equation, such as \( (k+2)^2 = 36 \) or \( k^2 + 4k - 32 = 0 \).
* **M1**: Solves their quadratic equation or takes square roots to find values for \( k \).
* **A0.5**: Correctly states \( k = 4 \) only and explicitly rejects the negative solution \( k = -8 \) based on \( k > 0 \).

(a) For the quadratic equation to have real roots, its discriminant must be greater than or equal to zero. That is, \( b^2 - 4ac \ge 0 \).
Here, \( a = 1 \), \( b = 2k+3 \), and \( c = 2k^2 - k + 1 \).
Substituting these into the discriminant: \( (2k+3)^2 - 4(1)(2k^2 - k + 1) \ge 0 \).
Expanding the terms: \( 4k^2 + 12k + 9 - (8k^2 - 4k + 4) \ge 0 \).
Simplifying: \( -4k^2 + 16k + 5 \ge 0 \).
Multiplying by \(-1\) and reversing the inequality sign gives: \( 4k^2 - 16k - 5 \le 0 \) (as required).

(b) To solve the quadratic inequality, we first find the critical values by solving the equation \( 4k^2 - 16k - 5 = 0 \).
Using the quadratic formula: \( k = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(4)(-5)}}{2(4)} \).
\( k = \frac{16 \pm \sqrt{256 + 80}}{8} = \frac{16 \pm \sqrt{336}}{8} \).
Since \( \sqrt{336} = \sqrt{16 \times 21} = 4\sqrt{21} \), we have:
\( k = \frac{16 \pm 4\sqrt{21}}{8} = \frac{4 \pm \sqrt{21}}{2} \).
Since the inequality is \( 4k^2 - 16k - 5 \le 0 \), the set of possible values for \( k \) lies between the two critical values.
Thus, \( \frac{4 - \sqrt{21}}{2} \le k \le \frac{4 + \sqrt{21}}{2} \).

(a)
M1: Attempts to use the discriminant \( b^2 - 4ac \) with correct identification of \( a \), \( b \), and \( c \).
A1: Correct expansion of \( (2k+3)^2 = 4k^2 + 12k + 9 \) or \( -4(2k^2-k+1) = -8k^2 + 4k - 4 \).
M1: Sets their discriminant expression \( \ge 0 \) and attempts to simplify to the given form.
A1*: Fully correct proof with no errors leading to \( 4k^2 - 16k - 5 \le 0 \).

(b)
M1: Attempts to find the critical values by solving \( 4k^2 - 16k - 5 = 0 \) using a valid method (formula or completing the square).
A1: Obtains \( \sqrt{336} \) or simplified surd equivalent.
A1: Correct critical values in exact form, e.g., \( \frac{4 \pm \sqrt{21}}{2} \) or equivalent.
M1: Chooses the inside region for their critical values.
A1: Fully correct inequality: \( \frac{4 - \sqrt{21}}{2} \le k \le \frac{4 + \sqrt{21}}{2} \) (accept interval notation).

(a) Substitute \( y = 3x - 1 \) into the second equation:
\( x^2 + (3x - 1)^2 = 5 \).
Expand and simplify:
\( x^2 + 9x^2 - 6x + 1 = 5 \)
\( 10x^2 - 6x - 4 = 0 \).
Divide the entire equation by 2:
\( 5x^2 - 3x - 2 = 0 \).
Factor the quadratic equation:
\( (5x + 2)(x - 1) = 0 \).
This gives \( x = 1 \) or \( x = -\frac{2}{5} \).
Substitute these values back into \( y = 3x - 1 \):
If \( x = 1 \), then \( y = 3(1) - 1 = 2 \).
If \( x = -\frac{2}{5} \), then \( y = 3\left(-\frac{2}{5}\right) - 1 = -\frac{11}{5} \).
So the solutions are \( (1, 2) \) and \( \left(-\frac{2}{5}, -\frac{11}{5}\right) \).

(b) Substitute \( y = 3x + c \) into the circle equation:
\( x^2 + (3x + c)^2 = 5 \)
\( x^2 + 9x^2 + 6cx + c^2 = 5 \)
\( 10x^2 + 6cx + (c^2 - 5) = 0 \).
For the line to be a tangent, the quadratic must have equal roots, so the discriminant \( b^2 - 4ac = 0 \).
Here, \( a = 10 \), \( b = 6c \), and \( c' = c^2 - 5 \).
\( (6c)^2 - 4(10)(c^2 - 5) = 0 \)
\( 36c^2 - 40c^2 + 200 = 0 \)
\( -4c^2 + 200 = 0 \)
\( 4c^2 = 200 \implies c^2 = 50 \).
Thus, \( c = \pm \sqrt{50} = \pm 5\sqrt{2} \).

(a)
M1: Substitutes \( y = 3x - 1 \) into \( x^2 + y^2 = 5 \).
A1: Obtains a correct 3-term quadratic, e.g., \( 10x^2 - 6x - 4 = 0 \) or \( 5x^2 - 3x - 2 = 0 \).
M1: Solves their quadratic to find two values for \( x \).
A1: Correct \( x \) values: \( x = 1 \) and \( x = -\frac{2}{5} \) (or \(-0.4\)).
A1: Correct corresponding \( y \) values, paired correctly: \( (1, 2) \) and \( \left(-\frac{2}{5}, -\frac{11}{5}\right) \).

(b)
M1: Substitutes \( y = 3x + c \) into \( x^2 + y^2 = 5 \) and expands to form a 3-term quadratic in \( x \).
A1: Correct quadratic \( 10x^2 + 6cx + (c^2 - 5) = 0 \).
M1: Sets the discriminant \( b^2 - 4ac = 0 \) for their quadratic.
A1: Simplifies to find \( c = \pm 5\sqrt{2} \) (both positive and negative values required for full marks).

(a) To sketch the curve \( y = (x-3)^2(x+1) \):
- The \( x \)-intercepts occur where \( y = 0 \), giving \( x = 3 \) (a repeated root, so the curve touches the \( x \)-axis) and \( x = -1 \) (a simple root, so the curve crosses the \( x \)-axis).
- The \( y \)-intercept occurs where \( x = 0 \), giving \( y = (-3)^2(1) = 9 \). So the curve passes through \( (0, 9) \).
- The curve is a cubic with a positive \( x^3 \) coefficient, so it starts in the third quadrant and ends in the first quadrant, with a local maximum in the second quadrant and a local minimum at \( (3, 0) \).

(b) The curve \( y = \text{f}(x) + k \) passes through the origin \( (0, 0) \).
Therefore, when \( x = 0 \), \( y = 0 \).
This gives \( \text{f}(0) + k = 0 \).
Since \( \text{f}(0) = 9 \), we have \( 9 + k = 0 \implies k = -9 \).

(c) The original curve \( y = \text{f}(x) \) has a repeated root at \( x = 3 \), which corresponds to a local minimum at the point \( (3, 0) \).
The curve \( y = \text{f}(x+c) \) represents a translation of the curve \( y = \text{f}(x) \) by the vector \( \begin{pmatrix} -c \\ 0 \end{pmatrix} \).
Under this translation, the local minimum at \( (3, 0) \) is mapped to \( (3-c, 0) \).
Since we are given that the new local minimum is at \( (1, 0) \), we set:
\( 3 - c = 1 \implies c = 2 \).

(a)
M1: A cubic-shaped curve with one local maximum and one local minimum.
A1: Curve touches the \( x \)-axis at \( x = 3 \) and crosses the \( x \)-axis at \( x = -1 \).
A1: Curve crosses the \( y \)-axis at \( (0, 9) \).
A1: Fully correct sketch with all key coordinates labeled.

(b)
M1: Uses the condition that \( x = 0 \implies y = 0 \) on \( y = \text{f}(x) + k \).
A1: Correctly deduces \( k = -9 \).

(c)
M1: Identifies that the original local minimum is at \( (3, 0) \).
M1: Identifies the translation as a horizontal shift of \( -c \) units, leading to the equation \( 3-c = 1 \).
A1: Correctly finds \( c = 2 \).

(a) First, rewrite the equation of the curve using fractional indices:
\(y = x^2 - 8x^{1/2} + 12\)

Differentiating with respect to \(x\):
\(\dfrac{\text{d}y}{\text{d}x} = 2x - 8\left(\dfrac{1}{2}\right)x^{-1/2}\)
\(\dfrac{\text{d}y}{\text{d}x} = 2x - 4x^{-1/2}\) (or \(2x - \dfrac{4}{\sqrt{x}}\))

(b) At the point \(P\), \(x = 4\).
Substitute \(x = 4\) into the derivative to find the gradient of the tangent, \(m\):
\(m = 2(4) - \dfrac{4}{\sqrt{4}} = 8 - 2 = 6\)

The equation of the tangent line passing through \(P(4, 12)\) with gradient \(6\) is:
\(y - 12 = 6(x - 4)\)
\(y - 12 = 6x - 24\)
\(y = 6x - 12\)

(c) The tangent line crosses the \(x\)-axis at \(A\) (where \(y = 0\)):
\(0 = 6x - 12 \implies x = 2\)
So, \(A\) has coordinates \((2, 0)\).

The tangent line crosses the \(y\)-axis at \(B\) (where \(x = 0\)):
\(y = -12\)
So, \(B\) has coordinates \((0, -12)\).

The area of triangle \(OAB\) is:
\(\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times OA \times OB = \dfrac{1}{2} \times 2 \times 12 = 12\)

(a)
* **M1**: For an attempt to differentiate: decreasing the power of at least one term by 1 (e.g., \(x^2 \to ax\) or \(x^{1/2} \to bx^{-1/2}\)).
* **A1**: Any one term correct: \(2x\) or \(-4x^{-1/2}\).
* **A1**: Fully correct simplified derivative: \(\dfrac{\text{d}y}{\text{d}x} = 2x - 4x^{-1/2}\) (or equivalent).

(b)
* **M1**: Substitutes \(x = 4\) into their \(\dfrac{\text{d}y}{\text{d}x}\) to find a numerical gradient.
* **M1**: Employs their numerical gradient and the coordinates of \(P(4, 12)\) to form a linear equation.
* **A1**: \(y = 6x - 12\) (must be in the form \(y = mx + c\)).

(c)
* **M1**: Uses a correct method to find the \(x\)-intercept and \(y\)-intercept of their tangent line (by setting \(y = 0\) and \(x = 0\) respectively) and applies \(\dfrac{1}{2} \times |x\text{-intercept}| \times |y\text{-intercept}|\).
* **A1**: \(\text{Area} = 12\) (must be positive, follow-through only allowed if tangent is of the form \(y = mx + c\) with \(mc \neq 0\)).

(a) In the right-angled triangle \(OAD\), we have \(\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OA}{OD}\). Since \(OA = r\) and \(OD = OB + BD = r + 8\), we get \(\cos\theta = \frac{r}{r+8}\). Multiplying both sides by \(r+8\) gives \(r\cos\theta + 8\cos\theta = r\). Rearranging terms to group \(r\) on one side: \(r(1 - \cos\theta) = 8\cos\theta\). Dividing by \(1 - \cos\theta\), we obtain the required result: \(r = \frac{8\cos\theta}{1 - \cos\theta}\).

(b) (i) Substituting \(\theta = \frac{\pi}{3}\) into the formula from part (a): Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), we have \(r = \frac{8(1/2)}{1 - 1/2} = \frac{4}{1/2} = 8\) cm.
(ii) The arc length of \(AB\) is given by \(s = r\theta\). Using \(r = 8\) and \(\theta = \frac{\pi}{3}\), the exact length is \(s = 8 \times \frac{\pi}{3} = \frac{8\pi}{3}\) cm.

(c) The perimeter of the shaded region \(R\) is given by \(P = \text{arc } AB + BD + AD\). We already have \(\text{arc } AB = \frac{8\pi}{3}\) and \(BD = 8\). In the right-angled triangle \(OAD\), we can find \(AD\) using trigonometry: \(AD = OA \tan\theta = 8 \tan\left(\frac{\pi}{3}\right) = 8\sqrt{3}\) cm (or by using Pythagoras: \(AD = \sqrt{OD^2 - OA^2} = \sqrt{16^2 - 8^2} = \sqrt{192} = 8\sqrt{3}\)). Thus, the perimeter is \(P = \frac{8}{3}\pi + 8\sqrt{3} + 8\) cm, so \(a = \frac{8}{3}\), \(b = 8\), and \(c = 8\).

(d) The area of the shaded region \(R\) is the area of the right-angled triangle \(OAD\) minus the area of the sector \(OAB\). Area of triangle \(OAD = \frac{1}{2} \times OA \times AD = \frac{1}{2} \times 8 \times 8\sqrt{3} = 32\sqrt{3}\) cm\(^2\). Area of sector \(OAB = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 8^2 \times \frac{\pi}{3} = \frac{32\pi}{3}\) cm\(^2\). Therefore, the exact area of the shaded region is \(32\sqrt{3} - \frac{32\pi}{3}\) cm\(^2\).

(a) M1: Realises that in the right-angled triangle \(OAD\), \(\cos\theta = \frac{OA}{OD}\) or \(\cos\theta = \frac{r}{r+8}\).
A1: Correct substitution of \(OA = r\) and \(OD = r + 8\) leading to \(\cos\theta = \frac{r}{r+8}\).
M1: Correct algebraic steps to make \(r\) the subject, e.g., expanding \(r\cos\theta + 8\cos\theta = r\) and factorising.
A1*: Fully correct proof with no errors or omissions.

(b)(i) M1: Substitutes \(\theta = \frac{\pi}{3}\) into the formula or uses \(\cos\left(\frac{\pi}{3}\right) = 0.5\) in \(\cos\theta = \frac{r}{r+8}\).
A1: Correct value of \(r = 8\).
(b)(ii) M1: Correct use of the formula for arc length, \(s = r\theta\), with their \(r\) and \(\theta = \frac{\pi}{3}\).
A1: Exact arc length \(\frac{8\pi}{3}\) (or equivalent).

(c) M1: Attempts to find the length of \(AD\) using Pythagoras' theorem or trigonometry (e.g., \(8\tan(\pi/3)\) or \(\sqrt{16^2 - 8^2}\)).
M1: Formulates the perimeter of the region as \(\text{arc } AB + BD + AD\) using their values.
A1: Fully correct simplified expression \(\frac{8}{3}\pi + 8\sqrt{3} + 8\) (accept equivalent exact forms, e.g., \(a = 8/3\), \(b = 8\), \(c = 8\)).

(d) M1: Area of region \(R\) = Area of triangle - Area of sector. Attempts to find both areas using \(\frac{1}{2}ab\sin C\) or \(\frac{1}{2}bh\) for the triangle, and \(\frac{1}{2}r^2\theta\) for the sector.
A1: Correct exact area of \(32\sqrt{3} - \frac{32\pi}{3}\) (or equivalent factored exact form, e.g., \(32\left(\sqrt{3} - \frac{\pi}{3}\right)\)).

(a) The terms of the series are \(u_1 = 5 \times 3^{-1} = \frac{5}{3}\), \(u_2 = 5 \times 3^{-2} = \frac{5}{9}\), etc. This is a geometric series with first term \(a = \frac{5}{3}\) and common ratio \(r = \frac{1}{3}\). The sum to infinity is given by \(S_{\infty} = \frac{a}{1 - r} = \frac{5/3}{1 - 1/3} = \frac{5/3}{2/3} = \frac{5}{2} = 2.5\). (b) The sum of the first \(K\) terms is \(S_K = \frac{a(1 - r^K)}{1 - r} = 2.5\left(1 - \left(\frac{1}{3}\right)^K\right)\). We want \(2.5\left(1 - \left(\frac{1}{3}\right)^K\right) > 2.499\). Dividing by 2.5 gives \(1 - \left(\frac{1}{3}\right)^K > 0.9996\), which simplifies to \(\left(\frac{1}{3}\right)^K < 0.0004\). This means \(3^K > 2500\). Taking logarithms on both sides: \(K \log(3) > \log(2500) \implies K > \frac{\log(2500)}{\log(3)} \approx 7.12\). Since \(K\) must be an integer, the smallest value of \(K\) is 8.

(a) M1: Attempts to find the first term and the common ratio of the geometric series. A1: Correct sum to infinity of 2.5 or 5/2. (b) M1: Uses the geometric sum formula \(S_K = \frac{a(1-r^K)}{1-r}\) with their \(a\) and \(r\) to set up an inequality or equation with 2.499. M1: Solves the inequality using logarithms to find a critical value for \(K\). A1: Correct value of \(K = 8\).

(a) The sum of the first \(n\) terms of an arithmetic series is given by \(S_n = \frac{n}{2}[2a + (n-1)d]\). For \(n=8\), we have \(S_8 = \frac{8}{2}[2a + 7d] = 152 \implies 4[2a + 7d] = 152 \implies 2a + 7d = 38\), as required. (b) For \(n=20\), we have \(S_{20} = \frac{20}{2}[2a + 19d] = 860 \implies 10[2a + 19d] = 860 \implies 2a + 19d = 86\). Subtracting the equation from part (a) from this equation: \((2a + 19d) - (2a + 7d) = 86 - 38 \implies 12d = 48 \implies d = 4\). Substituting \(d = 4\) back into the first equation: \(2a + 7(4) = 38 \implies 2a + 28 = 38 \implies 2a = 10 \implies a = 5\). Therefore, \(a = 5\) and \(d = 4\).

(a) M1: Applies the sum formula for an arithmetic series with \(n=8\) and sets it equal to 152. A1*: Correctly simplifies to the given equation \(2a + 7d = 38\) with no errors shown. (b) M1: Applies the sum formula for \(n=20\), sets it equal to 860, and simplifies to obtain a second equation in \(a\) and \(d\). M1: Solves the simultaneous equations to find a value for \(d\) or \(a\). A1: Both \(a = 5\) and \(d = 4\) correct.

The binomial expansion of \((1 + kx)^n\) is given by \(1 + nkx + \frac{n(n-1)}{2}(kx)^2 + \dots\). The coefficient of \(x\) is \(nk = 24\). The coefficient of \(x^2\) is \(\frac{n(n-1)}{2}k^2 = 252\). From the first equation, we have \(k = \frac{24}{n}\). Substituting this into the second equation: \(\frac{n(n-1)}{2} \left(\frac{24}{n}\right)^2 = 252 \implies \frac{n(n-1)}{2} \times \frac{576}{n^2} = 252 \implies \frac{288(n-1)}{n} = 252 \implies 288n - 288 = 252n \implies 36n = 288 \implies n = 8\). Substituting \(n = 8\) back into the expression for \(k\): \(k = \frac{24}{8} = 3\). Thus, \(n = 8\) and \(k = 3\).

M1: Identifies the coefficient of \(x\) as \(nk\) and the coefficient of \(x^2\) as \(\frac{n(n-1)}{2}k^2\), and sets up two equations. M1: Attempts to eliminate one variable (either \(n\) or \(k\)) from their equations. A1: Obtains a correct single-variable equation, e.g., \(\frac{288(n-1)}{n} = 252\) or equivalent. dM1: Solves their equation to find a positive integer value for \(n\) or a value for \(k\) (dependent on the first M mark). A1: Both \(n = 8\) and \(k = 3\) correct.

**Part (a)**\
\
The volume \(V\) of the cuboid is given by:\
\[V = x^2 h\]\
Given that the volume is \(400\text{ cm}^3\):\
\[x^2 h = 400 \implies h = \frac{400}{x^2}\]\
\
The box has an open top, so the surface area consists of:\
- The square base: \(x^2\)\
- Four vertical outer faces: \(4 \times (x \times h) = 4xh\)\
- One internal vertical partition: \(x \times h = xh\)\
\
Thus, the total surface area \(S\) is:\
\[S = x^2 + 4xh + xh = x^2 + 5xh\]\
\
Substituting \(h = \frac{400}{x^2}\) into this expression:\
\[S = x^2 + 5x\left(\frac{400}{x^2}\right)\]\[S = x^2 + \frac{2000}{x}\]\
*(as required)*\
\
**Part (b)**\
\
To find the minimum value of \(S\), we first differentiate \(S\) with respect to \(x\):\
\[S = x^2 + 2000x^{-1}\]\[\frac{dS}{dx} = 2x - 2000x^{-2} = 2x - \frac{2000}{x^2}\]\
\
Setting \
\[\frac{dS}{dx} = 0\]\
\[2x - \frac{2000}{x^2} = 0 \implies 2x = \frac{2000}{x^2}\]\[2x^3 = 2000 \implies x^3 = 1000 \implies x = 10\]\
\
To find the minimum surface area, substitute \(x = 10\) back into the expression for \(S\):\
\[S = 10^2 + \frac{2000}{10} = 100 + 200 = 300\text{ cm}^2\]\
\
To justify that this value is a minimum, we find the second derivative:\
\[\frac{d^2S}{dx^2} = 2 - 2000(-2)x^{-3} = 2 + \frac{4000}{x^3}\]\
\
Substituting \(x = 10\) into the second derivative:\
\[\frac{d^2S}{dx^2} = 2 + \frac{4000}{10^3} = 2 + 4 = 6\]\
\
Since \
\[\frac{d^2S}{dx^2} = 6 > 0\]\
the value of \(S\) is indeed a minimum.

**Part (a)**\
* **M1**: Sets up a volume equation \(x^2 h = 400\) and makes \(h\) the subject of the formula to get \(h = \frac{400}{x^2}\).\
* **M1**: Formulates a correct expression for the total surface area \(S = x^2 + 5xh\).\
* **A1**: Substitutes \(h\) into the surface area equation to get \(S = x^2 + 5x\left(\frac{400}{x^2}\right)\).\
* **A1***: Correctly simplifies to the given expression \(S = x^2 + \frac{2000}{x}\) with no errors seen. (This is a show-that question, so full working must be shown).\
\
**Part (b)**\
* **M1**: Attempts to differentiate \(S\) to obtain \(\frac{dS}{dx} = Ax \pm Bx^{-2}\) where \(A, B \neq 0\).\
* **A1**: Correct derivative: \(\frac{dS}{dx} = 2x - \frac{2000}{x^2}\) (allow unsimplified form).\
* **M1**: Sets their derivative equal to \(0\) and attempts to solve for \(x^3\) or \(x\).\
* **A1**: Obtains \(x = 10\).\
* **M1**: Differentiates again to find \(\frac{d^2S}{dx^2}\) and evaluates it at their \(x\) value (or considers the sign of the derivative on either side of \(x=10\)), and states that \(\frac{d^2S}{dx^2} > 0\) means a minimum.\
* **A1**: Correct minimum value of \(S = 300\) (units not required).

(a) Since \( (x - 2) \) is a factor of \( f(x) \), by the Factor Theorem:
\( f(2) = 0 \)
\( f(2) = 2(2)^3 + a(2)^2 + b(2) - 10 = 0 \)
\( 16 + 4a + 2b - 10 = 0 \)
\( 4a + 2b = -6 \)
\( 2a + b = -3 \) (Equation 1)

By the Remainder Theorem, since dividing by \( (x + 1) \) gives a remainder of \( -12 \):
\( f(-1) = -12 \)
\( f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) - 10 = -12 \)
\( -2 + a - b - 10 = -12 \)
\( a - b - 12 = -12 \)
\( a - b = 0 \implies a = b \) (Equation 2)

Substitute Equation 2 into Equation 1:
\( 2a + a = -3 \implies 3a = -3 \implies a = -1 \)
Since \( a = b \), we have \( b = -1 \).

(b) Using \( a = -1 \) and \( b = -1 \), the cubic function is:
\( f(x) = 2x^3 - x^2 - x - 10 \)

Since \( (x - 2) \) is a factor of \( f(x) \), we can write:
\( 2x^3 - x^2 - x - 10 = (x - 2)(2x^2 + px + q) \)
By expansion:
\( (x - 2)(2x^2 + px + q) = 2x^3 + (p - 4)x^2 + (q - 2p)x - 2q \)
Comparing coefficients:
For \( x^2 \): \( p - 4 = -1 \implies p = 3 \)
For the constant term: \( -2q = -10 \implies q = 5 \)

So, \( f(x) = (x - 2)(2x^2 + 3x + 5) \).

To show that \( x = 2 \) is the only real root, we must show that the quadratic equation \( 2x^2 + 3x + 5 = 0 \) has no real roots.
Calculating the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = (3)^2 - 4(2)(5) = 9 - 40 = -31 \)

Since \( \Delta < 0 \), the quadratic equation has no real roots. Therefore, \( x = 2 \) is the only real root of the equation \( f(x) = 0 \).

Part (a):
* M1: Attempts \( f(2) = 0 \) to get an equation in terms of \( a \) and \( b \).
* M1: Attempts \( f(-1) = -12 \) to get a second equation in terms of \( a \) and \( b \).
* A1: Obtains two correct simplified equations, e.g., \( 2a + b = -3 \) and \( a - b = 0 \) (or equivalent).
* M1: Solves their simultaneous equations to find values for \( a \) and \( b \).
* A1: Correct values: \( a = -1 \) and \( b = -1 \).

Part (b):
* M1: Attempts algebraic division or coefficient matching to find the quadratic factor of \( f(x) \).
* A1: Correct quadratic factor \( 2x^2 + 3x + 5 \).
* M1: Calculates the discriminant \( b^2 - 4ac \) of their quadratic factor, or attempts to complete the square.
* A1: Correctly finds \( \Delta = -31 \), states that \( -31 < 0 \) (or equivalent justification), and concludes that \( x = 2 \) is the only real root.

(a) Substitute the coordinates of \(P(1, 1)\) into the equation of the circle \(C\): \(1^2 + 1^2 - 10(1) + 4(1) + k = 0 \implies 1 + 1 - 10 + 4 + k = 0 \implies -4 + k = 0 \implies k = 4\). With \(k = 4\), the equation becomes \(x^2 + y^2 - 10x + 4y + 4 = 0\). Complete the square for both \(x\) and \(y\): \((x - 5)^2 - 25 + (y + 2)^2 - 4 + 4 = 0 \implies (x - 5)^2 + (y + 2)^2 = 25\). Therefore, the centre of \(C\) is \((5, -2)\) and the radius is \(\sqrt{25} = 5\). (b) Method 1 (Algebraic): Rearrange the line equation to make \(x\) the subject: \(3x = 4y - c \implies x = \frac{4y - c}{3}\). Substitute this into the equation of the circle \((x-5)^2 + (y+2)^2 = 25\): \(\left(\frac{4y - c}{3} - 5\right)^2 + (y+2)^2 = 25 \implies \left(\frac{4y - (c + 15)}{3}\right)^2 + (y+2)^2 = 25\). Expanding this gives: \(\frac{16y^2 - 8(c + 15)y + (c + 15)^2}{9} + y^2 + 4y + 4 = 25\). Multiply the entire equation by 9: \(16y^2 - 8(c + 15)y + (c + 15)^2 + 9(y^2 + 4y + 4) = 225 \implies 25y^2 + (36 - 8c - 120)y + c^2 + 30c + 225 + 36 - 225 = 0 \implies 25y^2 - (8c + 84)y + (c^2 + 30c + 36) = 0\). Since the line \(l\) is a tangent to the circle, the discriminant of this quadratic equation must be zero (representing exactly one point of intersection): \(\Delta = [-(8c + 84)]^2 - 4(25)(c^2 + 30c + 36) = 0 \implies (8c + 84)^2 - 100(c^2 + 30c + 36) = 0\). Dividing the entire equation by 4: \((4c + 42)^2 - 25(c^2 + 30c + 36) = 0 \implies 16c^2 + 336c + 1764 - 25c^2 - 750c - 900 = 0 \implies -9c^2 - 414c + 864 = 0\). Dividing by \(-9\) yields: \(c^2 + 46c - 96 = 0\). Factorising this gives: \((c + 48)(c - 2) = 0\), which results in \(c = -48\) or \(c = 2\). Method 2 (Geometric): The perpendicular distance from the centre of the circle \((5, -2)\) to the line \(3x - 4y + c = 0\) must equal the radius of the circle, which is \(5\). Using the perpendicular distance formula: \(\frac{|3(5) - 4(-2) + c|}{\sqrt{3^2 + (-4)^2}} = 5 \implies \frac{|15 + 8 + c|}{5} = 5 \implies |23 + c| = 25\). This gives two possible equations: \(23 + c = 25 \implies c = 2\) or \(23 + c = -25 \implies c = -48\).

Part (a): [M1] Substituting \(x=1, y=1\) into the circle equation and attempting to solve for \(k\). [A1] Finding the correct value of \(k = 4\). [A1] Correctly completing the square to find both the centre \((5, -2)\) and the radius \(5\) (or \(r^2 = 25\)). Part (b): [M1] Setting up a method to find the relationship for tangency, either by substituting \(x\) or \(y\) from the line equation into the circle equation to form a quadratic, or by using the perpendicular distance formula from the centre to the line. [A1] Obtaining a correct simplified quadratic equation in \(c\), e.g., \(c^2 + 46c - 96 = 0\), or a correct distance equation, e.g., \(|23 + c| = 25\). [M1] A valid attempt to solve their quadratic equation or absolute value equation for \(c\). [A1] One correct value of \(c\) (either \(2\) or \(-48\)). [A1] Both correct values of \(c = 2\) and \(c = -48\).

**(a)**
We can rewrite the first term of the equation using index laws:
\( 2^{2x+3} = 2^3 \times 2^{2x} = 8(2^x)^2 \)
Substituting this into the original equation gives:
\( 8(2^x)^2 - 9(2^x) + 1 = 0 \)
Let \( y = 2^x \). The equation becomes a quadratic in \( y \):
\( 8y^2 - 9y + 1 = 0 \)
Factorising the quadratic:
\( (8y - 1)(y - 1) = 0 \)
This gives:
\( y = \frac{1}{8} \) or \( y = 1 \).
Now substitute back \( y = 2^x \):
- For \( y = \frac{1}{8} \):
\( 2^x = \frac{1}{8} \implies 2^x = 2^{-3} \implies x = -3 \)
- For \( y = 1 \):
\( 2^x = 1 \implies 2^x = 2^0 \implies x = 0 \)
The solutions are \( x = -3 \) and \( x = 0 \).

**(b)**
Using the power law of logarithms, \( 2\log_2 w = \log_2(w^2) \).
The equation becomes:
\( \log_2(3w + 5) - \log_2(w^2) = 3 \)
Using the subtraction law of logarithms:
\( \log_2\left(\frac{3w + 5}{w^2}\right) = 3 \)
Rewriting in exponential form:
\( \frac{3w + 5}{w^2} = 2^3 \implies \frac{3w + 5}{w^2} = 8 \)
Multiply both sides by \( w^2 \):
\( 3w + 5 = 8w^2 \implies 8w^2 - 3w - 5 = 0 \)
Factorising this quadratic:
\( (8w + 5)(w - 1) = 0 \)
This gives:
\( w = -\frac{5}{8} \) or \( w = 1 \).
Since the original equation contains \( \log_2 w \), we must have \( w > 0 \).
Therefore, \( w = -\frac{5}{8} \) is rejected.
The only valid solution is: \( w = 1 \).

**(a)**
* **M1**: Attempts to write \( 2^{2x+3} \) in the form \( k(2^x)^2 \) where \( k \) is a constant. Usually \( 8(2^x)^2 \) or \( 8y^2 \).
* **M1**: Attempts to solve the quadratic equation \( 8y^2 - 9y + 1 = 0 \) by factorisation, completing the square, or formula.
* **A1**: Obtains correct values of \( y = \frac{1}{8} \) and \( y = 1 \) (or equivalent).
* **A1**: Obtains both correct integer solutions \( x = -3 \) and \( x = 0 \).

**(b)**
* **M1**: Uses the power law of logarithms to write \( 2\log_2 w \) as \( \log_2(w^2) \).
* **M1**: Uses the subtraction law of logarithms to write the left-hand side as a single logarithm.
* **A1**: Correctly removes logs to obtain \( \frac{3w + 5}{w^2} = 8 \) or a correct equivalent quadratic equation such as \( 8w^2 - 3w - 5 = 0 \).
* **A1**: Obtains \( w = 1 \) and explicitly rejects \( w = -\frac{5}{8} \) with a suitable reason (e.g. \( w > 0 \) or \( \log_2 w \) is undefined for negative values).

Method 1: Completing the Square. 1. Rearrange the inequality to bring all terms to one side: \(x^2 + 2y^2 + 1 - 2y(x + 1) \ge 0\). 2. Expand the brackets: \(x^2 + 2y^2 + 1 - 2xy - 2y \ge 0\). 3. Rearrange and group the terms into two perfect squares: \((x^2 - 2xy + y^2) + (y^2 - 2y + 1) \ge 0\). 4. Factorise both parts to express as a sum of squares: \((x - y)^2 + (y - 1)^2 \ge 0\). 5. Since the square of any real number is non-negative, \((x - y)^2 \ge 0\) and \((y - 1)^2 \ge 0\) for all real numbers \(x\) and \(y\). Therefore, their sum \((x - y)^2 + (y - 1)^2 \ge 0\) must also be non-negative. This completes the proof. Method 2: Discriminant of a Quadratic. 1. Rearrange the inequality as a quadratic expression in \(x\): \(x^2 - (2y)x + (2y^2 - 2y + 1) \ge 0\). 2. For this quadratic to be greater than or equal to 0 for all real \(x\), we require its discriminant \(\Delta = B^2 - 4AC \le 0\) (since the coefficient of \(x^2\) is \(1 > 0\)). 3. Calculate the discriminant: \(\Delta = (-2y)^2 - 4(1)(2y^2 - 2y + 1) = 4y^2 - 8y^2 + 8y - 4 = -4y^2 + 8y - 4\). 4. Factorise to simplify: \(\Delta = -4(y^2 - 2y + 1) = -4(y - 1)^2\). 5. Since \((y - 1)^2 \ge 0\) for all real \(y\), we have \(-4(y - 1)^2 \le 0\), meaning \(\Delta \le 0\) for all real \(y\). Because the discriminant is never positive, the quadratic expression is always non-negative, which completes the proof.

Method 1: M1: For rearranging terms and expanding brackets to obtain \(x^2 + 2y^2 + 1 - 2xy - 2y \ge 0\) or equivalent. M1: For grouping terms into two quadratic parts: \((x^2 - 2xy + y^2)\) and \((y^2 - 2y + 1)\). M1: For attempting to factorise both parts into squared terms. A1: For correctly writing the expression as \((x - y)^2 + (y - 1)^2\). A1: For a complete and correct conclusion, explaining that because squares of real numbers are non-negative, the sum of two squares is non-negative, hence the inequality is proven. Method 2: M1: For rearranging the inequality as a quadratic in \(x\): \(x^2 - (2y)x + (2y^2 - 2y + 1) \ge 0\). M1: For identifying that for the expression to be non-negative for all real \(x\), the discriminant \(\Delta\) must satisfy \(\Delta \le 0\). M1: For attempting to find the discriminant: \(\Delta = (-2y)^2 - 4(1)(2y^2 - 2y + 1)\). A1: For simplifying the discriminant correctly to \(-4(y - 1)^2\). A1: For concluding that since \((y - 1)^2 \ge 0\), then \(-4(y - 1)^2 \le 0\), and hence \(\Delta \le 0\), completing the proof.

**(a)**
Equate the equations of the curve and the line:
\[ x^2 - 5x + 6 = 2x - 4 \]

Rearrange to form a 3-term quadratic equation:
\[ x^2 - 7x + 10 = 0 \]

Factorise the quadratic:
\[ (x - 2)(x - 5) = 0 \]

So, the \( x \)-coordinates of the intersection points are:
\[ x = 2 \quad \text{and} \quad x = 5 \]

Substitute these \( x \)-values back into \( L_1 \) (or \( C \)) to find the corresponding \( y \)-coordinates:
- For \( x = 2 \): \( y = 2(2) - 4 = 0 \)
- For \( x = 5 \): \( y = 2(5) - 4 = 6 \)

Thus, the coordinates of the points of intersection are \( (2, 0) \) and \( (5, 6) \).

**(b)**
The line lies above the curve for the interval \( x \in [2, 5] \). The area \( R \) is given by:
\[ \text{Area} = \int_{2}^{5} \left( (2x - 4) - (x^2 - 5x + 6) \right) \text{d}x \]
\[ = \int_{2}^{5} (-x^2 + 7x - 10) \text{d}x \]

Integrate each term with respect to \( x \):
\[ = \left[ -\frac{x^3}{3} + \frac{7x^2}{2} - 10x \right]_{2}^{5} \]

Substitute the upper limit \( x = 5 \):
\[ \left( -\frac{5^3}{3} + \frac{7(5^2)}{2} - 10(5) \right) = -\frac{125}{3} + \frac{175}{2} - 50 = -\frac{25}{6} \]

Substitute the lower limit \( x = 2 \):
\[ \left( -\frac{2^3}{3} + \frac{7(2^2)}{2} - 10(2) \right) = -\frac{8}{3} + 14 - 20 = -\frac{26}{3} \]

Subtract the lower limit value from the upper limit value:
\[ \text{Area} = -\frac{25}{6} - \left(-\frac{26}{3}\right) = -\frac{25}{6} + \frac{52}{6} = \frac{27}{6} = \frac{9}{2} \text{ or } 4.5 \]

**(c)**
Equate the equations of the curve and the line \( L_2 \):
\[ x^2 - 5x + 6 = 2x + c \]
\[ x^2 - 7x + (6 - c) = 0 \]

For the line and the curve to have no points of intersection, the discriminant of this quadratic equation must be strictly negative:
\[ b^2 - 4ac < 0 \]
\[ (-7)^2 - 4(1)(6 - c) < 0 \]
\[ 49 - 24 + 4c < 0 \]
\[ 25 + 4c < 0 \]
\[ c < -\frac{25}{4} \quad (\text{or } c < -6.25) \]

**(a)**
- **M1**: Sets up an equation by equating the curve and the line, attempting to collect terms to form a quadratic equation equal to 0.
- **M1**: Solves the resulting 3-term quadratic equation by factorisation, completing the square, or using the quadratic formula.
- **A1**: Finding one correct coordinate pair, either \( (2, 0) \) or \( (5, 6) \).
- **A1**: Finding both correct coordinate pairs: \( (2, 0) \) and \( (5, 6) \).

**(b)**
- **M1**: Sets up a correct integral for the area with the limits \( 2 \) and \( 5 \) found in part (a). (Alternatively, calculates the area of the trapezium formed under \( L_1 \) and subtracts the integrated area under curve \( C \) between 2 and 5).
- **M1**: Integrates a quadratic expression of the form \( \pm x^2 + kx + c \), raising the power of at least two terms by 1.
- **A1**: Fully correct integrated expression: \( -\frac{x^3}{3} + \frac{7x^2}{2} - 10x \) (or equivalent if integrated separately).
- **M1**: Substitutes limits of \( 5 \) and \( 2 \) into their integrated expression (must show substitution or clear evidence of substitution).
- **A1**: Correct intermediate evaluations, e.g., obtaining \( -\frac{25}{6} \) and \( -\frac{26}{3} \) (or equivalent if split).
- **A1**: Exact final area of \( \frac{9}{2} \) or \( 4.5 \).

**(c)**
- **M1**: Equates the curve and \( L_2 \) to form a quadratic in terms of \( x \) and \( c \), and sets the discriminant \( b^2 - 4ac < 0 \).
- **A1**: Correct range: \( c < -\frac{25}{4} \) (or \( c < -6.25 \)).

(a) Given the equation: \( 2\sin \theta \tan \theta = 5 - 4\cos \theta \)

Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\( 2\sin \theta \left( \frac{\sin \theta}{\cos \theta} \right) = 5 - 4\cos \theta \)

\( \frac{2\sin^2 \theta}{\cos \theta} = 5 - 4\cos \theta \)

Multiplying both sides by \( \cos \theta \):
\( 2\sin^2 \theta = 5\cos \theta - 4\cos^2 \theta \)

Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( 2(1 - \cos^2 \theta) = 5\cos \theta - 4\cos^2 \theta \)

\( 2 - 2\cos^2 \theta = 5\cos \theta - 4\cos^2 \theta \)

Rearranging all terms to the left-hand side:
\( 2\cos^2 \theta - 5\cos \theta + 2 = 0 \) (as required).

(b) Substituting \( \theta = 2x \) into the result from part (a) gives:
\( 2\cos^2(2x) - 5\cos(2x) + 2 = 0 \)

Factorising the quadratic equation in terms of \( \cos(2x) \):
\( (2\cos(2x) - 1)(\cos(2x) - 2) = 0 \)

This gives:
\( \cos(2x) = \frac{1}{2} \) or \( \cos(2x) = 2 \)

Since \( -1 \le \cos(2x) \le 1 \), the equation \( \cos(2x) = 2 \) has no real solutions.

We solve \( \cos(2x) = \frac{1}{2} \) for the interval \( 0 \le x \le 2\pi \), which implies \( 0 \le 2x \le 4\pi \).

The principal value for \( 2x \) is:
\( 2x = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \)

Finding other values of \( 2x \) in the interval \( [0, 4\pi] \):
\( 2x = \frac{\pi}{3}, \quad 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}, \quad 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}, \quad 4\pi - \frac{\pi}{3} = \frac{11\pi}{3} \)

Dividing by 2 to find \( x \):
\( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \)

Part (a):
- M1: Uses \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and multiplies through by \( \cos \theta \) to obtain a form with \( \sin^2 \theta \) and \( \cos \theta \).
- M1: Applies the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to form a quadratic equation in terms of \( \cos \theta \) only.
- A1*: Fully correct proof with no errors or omissions, leading to the given quadratic equation.

Part (b):
- M1: Recognises the link to part (a) and attempts to solve the quadratic equation to find a value for \( \cos(2x) \).
- A1: Identifies that \( \cos(2x) = 2 \) has no solutions and deduces \( \cos(2x) = \frac{1}{2} \).
- M1: Finds at least one correct value of \( 2x \) or \( x \) in radians, or uses a correct method to find multiple solutions in the interval.
- A1: Any two correct solutions for \( x \) from \( \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \).
- A1: All four correct solutions and no extra solutions in the range.

### Part (a)

1. Since the sum of all probabilities must equal 1:
\[ a + b + b + 0.2 + c = 1 \implies a + 2b + c = 0.8 \quad \text{--- [Equation 1]} \]

2. We are given \(P(X \ge 3) = 0.55\):
\[ P(X = 3) + P(X = 4) + P(X = 5) = 0.55 \implies b + 0.2 + c = 0.55 \implies b + c = 0.35 \quad \text{--- [Equation 2]} \]

Substituting [Equation 2] into [Equation 1]:
\[ a + b + (b + c) = 0.8 \implies a + b + 0.35 = 0.8 \implies a + b = 0.45 \quad \text{--- [Equation 3]} \]

3. We are given \(E(X) = 2.9\):
\[ E(X) = 1(a) + 2(b) + 3(b) + 4(0.2) + 5(c) = 2.9 \]
\[ a + 5b + 0.8 + 5c = 2.9 \implies a + 5b + 5c = 2.1 \implies a + 5(b + c) = 2.1 \quad \text{--- [Equation 4]} \]

Substitute \(b + c = 0.35\) into [Equation 4]:
\[ a + 5(0.35) = 2.1 \implies a + 1.75 = 2.1 \implies a = 0.35 \]

Using [Equation 3]:
\[ 0.35 + b = 0.45 \implies b = 0.10 \]

Using [Equation 2]:
\[ 0.10 + c = 0.35 \implies c = 0.25 \]

So, \(a = 0.35\), \(b = 0.10\), and \(c = 0.25\).

---

### Part (b)

First, find \(E(X^2)\):
\[ E(X^2) = \sum x^2 P(X = x) = 1^2(a) + 2^2(b) + 3^2(b) + 4^2(0.2) + 5^2(c) \]
\[ E(X^2) = 1(0.35) + 4(0.10) + 9(0.10) + 16(0.20) + 25(0.25) \]
\[ E(X^2) = 0.35 + 0.40 + 0.90 + 3.20 + 6.25 = 11.1 \]

Now, calculate the variance:
\[ Var(X) = E(X^2) - [E(X)]^2 \]
\[ Var(X) = 11.1 - (2.9)^2 \]
\[ Var(X) = 11.1 - 8.41 = 2.69 \]

---

### Part (c)

(i) Using the linearity of expectation:
\[ E(Y) = E(3 - 2X) = 3 - 2E(X) \]
\[ E(Y) = 3 - 2(2.9) = 3 - 5.8 = -2.8 \]

(ii) Using the variance properties:
\[ Var(Y) = Var(3 - 2X) = (-2)^2 Var(X) = 4 \times 2.69 = 10.76 \]

(iii) Setting up the inequality for \(Y\):
\[ P(Y < -1) = P(3 - 2X < -1) \]
\[ P(-2X < -4) \]
\[ P(X > 2) \]

Since \(X\) only takes discrete integer values:
\[ P(X > 2) = P(X \ge 3) = 0.55 \]

**Part (a)**
* **M1**: Sets up a probability sum equation: \(a + 2b + c + 0.2 = 1\) (or simplified to \(a + 2b + c = 0.8\)).
* **M1**: Sets up an equation using \(P(X \ge 3) = 0.55\): \(b + c + 0.2 = 0.55\) (or simplified to \(b + c = 0.35\)).
* **M1**: Sets up an equation for expectation: \(1(a) + 2(b) + 3(b) + 4(0.2) + 5(c) = 2.9\) (or simplified to \(a + 5b + 5c = 2.1\)).
* **M1**: A complete algebraic method to solve their equations simultaneously to find at least one of \(a\), \(b\), or \(c\).
* **A1**: Both \(a = 0.35\) and \(b = 0.10\) correct.
* **A1**: \(c = 0.25\) correct.

**Part (b)**
* **M1**: Attempting to find \(E(X^2)\) with at least three terms of the form \(x^2 P(X=x)\) correct.
* **A1**: \(E(X^2) = 11.1\) (or correct equivalent).
* **M1**: Correctly applies \(Var(X) = E(X^2) - [E(X)]^2\) using their \(E(X^2)\) and \(E(X) = 2.9\).
* **A1**: \(Var(X) = 2.69\).

**Part (c)**
* **B1**: For \(E(Y) = -2.8\) (or correct exact equivalent, accept follow-through from their \(E(X)\) if working is shown).
* **M1**: Correct use of the variance property: \(Var(Y) = (-2)^2 Var(X)\).
* **A1**: For \(Var(Y) = 10.76\) (accept follow-through from their \(Var(X)\), i.e., \(4 \times Var(X)\)).
* **M1**: Realising that \(P(Y < -1)\) is equivalent to \(P(X > 2)\) or \(P(X \ge 3)\).
* **A1**: \(P(Y < -1) = 0.55\).

(a) The total number of observations is \(n = 80\). The median corresponds to the 40th observation.
Cumulative frequencies are:
- Up to 149.5 (upper bound of first class): 8
- Up to 199.5: 23
- Up to 219.5: 41
Since the 40th observation lies in the class \(200 - 219\) (class boundaries \(199.5\) to \(219.5\)), we interpolate:
\(\text{Median} = 199.5 + \frac{40 - 23}{18} \times (219.5 - 199.5) = 199.5 + \frac{17}{18} \times 20 \approx 218.39\text{ hours}\) (or 218 hours to 3 s.f.).

(b) Using midpoints (\(x\)) of the classes: \(124.5\), \(174.5\), \(209.5\), \(229.5\), \(269.5\), and \(349.5\).
\(\sum f x = (8 \times 124.5) + (15 \times 174.5) + (18 \times 209.5) + (22 \times 229.5) + (12 \times 269.5) + (5 \times 349.5) = 996 + 2617.5 + 3771 + 5049 + 3234 + 1747.5 = 17415\)
\(\text{Mean } \bar{x} = \frac{17415}{80} = 217.6875 \approx 218\text{ hours}\) (or 217.7 to 4 s.f.).

To find the standard deviation, we first find \(\sum f x^2\):
\(\sum f x^2 = (8 \times 124.5^2) + (15 \times 174.5^2) + (18 \times 209.5^2) + (22 \times 229.5^2) + (12 \times 269.5^2) + (5 \times 349.5^2)\)
\(\sum f x^2 = 124002 + 456753.75 + 790024.5 + 1158745.5 + 871563 + 610751.25 = 4011840\)
Using population standard deviation formula:
\(\sigma = \sqrt{\frac{\sum f x^2}{n} - \bar{x}^2} = \sqrt{\frac{4011840}{80} - 217.6875^2} = \sqrt{50148 - 47387.8477} = \sqrt{2760.1523} \approx 52.5\text{ hours}\)
Using sample standard deviation formula:
\(s = \sqrt{\frac{\sum f x^2 - \frac{(\sum f x)^2}{n}}{n - 1}} = \sqrt{\frac{4011840 - 3791027.81}{79}} = \sqrt{\frac{220812.19}{79}} \approx 52.9\text{ hours}\).

(c) For class \(200 - 219\):
Class width \(w_1 = 20\), frequency \(f_1 = 18\), frequency density \(fd_1 = \frac{18}{20} = 0.9\).
This is represented by a bar of width \(W_1 = 1.5\text{ cm}\) and height \(H_1 = 5.4\text{ cm}\).
For class \(240 - 299\):
Class width \(w_2 = 60\), frequency \(f_2 = 12\), frequency density \(fd_2 = \frac{12}{60} = 0.2\).
Width of new bar: \(W_2 = W_1 \times \frac{w_2}{w_1} = 1.5 \times \frac{60}{20} = 4.5\text{ cm}\).
Height of new bar: \(H_2 = H_1 \times \frac{fd_2}{fd_1} = 5.4 \times \frac{0.2}{0.9} = 1.2\text{ cm}\).

(a)
- M1: For identifying class 200 - 219 and setting up a correct interpolation expression using boundaries 199.5 and 219.5.
- A1: Correctly substituted fraction \(\frac{40-23}{18}\) or equivalent.
- A1: 218 or 218.4 (awarded for correct final answer).

(b)
- M1: Standard calculation for mean, using midpoints.
- A1: Mean = 218 or 217.7.
- M1: Correct formula for standard deviation or variance with their values substituted.
- A1: Standard deviation = 52.5 or 52.9.

(c)
- M1: Attempt to use area or frequency density relationships.
- A1: Width = 4.5 cm.
- A1: Height = 1.2 cm.

(a) Using Method 1 (fractional parts rounded up):
- Median: position \(25 \times 0.5 = 12.5\), which rounds up to the 13th term. \(Q_2 = 32\).
- Lower quartile \(Q_1\): position \(25 \times 0.25 = 6.25\), which rounds up to the 7th term. \(Q_1 = 23\).
- Upper quartile \(Q_3\): position \(25 \times 0.75 = 18.75\), which rounds up to the 19th term. \(Q_3 = 42\).

Using Method 2 (interpolation/formulaic):
- Median: 13th term. \(Q_2 = 32\).
- Lower quartile \(Q_1\): position \(\frac{25+1}{4} = 6.5\)th term. \(Q_1 = \frac{22+23}{2} = 22.5\).
- Upper quartile \(Q_3\): position \(\frac{3(25+1)}{4} = 19.5\)th term. \(Q_3 = \frac{42+45}{2} = 43.5\).

(b) Using Method 1:
- \(\text{IQR} = Q_3 - Q_1 = 42 - 23 = 19\).
- Lower outlier boundary: \(Q_1 - 1.5 \times \text{IQR} = 23 - 1.5(19) = -5.5\).
- Upper outlier boundary: \(Q_3 + 1.5 \times \text{IQR} = 42 + 1.5(19) = 70.5\).
Since \(80 > 70.5\), 80 is an outlier. The next largest value is 54, which is less than 70.5. Since the minimum value 12 is greater than \(-5.5\), there are no lower outliers. Thus, 80 is the only outlier.

Using Method 2:
- \(\text{IQR} = 43.5 - 22.5 = 21\).
- Lower outlier boundary: \(22.5 - 1.5(21) = -9\).
- Upper outlier boundary: \(43.5 + 1.5(21) = 75\).
Since \(80 > 75\), 80 is an outlier. The next largest value is 54, which is less than 75. Minimum 12 is greater than \(-9\). Thus, 80 is the only outlier.

(c) Comparison of medians:
- Driver B has a higher median run time (35 minutes) compared to Driver A (32 minutes). This indicates that Driver B generally has longer daily run times than Driver A.

Comparison of spread (using IQR):
- Driver A has a larger interquartile range (19 or 21 minutes) compared to Driver B (12 minutes). This indicates that Driver A's daily run times are more variable/less consistent than Driver B's.

(a)
- B1: Median = 32.
- B1: \(Q_1 = 23\) or \(22.5\).
- B1: \(Q_3 = 42\) or \(43.5\).

(b)
- M1: For calculating \(\text{IQR}\) and attempting to find at least one boundary.
- A1: Showing upper boundary is 70.5 (or 75).
- A1: Concluding 80 is an outlier because \(80 > \text{boundary}\) and stating why it is the only one (e.g. 54 is not an outlier).

(c)
- B1: Direct numerical comparison of medians (e.g., 35 vs 32).
- B1: Contextual interpretation (e.g., Driver B's runs are generally longer).
- B1: Direct numerical comparison of IQRs (e.g., 19/21 vs 12).
- B1: Contextual interpretation (e.g., Driver A's run times are more variable).

Let \(A\), \(D\), and \(M\) represent the sets of students participating in Art, Drama, and Music respectively.

(a) To draw the Venn diagram, we determine the number of students in each mutually exclusive region from the intersection outwards:
- \(A \cap D \cap M = 4\)
- Only \(A \cap D = 12 - 4 = 8\)
- Only \(D \cap M = 15 - 4 = 11\)
- Only \(A \cap M = 10 - 4 = 6\)
- Only \(A = 45 - (8 + 4 + 6) = 27\)
- Only \(D = 40 - (8 + 4 + 11) = 17\)
- Only \(M = 35 - (6 + 4 + 11) = 14\)
- Outside all three circles: \(100 - (27 + 17 + 14 + 8 + 11 + 6 + 4) = 100 - 87 = 13\)

(b) The number of students participating in at least two activities is:
\(8 + 6 + 11 + 4 = 29\)

Probability \(= \frac{29}{100} = 0.29\)

(c) We want to find \(\text{P}(M \cap D' | A) = \frac{\text{P}(A \cap M \cap D')}{\text{P}(A)}\).
From the Venn diagram, the number of students in \(A \cap M \cap D'\) (Art and Music, but not Drama) is 6.
The total number of students in \(A\) is 45.

\(\text{P}(M \cap D' | A) = \frac{6}{45} = \frac{2}{15} \approx 0.133\) (3 s.f.)

(a)
- M1: For drawing three intersecting circles inside a bounding box.
- A1: For correctly placing the inner intersection 4, and at least two of the two-set-only intersections (8, 6, or 11).
- A1: For fully correct Venn diagram with all numbers in regions correct (27, 17, 14, 8, 6, 11, 4, and 13 outside).

(b)
- M1: For attempting to sum the four intersection regions: \(8 + 6 + 11 + 4\) or equivalent.
- A1: For \(0.29\) or \(\frac{29}{100}\).

(c)
- M1: For identifying the conditional probability fraction with denominator 45 (or \(0.45\)) or selecting the numerator 6 (or \(0.06\)).
- A1: For \(\frac{2}{15}\) (or \(\frac{6}{45}\) or \(0.133\) to 3 s.f.).

(a) The tree diagram should have three main branches for the machines:
- Branch \(X\) with probability \(0.40\)
- Branch \(Y\) with probability \(0.35\)
- Branch \(Z\) with probability \(0.25\)

Each of these branches splits into Defective (\(D\)) and Not Defective (\(D'\)):
- From \(X\): \(D\) has probability \(0.02\), \(D'\) has probability \(0.98\)
- From \(Y\): \(D\) has probability \(0.03\), \(D'\) has probability \(0.97\)
- From \(Z\): \(D\) has probability \(0.04\), \(D'\) has probability \(0.96\)

(b) To find the probability that the component is defective:
\(\text{P}(D) = \text{P}(X \cap D) + \text{P}(Y \cap D) + \text{P}(Z \cap D)\)
\(\text{P}(D) = (0.40 \times 0.02) + (0.35 \times 0.03) + (0.25 \times 0.04)\)
\(\text{P}(D) = 0.008 + 0.0105 + 0.010 = 0.0285\)

(c) Using Bayes' Theorem / Conditional Probability:
\(\text{P}(Y | D) = \frac{\text{P}(Y \cap D)}{\text{P}(D)}\)
\(\text{P}(Y | D) = \frac{0.35 \times 0.03}{0.0285} = \frac{0.0105}{0.0285} = \frac{105}{285} = \frac{7}{19} \approx 0.368\) (to 3 s.f.)

(a)
- M1: For a tree diagram structure showing three initial branches followed by two secondary branches on each.
- A1: For correct probabilities on the first branches (0.40, 0.35, 0.25).
- A1: For correct conditional probabilities on the second branches (0.02, 0.98; 0.03, 0.97; 0.04, 0.96).

(b)
- M1: For attempting to find the sum of products of branch probabilities: \((0.40 \times 0.02) + (0.35 \times 0.03) + (0.25 \times 0.04)\).
- A1: For \(0.0285\) (or equivalent fraction, e.g., \(\frac{57}{2000}\)).

(c)
- M1: For identifying the conditional probability formula and setting up the fraction: \(\frac{\text{P}(Y \cap D)}{\text{P}(D)}\).
- A1: For substituting correct values: \(\frac{0.0105}{0.0285}\) (or using their answer from (b) as denominator and \(0.0105\) as numerator).
- A1: For \(\frac{7}{19}\) or \(0.368\) (awrt 0.368).

### Part (a)
Let \(W \sim \text{N}(\mu, \sigma^2)\).

We are given:
\(P(W < 100) = 0.05\)
\(P(W > 160) = 0.10\)

Using the standard normal cumulative distribution tables:
- For \(P(Z < z_1) = 0.05\), we find \(z_1 = -1.6449\) (accept \(-1.645\) or \(-1.64\))
- For \(P(Z > z_2) = 0.10\), we find \(z_2 = 1.2816\) (accept \(1.28\))

Setting up the standardisation equations:
\[ \frac{100 - \mu}{\sigma} = -1.6449 \implies \mu - 1.6449\sigma = 100 \quad (1) \]
\[ \frac{160 - \mu}{\sigma} = 1.2816 \implies \mu + 1.2816\sigma = 160 \quad (2) \]

Subtracting equation (1) from equation (2):
\[ 2.9265\sigma = 60 \implies \sigma \approx 20.5023 \implies \sigma = 20.5 \text{ (to 1 d.p.)} \]

Substituting \(\sigma\) back into equation (2):
\[ \mu = 160 - 1.2816(20.5023) \approx 133.724 \implies \mu = 133.7 \text{ (to 1 d.p.)} \]

---

### Part (b)
Using \(\mu = 133.7\) and \(\sigma = 20.5\):

We first find \(P(W > 120)\):
\[ Z = \frac{120 - 133.7}{20.5} = -0.6683 \]

Using standard normal tables (with \(Z = -0.67\)):
\[ P(W > 120) = P(Z > -0.67) = P(Z < 0.67) = 0.7486 \]
(Alternatively, using a calculator directly: \(P(W > 120) \approx 0.7480\))

Since the three pears are chosen at random, their weights are independent. The probability that all three weigh more than \(120\text{ g}\) is:
- Using tables: \(0.7486^3 \approx 0.420\) (to 3 s.f.)
- Using calculator: \(0.7480^3 \approx 0.419\) (to 3 s.f.)

---

### Part (c)
We require the conditional probability \(P(W < 170 \mid W > 150) = \frac{P(150 < W < 170)}{P(W > 150)}\).

1. Find \(P(W > 150)\):
\[ Z_1 = \frac{150 - 133.7}{20.5} = 0.7951 \]
- Using tables (with \(Z = 0.80\)): \(P(W > 150) = 1 - \Phi(0.80) = 1 - 0.7881 = 0.2119\)
- Using calculator / interpolation: \(P(W > 150) \approx 0.2133\)

2. Find \(P(W < 170)\):
\[ Z_2 = \frac{170 - 133.7}{20.5} = 1.7707 \]
- Using tables (with \(Z = 1.77\)): \(P(W < 170) = \Phi(1.77) = 0.9616\)
- Using calculator / interpolation: \(P(W < 170) \approx 0.9617\)

3. Calculate the numerator \(P(150 < W < 170)\):
- Using tables: \(P(150 < W < 170) = 0.9616 - (1 - 0.2119) = 0.9616 - 0.7881 = 0.1735\)
- Using calculator: \(P(150 < W < 170) = 0.9617 - 0.7867 = 0.1750\)

4. Calculate the conditional probability:
- Using tables: \(\frac{0.1735}{0.2119} \approx 0.819\) (to 3 s.f.)
- Using calculator: \(\frac{0.1750}{0.2133} \approx 0.820\) (to 3 s.f.)

### Part (a)
- **B1**: For finding \(z = 1.2816\) (or \(1.28\)) associated with the upper \(10\%\).
- **B1**: For finding \(z = -1.6449\) (or \(-1.645\) or \(-1.64\)) associated with the lower \(5\%\).
- **M1**: For attempting to set up two simultaneous standardisation equations with correct signs and equating to their \(z\)-values.
- **A1**: Two correct equations in \(\mu\) and \(\sigma\) (e.g. \(\mu - 1.6449\sigma = 100\) and \(\mu + 1.2816\sigma = 160\) or equivalent).
- **M1**: Solving their simultaneous equations to find a value for \(\sigma\) or \(\mu\).
- **A1**: \(\mu = 133.7\) and \(\sigma = 20.5\) (both values must be correctly rounded to 1 decimal place).

### Part (b)
- **M1**: For standardising \(120\) with \(133.7\) and \(20.5\).
- **A1**: For finding \(P(W > 120)\) in the range \([0.748, 0.750]\).
- **M1**: For calculating \(p^3\) where \(p\) is their \(P(W > 120)\).
- **A1**: For an answer of \(0.419\) or \(0.420\) (accept any answer in the range \([0.418, 0.421]\)).

### Part (c)
- **M1**: For writing or using a correct conditional probability ratio, e.g. \(\frac{P(150 < W < 170)}{P(W > 150)}\).
- **A1**: For finding both \(P(150 < W < 170)\) and \(P(W > 150)\) (numerator in the range \([0.173, 0.175]\) and denominator in the range \([0.211, 0.214]\)).
- **A1**: For a final answer in the range \([0.818, 0.821]\).

**(a)**
Using the formula for the sum of squares:
\[S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 4200 - \frac{200^2}{10} = 4200 - 4000 = 200 \quad \text{(as shown)}\]

Now calculate \(S_{yy}\):
\[S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 233000 - \frac{1500^2}{10} = 233000 - 225000 = 8000\]

Now calculate \(S_{xy}\):
\[S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 31100 - \frac{200 \times 1500}{10} = 31100 - 30000 = 1100\]

**(b)**
The product-moment correlation coefficient \(r\) is given by:
\[r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{1100}{\sqrt{200 \times 8000}} = \frac{1100}{\sqrt{1600000}} = \frac{1100}{1264.911} \approx 0.870 \quad \text{(to 3 s.f.)}\]

**(c)**
Yes, because \(r = 0.870\) is positive (and close to 1), indicating a strong positive correlation, which supports the claim that higher temperatures are associated with higher sales.

**(d)**
The gradient \(b\) of the regression line is:
\[b = \frac{S_{xy}}{S_{xx}} = \frac{1100}{200} = 5.5\]

The mean values of \(x\) and \(y\) are:
\[\bar{x} = \frac{200}{10} = 20\]
\[\bar{y} = \frac{1500}{10} = 150\]

The intercept \(a\) is:
\[a = \bar{y} - b\bar{x} = 150 - 5.5(20) = 150 - 110 = 40\]

Therefore, the regression line equation is:
\[y = 40 + 5.5x\]

**(e)**
Substituting \(x = 25\) into the regression line:
\[y = 40 + 5.5(25) = 40 + 137.5 = 177.5\text{ litres}\]

Since \(25^\circ\text{C}\) lies within the range of temperatures recorded during the study (\(14^\circ\text{C}\) to \(28^\circ\text{C}\)), this is interpolation, so the estimate is likely to be reliable.

**(a)**
- **M1**: For a correct method to find either \(S_{yy}\) or \(S_{xy}\).
- **A1**: \(S_{yy} = 8000\) (c.a.o.)
- **A1**: \(S_{xy} = 1100\) (c.a.o.)

**(b)**
- **M1**: For substituting their values into a correct PMCC formula.
- **A1**: \(r = 0.870\) or \(0.87\) (accept awrt 0.870).

**(c)**
- **B1**: Concluding 'Yes' with a reason based on a positive correlation (or strong positive correlation).

**(d)**
- **M1**: For a correct calculation of \(b = S_{xy}/S_{xx}\) using their values.
- **A1**: \(b = 5.5\) (or equivalent fraction).
- **M1**: For a correct method to find \(a\) using their \(b\), \(\bar{x}\), and \(\bar{y}\).
- **A1**: Correct equation in the form \(y = 40 + 5.5x\) (or equivalent fraction/decimals, but must have \(y\) and \(x\)).

**(e)**
- **M1**: For substituting \(x = 25\) into their regression line equation.
- **A1**: Obtaining \(177.5\) (litres) and stating that the estimate is reliable because \(25\) is within the range of the data (interpolation).