2023 Edexcel IAL Mathematics (YMA01) Practice Paper with Answers

(a) Sum of items = \(15 + 27 + 12 + 19 + 8 + 22 + 14 + 18 = 135\).
Lower bound = \(\lceil 135 / 40 \rceil = \lceil 3.375 \rceil = 4\).

(b) Quick Sort (descending order):
Initial list: \([15, 27, 12, 19, 8, 22, 14, 18]\)
Length 8: Left middle is 19 (4th element). Pivot is 19.
- Pass 1 (comparing items to 19: larger to left, smaller to right):
\(27, 22, [19], 15, 12, 8, 14, 18\)

Now sublists are \([27, 22]\) and \([15, 12, 8, 14, 18]\).
For \([27, 22]\), length 2, pivot is 27.
For \([15, 12, 8, 14, 18]\), length 5, pivot is 8 (3rd element).
- Pass 2:
\([27], 22, [19], 15, 12, 14, 18, [8]\)
Note: 22 and 8 are also now selected as pivots as they are single-element sublists/pivots.

Now remaining sublist to sort is \([15, 12, 14, 18]\), length 4, left middle is 12 (2nd element). Pivot is 12.
- Pass 3:
\(27, 22, 19, 15, 14, 18, [12], 8\)

Now remaining sublist to sort is \([15, 14, 18]\), length 3, pivot is 14 (2nd element).
- Pass 4:
\(27, 22, 19, 15, 18, [14], 12, 8\)

Now remaining sublist to sort is \([15, 18]\), length 2, pivot is 15 (1st element).
- Pass 5:
\(27, 22, 19, 18, [15], 14, 12, 8\)
All elements have been pivots. Sorted list is:
\(27, 22, 19, 18, 15, 14, 12, 8\)

(c) First-Fit Decreasing:
- Item 27: Bin 1 (space left: 13)
- Item 22: Bin 2 (space left: 18)
- Item 19: Bin 3 (space left: 21)
- Item 18: Bin 2 (space left: 0)
- Item 15: Bin 3 (space left: 6)
- Item 14: Bin 4 (space left: 26)
- Item 12: Bin 1 (space left: 1)
- Item 8: Bin 4 (space left: 18)

Final Packing:
Bin 1: 27, 12
Bin 2: 22, 18
Bin 3: 19, 15
Bin 4: 14, 8

(d) The packing is optimal because the number of bins used (4) is equal to the calculated lower bound (4). It is impossible to use fewer than 4 bins.

(a) M1: For sum of weights divided by 40.
A1: For correct lower bound of 4 (must show working).

(b) M1: First pass complete, 19 correctly placed as pivot with larger items to the left and smaller to the right.
A1: Second pass complete with pivots 27 and 8 identified correctly.
A1: Third pass complete with pivot 12 identified correctly.
A1: Fourth and fifth passes complete with pivots 14 and 15 identified correctly.
A1: Final sorted list fully correct and all pivots shown.

(c) M1: First 4 items (27, 22, 19, 18) placed correctly (18 must go into Bin 2).
A1: Item 15 and 14 placed correctly.
A1: Items 12 and 8 placed correctly, completing the four bins.
(d) B1: States 'Optimal'.
B1: Gives a valid reason referencing the lower bound of 4 calculated in part (a).

(a) Applying Dijkstra's algorithm:
- Vertex A: Final Label = 0, Order = 1
- Vertex B: Working Values: 12. Final Label = 12, Order = 2
- Vertex D: Working Values: 15. Final Label = 15, Order = 3
- Vertex C: Working Values: 20. Final Label = 20, Order = 4
- Vertex E: Working Values: 26, 21. Final Label = 21, Order = 5
- Vertex F: Working Values: 37, 29. Final Label = 29, Order = 6
- Vertex G: Working Values: 35, 33. Final Label = 33, Order = 7

Shortest path: \(A \rightarrow D \rightarrow E \rightarrow G\)
Length of shortest path = 33.

(b) To find the shortest route from \(A\) to \(G\) passing through \(C\):
- Shortest path \(A \rightarrow C\) is \(A \rightarrow B \rightarrow C\) of length \(12 + 8 = 20\).
- Shortest path \(C \rightarrow G\) is \(C \rightarrow E \rightarrow G\) of length \(11 + 12 = 23\).
- Combining these gives the route: \(A \rightarrow B \rightarrow C \rightarrow E \rightarrow G\) of length \(20 + 23 = 43\).

(c) Sum of all arc weights in the network:
\(12 + 15 + 8 + 10 + 14 + 9 + 11 + 17 + 6 + 20 + 8 + 12 + 13 = 155\).

Identify the degrees of all vertices:
- \(A\): 2, \(B\): 4, \(C\): 4, \(D\): 5 (Odd), \(E\): 5 (Odd), \(F\): 3 (Odd), \(G\): 3 (Odd).
Odd vertices are \(D, E, F, G\).

Form pairings of these four odd vertices:
1. \(DE\) and \(FG\):
- Shortest path \(DE = 6\)
- Shortest path \(FG = 13\)
- Total = \(6 + 13 = 19\)

2. \(DF\) and \(EG\):
- Shortest path \(DF = 14\) (via \(D-E-F\))
- Shortest path \(EG = 12\)
- Total = \(14 + 12 = 26\)

3. \(DG\) and \(EF\):
- Shortest path \(DG = 18\) (via \(D-E-G\))
- Shortest path \(EF = 8\)
- Total = \(18 + 8 = 26\)

Minimum pairing is \(DE + FG = 19\).
Length of shortest inspection route = \(155 + 19 = 174\).

(a) M1: Standard Dijkstra table format with working values and final labels at vertices A, B, and D.
A1: Correct order of labelling and final labels for A, B, D.
M1: Standard Dijkstra values computed for C, E, F.
A1: Correct final labels for C, E, F.
A1: Correct final label for G (33) and order of labelling (7).
A1: Correct path A-D-E-G and length of 33.

(b) M1: Evaluates paths from A to C and C to G.
A1: Correct route A-B-C-E-G.
A0.5: Correct length of 43.

(c) M1: Calculates the sum of all arc weights (155) and identifies the four odd vertices (D, E, F, G).
M1: Identifies the three possible pairings of the odd vertices and calculates their totals.
A1: Correctly identifies the minimum pairing as DE + FG with weight 19.
A1: Correct final route inspection length of 174.

(a) Formulation of the linear programming problem:
Let \(x\) be the number of Royal Chocolate cakes and \(y\) be the number of Imperial Vanilla cakes.

- Flour constraint: \(200x + 300y \le 2400 \implies 2x + 3y \le 24\)
- Eggs constraint: \(3x + 2y \le 24\)
- Sugar constraint: \(200x + 100y \le 1400 \implies 2x + y \le 14\)
- Minimum Imperial Vanilla constraint: \(y \ge 2\)
- Non-negativity constraint: \(x \ge 0\)

Objective function:
Maximize \(P = 4x + 5y\)

(b) Draw the boundary lines on a coordinate grid:
1. \(2x + 3y = 24\): passes through \((0, 8)\) and \((12, 0)\).
2. \(3x + 2y = 24\): passes through \((0, 12)\) and \((8, 0)\).
3. \(2x + y = 14\): passes through \((0, 14)\) and \((7, 0)\).
4. \(y = 2\): horizontal line.

The feasible region \(R\) is the region bounded by vertices:
- \((0, 2)\)
- \((6, 2)\) [intersection of \(y = 2\) and \(2x + y = 14\)]
- \((4.5, 5)\) [intersection of \(2x + y = 14\) and \(2x + 3y = 24\)]
- \((0, 8)\) [intersection of \(x = 0\) and \(2x + 3y = 24\)]

(c) Objective Line Method:
Draw a profit line, e.g., \(4x + 5y = 20\), which passes through \((5, 0)\) and \((0, 4)\).
Moving this line parallel to itself in the direction of increasing profit, the last point in \(R\) it touches is the vertex \((4.5, 5)\).

At this vertex:
\(P = 4(4.5) + 5(5) = 18 + 25 = 43\).

(d) Integer Solution:
Since \(x\) and \(y\) must be integers, we test integer points inside \(R\) close to \((4.5, 5)\):
- \((4, 5)\): \(2(4)+3(5)=23 \le 24\); \(3(4)+2(5)=22 \le 24\); \(2(4)+5=13 \le 14\). (In region) -> \(P = 4(4) + 5(5) = 41\).
- \((3, 6)\): \(2(3)+3(6)=24 \le 24\); \(3(3)+2(6)=21 \le 24\); \(2(3)+6=12 \le 14\). (In region) -> \(P = 4(3) + 5(6) = 42\).
- \((5, 4)\): \(2(5)+3(4)=22 \le 24\); \(3(5)+2(4)=23 \le 24\); \(2(5)+4=14 \le 14\). (In region) -> \(P = 4(5) + 5(4) = 40\).
- \((4, 6)\): \(2(4)+3(6)=26 > 24\) (Outside region).

The maximum integer profit is £42, which is achieved by producing 3 Royal Chocolate cakes and 6 Imperial Vanilla cakes.

(a) B1: Correct simplified flour constraint: \(2x + 3y \le 24\) and egg constraint: \(3x + 2y \le 24\).
B1: Correct sugar constraint: \(2x + y \le 14\) and other constraints: \(y \ge 2, x \ge 0\).
B1: Correct objective function: Maximize \(P = 4x + 5y\).

(b) M1: At least three boundary lines drawn correctly.
A1: All four boundary lines drawn correctly.
M1: Shading of incorrect regions (or clear selection of the correct region).
A1.5: Feasible region R clearly and correctly identified.

(c) M1: Draw or clearly show a valid objective line with a negative gradient of \(-0.8\).
A1: Identify \((4.5, 5)\) as the optimal vertex.
A1: Correct maximum profit of £43.

(d) M1: Test at least two integer coordinates near the optimal vertex.
A1: Identify \((3, 6)\) as the optimal integer solution and state the maximum profit of £42.

Part (a): Dijkstra's algorithm trace: Node A: final label 0, order 1. Temporary labels from A: B = 12, C = 15. Node B: final label 12, order 2. Temporary labels from B: D = 30, E = 22. Node C: final label 15, order 3. Temporary labels from C: D = min(30, 15+11) = 26, F = 29. Node E: final label 22, order 4. Temporary labels from E: G = 35. Node D: final label 26, order 5. Temporary labels from D: F = min(29, 26+9) = 29, G = min(35, 26+16) = 35. Node F: final label 29, order 6. Temporary labels from F: G = min(35, 29+12) = 35. Node G: final label 35, order 7. The shortest path is A - B - E - G with length 35. Part (b): Prim's algorithm starting at A: 1. Add B via edge AB (weight 12). 2. Add E via edge BE (weight 10). 3. Add D via edge ED (weight 8). 4. Add F via edge DF (weight 9). 5. Add C via edge CD (weight 11). 6. Add G via edge FG (weight 12). Vertices added in order: A, B, E, D, F, C, G. Edges in MST: AB, BE, ED, DF, CD, FG. Part (c): Total weight of the minimum spanning tree is 12 + 10 + 8 + 9 + 11 + 12 = 62. Part (d): Prim's algorithm relies on choosing the minimum undirected edge connecting any vertex in the tree to any vertex not in the tree. In a directed graph, edges have directions, meaning connectivity is asymmetric. Choosing a minimum incoming or outgoing edge without regard to direction can result in a subgraph that is not connected or does not form a valid directed spanning tree (arborescence) from the root.

Part (a): M1 for a systematic Dijkstra's algorithm run showing working at each vertex. A1 for correct final labels and order of labeling. A1 for identifying the correct shortest path A - B - E - G. A1 for the correct length of 35. Part (b): M1 for starting at A and selecting the correct first edge AB. A1 for the correct sequence of next three vertices (E, D, F). A1 for completing the MST with the correct order of all vertices (A, B, E, D, F, C, G) and listing the edges. Part (c): B1 for correct total weight of 62. Part (d): M1 for stating that directions affect the validity of connectivity. A1 for explaining that undirected selection in Prim's fails because a directed path must exist from the root to all other vertices, which is not guaranteed by standard Prim's.

Part (a): Let P be the profit. Objective function: Maximize P = 40x + 50y. Constraints: Clay: 2x + y <= 18. Kiln time: x + 3y <= 24. Minimum bowls: x >= 2. Minimum vases: y >= 3. Ratio constraint: y <= 2x. Part (b): Plot the boundary lines: 2x + y = 18 (crosses (0,18) and (9,0)), x + 3y = 24 (crosses (0,8) and (24,0)), x = 2, y = 3, y = 2x. The feasible region R is bounded by the vertices: (2, 3), (7.5, 3), (6, 6), (24/7, 48/7), and (2, 4). Part (c): Testing the vertices: At (2, 3), P = 40(2) + 50(3) = 230. At (7.5, 3), P = 40(7.5) + 50(3) = 450. At (6, 6), P = 40(6) + 50(6) = 540. At (24/7, 48/7), P = 40(24/7) + 50(48/7) = 3360/7 approx 480. At (2, 4), P = 40(2) + 50(4) = 280. The optimal vertex is (6, 6) and the maximum profit is 540 pounds.

Part (a): B1 for correct objective function P = 40x + 50y. M1 for setting up the resource constraints 2x + y <= 18 and x + 3y <= 24. A1 for the remaining constraints x >= 2, y >= 3, and y <= 2x. Part (b): M1 for plotting at least three boundary lines correctly. A1 for plotting all five boundary lines correctly. A2 for clearly identifying and labeling the feasible region R (e.g. by shading out the unwanted regions). Part (c): M1 for a valid method to find the optimal point (either drawing an objective line with gradient -0.8 or testing at least three vertices of R). A1 for identifying the optimal vertex (6, 6). A1 for calculating the maximum profit of 540 pounds.

Part (a): The network has 8 nodes. Node 1 is the start. Activities A, B, C start from Node 1. A ends at Node 2, B ends at Node 3, C ends at Node 4. A dummy activity D1 goes from Node 3 to Node 4 to show that F depends on B and C. F starts at Node 4 and ends at Node 7. E starts at Node 3 and ends at Node 5. D starts at Node 2 and ends at Node 5. H starts at Node 3 and ends at Node 6. G starts at Node 5 and ends at Node 6. I starts at Node 7 and ends at Node 6. J starts at Node 6 and ends at Node 8 (Finish). Part (b): Forward pass: EET(1) = 0. EET(2) = 5. EET(3) = 4. EET(4) = max(3, EET(3)+0) = 4. EET(5) = max(EET(2)+6, EET(3)+7) = max(11, 11) = 11. EET(7) = EET(4)+4 = 8. EET(6) = max(EET(5)+5, EET(3)+8, EET(7)+3) = max(16, 12, 11) = 16. EET(8) = EET(6)+6 = 22. Backward pass: LET(8) = 22. LET(6) = 22-6 = 16. LET(7) = 16-3 = 13. LET(5) = 16-5 = 11. LET(4) = 13-4 = 9. LET(2) = 11-6 = 5. LET(3) = min(11-7, 16-8, 9-0) = min(4, 8, 9) = 4. LET(1) = min(5-5, 4-4, 9-3) = 0. Part (c): Critical activities have zero float. They are A, B, D, E, G, J. The minimum project completion time is 22. Part (d): Total Float = LET(end) - EET(start) - duration. For H: LET(6) - EET(3) - 8 = 16 - 4 - 8 = 4. For F: LET(7) - EET(4) - 4 = 13 - 4 - 4 = 5.

Part (a): M1 for drawing a valid activity network with a start and finish node. A1 for correct placement of A, B, C, D, E, F. A1 for correctly incorporating the dummy activity from Node 3 to Node 4. A1 for correct placement of G, H, I, J with correct dependencies. Part (b): M1 for a forward pass showing systematic addition. A1 for all correct Early Event Times. M1 for a backward pass showing systematic subtraction. A1 for all correct Late Event Times. Part (c): B1 for listing all critical activities (A, B, D, E, G, J). B1 for stating the minimum completion time of 22. Part (d): M1 for the formula for total float. A1 for correct floats: H = 4 and F = 5.

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

(a) For particle \(A\), resolving forces perpendicular to the inclined plane:
\(R = 3mg\cos\alpha = 3mg(0.8) = 2.4mg\)

Since \(A\) moves up the plane, the frictional force \(F\) acts down the plane:
\(F = \mu R = \frac{1}{3}(2.4mg) = 0.8mg\)

Equation of motion for \(A\) parallel to the plane (upwards):
\(T - F - 3mg\sin\alpha = 3ma\)
\(T - 0.8mg - 3mg(0.6) = 3ma \implies T - 2.6mg = 3ma\)

Equation of motion for \(B\) moving downwards:
\(kmg - T = kma\)

(b) Adding the two equations of motion:
\((T - 2.6mg) + (kmg - T) = 3ma + kma\)
\(m(k - 2.6)g = m(k + 3)a\)

Dividing both sides by \(m(k+3)\):
\(a = \frac{k - 2.6}{k + 3} g\) (as required).

(c) Since the acceleration \(a\) is constant, we can use the equation of motion \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = h\), and \(a = \frac{k - 2.6}{k + 3} g\):
\(V^2 = 2 \left(\frac{k - 2.6}{k + 3} g\right) h\)
\(V = \sqrt{\frac{2(k - 2.6)gh}{k + 3}}\)

(d) The inextensibility of the string means that both particles \(A\) and \(B\) move with the same acceleration (they have the same magnitude of acceleration).

(a)
- M1: Attempt to find the normal reaction \(R = 3mg\cos\alpha\) and use \(F = \mu R\).
- A1: Correct equation of motion for particle \(A\): \(T - 2.6mg = 3ma\) (or equivalent).
- A1: Correct equation of motion for particle \(B\): \(kmg - T = kma\).

(b)
- M1: Attempt to solve the equations simultaneously by adding them to eliminate \(T\).
- A1: Obtain \((k+3)ma = (k - 2.6)mg\).
- A1: Clearly show the division by \(m\).
- A1: Fully correct proof showing all intermediate steps leading to \(a = \frac{k - 2.6}{k + 3} g\).

(c)
- M1: Use of \(v^2 = u^2 + 2as\) with \(u = 0\) and their expression for \(a\).
- A1: Correct expression for \(V\) in terms of \(g, h,\) and \(k\): \(V = \sqrt{\frac{2(k - 2.6)gh}{k + 3}}\).

(d)
- B2: States clearly that the acceleration of both particles has the same magnitude (or they move with the same speed at any instant). (Award B1 if only 'same speed' or 'same acceleration' is mentioned without relating it to both particles).

(a) Using the formula \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) for constant velocity:
\(\mathbf{r}_P = (2\mathbf{i} - \mathbf{j}) + (-3\mathbf{i} + 4\mathbf{j})t\)
\(\mathbf{r}_P = (2 - 3t)\mathbf{i} + (-1 + 4t)\mathbf{j}\)

(b) The position vector of \(Q\) at time \(t\) is:
\(\mathbf{r}_Q = (-\mathbf{i} + 5\mathbf{j}) + (u\mathbf{i} + v\mathbf{j})t = (-1 + ut)\mathbf{i} + (5 + vt)\mathbf{j}\)

At the collision time \(t = 2\), the position vectors must be equal (\(\mathbf{r}_P(2) = \mathbf{r}_Q(2)\)):
\(\mathbf{r}_P(2) = (2 - 3(2))\mathbf{i} + (-1 + 4(2))\mathbf{j} = -4\mathbf{i} + 7\mathbf{j}\)
\(\mathbf{r}_Q(2) = (-1 + 2u)\mathbf{i} + (5 + 2v)\mathbf{j}\)

Equating components:
(i) \(-1 + 2u = -4 \implies 2u = -3 \implies u = -\frac{3}{2}\) (as required).
(ii) \(5 + 2v = 7 \implies 2v = 2 \implies v = 1\).

(c) If \(\mathbf{v}_Q = -2\mathbf{i} + \mathbf{j}\), then at \(t = 3\):
\(\mathbf{r}_Q(3) = (-\mathbf{i} + 5\mathbf{j}) + 3(-2\mathbf{i} + \mathbf{j}) = (-1 - 6)\mathbf{i} + (5 + 3)\mathbf{j} = -7\mathbf{i} + 8\mathbf{j}\)

And for \(P\) at \(t = 3\):
\(\mathbf{r}_P(3) = (2 - 3(3))\mathbf{i} + (-1 + 4(3))\mathbf{j} = -7\mathbf{i} + 11\mathbf{j}\)

The displacement vector from \(Q\) to \(P\) at \(t = 3\) is:
\(\mathbf{r}_P(3) - \mathbf{r}_Q(3) = (-7\mathbf{i} + 11\mathbf{j}) - (-7\mathbf{i} + 8\mathbf{j}) = 3\mathbf{j}\)

The distance is the magnitude of this displacement vector:
\(\text{Distance} = |3\mathbf{j}| = 3\text{ km}\).

(a)
- M1: Uses \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) for \(P\).
- A1: Correct simplified vector: \((2 - 3t)\mathbf{i} + (-1 + 4t)\mathbf{j}\).

(b)
- M1: Substitutes \(t = 2\) into their expression for \(\mathbf{r}_P\) to find the position at collision.
- A1: Finds \(\mathbf{r}_P(2) = -4\mathbf{i} + 7\mathbf{j}\).
- M1: Equates components of \(\mathbf{r}_P(2)\) and \(\mathbf{r}_Q(2)\).
- A1: Shows clearly that \(u = -\frac{3}{2}\).
- A1: Correctly finds \(v = 1\).

(c)
- M1: Writes down the new position vector equation for \(Q\) at \(t = 3\).
- A1: Finds correct position vectors at \(t = 3\): \(\mathbf{r}_P(3) = -7\mathbf{i} + 11\mathbf{j}\) and \(\mathbf{r}_Q(3) = -7\mathbf{i} + 8\mathbf{j}\).
- M1: Calculates the difference vector \(\mathbf{r}_P - \mathbf{r}_Q\) or \(\mathbf{r}_Q - \mathbf{r}_P\).
- A1: Finds distance is \(3\text{ km}\).

(a) (i) Let the tension in the rope at \(A\) be \(T_A\) and the tension in the rope at \(C\) be \(T_C\).
We are given \(T_C = 3T_A\).

Resolving forces vertically for the equilibrium of the system:
\(T_A + T_C = 15g + 10g\)
\(T_A + 3T_A = 25g\)
\(4T_A = 25g \implies T_A = 6.25g\)

Taking \(g = 9.8\text{ m s}^{-2}\):
\(T_A = 6.25 \times 9.8 = 61.25\text{ N}\) (as required).

(ii) Since \(AB = 6\text{ m}\) and \(BC = 1.5\text{ m}\), the distance \(AC = 6 - 1.5 = 4.5\text{ m}\).
Taking moments about \(A\):
\(T_C \times AC = (15g \times d) + (10g \times AB)\)

Since \(T_C = 3T_A = 18.75g\), we substitute the known values:
\(18.75g \times 4.5 = 15gd + 10g \times 6\)

Divide throughout by \(g\):
\(18.75 \times 4.5 = 15d + 60\)
\(84.375 = 15d + 60\)
\(15d = 24.375 \implies d = 1.625\)

(b) By modeling the box as a particle, we assume that its entire weight acts through a single concentrated point at the end \(B\) of the plank, rather than being distributed over a section of the plank.

(a)(i)
- M1: For resolving forces vertically: \(T_A + T_C = 15g + 10g\).
- M1: Correct use of the relation \(T_C = 3T_A\).
- A1: Fully correct proof showing \(T_A = 61.25\text{ N}\).

(a)(ii)
- B1: Correctly identifies that \(AC = 4.5\text{ m}\).
- M1: Set up a moments equation about a chosen point (e.g., about \(A\) or \(C\)).
- A1: Correctly substituted moments equation (e.g., about \(A\): \(18.75g \times 4.5 = 15gd + 60g\)).
- M1: Solves the equation for \(d\).
- A1: Obtains \(d = 1.625\).

(b)
- B2: Clearly explains that the weight of the box acts at a single concentrated point (at the end \(B\)). (B1 for mentioning the weight acting at a single point without specifying \(B\) or the context).

Given:
$m_A = 2\text{ kg}$
$m_B = 3\text{ kg}$
$\tan \theta = \frac{3}{4} \implies \sin \theta = 0.6, \cos \theta = 0.8$
$\mu = 0.25$

(a) Let $R$ be the normal reaction force on $A$. Resolving perpendicular to the inclined plane:
\(R = m_A g \cos \theta = 2g \times 0.8 = 1.6g\)

The maximum frictional force $F$ is:
\(F = \mu R = 0.25 \times 1.6g = 0.4g\)

Let $T$ be the tension in the string and $a$ be the acceleration of the system.
For particle $A$ (moving up the plane):
\(T - m_A g \sin \theta - F = m_A a\)
\(T - 2g(0.6) - 0.4g = 2a\)
\(T - 1.6g = 2a \quad \text{--- (Equation 1)}\)

For particle $B$ (moving vertically downwards):
\(m_B g - T = m_B a\)
\(3g - T = 3a \quad \text{--- (Equation 2)}\)

Adding (Equation 1) and (Equation 2):
\(3g - 1.6g = 5a\)
\(1.4g = 5a\)
\(a = 0.28g = 0.28 \times 9.8 = 2.744\text{ m s}^{-2}\)

To 2 significant figures, \(a = 2.7\text{ m s}^{-2}\) (or \(2.74\text{ m s}^{-2}\) to 3 s.f.).

(b) Using (Equation 2):
\(T = 3g - 3a = 3(9.8 - 2.744) = 21.168\text{ N}\)

To 2 significant figures, \(T = 21\text{ N}\) (or \(21.2\text{ N}\) to 3 s.f.).

(c) For the motion of $B$ before it hits the ground:
\(u = 0\)
\(a = 2.744\text{ m s}^{-2}\)
\(s = 1.5\text{ m}\)

Using \(v^2 = u^2 + 2as\):
\(v^2 = 0 + 2 \times 2.744 \times 1.5 = 8.232\)
\(v = \sqrt{8.232} \approx 2.8691\text{ m s}^{-1}\)

To 2 significant figures, \(v = 2.9\text{ m s}^{-1}\) (or \(2.87\text{ m s}^{-1}\) to 3 s.f.).

Part (a)
* M1: Resolving forces on $A$ perpendicular to the plane to find $R = m_A g \cos \theta$ with $\cos \theta = 0.8$.
* A1: Correct $R = 1.6g$ or $15.68\text{ N}$.
* M1: Setting up the equation of motion for $A$ parallel to the plane: \(T - m_A g \sin \theta - \mu R = m_A a\).
* M1: Setting up the equation of motion for $B$: \(3g - T = 3a\).
* A1: Correct system of equations, e.g., \(T - 1.6g = 2a\) and \(3g - T = 3a\).
* A1: Correct acceleration \(a = 0.28g \approx 2.7\text{ m s}^{-2}\) or \(2.74\text{ m s}^{-2}\).

Part (b)
* M1: Substituting their $a$ into one of the equations of motion to find $T$.
* A1: Correct tension \(T \approx 21\text{ N}\) or \(21.2\text{ N}\).

Part (c)
* M1: Using a constant acceleration formula with their $a$, $u = 0$, and $s = 1.5$ to find $v$.
* A1: Correct substituted equation: \(v^2 = 2 \times 2.744 \times 1.5\).
* A1: Correct speed \(v \approx 2.9\text{ m s}^{-1}\) or \(2.87\text{ m s}^{-1}\).

(a) Since the force $\mathbf{F}$ is constant, the acceleration $\mathbf{a}$ of $P$ is constant.
\(\mathbf{a} = \frac{\mathbf{v} - \mathbf{u}}{t}\)
\(\mathbf{a} = \frac{(8\mathbf{i} + 5\mathbf{j}) - (2\mathbf{i} - 3\mathbf{j})}{4}\)
\(\mathbf{a} = \frac{6\mathbf{i} + 8\mathbf{j}}{4} = (1.5\mathbf{i} + 2\mathbf{j})\text{ m s}^{-2}\)

(b) Using Newton's second law:
\(\mathbf{F} = m \mathbf{a}\)
\(\mathbf{F} = 0.5 \times (1.5\mathbf{i} + 2\mathbf{j}) = (0.75\mathbf{i} + \mathbf{j})\text{ N}\)

The magnitude of $\mathbf{F}$ is:
\(|\mathbf{F}| = \sqrt{0.75^2 + 1^2} = \sqrt{0.5625 + 1} = \sqrt{1.5625} = 1.25\text{ N}\)

(c) At any time $t$, the velocity vector $\mathbf{v}(t)$ of $P$ is given by:
\(\mathbf{v}(t) = \mathbf{u} + \mathbf{a}t\)
\(\mathbf{v}(t) = (2\mathbf{i} - 3\mathbf{j}) + (1.5\mathbf{i} + 2\mathbf{j})t\)
\(\mathbf{v}(t) = (2 + 1.5t)\mathbf{i} + (-3 + 2t)\mathbf{j}\)

$P$ is moving parallel to the vector $\mathbf{i} + \mathbf{j}$ when the $\mathbf{i}$ and $\mathbf{j}$ components of velocity are equal and have the same sign:
\(2 + 1.5T = -3 + 2T\)
\(0.5T = 5 \implies T = 10\text{ s}\)

At $T = 10$, $\mathbf{v}(10) = 17\mathbf{i} + 17\mathbf{j}$, which is parallel to $\mathbf{i} + \mathbf{j}$ (since $17 > 0$).

The position vector $\mathbf{r}(t)$ at time $t$ is:
\(\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\)
At $T = 10\text{ s}$:
\(\mathbf{r}(10) = (10\mathbf{i} + 12\mathbf{j}) + (2\mathbf{i} - 3\mathbf{j})(10) + \frac{1}{2}(1.5\mathbf{i} + 2\mathbf{j})(10)^2\)
\(\mathbf{r}(10) = (10\mathbf{i} + 12\mathbf{j}) + (20\mathbf{i} - 30\mathbf{j}) + 50(1.5\mathbf{i} + 2\mathbf{j})\)
\(\mathbf{r}(10) = (10\mathbf{i} + 12\mathbf{j}) + (20\mathbf{i} - 30\mathbf{j}) + (75\mathbf{i} + 100\mathbf{j})\)
\(\mathbf{r}(10) = (10 + 20 + 75)\mathbf{i} + (12 - 30 + 100)\mathbf{j} = (105\mathbf{i} + 82\mathbf{j})\text{ m}\)

Part (a)
* M1: Attempting to use \(\mathbf{a} = \frac{\mathbf{v} - \mathbf{u}}{t}\).
* A1: Correct acceleration vector \((1.5\mathbf{i} + 2\mathbf{j})\text{ m s}^{-2}\).

Part (b)
* M1: Applying \(\mathbf{F} = m \mathbf{a}\) using their acceleration from part (a).
* M1: Calculating the magnitude of their force vector using Pythagoras' theorem.
* A1: Correct magnitude \(1.25\text{ N}\).

Part (c)
* M1: Setting up the velocity vector expression \(\mathbf{v}(t) = (2 + 1.5t)\mathbf{i} + (-3 + 2t)\mathbf{j}\).
* M1: Equating the components of velocity to find $T$: \(2 + 1.5T = -3 + 2T\).
* A1: Correct value of \(T = 10\).
* M1: Using a valid integration or constant acceleration formula for position, e.g., \(\mathbf{r}(T) = \mathbf{r}_0 + \mathbf{u}T + \frac{1}{2}\mathbf{a}T^2\).
* A1: Substituting their $T = 10$, $\mathbf{r}_0$, $\mathbf{u}$, and $\mathbf{a}$ into the formula.
* A1: Correct final position vector \((105\mathbf{i} + 82\mathbf{j})\text{ m}\).

(a) A correct speed-time graph starts at the origin \((0,0)\), goes up linearly with positive gradient to a maximum speed of $V$ at time $t_1$, remains constant at speed $V$ for duration $T$, and then goes down linearly with a steeper negative gradient to $0$ speed at total time $80$ seconds.

(b) Using \(v = u + at\):
For the acceleration phase:
\(V = 0 + 1.25 t_1 \implies t_1 = \frac{V}{1.25} = 0.8V\text{ seconds}\).

For the deceleration phase:
\(0 = V - 2.5 t_3 \implies t_3 = \frac{V}{2.5} = 0.4V\text{ seconds}\).

(c) The total time for the journey is $80$ seconds:
\(t_1 + T + t_3 = 80\)
\(0.8V + T + 0.4V = 80\)
\(T = 80 - 1.2V\)

The total distance traveled is represented by the area under the speed-time graph (area of a trapezium):
\(\text{Distance} = \frac{1}{2}(a + b)h\)
where $a = T = 80 - 1.2V$ (the parallel top side), $b = 80$ (the bottom side), and $h = V$ (the height).
\(\text{Distance} = \frac{1}{2} (80 - 1.2V + 80) V = 1860\)
\(\frac{1}{2} (160 - 1.2V) V = 1860\)
\((80 - 0.6V) V = 1860\)
\(80V - 0.6V^2 = 1860\)
\(0.6V^2 - 80V + 1860 = 0\)

Multiplying the entire equation by $5$ to obtain integer coefficients:
\(3V^2 - 400V + 9300 = 0\) (as required).

(d) Solving the quadratic equation using the quadratic formula:
\(V = \frac{-(-400) \pm \sqrt{(-400)^2 - 4 \times 3 \times 9300}}{2 \times 3}\)
\(V = \frac{400 \times \sqrt{160000 - 111600}}{6}\)
\(V = \frac{400 \pm \sqrt{48400}}{6}\)
\(V = \frac{400 \pm 220}{6}\)

This gives two possible mathematical values:
\(V = \frac{620}{6} \approx 103.3\text{ m s}^{-1}\) or \(V = \frac{180}{6} = 30\text{ m s}^{-1}\).

If \(V \approx 103.3\text{ m s}^{-1}\), the constant-speed duration is:
\(T = 80 - 1.2 \times 103.3 = -44\text{ seconds}\), which is impossible because duration must be non-negative.

Therefore, \(V = 30\text{ m s}^{-1}\).

Part (a)
* B1: Sketch showing three distinct phases: acceleration, constant speed, deceleration.
* B1: Labels of key features: $V$ on vertical axis, total time of $80$ seconds on horizontal axis.

Part (b)
* B1: Show that \(t_1 = 0.8V\) using basic kinematics.
* B1: Correct expression for deceleration time: \(t_3 = 0.4V\).

Part (c)
* M1: Finding an expression for the constant speed time \(T\) in terms of $V$, i.e., \(T = 80 - 1.2V\).
* M1: Using area under graph = total distance: e.g., \(\frac{1}{2}(T + 80)V = 1860\).
* A1: Correct substitution of their expression for $T$ into the area equation.
* M1: Expanding and simplifying to form a quadratic equation in $V$.
* A1: Correctly showing the printed equation: \(3V^2 - 400V + 9300 = 0\).

Part (d)
* M1: Solving the quadratic equation to find two roots: \(V = 30\) and \(V \approx 103\).
* A1: Rejecting \(103.3\) and choosing \(V = 30\text{ m s}^{-1}\) with a valid reason (e.g., \(V = 103.3\) yields negative constant-speed duration $T$).

**(a)**

First, we resolve the forces acting on particle \( A \) perpendicular to the inclined plane.
\[ R = 3mg \cos \alpha \]
Since \( \tan \alpha = \frac{3}{4} \), we have:
\[ \sin \alpha = \frac{3}{5} = 0.6 \quad \text{and} \quad \cos \alpha = \frac{4}{5} = 0.8 \]
Substituting this into the expression for \( R \):
\[ R = 3mg(0.8) = 2.4mg \]

The maximum friction force \( F \) acting on \( A \) is:
\[ F = \mu R = \frac{1}{3} (2.4mg) = 0.8mg \]

When the system is released, particle \( B \) moves downwards with acceleration \( a \) and particle \( A \) moves up the plane with acceleration \( a \).

Writing the equation of motion for particle \( A \) along the plane (taking the direction up the plane as positive):
\[ T - 3mg \sin \alpha - F = 3ma \]
\[ T - 3mg(0.6) - 0.8mg = 3ma \]
\[ T - 2.6mg = 3ma \quad \text{--- (Eq 1)} \]

Writing the equation of motion for particle \( B \) (taking downwards as positive):
\[ 5mg - T = 5ma \quad \text{--- (Eq 2)} \]

Adding (Eq 1) and (Eq 2):
\[ (5mg - T) + (T - 2.6mg) = 5ma + 3ma \]
\[ 2.4mg = 8ma \]
\[ a = 0.3g \]

Using \( g = 9.8\text{ m s}^{-2} \):
\[ a = 0.3 \times 9.8 = 2.94\text{ m s}^{-2} \]

---

**(b)**

Using (Eq 2) to find the tension \( T \):
\[ T = 5mg - 5ma \]
\[ T = 5m(g - 0.3g) \]
\[ T = 3.5mg \]

---

**(c)**

Before \( B \) hits the ground, it falls through a distance of \( 1.3\text{ m} \). Let the speed of both particles at the instant \( B \) hits the ground be \( v \).

Using \( v^2 = u^2 + 2as \):
\[ v^2 = 0 + 2(0.3g)(1.3) = 0.78g \]

At the instant \( B \) hits the ground, the string becomes slack and the tension \( T = 0 \).
Particle \( A \) is now moving up the plane under gravity and friction. Let its new acceleration (deceleration) be \( a' \).

Writing the equation of motion for \( A \) up the plane:
\[ -3mg \sin \alpha - F = 3ma' \]
\[ -1.8mg - 0.8mg = 3ma' \]
\[ -2.6mg = 3ma' \implies a' = -\frac{2.6}{3}g = -\frac{13}{15}g \]

Let \( s' \) be the additional distance traveled by \( A \) up the plane before it comes to instantaneous rest. Using \( v_{\text{final}}^2 = v_{\text{initial}}^2 + 2a's' \):
\[ 0 = 0.78g + 2\left(-\frac{13}{15}g\right)s' \]
\[ \frac{26}{15}g s' = 0.78g \]
\[ s' = \frac{0.78 \times 15}{26} = 0.45\text{ m} \]

Therefore, the total distance traveled by \( A \) from the instant of release is:
\[ s_{\text{total}} = 1.3 + s' = 1.3 + 0.45 = 1.75\text{ m} \]

**(a)**
* **M1**: For resolving forces perpendicular to the inclined plane to find \( R = 3mg \cos \alpha \) with correct trigonometry.
* **A1**: For correct \( R = 2.4mg \) and hence friction force \( F = 0.8mg \).
* **M1**: For setting up the equations of motion for both particles (at least one equation correct) and attempting to solve for \( a \).
* **A1**: For obtaining the correct acceleration \( a = 2.94\text{ m s}^{-2} \) (c.s.o.).

**(b)**
* **M1**: For substituting the value of \( a \) back into one of the equations of motion to find \( T \) in terms of \( m \) and \( g \).
* **A1**: For \( T = 3.5mg \) (or equivalent fraction \( \frac{7}{2}mg \)).

**(c)**
* **M1**: For using a constant acceleration equation to find \( v^2 \) (or \( v \)) just before \( B \) hits the ground.
* **A1**: For \( v^2 = 7.644 \) (or \( 0.78g \)).
* **M1**: For setting up the new equation of motion for \( A \) after \( B \) hits the ground to find the new deceleration \( a' \).
* **A1**: For finding the correct deceleration \( a' = -\frac{13}{15}g \approx -8.49\text{ m s}^{-2} \).
* **A1**: For calculating the correct additional distance \( s' = 0.45\text{ m} \) and the total distance as \( 1.75\text{ m} \) (accept \( 1.8\text{ m} \) to 2 s.f.).

(a) Let the velocities of \(A\) and \(B\) after the collision be \(v_A\) and \(v_B\) respectively.

By conservation of linear momentum:
\(2m(3u) + 3m(u) = 2m v_A + 3m v_B\)
\(9u = 2v_A + 3v_B\) --- (1)

By Newton's law of restitution:
\(v_B - v_A = e(3u - u)\)
\(v_B - v_A = 2eu \implies v_A = v_B - 2eu\) --- (2)

Substitute (2) into (1):
\(9u = 2(v_B - 2eu) + 3v_B\)
\(9u = 5v_B - 4eu\)
\(5v_B = 9u + 4eu \implies v_B = \frac{u}{5}(9 + 4e)\) (as required).

(b) First find \(v_A\):
\(v_A = v_B - 2eu = \frac{u}{5}(9 + 4e) - \frac{10eu}{5} = \frac{u}{5}(9 - 6e)\)

Initial kinetic energy of the system:
\(T_i = \frac{1}{2}(2m)(3u)^2 + \frac{1}{2}(3m)u^2 = 9mu^2 + 1.5mu^2 = 10.5mu^2\)

Final kinetic energy of the system:
\(T_f = \frac{1}{2}(2m)v_A^2 + \frac{1}{2}(3m)v_B^2 = mv_A^2 + 1.5mv_B^2\)
\(T_f = m \left[ \frac{u}{5}(9 - 6e) \right]^2 + 1.5m \left[ \frac{u}{5}(9 + 4e) \right]^2\)
\(T_f = \frac{mu^2}{50} \left[ 2(81 - 108e + 36e^2) + 3(81 + 72e + 16e^2) \right]\)
\(T_f = \frac{mu^2}{50} [ 162 - 216e + 72e^2 + 243 + 216e + 48e^2 ]\)
\(T_f = \frac{mu^2}{50} [ 405 + 120e^2 ] = \frac{mu^2}{10} (81 + 24e^2)\)

Loss in kinetic energy is given as \(\frac{9}{5}mu^2\):
\(\Delta T = T_i - T_f = \frac{21}{2}mu^2 - \frac{mu^2}{10}(81 + 24e^2) = \frac{9}{5}mu^2\)
\(\frac{105 - 81 - 24e^2}{10} = \frac{18}{10}\)
\(24 - 24e^2 = 18\)
\(24e^2 = 6 \implies e^2 = 0.25 \implies e = 0.5\) (since \(e > 0\)).

(a)
- M1: Attempts to write the conservation of linear momentum equation with correct terms.
- A1: Obtains a correct simplified equation: \(9u = 2v_A + 3v_B\).
- M1: Attempts to write the restitution equation with \(e\) and correct relative speeds.
- M1: Eliminates \(v_A\) to obtain an equation in \(v_B\) and \(e\).
- A1: Correctly shows the given expression for \(v_B\).

(b)
- B1: Correctly expresses \(v_A\) in terms of \(u\) and \(e\).
- B1: Correct initial Kinetic Energy of \(10.5mu^2\).
- M1: Sets up the expression for final Kinetic Energy in terms of \(m\), \(u\), and \(e\).
- A1: Obtains a correct simplified expression for final KE: \(\frac{mu^2}{10}(81+24e^2)\).
- M1: Equates the difference in KE to \(\frac{9}{5}mu^2\) and attempts to solve for \(e\).
- A1: Correct value of \(e = 0.5\).

(a) Mass of the car \(m = 1500\text{ kg}\).
Component of gravity acting down the slope:
\(W_{\parallel} = mg \sin\theta = 1500 \times 9.8 \times \frac{1}{49} = 300\text{ N}\)

Power of the engine \(P = 30000\text{ W}\).
At speed \(v = 15\text{ m s}^{-1}\), the tractive force \(F\) is:
\(F = \frac{P}{v} = \frac{30000}{15} = 2000\text{ N}\)

Using Newton's second law along the plane:
\(F - R - mg\sin\theta = ma\)
\(2000 - R - 300 = 1500 \times 0.4\)
\(1700 - R = 600 \implies R = 1100\).

(b) When traveling down the road, gravity acts down the slope, helping the motion.
At maximum speed \(V\), the acceleration is 0.
Let the new tractive force be \(F'\).
The equation of equilibrium down the plane is:
\(F' + mg\sin\theta = R\)
\(F' + 300 = 1100 \implies F' = 800\text{ N}\)

Since the power is still 30 kW:
\(F' = \frac{P}{V} \implies 800 = \frac{30000}{V}\)
\(V = \frac{30000}{800} = 37.5\text{ m s}^{-1}\).

(a)
- M1: Attempts to calculate the gravitational force component down the plane.
- A1: Correct gravitational force of \(300\text{ N\).
- M1: Uses \(P = Fv\) to find the tractive force at \(15\text{ m s}^{-1}\).
- A1: Correct tractive force of \(2000\text{ N\).
- M1: Sets up Newton's second law equation: \(F - R - mg\sin\theta = ma\).
- A1: Correct value of \(R = 1100\).

(b)
- M1: Realises that at maximum speed, acceleration is 0 and sets up a force balance equation down the plane.
- A1: Correct equation: \(F' + 300 = 1100\) (or equivalent).
- A1: Obtains \(F' = 800\text{ N}\).
- M1: Uses \(P = F' V\) to express \(V\) in terms of power and force.
- A1: Correct final answer of \(37.5\text{ m s}^{-1}\).

(a) Let \(R_A\) and \(F_A\) be the normal reaction and frictional force at the ground \(A\) respectively.
Since the ladder is in limiting equilibrium, \(F_A = \mu_1 R_A = 0.5 R_A\).
Let \(R_B\) and \(F_B\) be the normal reaction and frictional force at the wall \(B\) respectively.
Since the ladder is in limiting equilibrium, \(F_B = \mu_2 R_B = 0.4 R_B\).

Resolving horizontally:
\(F_A = R_B \implies 0.5 R_A = R_B\) --- (1)

Resolving vertically:
\(R_A + F_B = Mg \implies R_A + 0.4 R_B = Mg\) --- (2)

Substitute (1) into (2):
\(R_A + 0.4(0.5 R_A) = Mg \implies 1.2 R_A = Mg \implies R_A = \frac{5}{6}Mg\)
Then \(R_B = \frac{5}{12}Mg\) and \(F_B = \frac{1}{6}Mg\).

Taking moments about \(A\):
\(Mg(L \cos\theta) = R_B(2L \sin\theta) + F_B(2L \cos\theta)\)

Divide by \(L\cos\theta\):
\(Mg = 2 R_B \tan\theta + 2 F_B\)

Substitute \(R_B\) and \(F_B\):
\(Mg = 2 \left(\frac{5}{12}Mg\right) \tan\theta + 2 \left(\frac{1}{6}Mg\right)\)
\(1 = \frac{5}{6} \tan\theta + \frac{1}{3}\)
\(\frac{5}{6} \tan\theta = \frac{2}{3} \implies \tan\theta = \frac{2}{3} \times \frac{6}{5} = 0.8\) (as required).

(b) Let the person of mass \(2M\) be at a distance \(x = 2kL\) from \(A\), where \(k\) is the fraction of the ladder's length.
At the point of slipping, the friction forces are still limiting: \(F_A = 0.5 R_A\) and \(F_B = 0.4 R_B\).

Resolving horizontally:
\(F_A = R_B \implies 0.5 R_A = R_B\)

Resolving vertically:
\(R_A + F_B = Mg + 2Mg = 3Mg \implies R_A + 0.4 R_B = 3Mg\)
Using \(0.5 R_A = R_B\):
\(R_A + 0.2 R_A = 3Mg \implies 1.2 R_A = 3Mg \implies R_A = 2.5Mg\)
Thus \(R_B = 1.25Mg\) and \(F_B = 0.5Mg\).

Taking moments about \(A\):
\(Mg(L\cos\theta) + 2Mg(2kL\cos\theta) = R_B(2L\sin\theta) + F_B(2L\cos\theta)\)

Divide by \(MgL
\cos\theta\):
\(1 + 4k = \frac{R_B}{Mg} (2\tan\theta) + \frac{F_B}{Mg} (2)\)
Using \(\tan\theta = 0.8\), \(\frac{R_B}{Mg} = 1.25\), \(\frac{F_B}{Mg} = 0.5\):
\(1 + 4k = 1.25 \times (1.6) + 0.5 \times 2\)
\(1 + 4k = 2.0 + 1.0 = 3.0\)
\(4k = 2.0 \implies k = 0.5\).

So the person can climb a fraction of \(0.5\) (or half) of the length of the ladder.

(a)
- M1: Resolves forces horizontally to get a relation between \(F_A\) and \(R_B\).
- M1: Resolves forces vertically to get an equation in \(R_A\), \(F_B\), and \(Mg\).
- M1: Uses limiting friction relations to express \(R_A\), \(R_B\), \(F_B\) in terms of \(Mg\).
- A1: Correct expressions for \(R_A\), \(R_B\), \(F_B\) in terms of \(Mg\).
- M1: Takes moments about \(A\) (or any other point) with correct terms.
- A1: Correct moment equation.
- A1: Correctly shows \(\tan\theta = 0.8\).

(b)
- M1: Sets up the vertical force equation including the person's weight \(2Mg\).
- A1: Finds the new reactions \(R_A = 2.5Mg\) and \(R_B = 1.25Mg\).
- M1: Sets up the new moments equation about \(A\) with the person's weight term included.
- A1: Solves to find the fraction \(k = 0.5\).

(a) Let \(W\) be the power of the engine in watts, so \(W = 1000P\).

Component of weight down the slope: \(mg\sin\theta = 1000 \times 9.8 \times 0.08 = 784\text{ N}\).

At \(v = 10\text{ m/s}\), the driving force is \(F_1 = \frac{W}{10}\).
Applying Newton's second law up the slope: \(F_1 - R - mg\sin\theta = ma\)
\(\frac{W}{10} - R - 784 = 1000 \times 1.5\)
\(\frac{W}{10} - R = 2284\) [Equation 1]

At \(v = 25\text{ m/s}\), the car moves with constant speed, so acceleration is 0.
The driving force is \(F_2 = \frac{W}{25}\).
Applying Newton's second law: \(F_2 - R - mg\sin\theta = 0\)
\(\frac{W}{25} - R = 784\) [Equation 2]

Subtract Equation 2 from Equation 1:
\(\left(\frac{W}{10} - R\right) - \left(\frac{W}{25} - R\right) = 2284 - 784\)
\(\frac{3W}{50} = 1500 \implies W = 25000\text{ W}\).
Since \(W = 1000P\), we have \(P = 25\) as required.

Substituting \(W = 25000\) into Equation 2:
\(\frac{25000}{25} - R = 784 \implies 1000 - R = 784 \implies R = 216\).

(b) When the car is traveling down the hill at \(20\text{ m/s}\):
Driving force down the hill: \(F_{\text{down}} = \frac{25000}{20} = 1250\text{ N}\).
Component of weight down the hill: \(mg\sin\theta = 784\text{ N}\).
Resistance opposing motion: \(R = 216\text{ N}\).

Applying Newton's second law down the hill:
\(F_{\text{down}} + mg\sin\theta - R = ma\)
\(1250 + 784 - 216 = 1000a\)
\(1818 = 1000a \implies a = 1.818\text{ m/s}^2 \approx 1.82\text{ m/s}^2\) (3 s.f.)

(a)
- M1: Attempts to set up the equation of motion for v = 10 m/s with driving force, weight component, and resistance.
- A1: Correct equation: W/10 - R - 784 = 1500.
- M1: Attempts to set up the equation of motion for constant speed v = 25 m/s.
- A1: Correct equation: W/25 - R - 784 = 0.
- M1: Solves the simultaneous equations to find W or P.
- A1: Shows clearly that P = 25 (cso).
- A1: Finds R = 216.

(b)
- M1: Calculates the new driving force down the hill: 25000 / 20 = 1250 N.
- M1: Sets up the equation of motion down the hill: F_down + mg sin(theta) - R = 1000a with their R.
- A1: Correct equation: 1250 + 784 - 216 = 1000a.
- A1: a = 1.82 m/s^2 or 1.818 m/s^2.

(a) Let \(v_A\) and \(v_B\) be the velocities of \(A\) and \(B\) immediately after their collision.
By conservation of linear momentum:
\(3m u = 3m v_A + m v_B \implies 3v_A + v_B = 3u\) [Equation 1]

By Newton's law of restitution:
\(v_B - v_A = eu \implies v_A = v_B - eu\) [Equation 2]

Substitute Equation 2 into Equation 1:
\(3(v_B - eu) + v_B = 3u \implies 4v_B - 3eu = 3u \implies v_B = \frac{3}{4}u(1+e)\) [as required]

(b) From Equation 2, the velocity of \(A\) after the first collision is:
\(v_A = \frac{3}{4}u(1+e) - eu = \frac{1}{4}u(3-e)\).
Since \(e \le 1\), \(3-e \ge 2\), which means \(v_A \ge \frac{1}{2}u > 0\). thus \(A\) continues to move forward.

Now, \(B\) (mass \(m\), velocity \(v_B\)) collides with \(C\) (mass \(2m\), at rest).
Let \(w_B\) and \(w_C\) be the velocities of \(B\) and \(C\) after this second collision.
By conservation of linear momentum:
\(m v_B = m w_B + 2m w_C \implies w_B + 2w_C = v_B\) [Equation 3]

By Newton's law of restitution (with coefficient \(\frac{1}{2}\)):
\(w_C - w_B = \frac{1}{2}v_B \implies 2w_C - 2w_B = v_B\) [Equation 4]

Subtracting Equation 4 from Equation 3:
\(3w_B = 0 \implies w_B = 0\).

Since \(v_A = \frac{1}{4}u(3-e) > 0\) and \(w_B = 0\), we have \(v_A > w_B\), so particle \(A\) must collide with particle \(B\) again.

(a)
- M1: Applies conservation of linear momentum to the first collision.
- A1: Correct momentum equation: 3v_A + v_B = 3u.
- M1: Applies Newton's law of restitution to the first collision.
- A1: Correct restitution equation: v_B - v_A = eu.
- A1: Solves simultaneously to show v_B = 3/4 * u * (1+e) clearly (cso).

(b)
- M1: Expresses the velocity of A, v_A, in terms of u and e.
- A1: Correct expression: v_A = 1/4 * u * (3-e).
- M1: Applies conservation of linear momentum to the collision between B and C.
- M1: Applies Newton's law of restitution to the collision between B and C with e' = 1/2.
- A1: Solves to show w_B = 0.
- A1: Explains clearly that since v_A > 0 and w_B = 0, a further collision between A and B must occur.

(a) Let \(R_A\) and \(F_A\) be the normal reaction and friction force at \(A\), and \(R_B\) be the reaction force at \(B\).
Resolving vertically: \(R_A = Mg\).
Since the rod is on the point of slipping: \(F_A = \mu R_A = \mu Mg\).
Resolving horizontally: \(R_B = F_A = \mu Mg\).

Taking moments about \(A\):
\(Mg(d\cos\theta) = R_B(2a\sin\theta)\)
Substituting \(R_B = \mu Mg\):
\(Mg d \cos\theta = \mu Mg (2a \sin\theta) \implies d = 2a \mu \tan\theta\).
Since \(\tan\theta = \frac{4}{3}\):
\(d = 2a \mu \left(\frac{4}{3}\right) = \frac{8a\mu}{3} \implies \mu = \frac{3d}{8a}\) [as required].

(b) With \(d = a\) and \(\mu = \frac{3}{8}\), and an added particle of mass \(M\) at \(C\) (\(AC = 0.5a\)):
New vertical reaction: \(R'_A = Mg + Mg = 2Mg\).
New friction force at limit of slipping: \(F'_A = \mu R'_A = \frac{3}{8}(2Mg) = \frac{3}{4}Mg\).
New horizontal reaction at wall: \(R'_B = F'_A = \frac{3}{4}Mg\).

Taking moments about \(A\):
\(Mg (a \cos\phi) + Mg (0.5a \cos\phi) = R'_B (2a \sin\phi)\)
\(1.5 Mg a \cos\phi = \frac{3}{4} Mg (2a \sin\phi)\)
\(1.5 Mg a \cos\phi = 1.5 Mg a \sin\phi \implies \tan\phi = 1\).

(a)
- M1: Resolves forces vertically to find R_A = Mg.
- M1: Uses F_A = mu * R_A and resolves horizontally to find R_B = mu * Mg.
- M1: Attempts to take moments about A.
- A1: Correct moments equation: Mg * d * cos(theta) = R_B * 2a * sin(theta).
- M1: Substitutes R_B into the moments equation.
- A1: Substitutes tan(theta) = 4/3 correctly.
- A1: Obtains mu = 3d / 8a with no errors (cso).

(b)
- M1: Finds the new vertical reaction R'_A = 2Mg and the corresponding maximum friction F'_A = 3/4 * Mg.
- M1: Sets up the new moments equation about A including the particle's weight.
- A1: Correct moments equation: Mg * a * cos(phi) + Mg * 0.5a * cos(phi) = 3/4 * Mg * 2a * sin(phi).
- A1: Solves to find tan(phi) = 1.

### Part (a)

The forces acting on the rod \(AB\) are:
* The weight of the rod, \(Mg\), acting vertically downwards at the midpoint of the rod (at a distance of \(2a\) from \(A\)).
* The weight of the particle, \(2Mg\), acting vertically downwards at \(C\) (at a distance of \(x\) from \(A\)).
* The normal reaction of the ground at \(A\), \(R\), acting vertically upwards.
* The frictional force of the ground at \(A\), \(F\), acting horizontally towards the wall.
* The normal reaction of the smooth wall at \(B\), \(S\), acting horizontally away from the wall.

Resolving vertically for equilibrium:
\[ R - Mg - 2Mg = 0 \implies R = 3Mg \]

Since the rod is on the point of slipping, the friction is limiting:
\[ F = \mu R = \frac{1}{2} (3Mg) = 1.5Mg \]

Resolving horizontally for equilibrium:
\[ S - F = 0 \implies S = F = 1.5Mg \]

Taking moments about the point \(A\):
\[ Mg(2a\cos\theta) + 2Mg(x\cos\theta) = S(4a\sin\theta) \]

Divide the entire equation by \(Mg\cos\theta\):
\[ 2a + 2x = 4a \left(\frac{S}{Mg}\right) \tan\theta \]

We are given that \(\tan\theta = \frac{4}{3}\) and we found \(S = 1.5Mg\), which implies \(\frac{S}{Mg} = 1.5 = \frac{3}{2}\).
Substituting these values in:
\[ 2a + 2x = 4a \left(\frac{3}{2}\right) \left(\frac{4}{3}\right) \]
\[ 2a + 2x = 8a \]
\[ 2x = 6a \implies x = 3a \]

### Part (b)

The force exerted by the ground on the rod at \(A\) is the resultant of the normal reaction \(R\) and the friction force \(F\).
\[ R_{\text{ground}} = \sqrt{R^2 + F^2} \]

Substituting \(R = 3Mg\) and \(F = 1.5Mg\):
\[ R_{\text{ground}} = \sqrt{(3Mg)^2 + (1.5Mg)^2} \]
\[ R_{\text{ground}} = \sqrt{9 M^2 g^2 + 2.25 M^2 g^2} = \sqrt{11.25 M^2 g^2} \]
\[ R_{\text{ground}} = \frac{3\sqrt{5}}{2}Mg \approx 3.35Mg \text{ (to 3 s.f.)} \]

### Part (a)
* **M1**: Resolves vertically to find an equation relating \(R\), \(M\), and \(g\).
* **A1**: Correct expression for the normal reaction, \(R = 3Mg\).
* **B1**: Uses \(F = \mu R\) to find the limiting friction force, \(F = 1.5Mg\).
* **M1**: Resolves horizontally to relate the normal reaction at the wall, \(S\), to the friction force \(F\).
* **A1**: Correct expression for \(S\), i.e., \(S = 1.5Mg\).
* **M1**: Takes moments about \(A\) (or any other point) with a correct number of terms and correct trigonometric components.
* **A1**: Fully correct completion to show \(x = 3a\).

### Part (b)
* **M1**: Identifies that the force exerted by the ground is the resultant of \(R\) and \(F\), and sets up \(R_{\text{ground}} = \sqrt{R^2 + F^2}\).
* **M1**: Substitutes their values of \(R\) and \(F\) into \(\sqrt{R^2 + F^2}\).
* **A1.7**: Obtains the correct magnitude \(\frac{3\sqrt{5}}{2}Mg\) or \(3.35Mg\) (or \(3.4Mg\) if 2 s.f. is used).

Let \(\theta\) be the angle that the radius from \(O\) to \(P\) makes with the downward vertical. The radius of the horizontal circular path of \(P\) is \(r = R \sin \theta\).

(a) The forces acting on the particle are:
- The weight \(mg\) acting vertically downwards.
- The normal reaction \(N\) directed towards \(O\).

Resolving horizontally towards the centre of the circular path:
\[N \sin \theta = m r \omega^2\]
Since \(r = R \sin \theta\):
\[N \sin \theta = m R \sin \theta \omega^2\]
Since \(\sin \theta \ne 0\) for a circular path of non-zero radius:
\[N = m R \omega^2\] (as required).

(b) Resolving vertically:
\[N \cos \theta = mg\]
Substitute \(N = m R \omega^2\):
\[(m R \omega^2) \cos \theta = mg\]
\[\cos \theta = \frac{g}{R \omega^2}\]
The height \(h\) of the plane of the circular path above the lowest point of the bowl is:
\[h = R - R \cos \theta = R\left(1 - \frac{g}{R \omega^2}\right) = R - \frac{g}{\omega^2}\] (as required).

(c) For the particle to describe a horizontal circle on the inner surface of the bowl, we must have \(\theta > 0\), which requires:
\[\cos \theta < 1\]
\[\frac{g}{R \omega^2} < 1\]
\[\omega^2 > \frac{g}{R}\]
\[\omega > \sqrt{\frac{g}{R}}\]
Thus, the minimum angular speed is \(\sqrt{\frac{g}{R}}\).

**Part (a)**
- **M1**: Resolving horizontally with acceleration term \(m r \omega^2\) or \(m R \sin\theta \omega^2\).
- **A1**: Correct horizontal equation: \(N \sin \theta = m R \sin \theta \omega^2\).
- **M1**: Utilizing \(r = R \sin \theta\) and cancelling \(\sin \theta\).
- **A1**: Showing clearly that \(N = m R \omega^2\) with no errors.

**Part (b)**
- **M1**: Resolving vertically to obtain an equation containing \(N\), \(\theta\) and \(mg\).
- **A1**: Correct vertical equation: \(N \cos \theta = mg\).
- **M1**: Expressing the height above the lowest point as \(h = R(1 - \cos \theta)\) and substituting \(\cos \theta\).
- **A1**: Showing clearly that \(h = R - \frac{g}{\omega^2}\).

**Part (c)**
- **M1**: Stating or using the physical condition \(\cos \theta < 1\) (or equivalently \(h > 0\)).
- **M1**: Setting up the inequality \(\frac{g}{R \omega^2} < 1\) or \(R - \frac{g}{\omega^2} > 0\).
- **A1**: Correct final inequality \(\omega > \sqrt{\frac{g}{R}}\).

(a) At the equilibrium position, the extension of the string is:
\[e = 1.2 - 1.0 = 0.2\text{ m}\]
In equilibrium, the tension \(T_0\) in the string balances the weight of the particle:
\[T_0 = mg\]
By Hooke's Law:
\[T_0 = \frac{\lambda e}{l} = \frac{\lambda \times 0.2}{1.0} = 0.2\lambda\]
Therefore:
\[0.2\lambda = 0.5 \times 9.8 = 4.9\]
\[\lambda = 24.5\] (as required).

(b) Let \(x\) be the vertical displacement of \(P\) downwards from its equilibrium position.
The extension of the string at this position is \(0.2 + x\).
The tension in the string is:
\[T = \frac{24.5(0.2 + x)}{1.0} = 4.9 + 24.5x\]
Using Newton's Second Law downwards:
\[mg - T = m \ddot{x}\]
\[4.9 - (4.9 + 24.5x) = 0.5 \ddot{x}\]
\[-24.5x = 0.5 \ddot{x}\]
\[\ddot{x} = -49x\]
This is of the form \(\ddot{x} = -\omega^2 x\) with \(\omega^2 = 49\), so the motion is simple harmonic.
The period of the motion is:
\[T_{\text{period}} = \frac{2\pi}{\omega} = \frac{2\pi}{7}\text{ s}\] (as required).

(c) The amplitude of the motion is \(a = 0.15\text{ m\}}.
Since the maximum upward displacement from the equilibrium position is \)0.15\text{ m\}}, the extension of the string at the highest point of the motion is \(0.2 - 0.15 = 0.05\text{ m\}}.
Since \)0.05 > 0\), the string remains taut throughout the motion.
The maximum speed of the particle is given by:
\[v_{\text{max}} = a\omega = 0.15 \times 7 = 1.05\text{ m s}^{-1}\]

**Part (a)**
- **M1**: Finding the equilibrium extension \(e = 0.2\text{ m}\) and setting up the equilibrium equation \(T_0 = mg\).
- **M1**: Applying Hooke's Law \(T_0 = \frac{\lambda e}{l}\) and substituting \(g = 9.8\).
- **A1**: Finding \(\lambda = 24.5\) clearly with no errors.

**Part (b)**
- **M1**: Setting up the equation of motion \(mg - T = m \ddot{x}\) for a general displacement \(x\) from equilibrium.
- **A1**: Finding a correct expression for the tension \(T = 4.9 + 24.5x\).
- **M1**: Simplifying to obtain the SHM differential equation in the form \(\ddot{x} = -\omega^2 x\).
- **A1**: Correctly identifying \(\omega = 7\) and showing the period is \(\frac{2\pi}{7}\text{ s}\).

**Part (c)**
- **M1**: Identifying the amplitude \(a = 0.15\text{ m}\).
- **M1**: Checking or stating that the string remains taut (the extension at the highest point is \(0.05\text{ m} > 0\)).
- **M1**: Using \(v_{\text{max}} = a\omega\) with their values.
- **A1**: Calculating the maximum speed as \(1.05\text{ m s}^{-1}\).

(a) The volume \(V_S\) of the solid of revolution \(S\) is:
\[V_S = \pi \int_0^3 y^2 \, dx = \pi \int_0^3 9x \, dx = \pi \left[ \frac{9x^2}{2} \right]_0^3 = \frac{81\pi}{2}\]
The position of the centre of mass \(\bar{x}_S\) of \(S\) is given by:
\[V_S \bar{x}_S = \pi \int_0^3 x y^2 \, dx = \pi \int_0^3 9x^2 \, dx = \pi \left[ 3x^3 \right]_0^3 = 81\pi\]
Therefore:
\[\bar{x}_S = \frac{81\pi}{\frac{81\pi}{2}} = 2\] (as required).

(b) The flat circular face of \(S\) at \(x = 3\) has radius \(y(3) = 3\sqrt{3}\).
The cylinder \(C\) has radius \(3\sqrt{3}\) and length 4, and lies in \(3 \le x \le 7\).
The volume of the cylinder is:
\[V_C = \pi \times \text{radius}^2 \times \text{length} = \pi \times (3\sqrt{3})^2 \times 4 = 108\pi\]
The centre of mass of the cylinder is at the midpoint of its axis of symmetry:
\[\bar{x}_C = 3 + 2 = 5\]
Let \(\bar{X}\) be the distance of the centre of mass of the combined solid from \(O\):
\[(V_S + V_C)\bar{X} = V_S \bar{x}_S + V_C \bar{x}_C\]
\[\left(\frac{81\pi}{2} + 108\pi\right)\bar{X} = \frac{81\pi}{2}(2) + 108\pi(5)\]
\[\frac{297\pi}{2} \bar{X} = 81\pi + 540\pi = 621\pi\]
\[\bar{X} = \frac{621 \times 2}{297} = \frac{1242}{297} = \frac{46}{11} \approx 4.18\]

(c) The flat circular end of the cylinder is at \(x = 7\).
The centre of mass \(G\) of the combined solid lies on the axis of symmetry at a distance of \(\frac{46}{11}\) from \(O\).
The distance from the flat circular end at \(x = 7\) to \(G\) along the axis of symmetry is:
\[d = 7 - \bar{X} = 7 - \frac{46}{11} = \frac{31}{11} \approx 2.818\]
The point of suspension \(P\) lies on the circumference of this flat circular end of radius \(R = 3\sqrt{3}\).
In equilibrium, the line \(PG\) is vertical.
The angle \(\theta\) that the axis of symmetry makes with the vertical is given by:
\[\tan \theta = \frac{R}{d} = \frac{3\sqrt{3}}{\frac{31}{11}} = \frac{33\sqrt{3}}{31} \approx 1.8437\]
\[\theta = \arctan(1.8437) \approx 61.52^\circ\]
To the nearest degree, \(\theta = 62^\circ\).

**Part (a)**
- **M1**: Correct integration of \(\pi y^2\) to find the volume of \(S\) (limits not required for M mark).
- **A1**: Finding volume of \(S\) is \(\frac{81\pi}{2}\).
- **M1**: Integrating \(\pi x y^2\) to find the moment of \(S\).
- **A1**: Finding \(\bar{x}_S = 2\) with no errors shown.

**Part (b)**
- **B1**: Stating or finding the volume of the cylinder \(V_C = 108\pi\) and its centre of mass \(\bar{x}_C = 5\).
- **M1**: Using the composite body formula: \((V_S + V_C)\bar{X} = V_S \bar{x}_S + V_C \bar{x}_C\).
- **A1**: Substituting their correct values into the composite body formula.
- **A1**: Calculating \(\bar{X} = \frac{46}{11}\) (or \(4.18\) or better).

**Part (c)**
- **M1**: Finding the distance of \(G\) from the circular end \(x = 7\) as \(7 - \bar{X}\).
- **M1**: Using trigonometry \(\tan \theta = \frac{\text{radius}}{\text{distance}}\).
- **A1**: Correct angle \(62^\circ\) (accept \(61.5^\circ\)).

For part (a), let the extension of the string be \(x\). Initially, \(OA = 1.5l\), so the initial extension is \(x_0 = 1.5l - l = 0.5l\). By conservation of energy, the initial elastic potential energy (EPE) is given by \(V_{EPE} = \frac{\lambda x_0^2}{2l} = \frac{4mg (0.5l)^2}{2l} = 0.5mgl\). When the string first becomes slack, \(x = 0\), meaning EPE is zero and all energy is converted to kinetic energy: \(\frac{1}{2}mv^2 = 0.5mgl \implies v = \sqrt{gl}\). For part (b), while the string is taut (\(x > 0\)), the equation of motion is \(m\ddot{x} = -T = -\frac{4mgx}{l} \implies \ddot{x} = -\frac{4g}{l}x\). This is simple harmonic motion with \(\omega = 2\sqrt{\frac{g}{l}}\). The time \(t_1\) to go from \(x = 0.5l\) to \(x = 0\) is a quarter of a period: \(t_1 = \frac{\pi}{2\omega} = \frac{\pi}{4}\sqrt{\frac{l}{g}}\). Once the string is slack (\(x < 0\)), the particle moves at a constant speed \(v = \sqrt{gl}\) over a distance \(l\). The time \(t_2\) for this stage is \(t_2 = \frac{l}{v} = \sqrt{\frac{l}{g}}\). The total time is \(t = t_1 + t_2 = \left(\frac{\pi}{4} + 1\right)\sqrt{\frac{l}{g}}\).

Part (a): M1 for a conservation of energy equation relating initial EPE and final KE. A1 for correct EPE expression of \(\frac{4mg(0.5l)^2}{2l}\). A1 for equating EPE to KE: \(\frac{1}{2}mv^2 = 0.5mgl\). A1 for final speed \(\sqrt{gl}\). Part (b): M1 for setting up the equation of motion and identifying SHM. A1 for finding \(\omega = 2\sqrt{\frac{g}{l}}\). M1 for calculating the time \(t_1\) for the SHM phase. A1 for \(t_1 = \frac{\pi}{4}\sqrt{\frac{l}{g}}\). M1 for setting up \(t_2 = \frac{\text{distance}}{\text{speed}}\) for the constant velocity phase. A1 for \(t_2 = \sqrt{\frac{l}{g}}\). A1 for the correct total time \(\left(\frac{\pi}{4} + 1\right)\sqrt{\frac{l}{g}}\).

For part (a), let \(v\) be the speed of the particle when the string makes an angle \(\theta\) with the downward vertical. By conservation of energy: \(\frac{1}{2}mU^2 = \frac{1}{2}mv^2 + mgL(1 - \cos\theta)\), which gives \(v^2 = U^2 - 2gL(1 - \cos\theta)\). The equation of motion radially towards the center \(O\) is: \(T - mg\cos\theta = \frac{mv^2}{L}\). Substituting \(v^2\) gives: \(T = mg\cos\theta + \frac{m}{L}(U^2 - 2gL + 2gL\cos\theta) = \frac{mU^2}{L} - mg(2 - 3\cos\theta)\) as required. For part (b), the string becomes slack when \(T = 0\) at \(\theta = 120^\circ\). Substituting \(\cos 120^\circ = -0.5\) and \(T = 0\) gives: \(0 = \frac{mU^2}{L} - mg(2 - 3(-0.5)) \implies 0 = \frac{U^2}{L} - 3.5g \implies U^2 = 3.5gL = \frac{7}{2}gL \implies U = \sqrt{\frac{7gL}{2}}\).

Part (a): M1 for applying conservation of energy between the lowest point and the general position. A1 for a correct energy equation: \(\frac{1}{2}mU^2 = \frac{1}{2}mv^2 + mgL(1 - \cos\theta)\). M1 for writing the equation of motion along the radius. A1 for a correct radial equation: \(T - mg\cos\theta = \frac{mv^2}{L}\). A1 for fully simplifying to show the given expression for \(T\). Part (b): M1 for setting \(T = 0\) at \(\theta = 120^\circ\). A1 for stating \(\cos 120^\circ = -0.5\). M1 for solving the resulting equation for \(U^2\). A1 for obtaining \(U^2 = 3.5gL\) or equivalent. A1 for the final answer \(U = \sqrt{\frac{7gL}{2}}\).

For part (a), the volume \(V\) of the solid of revolution is \(V = \pi \int_0^h y^2 dx = \pi \int_0^h k^2 x^3 dx = \pi k^2 [\frac{x^4}{4}]_0^h = \frac{\pi k^2 h^4}{4}\). The x-coordinate of the centre of mass \(\bar{x}\) is given by \(\bar{x} = \frac{\pi \int_0^h x y^2 dx}{V}\). The numerator is \(\pi \int_0^h x(k^2 x^3) dx = \pi k^2 \int_0^h x^4 dx = \pi k^2 [\frac{x^5}{5}]_0^h = \frac{\pi k^2 h^5}{5}\). Thus, \(\bar{x} = \frac{\pi k^2 h^5 / 5}{\pi k^2 h^4 / 4} = \frac{4}{5}h\) as required. For part (b), we are given \(h = 5\text{ cm}\) and the radius of the circular base is \(5\text{ cm}\). At \(x = 5\), \(y = 5 \implies k (5)^{3/2} = 5 \implies k^2 = \frac{1}{5}\). Thus, the volume of \(S\) is \(V_S = \frac{\pi (1/5) (5^4)}{4} = \frac{125\pi}{4} = 31.25\pi\), and its centre of mass is at \(\bar{x}_S = \frac{4}{5}(5) = 4\text{ cm}\). The cylinder has radius \(5\text{ cm}\) and length \(6\text{ cm}\), so its volume is \(V_C = \pi (5^2)(6) = 150\pi\). Since it is joined to the base of \(S\) (at \(x = 5\)), the cylinder lies in the interval \(5 \le x \le 11\), and its centre of mass is at \(\bar{x}_C = 5 + 3 = 8\text{ cm}\). The distance \(\bar{X}\) of the centre of mass of the combined solid from the origin is given by \(\bar{X} = \frac{V_S \bar{x}_S + V_C \bar{x}_C}{V_S + V_C} = \frac{31.25\pi (4) + 150\pi (8)}{31.25\pi + 150\pi} = \frac{125\pi + 1200\pi}{181.25\pi} = \frac{1325}{181.25} = \frac{212}{29} \approx 7.31\text{ cm}\).

Part (a): M1 for setting up the integral for volume \(\pi \int y^2 dx\). A1 for finding the volume as \(\frac{\pi k^2 h^4}{4}\). M1 for setting up the integral for the moment \(\pi \int x y^2 dx\). A1 for finding the moment as \(\frac{\pi k^2 h^5}{5}\). M1 for dividing the moment by the volume. A1 for showing \(\bar{x} = \frac{4}{5}h\). Part (b): M1 for finding the value of \(k^2 = 0.2\) and calculating the volume of \(S\) as \(31.25\pi\) or \(\frac{125\pi}{4}\). A1 for finding the volume of the cylinder as \(150\pi\). M1 for locating the centre of mass of the cylinder at \(x = 8\). M1 for setting up the moments equation about the origin. A1 for obtaining the final distance \(\frac{212}{29}\text{ cm}\) (or 3 s.f. equivalent of \(7.31\text{ cm}\)).

Part (a) Let \(x\) be the extension of the string. When the particle has fallen a distance of \(l + x\), its loss in gravitational potential energy (GPE) is \(mg(l + x)\). The elastic potential energy (EPE) stored in the string is \(\frac{\lambda x^2}{2l}\), where \(\lambda = 4mg\). At the point of instantaneous rest, the kinetic energy (KE) of the particle is zero. By conservation of energy: Loss in GPE = Gain in EPE, which gives \(mg(l + x) = \frac{4mg x^2}{2l}\). Simplifying this, we get \(l + x = \frac{2x^2}{l}\), which rearranges to the quadratic equation \(2x^2 - lx - l^2 = 0\). Factoring the quadratic gives \((2x + l)(x - l) = 0\). Since the extension \(x\) must be positive, we find \(x = l\). Therefore, the total distance fallen by \(P\) before it first comes to instantaneous rest is \(l + x = 2l\). Part (b) The maximum speed of \(P\) occurs when its acceleration is zero, which is at the equilibrium position where the upward tension equals the downward weight. This gives \(T = mg \implies \frac{\lambda x_0}{l} = mg\). Substituting \(\lambda = 4mg\), we have \(\frac{4mg x_0}{l} = mg \implies x_0 = \frac{1}{4}l\). Applying conservation of energy between the release point \(O\) and the position of maximum speed: Loss in GPE = Gain in KE + Gain in EPE, which gives \(mg(l + x_0) = \frac{1}{2}mv^2 + \frac{\lambda x_0^2}{2l}\). Substituting \(x_0 = \frac{1}{4}l\) and \(\lambda = 4mg\) into the equation gives \(mg\left(l + \frac{1}{4}l\right) = \frac{1}{2}mv^2 + \frac{4mg\left(\frac{1}{4}l\right)^2}{2l}\), which simplifies to \(\frac{5}{4}mgl = \frac{1}{2}mv^2 + \frac{1}{8}mgl\). Rearranging for \(v^2\), we get \(\frac{1}{2}mv^2 = \frac{9}{8}mgl \implies v^2 = \frac{9}{4}gl\). Taking the square root gives the maximum speed \(v = \frac{3}{2}\sqrt{gl}\).

Part (a): M1: Attempting to set up an energy equation of the form \(mg(l + x) = \frac{\lambda x^2}{2l}\). A1: Correct energy equation with \(\lambda = 4mg\) substituted: \(mg(l + x) = \frac{4mg x^2}{2l}\) (or equivalent). M1: Solving the resulting quadratic equation in \(x\). A1: Obtaining the positive extension \(x = l\). A1: Finding the total distance fallen as \(2l\). Part (b): M1: Realising that maximum speed occurs at the equilibrium position where \(T = mg\). A1: Correctly finding the extension at this position as \(x_0 = \frac{1}{4}l\). M1: Setting up a three-term energy equation relating the release point to this equilibrium position. A1: Correct energy equation: \(mg(l + \frac{1}{4}l) = \frac{1}{2}mv^2 + \frac{1}{8}mgl\) (or equivalent). M1: Solving for \(v\) or \(v^2\) from their equation. A1: Correctly obtaining \(v = \frac{3}{2}\sqrt{gl}\) (or \(1.5\sqrt{gl}\)).

(a) The equation of \(l_1\) is \(2x - y + 1 = 0 \implies y = 2x + 1\).
Therefore, the gradient of \(l_1\) is \(m_1 = 2\).
Since \(l_2\) is perpendicular to \(l_1\), its gradient is \(m_2 = -\frac{1}{m_1} = -\frac{1}{2}\).
Using the equation of a straight line with gradient \(-\frac{1}{2}\) passing through \(A(4, 4)\):
\(y - 4 = -\frac{1}{2}(x - 4)\)
\(2(y - 4) = -(x - 4)\)
\(2y - 8 = -x + 4\)
\(x + 2y - 12 = 0\).

(b) To find the coordinates of \(P\), we solve the equations of \(l_1\) and \(l_2\) simultaneously:
From \(l_1\): \(y = 2x + 1\).
Substitute into the equation of \(l_2\):
\(x + 2(2x + 1) - 12 = 0\)
\(x + 4x + 2 - 12 = 0\)
\(5x - 10 = 0 \implies x = 2\).
Substitute \(x = 2\) back into \(y = 2x + 1\):
\(y = 2(2) + 1 = 5\).
So the coordinates of \(P\) are \((2, 5)\).

(a)
M1: Identifies the gradient of \(l_1\) as \(2\).
M1: Uses the perpendicular gradient rule \(m_2 = -1/m_1\) and writes the equation of the line passing through \((4, 4)\) with gradient \(-\frac{1}{2}\).
A1: Obtains a correct equation in the form \(ax + by + c = 0\) with integer coefficients (e.g. \(x + 2y - 12 = 0\) or \(-x - 2y + 12 = 0\)).

(b)
M1: Attempts to solve the simultaneous equations for \(l_1\) and \(l_2\).
A1: Find one correct coordinate (either \(x = 2\) or \(y = 5\)).
A1.8: Both coordinates correct, \(P(2, 5)\) or \(x = 2, y = 5\).

(a) Express \(\mathrm{f}'(x)\) in index form:
\(\mathrm{f}'(x) = 5x^{3/2} - 4x^{-1/2}\).
Integrate \(\mathrm{f}'(x)\) to find \(\mathrm{f}(x)\):
\(\mathrm{f}(x) = \int \left(5x^{3/2} - 4x^{-1/2}\right) \mathrm{d}x\)
\(\mathrm{f}(x) = \frac{5}{5/2}x^{5/2} - \frac{4}{1/2}x^{1/2} + C\)
\(\mathrm{f}(x) = 2x^{5/2} - 8x^{1/2} + C\).
Substitute the coordinates of \(P(4, 30)\) into the equation:
\(30 = 2(4)^{5/2} - 8(4)^{1/2} + C\)
\(30 = 2(32) - 8(2) + C\)
\(30 = 64 - 16 + C\)
\(30 = 48 + C \implies C = -18\).
So, \(\mathrm{f}(x) = 2x^{5/2} - 8x^{1/2} - 18\).

(b) The gradient of the tangent to \(C\) at \(P(4, 30)\) is given by \(\mathrm{f}'(4)\):
\(\mathrm{f}'(4) = 5(4)\sqrt{4} - \frac{4}{\sqrt{4}} = 40 - 2 = 38\).
Using the point-slope form with gradient \(38\) and point \((4, 30)\):
\(y - 30 = 38(x - 4)\)
\(y - 30 = 38x - 152 \implies y = 38x - 122\).

(a)
M1: Integrates \(\mathrm{f}'(x)\) by increasing at least one fractional power by \(1\).
A1: Integrates at least one term correctly: \(\frac{5}{5/2}x^{5/2} = 2x^{5/2}\) or \(-\frac{4}{1/2}x^{1/2} = -8x^{1/2}\).
A1: Fully correct integrated terms, including the constant of integration \(+ C\): \(\mathrm{f}(x) = 2x^{5/2} - 8x^{1/2} + C\).
M1: Substitutes \(x = 4\) and \(y = 30\) into their integrated expression to find \(C\).
A0.8: Correct function \(\mathrm{f}(x) = 2x^{5/2} - 8x^{1/2} - 18\).

(b)
M1: Substitutes \(x = 4\) into \(\mathrm{f}'(x)\) to find the gradient \(m = 38\), and attempts to write the equation of the tangent.
A1: Correct equation of the tangent: \(y = 38x - 122\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation in terms of \(\cos \theta\):
\(6(1 - \cos^2 \theta) + 5 \cos \theta - 5 = 0\)
\(6 - 6 \cos^2 \theta + 5 \cos \theta - 5 = 0\)
\(-6 \cos^2 \theta + 5 \cos \theta + 1 = 0\)
\(6 \cos^2 \theta - 5 \cos \theta - 1 = 0\).

Factorise the quadratic equation:
\((6 \cos \theta + 1)(\cos \theta - 1) = 0\).

This gives two possible solutions for \(\cos \theta\):
1) \(\cos \theta = 1\)
In the interval \(0 \le \theta < 360^\circ\), this gives:
\(\theta = 0^\circ\).

2) \(\cos \theta = -\frac{1}{6}\)
The principal value is \(\theta = \arccos\left(-\frac{1}{6}\right) \approx 99.59^\circ\).
The second solution in the given interval is:
\(\theta = 360^\circ - 99.59^\circ \approx 260.41^\circ\).

Rounding to one decimal place, the solutions are:
\(\theta = 0^\circ, 99.6^\circ, 260.4^\circ\).

M1: Substitutes \(\sin^2\theta = 1 - \cos^2\theta\) to form a quadratic equation in \(\cos\theta\).
A1: Obtains the correct quadratic equation: \(6 \cos^2 \theta - 5 \cos \theta - 1 = 0\) (or equivalent).
M1: Solves their quadratic equation in \(\cos \theta\) by factorising or using the quadratic formula.
A1: Obtains both \(\cos \theta = 1\) and \(\cos \theta = -\frac{1}{6}\).
B1: Identifies \(\theta = 0^\circ\) (allow \(0\); ignore inclusion of \(360^\circ\)).
M1: Uses a correct method to find at least one value of \(\theta\) in the range from \(\cos \theta = -\frac{1}{6}\).
A0.8: Both \(\theta = 99.6^\circ\) and \(\theta = 260.4^\circ\) correct (to 1 d.p.) and no extra solutions in range.

\( \textbf{Part (a)} \)

Rearranging the equation of \( l_1 \) into the form \( y = mx + c \):
\[ 2y = 3x + 8 \implies y = \frac{3}{2}x + 4 \]
So the gradient of \( l_1 \) is \( m_1 = \frac{3}{2} \).

Since \( l_2 \) is perpendicular to \( l_1 \), its gradient \( m_2 \) satisfies:
\[ m_2 = -\frac{1}{m_1} = -\frac{2}{3} \]

Using the point-gradient formula with \( A(-2, 1) \):
\[ y - 1 = -\frac{2}{3}(x + 2) \]
Multiply both sides by 3:
\[ 3(y - 1) = -2(x + 2) \]
\[ 3y - 3 = -2x - 4 \]
Rearranging into the form \( ax + by + c = 0 \):
\[ 2x + 3y + 1 = 0 \]

\( \textbf{Part (b)} \)

To find the coordinates of \( B \), we set \( x = 0 \) in the equation of \( l_1 \):
\[ 3(0) - 2y + 8 = 0 \implies 2y = 8 \implies y = 4 \]
So \( B \) has coordinates \( (0, 4) \).

To find the coordinates of \( C \), we set \( x = 0 \) in the equation of \( l_2 \):
\[ 2(0) + 3y + 1 = 0 \implies 3y = -1 \implies y = -\frac{1}{3} \]
So \( C \) has coordinates \( \left(0, -\frac{1}{3}\right) \).

The length of the vertical base \( BC \) is:
\[ BC = y_B - y_C = 4 - \left(-\frac{1}{3}\right) = \frac{13}{3} \]

The perpendicular height of the triangle \( ABC \) from the line \( BC \) (the \( y \)-axis) to the vertex \( A(-2, 1) \) is the absolute value of the \( x \)-coordinate of \( A \):
\[ \text{height} = |-2| = 2 \]

Therefore, the area of triangle \( ABC \) is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{13}{3} \times 2 = \frac{13}{3} \]

\( \textbf{Part (a)} \)
- **M1**: Attempts to find the gradient of \( l_1 \) and uses \( m_1 m_2 = -1 \) to find the gradient of \( l_2 \).
- **M1**: Correct straight line equation method using their perpendicular gradient and the point \( (-2, 1) \).
- **A1**: Correct equation in the required form, e.g. \( 2x + 3y + 1 = 0 \) (or any non-zero integer multiple of this equation).

\( \textbf{Part (b)} \)
- **B1**: Identifies the \( y \)-coordinates of \( B \) and \( C \) correctly: \( y = 4 \) and \( y = -\frac{1}{3} \).
- **M1**: Correct method to find the length of \( BC \), i.e., subtracting the \( y \)-coordinates: \( 4 - (-\frac{1}{3}) \).
- **M1**: Correct method for the area of the triangle using \( \frac{1}{2} \times \text{their } BC \times 2 \).
- **A1**: Correct exact area of \( \frac{13}{3} \) (or \( 4\frac{1}{3} \) or recurring \( 4.\dot{3} \)).

\( \textbf{Part (a)} \)

Differentiating \( y = x^3 - 5x^2 + 3x + 7 \) term by term with respect to \( x \):
\[ \frac{\text{d}y}{\text{d}x} = 3x^2 - 10x + 3 \]

\( \textbf{Part (b)} \)

For \( y \) to be a decreasing function, we need:
\[ \frac{\text{d}y}{\text{d}x} < 0 \implies 3x^2 - 10x + 3 < 0 \]

Factorising the quadratic expression:
\[ (3x - 1)(x - 3) < 0 \]

The critical values are \( x = \frac{1}{3} \) and \( x = 3 \).
Since the inequality is \( < 0 \), the solution is the region between the critical values:
\[ \frac{1}{3} < x < 3 \]
(Note: \( \frac{1}{3} \le x \le 3 \) is also accepted).

\( \textbf{Part (c)} \)

First, find the gradient of the tangent at \( P \) (where \( x = 1 \)):
\[ \left.\frac{\text{d}y}{\text{d}x}\right|_{x=1} = 3(1)^2 - 10(1) + 3 = 3 - 10 + 3 = -4 \]

Since the tangent at \( Q \) is parallel to the tangent at \( P \), the gradient at \( Q \) must also be \( -4 \):
\[ 3x^2 - 10x + 3 = -4 \]
\[ 3x^2 - 10x + 7 = 0 \]

Factorising this quadratic equation:
\[ (3x - 7)(x - 1) = 0 \]

This gives \( x = \frac{7}{3} \) or \( x = 1 \).
Since \( Q \) is distinct from \( P \), the \( x \)-coordinate of \( Q \) is \( x = \frac{7}{3} \).

\( \textbf{Part (a)} \)
- **M1**: Differentiates to find at least two terms correct of the form \( A x^2 + B x + C \).
- **A1**: Fully correct derivative: \( 3x^2 - 10x + 3 \).

\( \textbf{Part (b)} \)
- **M1**: Sets their \( \frac{\text{d}y}{\text{d}x} < 0 \) (or \( \le 0 \)) and attempts to find critical values by factorising or using the quadratic formula.
- **M1**: Solves the quadratic inequality to find an inside interval (e.g. \( a < x < b \)) for their critical values.
- **A1**: Correct set of values: \( \frac{1}{3} < x < 3 \) (or \( \left(\frac{1}{3}, 3\right) \) or \( \frac{1}{3} \le x \le 3 \)).

\( \textbf{Part (c)} \)
- **M1**: Substitutes \( x = 1 \) into their \( \frac{\text{d}y}{\text{d}x} \) to find the gradient at \( P \).
- **M1**: Sets their \( \frac{\text{d}y}{\text{d}x} \) equal to their gradient at \( P \) and attempts to solve the resulting quadratic equation.
- **A1**: Obtains \( x = \frac{7}{3} \) (or \( 2.33 \) rounded to 3 significant figures, accept \( 2\frac{1}{3} \)), rejecting \( x = 1 \).

Using the trigonometric identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we can substitute this into the equation:
\[ 6(1 - \cos^2 \theta) - \cos \theta - 4 = 0 \]
\[ 6 - 6\cos^2 \theta - \cos \theta - 4 = 0 \]
\[ -6\cos^2 \theta - \cos \theta + 2 = 0 \]

Multiply by \( -1 \) to write as a standard quadratic equation:
\[ 6\cos^2 \theta + \cos \theta - 2 = 0 \]

Let \( u = \cos \theta \). Then:
\[ 6u^2 + u - 2 = 0 \]

We can factorise this quadratic:
\[ (3u + 2)(2u - 1) = 0 \]

This gives two possible values for \( \cos \theta \):
\[ \cos \theta = -\frac{2}{3} \quad \text{or} \quad \cos \theta = \frac{1}{2} \]

**Case 1: \( \cos \theta = \frac{1}{2} \)**
Within the range \( 0 \le \theta < 360^\circ \):
\[ \theta = 60^\circ \]
The second solution is:
\[ \theta = 360^\circ - 60^\circ = 300^\circ \]

**Case 2: \( \cos \theta = -\frac{2}{3} \)**
Using a calculator to find the principal value:
\[ \theta = \arccos\left(-\frac{2}{3}\right) \approx 131.81^\circ \]
The second solution is:
\[ \theta = 360^\circ - 131.81^\circ \approx 228.19^\circ \]

Rounding all solutions to one decimal place, we obtain:
\[ \theta = 60^\circ, 131.8^\circ, 228.2^\circ, 300^\circ \]

- **M1**: Uses the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to form an equation in terms of \( \cos \theta \) only.
- **A1**: Correct 3-term quadratic equation, e.g. \( 6\cos^2 \theta + \cos \theta - 2 = 0 \).
- **M1**: Solves their quadratic equation to find two values for \( \cos \theta \) (must show working if using a calculator, or show factorisation/formula usage).
- **A1**: Identifies both correct values: \( \cos \theta = \frac{1}{2} \) and \( \cos \theta = -\frac{2}{3} \).
- **M1**: Finds at least one correct angle for \( \cos \theta = -\frac{2}{3} \), showing a correct method (e.g. \( \arccos(-2/3) \) or \( 180^\circ - \arccos(2/3) \)).
- **A1**: All four correct angles: \( \theta = 60^\circ \), \( 131.8^\circ \), \( 228.2^\circ \), \( 300^\circ \) (with no other values within the range). Allow minor rounding errors if working is shown.

**(a)** First, find the gradient of the line \(l_1\) passing through \(A(-2, 5)\) and \(B(4, -3)\): \(m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3}\). Now, use the equation of a straight line \(y - y_1 = m_1(x - x_1)\) with point \(A(-2, 5)\): \(y - 5 = -\frac{4}{3}(x + 2)\). Multiply both sides by 3: \(3(y - 5) = -4(x + 2)\) which simplifies to \(3y - 15 = -4x - 8\). Rearranging gives: \(4x + 3y - 7 = 0\). **(b)** First, find the midpoint \(M\) of the line segment \(AB\): \(M = \left(\frac{-2 + 4}{2}, \frac{5 - 3}{2}\right) = (1, 1)\). Since \(l_2\) is perpendicular to \(l_1\), its gradient \(m_2\) satisfies: \(m_2 = -\frac{1}{m_1} = -\frac{1}{-4/3} = \frac{3}{4}\). Using the point-slope form with the midpoint \(M(1, 1)\) and gradient \(m_2 = \frac{3}{4}\): \(y - 1 = \frac{3}{4}(x - 1)\) which simplifies to \(y = \frac{3}{4}x - \frac{3}{4} + 1\), leading to \(y = \frac{3}{4}x + \frac{1}{4}\).

**(a)** * **M1**: Attempt to find the gradient of \(l_1\) using \((y_2 - y_1)/(x_2 - x_1)\). * **M1**: Correct method to find the equation of \(l_1\) using their gradient and either point \(A\) or point \(B\). * **A1**: Correct equation in the form \(ax + by + c = 0\) with integer coefficients, e.g. \(4x + 3y - 7 = 0\) (or any non-zero integer multiple). **(b)** * **M1**: Attempt to find the midpoint of \(AB\) by averaging coordinates. * **B1**: Midpoint is \((1, 1)\). * **M1**: Uses the perpendicular gradient rule \(m_2 = -1/m_1\) with their gradient from part (a) and attempts to find the equation of the line passing through their midpoint. * **A1**: Correct equation in the form \(y = mx + c\), i.e. \(y = \frac{3}{4}x + \frac{1}{4}\).

First, rewrite the derivative \(\frac{\text{d}y}{\text{d}x}\) in index form by dividing each term in the numerator by \(\sqrt{x} = x^{1/2}\): \(\frac{\text{d}y}{\text{d}x} = \frac{3x^2}{x^{1/2}} - \frac{8}{x^{1/2}} = 3x^{3/2} - 8x^{-1/2}\). Now, integrate each term with respect to \(x\) to find \(y\): \(y = \int (3x^{3/2} - 8x^{-1/2}) \text{d}x = \frac{3}{5/2}x^{5/2} - \frac{8}{1/2}x^{1/2} + C = \frac{6}{5}x^{5/2} - 16x^{1/2} + C\). We are given that the curve passes through the point \((4, 15)\). Substitute \(x = 4\) and \(y = 15\) to find the constant of integration \(C\): \(15 = \frac{6}{5}(4)^{5/2} - 16(4)^{1/2} + C\). Since \(4^{5/2} = 32\) and \(4^{1/2} = 2\), we have: \(15 = \frac{6}{5}(32) - 16(2) + C \implies 15 = 38.4 - 32 + C \implies 15 = 6.4 + C \implies C = 8.6 = \frac{43}{5}\). Thus, the equation of the curve \(C\) is: \(y = \frac{6}{5}x^{5/2} - 16x^{1/2} + \frac{43}{5}\).

* **M1**: Attempt to expand/split the fraction into index form with at least one correct index. * **M1**: Attempt to integrate at least one term of the form \(Ax^k\) where \(k \ne -1\). * **A1**: One correct integrated term, either \(\frac{6}{5}x^{5/2}\) or \(-16x^{1/2}\). * **A1**: Both integrated terms correct (constant of integration not required for this mark). * **M1**: Substitutes \(x = 4\) and \(y = 15\) into an integrated expression containing a constant of integration \(C\) to find its value. * **A1**: Correct value of \(C = 8.6\) or \(\frac{43}{5}\). * **A1**: Correct simplified equation \(y = \frac{6}{5}x^{5/2} - 16x^{1/2} + \frac{43}{5}\) or equivalent.

Use the identity \(\sin^2\theta = 1 - \cos^2\theta\) to rewrite the equation in terms of \(\cos\theta\) only: \(3(1 - \cos^2\theta) - 5\cos\theta - 1 = 0 \implies 3 - 3\cos^2\theta - 5\cos\theta - 1 = 0 \implies 3\cos^2\theta + 5\cos\theta - 2 = 0\). Factor the quadratic equation: \((3\cos\theta - 1)(\cos\theta + 2) = 0\). This yields: \(\cos\theta = \frac{1}{3}\) or \(\cos\theta = -2\). Since \(\cos\theta = -2\) has no real solutions, we solve \(\cos\theta = \frac{1}{3}\) for \(0 \le \theta < 360^\circ\). The principal solution is \(\theta = \arccos(1/3) \approx 70.5288^\circ \approx 70.5^\circ\). The second solution is \(\theta = 360^\circ - 70.5288^\circ \approx 289.4712^\circ \approx 289.5^\circ\). Therefore, the solutions are \(\theta = 70.5^\circ\) and \(\theta = 289.5^\circ\).

* **M1**: Uses the identity \(\sin^2\theta = 1 - \cos^2\theta\) to form an equation in terms of \(\cos\theta\) only. * **A1**: Correct quadratic equation, e.g., \(3\cos^2\theta + 5\cos\theta - 2 = 0\). * **M1**: Attempt to solve their quadratic equation in \(\cos\theta\) by factoring, formula, or calculator. * **A1**: Correct value \(\cos\theta = \frac{1}{3}\). * **B1**: Explains or shows that \(\cos\theta = -2\) has no solutions. * **M1**: Finds one correct solution for \(\theta\) in the range using \(\arccos(\text{their positive root})\). * **A1**: Both \(\theta = 70.5^\circ\) and \(\theta = 289.5^\circ\) and no other solutions in the interval.

(a) First, find the gradient of \(l_1\):
\[m_1 = \frac{7 - 2}{6 - (-4)} = \frac{5}{10} = \frac{1}{2}\]
Now, use the line equation formula with point \(A(-4, 2)\):
\[y - 2 = \frac{1}{2}(x - (-4))\]
\[2(y - 2) = x + 4\]
\[2y - 4 = x + 4\]
Rearranging into the form \(ax + by + c = 0\):
\[x - 2y + 8 = 0\]
(Any integer multiple, such as \(2y - x - 8 = 0\), is also acceptable).

(b) Since \(l_2\) is perpendicular to \(l_1\), its gradient is the negative reciprocal:
\[m_2 = -\frac{1}{m_1} = -2\]
Using the point \(C(4, 1)\), the equation of \(l_2\) is:
\[y - 1 = -2(x - 4)\]
\[y = -2x + 9\]
To find the point of intersection, substitute \(y = -2x + 9\) into the equation for \(l_1\):
\[x - 2(-2x + 9) + 8 = 0\]
\[x + 4x - 18 + 8 = 0\]
\[5x - 10 = 0\]
\[5x = 10 \implies x = 2\]
Substitute \(x = 2\) back into the equation for \(l_2\):
\[y = -2(2) + 9 = 5\]
Thus, the coordinates of the point of intersection are \((2, 5)\).

(a)
- M1: Attempt to find the gradient of \(l_1\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
- M1: Uses their gradient and one of the points \(A\) or \(B\) to form a linear equation.
- A1: Correct equation in the form \(ax + by + c = 0\) with integer coefficients, e.g., \(x - 2y + 8 = 0\) or \(-x + 2y - 8 = 0\).

(b)
- M1: Identifies the gradient of \(l_2\) as \(-2\) (the negative reciprocal of their gradient from part a) and attempts to find the equation of \(l_2\) using point \(C(4, 1)\).
- A1: Correct equation for \(l_2\), e.g., \(y = -2x + 9\) or \(2x + y - 9 = 0\).
- M1: Attempts to solve the simultaneous equations for \(l_1\) and \(l_2\) to find \(x\) or \(y\).
- A1: Correct coordinates \((2, 5)\) written as an ordered pair or explicitly stated.

(a) We use the trigonometric identity:
\[\sin^2 x = 1 - \cos^2 x\]
Substitute this into the given equation:
\[6(1 - \cos^2 x) - 5 \cos x - 5 = 0\]
\[6 - 6 \cos^2 x - 5 \cos x - 5 = 0\]
\[-6 \cos^2 x - 5 \cos x + 1 = 0\]
Multiply the entire equation by \(-1\):
\[6 \cos^2 x + 5 \cos x - 1 = 0\]
which is the required form.

(b) We solve the quadratic equation:
\[6 \cos^2 x + 5 \cos x - 1 = 0\]
Factorising the quadratic expression:
\[(6 \cos x - 1)(\cos x + 1) = 0\]
This gives two possible solutions for \(\cos x\):
\[\cos x = \frac{1}{6} \quad \text{or} \quad \cos x = -1\]
Case 1: \(\cos x = \frac{1}{6}\)
\[x = \arccos\left(\frac{1}{6}\right) \approx 80.4059^\circ \approx 80.4^\circ\]
The other solution in the range \(0 \le x < 360^\circ\) is:
\[x = 360^\circ - 80.4059^\circ \approx 279.5941^\circ \approx 279.6^\circ\]

Case 2: \(\cos x = -1\)
\[x = 180^\circ\]

Combining all solutions within the range, we get:
\[x = 80.4^\circ, 180^\circ, 279.6^\circ\]

(a)
- M1: Uses \(\sin^2 x = 1 - \cos^2 x\) to express the equation solely in terms of \(\cos x\).
- A1*: Completes the algebraic steps correctly with no errors or omissions to reach the given result.

(b)
- M1: Solves the quadratic equation to obtain two values for \(\cos x\). Accept factorisation, quadratic formula, or completing the square.
- A1: Obtains \(\cos x = \frac{1}{6}\) and \(\cos x = -1\).
- B1: Obtains \(x = 180^\circ\).
- M1: Calculates one correct angle from \(\cos x = \frac{1}{6}\) (e.g. \(x \approx 80.4^\circ\)), and attempts to find a second angle in the fourth quadrant by calculating \(360^\circ - \text{their primary angle}\).
- A1: Obtains \(80.4^\circ\) and \(279.6^\circ\) (must be rounded to 1 decimal place). Deduct 1 mark if there are extra solutions within the range.

**(a)**
Using the formula for the sum to infinity of a geometric series:
\[ S_{\infty} = \frac{a}{1-r} = 9 \implies a = 9(1-r) \]

Using the formula for the sum of the first two terms:
\[ S_2 = a + ar = a(1+r) = 5 \]

Substitute the expression for \( a \) from the sum to infinity into the sum of the first two terms:
\[ 9(1-r)(1+r) = 5 \]

Since \( (1-r)(1+r) = 1 - r^2 \):
\[ 9(1 - r^2) = 5 \]
\[ 9 - 9r^2 = 5 \]
\[ 9r^2 = 4 \] (as required).

**(b)**
Solve the equation for \( r \):
\[ r^2 = \frac{4}{9} \implies r = \pm\frac{2}{3} \]

Both values are valid since \( |r| < 1 \).

Find the corresponding values of \( a \) using \( a = 9(1-r) \):
- For \( r = \frac{2}{3} \):
\[ a = 9\left(1 - \frac{2}{3}\right) = 9\left(\frac{1}{3}\right) = 3 \]
- For \( r = -\frac{2}{3} \):
\[ a = 9\left(1 - \left(-\frac{2}{3}\right)\right) = 9\left(\frac{5}{3}\right) = 15 \]

So the two possible pairs of values are \( r = \frac{2}{3}, a = 3 \) and \( r = -\frac{2}{3}, a = 15 \).

**(a)**
- **M1**: Uses \( S_{\infty} = 9 \) to write \( a \) in terms of \( r \), i.e., \( a = 9(1-r) \) or equivalent.
- **M1**: Writes a correct equation for \( S_2 \), i.e., \( a + ar = 5 \) or \( a(1+r) = 5 \).
- **dM1**: Substitutes \( a \) into their \( S_2 \) equation to form a single equation in \( r \) (dependent on at least one of the previous M marks).
- **A1**: Obtains \( 9r^2 = 4 \) correctly with no errors in working.

**(b)**
- **M1**: Solves \( 9r^2 = 4 \) to find at least one value of \( r \).
- **A1**: Both \( r = \frac{2}{3} \) and \( r = -\frac{2}{3} \) (or written as \( \pm\frac{2}{3} \)).
- **M1**: Attempts to find at least one value of \( a \) using a valid equation.
- **A1**: Both correct pairs: \( r = \frac{2}{3}, a = 3 \) and \( r = -\frac{2}{3}, a = 15 \).

**(a)**
Use the change of base rule to express \( \log_4(x+2) \) in base 2:
\[ \log_4(x+2) = \frac{\log_2(x+2)}{\log_2 4} = \frac{\log_2(x+2)}{2} = \frac{1}{2}\log_2(x+2) \]

Substitute this back into the original equation:
\[ \log_2(x-6) - \frac{1}{2}\log_2(x+2) = 1 \]

Multiply the entire equation by 2:
\[ 2\log_2(x-6) - \log_2(x+2) = 2 \]

Apply the power law of logarithms:
\[ \log_2((x-6)^2) - \log_2(x+2) = 2 \]

Apply the division law of logarithms:
\[ \log_2\left( \frac{(x-6)^2}{x+2} \right) = 2 \]

Convert the logarithmic equation into exponential form:
\[ \frac{(x-6)^2}{x+2} = 2^2 \]
\[ \frac{(x-6)^2}{x+2} = 4 \]

Multiply both sides by \( (x+2) \):
\[ (x-6)^2 = 4(x+2) \]

Expand both sides:
\[ x^2 - 12x + 36 = 4x + 8 \]

Rearrange to form a quadratic equation:
\[ x^2 - 16x + 28 = 0 \] (as required).

**(b)**
Solve the quadratic equation:
\[ x^2 - 16x + 28 = 0 \implies (x-14)(x-2) = 0 \implies x = 14 \text{ or } x = 2 \]

Now check constraints for the logarithms in the original equation:
- For \( \log_2(x-6) \) to be defined, we require \( x - 6 > 0 \implies x > 6 \).
- For \( \log_4(x+2) \) to be defined, we require \( x + 2 > 0 \implies x > -2 \).

Since \( x = 2 \) does not satisfy \( x > 6 \), we must reject \( x = 2 \).

Thus, the only valid solution is \( x = 14 \).

**(a)**
- **M1**: Uses the change of base formula correctly to write \( \log_4(x+2) \) as \( \frac{1}{2}\log_2(x+2) \) or equivalent transition.
- **M1**: Combines the logarithms using logarithm laws after handling coefficients (e.g., obtaining \( \log_2\left(\frac{(x-6)^2}{x+2}\right) \) or equivalent).
- **M1**: Correctly removes the logarithms by converting to base-2 index form (e.g., \( \frac{(x-6)^2}{x+2} = 2^2 \)).
- **M1**: Multiplies out and expands \( (x-6)^2 \) to obtain a quadratic expression.
- **A1**: Correctly shows \( x^2 - 16x + 28 = 0 \) with all intermediate steps clearly laid out.

**(b)**
- **M1**: Attempts to solve the quadratic \( x^2 - 16x + 28 = 0 \) by factoring, completing the square, or using the quadratic formula.
- **A1**: Identifies both potential solutions: \( x = 14 \) and \( x = 2 \).
- **B1**: Explains why \( x = 2 \) is rejected (e.g., because \( x-6 > 0 \) is required) and concludes that the only valid solution is \( x = 14 \).

**(a)**
To find the points of intersection, equate the two equations:
\[ 8x - 2x^2 = 2x \]

Rearrange to form a quadratic equation equal to zero:
\[ 2x^2 - 6x = 0 \]
\[ 2x(x - 3) = 0 \]

This gives \( x = 0 \) or \( x = 3 \).

Substitute these values back into \( y = 2x \) to find the corresponding \( y \)-coordinates:
- For \( x = 0 \): \( y = 2(0) = 0 \implies (0, 0) \)
- For \( x = 3 \): \( y = 2(3) = 6 \implies (3, 6) \)

So the points of intersection are \( (0, 0) \) and \( (3, 6) \).

**(b)**
The finite region is bounded between \( x = 0 \) and \( x = 3 \). In this interval, the curve \( C \) lies above the line \( L \).

The area \( A \) of the region is given by the definite integral:
\[ A = \int_{0}^{3} ((8x - 2x^2) - 2x) \, dx \]
\[ A = \int_{0}^{3} (6x - 2x^2) \, dx \]

Integrate each term with respect to \( x \):
\[ A = \left[ 3x^2 - \frac{2}{3}x^3 \right]_{0}^{3} \]

Evaluate at the upper limit \( x = 3 \):
\[ \left( 3(3)^2 - \frac{2}{3}(3)^3 \right) = 27 - 18 = 9 \]

Evaluate at the lower limit \( x = 0 \):
\[ 3(0)^2 - \frac{2}{3}(0)^3 = 0 \]

Subtract the lower limit value from the upper limit value:
\[ A = 9 - 0 = 9 \]

So the exact area of the finite region is 9.

**(a)**
- **M1**: Equates the curve and line equations to set up a quadratic equation in \( x \).
- **A1**: Solves the quadratic to find both \( x = 0 \) and \( x = 3 \).
- **A1**: Obtains both correct coordinate pairs: \( (0, 0) \) and \( (3, 6) \).

**(b)**
- **M1**: Sets up a definite integral of the form \( \int_{a}^{b} (y_C - y_L) \, dx \) with their limits \( a \) and \( b \) from part (a).
- **A1**: Obtains a correct simplified integrand, e.g., \( \int (6x - 2x^2) \, dx \) (condone missing limits or \( dx \) for this mark).
- **M1**: Integrates to find an expression of the form \( px^2 - qx^3 \) where \( p, q \neq 0 \).
- **A1**: Correct integrated expression: \( 3x^2 - \frac{2}{3}x^3 \) (or equivalent).
- **A1**: Substitutes limits \( 3 \) and \( 0 \) correctly to obtain \( 9 \) (the final answer must be supported by clear integration steps).

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

Converting the equation from logarithmic to exponential form:
\(\frac{2x+1}{x-2} = 3^2\)
\(\frac{2x+1}{x-2} = 9\)

Solving the linear equation:
\(2x + 1 = 9(x - 2)\)
\(2x + 1 = 9x - 18\)
\(7x = 19\)
\(x = \frac{19}{7}\)

(Since \(x = \frac{19}{7} \approx 2.71 > 2\), the terms inside the logarithms remain positive, so this is a valid solution).

(b) Rewrite the given equation as:
\(2 \cdot (2^y)^2 - 15(2^y) - 8 = 0\)

Let \(u = 2^y\). The equation becomes:
\(2u^2 - 15u - 8 = 0\)

Factorising the quadratic equation:
\((2u + 1)(u - 8) = 0\)

This gives:
\(u = -\frac{1}{2}\) or \(u = 8\)

Since \(u = 2^y\) and \(2^y > 0\) for all real \(y\), the solution \(2^y = -\frac{1}{2}\) has no real solutions and is rejected.

Therefore:
\(2^y = 8\)
\(2^y = 2^3 \implies y = 3\)

**Part (a)**
* **M1**: Uses the subtraction law of logarithms correctly to obtain \(\log_3\left(\frac{2x+1}{x-2}\right) = 2\).
* **M1**: Converts the logarithmic equation into the correct linear form \(\frac{2x+1}{x-2} = 3^2\) or \(\frac{2x+1}{x-2} = 9\).
* **A1**: Correct exact solution \(x = \frac{19}{7}\) (or equivalent, e.g., \(2\frac{5}{7}\)). Do not accept decimals unless rounded to 3 significant figures after showing the exact form.

**Part (b)**
* **M1**: Writes the equation in the form \(2(2^y)^2 - 15(2^y) - 8 = 0\) or uses a suitable substitution such as \(u = 2^y\) to form \(2u^2 - 15u - 8 = 0\).
* **M1**: Solves the quadratic equation to find two values for \(2^y\) (or \(u\)), e.g., by factorising to \((2u + 1)(u - 8) = 0\) or using the quadratic formula.
* **A1**: Identifies \(2^y = 8\) and rejects the negative root with a valid reason (e.g., \(2^y > 0\) or \(\log_2(-1/2)\) is undefined).
* **A1.5**: Deduces that \(y = 3\) is the only real solution.

(a) To find the intersection points, equate the two equations:
\(9x - x^3 = 5x\)
\(x^3 - 4x = 0\)
\(x(x^2 - 4) = 0\)
\(x(x - 2)(x + 2) = 0\)

This gives \(x = -2\), \(x = 0\), and \(x = 2\).
For the first quadrant (including the origin, where \(x \ge 0\) and \(y \ge 0\)):
When \(x = 0\), \(y = 5(0) = 0 \implies (0, 0)\)
When \(x = 2\), \(y = 5(2) = 10 \implies (2, 10)\)

The coordinates of the intersection points in the first quadrant are \((0, 0)\) and \((2, 10)\).

(b) The region \(R\) is bounded above by the curve \(y = 9x - x^3\) and below by the line \(y = 5x\) between \(x = 0\) and \(x = 2\).

The area \(A\) of the region \(R\) is given by:
\(A = \int_{0}^{2} ((9x - x^3) - 5x) \, dx\)
\(A = \int_{0}^{2} (4x - x^3) \, dx\)

Integrating term by term:
\(A = \left[ 2x^2 - \frac{1}{4}x^4 \right]_{0}^{2}\)

Substituting the upper and lower limits:
\(A = \left( 2(2)^2 - \frac{1}{4}(2)^4 \right) - (0)\)
\(A = \left( 8 - 4 \right) - 0\)
\(A = 4\)

Alternatively, using the area under the curve minus the area of the triangle under the line:
\(\int_{0}^{2} (9x - x^3) \, dx = \left[ \frac{9}{2}x^2 - \frac{1}{4}x^4 \right]_{0}^{2} = \left( 18 - 4 \right) - 0 = 14\)
Area of the triangle: \(\frac{1}{2} \times 2 \times 10 = 10\)
Area \(R = 14 - 10 = 4\).

**Part (a)**
* **M1**: Equates \(9x - x^3\) and \(5x\) and attempts to factorise or solve the resulting cubic equation.
* **A1**: Correctly identifies the \(x\)-coordinates of interest as \(x = 0\) and \(x = 2\). (Do not penalise if \(x = -2\) is also written, provided it is not selected as part of the final answer in the first quadrant).
* **A0.5**: Finds the corresponding coordinates \((0, 0)\) and \((2, 10)\) correctly.

**Part (b)**
* **M1**: Sets up a correct integral for the area of region \(R\), either \(\int_{0}^{2} (4x - x^3) \, dx\) or \(\int_{0}^{2} (9x - x^3) \, dx - \frac{1}{2}(2)(10)\).
* **M1**: Integrates the terms correctly, increasing the power by 1 and dividing by the new power (at least one term integrated correctly).
* **A1**: Correct integrated expression: \(\left[ 2x^2 - \frac{1}{4}x^4 \right]\) (or \(\left[ \frac{9}{2}x^2 - \frac{1}{4}x^4 \right]\)).
* **M1**: Substitutes both limits \(2\) and \(0\) into their integrated expression and subtracts.
* **A1**: Obtains the correct exact area of \(4\).

(a) We use the trigonometric identity:
\(\cos^2 \theta = 1 - \sin^2 \theta\)

Substituting this into the given equation:
\(5(1 - \sin^2 \theta) + 3 \sin \theta - 3 = 0\)

Expanding the brackets:
\(5 - 5 \sin^2 \theta + 3 \sin \theta - 3 = 0\)

Simplifying and rearranging:
\(-5 \sin^2 \theta + 3 \sin \theta + 2 = 0\)

Multiplying the entire equation by \(-1\):
\(5 \sin^2 \theta - 3 \sin \theta - 2 = 0\) (as required)

(b) To solve \(5 \sin^2 \theta - 3 \sin \theta - 2 = 0\), let \(y = \sin \theta\):
\(5y^2 - 3y - 2 = 0\)

Factorising the quadratic equation:
\((5y + 2)(y - 1) = 0\)

This gives:
\(\sin \theta = -\frac{2}{5} = -0.4\) or \(\sin \theta = 1\)

Case 1: \(\sin \theta = 1\)
Within the interval \(0 \le \theta < 360^\circ\):
\(\theta = 90^\circ\)

Case 2: \(\sin \theta = -0.4\)
The principal value is:
\(\theta = \arcsin(-0.4) \approx -23.58^\circ\)

Since \(\sin \theta\) is negative, the solutions lie in the 3rd and 4th quadrants:
\(\theta_1 = 180^\circ - (-23.58^\circ) = 203.58^\circ \approx 203.6^\circ\)
\(\theta_2 = 360^\circ + (-23.58^\circ) = 336.42^\circ \approx 336.4^\circ\)

Thus, the set of solutions in the range \(0 \le \theta < 360^\circ\) is:
\(\theta = 90^\circ, 203.6^\circ, 336.4^\circ\)

**Part (a)**
* **M1**: Substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) into the given equation.
* **A1**: Expands and collects terms to obtain \(-5 \sin^2 \theta + 3 \sin \theta + 2 = 0\) or equivalent.
* **A0.5**: Shows a complete and convincing algebraic path to the final requested equation \(5 \sin^2 \theta - 3 \sin \theta - 2 = 0\) with no errors.

**Part (b)**
* **M1**: Solves the quadratic equation to find two values for \(\sin \theta\), e.g., by factorising to \((5\sin\theta + 2)(\sin\theta - 1) = 0\).
* **A1**: Identifies both correct values \(\sin \theta = 1\) and \(\sin \theta = -0.4\) (or \(-\frac{2}{5}\)).
* **B1**: Identifies \(\theta = 90^\circ\) as one correct solution.
* **M1**: Uses a correct method to find at least one value for \(\theta\) in the range \(0 \le \theta < 360^\circ\) corresponding to \(\sin \theta = -0.4\) (e.g. \(180^\circ - \arcsin(-0.4)\) or \(360^\circ + \arcsin(-0.4)\)).
* **A1**: Obtains both \(\theta \approx 203.6^\circ\) and \(\theta \approx 336.4^\circ\) and no other solutions in the interval. (Subtract 1 mark if extra solutions are given in the range, or if the answers are not rounded to 1 decimal place).

(a)
Starting with the equation:
\( 2\log_3(x-2) - \log_3(x-4) = 2 \)

Apply the power law of logarithms to the first term:
\( \log_3(x-2)^2 - \log_3(x-4) = 2 \)

Apply the subtraction law of logarithms to combine the terms:
\( \log_3\left(\frac{(x-2)^2}{x-4}\right) = 2 \)

Rewrite in exponential form (base 3):
\( \frac{(x-2)^2}{x-4} = 3^2 \)
\( \frac{x^2 - 4x + 4}{x-4} = 9 \)

Multiply both sides by \( (x-4) \):
\( x^2 - 4x + 4 = 9(x-4) \)
\( x^2 - 4x + 4 = 9x - 36 \)

Rearrange to form a quadratic equal to zero:
\( x^2 - 13x + 40 = 0 \) (as required)

(b)
Solve the quadratic equation:
\( x^2 - 13x + 40 = 0 \)
\( (x-5)(x-8) = 0 \)
So \( x = 5 \) or \( x = 8 \).

Validity check:
For the original terms \( \log_3(x-2) \) and \( \log_3(x-4) \) to be defined, their arguments must be strictly positive:
\( x - 2 > 0 \implies x > 2 \)
\( x - 4 > 0 \implies x > 4 \)

Taking the intersection of these conditions, we must have \( x > 4 \).
Since both \( x = 5 \) and \( x = 8 \) satisfy \( x > 4 \), both solutions are valid.

Part (a) [5 Marks]:
- M1: Applies the power law of logarithms to write \( 2\log_3(x-2) \) as \( \log_3(x-2)^2 \).
- M1: Applies the division law of logarithms to combine the terms into \( \log_3\left(\frac{(x-2)^2}{x-4}\right) \).
- M1: Removes the logarithm correctly to get \( \frac{(x-2)^2}{x-4} = 3^2 \) or \( 9 \).
- M1: Expands the numerator to \( x^2 - 4x + 4 \) and multiplies by \( (x-4) \).
- A1*: Fully correct proof with no errors, leading to the given quadratic equation \( x^2 - 13x + 40 = 0 \).

Part (b) [2.5 Marks]:
- M1: Attempts to solve the quadratic equation \( x^2 - 13x + 40 = 0 \) by factorisation, completing the square, or using the quadratic formula.
- A1: Finds both solutions \( x = 5 \) and \( x = 8 \).
- B0.5: Identifies that both solutions are valid because they satisfy \( x > 4 \) (making the arguments of the original logarithms positive).

(a)
Equate the equations of the curve and the line to find their intersection points:
\( x^2 - 5x + 9 = x + 4 \)
\( x^2 - 6x + 5 = 0 \)
\( (x-1)(x-5) = 0 \)
So \( x = 1 \) or \( x = 5 \).

For \( x = 1 \):
\( y = 1 + 4 = 5 \implies P(1, 5) \)

For \( x = 5 \):
\( y = 5 + 4 = 9 \implies Q(5, 9) \)

(b)
To find the area of the bounded region, integrate the difference between the upper function (the line) and the lower function (the curve) between the limits \( x = 1 \) and \( x = 5 \):
\( \text{Area} = \int_{1}^{5} \left( (x + 4) - (x^2 - 5x + 9) \right) \, dx \)
\( \text{Area} = \int_{1}^{5} (-x^2 + 6x - 5) \, dx \)

Integrate term by term:
\( \int (-x^2 + 6x - 5) \, dx = \left[ -\frac{1}{3}x^3 + 3x^2 - 5x \right]_{1}^{5} \)

Substitute the upper limit \( x = 5 \):
\( \left( -\frac{5^3}{3} + 3(5^2) - 5(5) \right) = -\frac{125}{3} + 75 - 25 = 50 - \frac{125}{3} = \frac{25}{3} \)

Substitute the lower limit \( x = 1 \):
\( \left( -\frac{1^3}{3} + 3(1^2) - 5(1) \right) = -\frac{1}{3} + 3 - 5 = -2 - \frac{1}{3} = -\frac{7}{3} \)

Calculate the area:
\( \text{Area} = \frac{25}{3} - \left(-\frac{7}{3}\right) = \frac{32}{3} \)

Part (a) [3 Marks]:
- M1: Equates the curve and line equations to form a quadratic equation in \( x \).
- A1: Solves the quadratic to find \( x = 1 \) and \( x = 5 \) (or finds both \( y \) values \( y = 5 \) and \( y = 9 \)).
- A1: Correctly identifies the coordinates of the two points as \( (1, 5) \) and \( (5, 9) \) (in any order).

Part (b) [4.5 Marks]:
- M1: Sets up a correct integral expression for the area with their limits from part (a).
- M1: Integrates their expression term by term, with at least one term correct.
- A1: Fully correct integration: \( \left[ -\frac{1}{3}x^3 + 3x^2 - 5x \right] \).
- dM1: Substitutes both of their limits from part (a) into their integrated expression and subtracts the lower limit value from the upper limit value. (Dependent on both previous M marks).
- A0.5: Obtains the correct final area of \( \frac{32}{3} \) (or \( 10\frac{2}{3} \) or exact equivalent).

(a)
Using the identity \( \cos^2(2\theta) = 1 - \sin^2(2\theta) \):
\( 6\left(1 - \sin^2(2\theta)\right) - \sin(2\theta) - 5 = 0 \)
\( 6 - 6\sin^2(2\theta) - \sin(2\theta) - 5 = 0 \)
\( 1 - 6\sin^2(2\theta) - \sin(2\theta) = 0 \)

Multiply through by \( -1 \):
\( 6\sin^2(2\theta) + \sin(2\theta) - 1 = 0 \) (as required)

(b)
Let \( y = 2\theta \).
Since \( 0 \le \theta < 180^\circ \), the range for \( y \) is \( 0 \le y < 360^\circ \).

Solve the quadratic equation:
\( 6\sin^2 y + \sin y - 1 = 0 \)
\( (3\sin y - 1)(2\sin y + 1) = 0 \)

This gives:
\( \sin y = \frac{1}{3} \) or \( \sin y = -\frac{1}{2} \)

For \( \sin y = \frac{1}{3} \):
\( y = \arcsin\left(\frac{1}{3}\right) \approx 19.4712^\circ \)
And the second solution in range is:
\( y = 180^\circ - 19.4712^\circ \approx 160.5288^\circ \)

For \( \sin y = -\frac{1}{2} \):
\( y = 180^\circ - (-30^\circ) = 210^\circ \)
And the second solution in range is:
\( y = 360^\circ - 30^\circ = 330^\circ \)

So the possible values for \( y = 2\theta \) are:
\( 2\theta = 19.4712^\circ, \quad 160.5288^\circ, \quad 210^\circ, \quad 330^\circ \)

Divide each by 2 to find \( \theta \):
\( \theta = 9.7356^\circ, \quad 80.2644^\circ, \quad 105^\circ, \quad 165^\circ \)

Rounding to 1 decimal place where appropriate:
\( \theta = 9.7^\circ, \quad 80.3^\circ, \quad 105^\circ, \quad 165^\circ \)

Part (a) [2 Marks]:
- M1: Uses the identity \( \cos^2(2\theta) = 1 - \sin^2(2\theta) \) to substitute into the given equation.
- A1*: Correctly rearranges terms to obtain the given quadratic equation \( 6\sin^2(2\theta) + \sin(2\theta) - 1 = 0 \).

Part (b) [5.5 Marks]:
- M1: Solves the quadratic to find \( \sin(2\theta) = \frac{1}{3} \) and \( \sin(2\theta) = -\frac{1}{2} \).
- M1: Finds at least one correct principal value for \( 2\theta \) (e.g. \( 19.5^\circ \) or \( 210^\circ \) or \( -30^\circ \)).
- A1: Finds at least two correct values for \( 2\theta \) from the set \( \{19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\} \).
- dM1: Halves their values of \( 2\theta \) to find values for \( \theta \). (Dependent on the first M1).
- A1: Finds any two of the final correct answers for \( \theta \) (e.g. \( 9.7^\circ \) or \( 80.3^\circ \) or \( 105^\circ \) or \( 165^\circ \)).
- A0.5: Finds all four correct answers: \( \theta = 9.7^\circ, \, 80.3^\circ, \, 105^\circ, \, 165^\circ \). Deduct 0.5 marks if there are any extra incorrect solutions within the range.

(a) Given the equation:
\[\log_2(y - 1) - \log_2(x) = 2\]

Using the subtraction law of logarithms:
\[\log_2\left(\frac{y - 1}{x}\right) = 2\]

Removing the logarithm by raising both sides as powers of 2:
\[\frac{y - 1}{x} = 2^2\]
\[\frac{y - 1}{x} = 4\]

Multiplying both sides by \(x\):
\[y - 1 = 4x\]
\[y = 4x + 1\quad \text{(as required)}\]

(b) Given the second equation:
\[2^y = 8^{x+1}\]

Since \(8 = 2^3\), we can express both sides with base 2:
\[2^y = (2^3)^{x+1}\]
\[2^y = 2^{3(x+1)}\]
\[2^y = 2^{3x+3}\]

Equating the exponents:
\[y = 3x + 3\]

We now have two simultaneous linear equations:
1) \(y = 4x + 1\)
2) \(y = 3x + 3\)

Equating these two expressions for \(y\):
\[4x + 1 = 3x + 3\]
\[x = 2\]

Substituting \(x = 2\) back into the first equation:
\[y = 4(2) + 1 = 9\]

Thus, the solutions are \(x = 2\) and \(y = 9\).

(a)
- M1: Applies the subtraction law of logarithms to write \(\log_2(y-1) - \log_2(x)\) as \(\log_2\left(\frac{y-1}{x}\right)\), or writes \(\log_2(y-1) = 2 + \log_2(x)\) and combines to \(\log_2(4x)\).
- M1: Eliminates logs correctly to obtain \(\frac{y-1}{x} = 4\) or \(y - 1 = 4x\).
- A1*: Correctly completes the proof to show \(y = 4x + 1\) with no errors or omissions.

(b)
- M1: Expresses \(8^{x+1}\) as \(2^{3x+3}\), or takes \(\log_2\) of both sides to obtain \(y = (x+1)\log_2(8)\).
- A1: Deduces the linear equation \(y = 3x + 3\) or equivalent.
- M1: Equates their two expressions for \(y\) (or substitutes one into the other) to form a single equation in one variable and attempts to solve for \(x\) or \(y\).
- A1: Correct values of \(x = 2\) and \(y = 9\). Both must be present and correct.

**(a)**
For \(x > 3\), \(\text{f}(x) = \frac{2(x-3)+11}{x-3} = 2 + \frac{11}{x-3}\).
Since \(x > 3\), \(x-3 > 0\), so \(\frac{11}{x-3} > 0\).
Therefore, the range of \(\text{f}\) is \(\text{f}(x) > 2\) (or \(y > 2\)).

**(b)**
Let \(y = \frac{2x+5}{x-3}\).
Multiply by \(x-3\):
\(y(x-3) = 2x+5 \implies yx - 3y = 2x+5\)
Rearrange to make \(x\) the subject:
\(yx - 2x = 3y+5 \implies x(y-2) = 3y+5 \implies x = \frac{3y+5}{y-2}\)
Therefore, the inverse function is:
$$\text{f}^{-1}(x) = \frac{3x+5}{x-2}$$
The domain of \(\text{f}^{-1}\) is the range of \(\text{f}\), which is \(x > 2\).

**(c)**
Given \(\text{fg}(x) = \frac{7}{2}\), we have:
$$\text{f}(\text{g}(x)) = \frac{2(x^2+2)+5}{(x^2+2)-3} = \frac{2x^2+9}{x^2-1}$$
Set this equal to \(\frac{7}{2}\):
$$\frac{2x^2+9}{x^2-1} = \frac{7}{2}$$
\(2(2x^2+9) = 7(x^2-1) \implies 4x^2+18 = 7x^2-7 \implies 3x^2 = 25 \implies x^2 = \frac{25}{3}\)
Since the domain of \(\text{g}\) is \(x \ge 1\), we take the positive square root:
\(x = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}\).
This value is valid as \(\text{g}\left(\frac{5\sqrt{3}}{3}\right) = \frac{25}{3} + 2 = \frac{31}{3} > 3\), which is in the domain of \(\text{f}\).

**(a)**
- **M1**: Attempts to find the horizontal asymptote or considers the limits of \(f(x)\) as \(x \to \infty\) and \(x \to 3\).
- **A1**: Correct range: \(\text{f}(x) > 2\) (or \(y > 2\)). Reject \(\text{f}(x) \ge 2\).

**(b)**
- **M1**: Attempts to make \(x\) the subject, e.g., multiplying through by \(x-3\) and expanding.
- **A1**: Correctly expresses \(x\) in terms of \(y\): \(x = \frac{3y+5}{y-2}\) (or vice versa).
- **A1**: Fully correct inverse function with its domain: \(\text{f}^{-1}(x) = \frac{3x+5}{x-2}\) with domain \(x > 2\).

**(c)**
- **M1**: Attempts to form the composite function \(\text{fg}(x)\) or sets \(\text{g}(x) = \text{f}^{-1}\left(\frac{7}{2}\right)\).
- **A1**: Obtains a correct equation in \(x^2\), e.g., \(3x^2 = 25\) or \(x^2+2 = \frac{31}{3}\).
- **A1**: Correct exact value of \(x = \frac{5\sqrt{3}}{3}\) (or equivalent exact surd). Reject negative root.

**(a)**
Using the product rule with \(u = 2x^2 - 3x\) and \(v = \text{e}^{-x}\):
$$\frac{\text{d}u}{\text{d}x} = 4x - 3$$
$$\frac{\text{d}v}{\text{d}x} = -\text{e}^{-x}$$
Then:
$$\frac{\text{d}y}{\text{d}x} = (2x^2-3x)(-\text{e}^{-x}) + (4x-3)\text{e}^{-x}$$
$$\frac{\text{d}y}{\text{d}x} = (-2x^2 + 3x + 4x - 3)\text{e}^{-x} = (-2x^2 + 7x - 3)\text{e}^{-x}$$

**(b)**
At stationary points, \(\frac{\text{d}y}{\text{d}x} = 0\).
Since \(\text{e}^{-x} \neq 0\) for all real \(x\):
$$-2x^2 + 7x - 3 = 0 \implies 2x^2 - 7x + 3 = 0$$
$$(2x - 1)(x - 3) = 0$$
Which gives \(x = \frac{1}{2}\) and \(x = 3\).

To find the corresponding \(y\)-coordinates:
For \(x = \frac{1}{2}\):
$$y = \left(2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right)\right)\text{e}^{-1/2} = \left(\frac{1}{2} - \frac{3}{2}\right)\text{e}^{-1/2} = -\text{e}^{-1/2} = -\frac{1}{\sqrt{\text{e}}}$$

For \(x = 3\):
$$y = (2(3)^2 - 3(3))\text{e}^{-3} = 9\text{e}^{-3}$$

So the exact coordinates of the stationary points are \(\left(\frac{1}{2}, -\text{e}^{-1/2}\right)\) and \(\left(3, 9\text{e}^{-3}\right)\).

**(a)**
- **M1**: Applies the product rule. Award for the form \(A(4x-3)\text{e}^{-x} + B(2x^2-3x)\text{e}^{-x}\) where \(A\) and \(B\) are non-zero constants.
- **A1**: Correct unsimplified derivative: \((4x-3)\text{e}^{-x} - (2x^2-3x)\text{e}^{-x}\).
- **A1**: Fully simplified correct derivative: \((-2x^2+7x-3)\text{e}^{-x}\).

**(b)**
- **M1**: Sets their \(\frac{\text{d}y}{\text{d}x} = 0\), leading to a 3-term quadratic in \(x\) and attempts to solve.
- **A1**: Correct \(x\)-coordinates: \(x = 0.5\) and \(x = 3\).
- **M1**: Substitutes at least one of their \(x\) values back into the original curve equation to find \(y\).
- **A1**: Correct exact coordinates for both points: \(\left(\frac{1}{2}, -\text{e}^{-1/2}\right)\) (or equivalent) and \(\left(3, 9\text{e}^{-3}\right)\).

**(a)**
Using the expansion \(R\sin(2\theta - \alpha) = R\sin 2\theta\cos\alpha - R\cos 2\theta\sin\alpha\):
Comparing coefficients with \(4\sin 2\theta - 3\cos 2\theta\) gives:
\(R\cos\alpha = 4\) and \(R\sin\alpha = 3\).
Squaring and adding:
\(R^2 = 4^2 + 3^2 = 25 \implies R = 5\).
Dividing the equations:
\(\tan\alpha = \frac{3}{4} \implies \alpha = \arctan(0.75) \approx 0.64350 \approx 0.644\) radians.
Therefore, \(4\sin 2\theta - 3\cos 2\theta = 5\sin(2\theta - 0.644)\).

**(b)**
Using the result from part (a):
$$5\sin(2\theta - 0.6435) = 2 \implies \sin(2\theta - 0.6435) = 0.4$$
Let \(X = 2\theta - 0.6435\).
Since \(0 \le \theta < \pi\), we have \(0 \le 2\theta < 2\pi\), which means \(-0.6435 \le X < 2\pi - 0.6435 \approx 5.640\).

Solving \(\sin X = 0.4\):
\(X_1 = \arcsin(0.4) \approx 0.4115\) radians.
\(X_2 = \pi - 0.4115 \approx 2.7301\) radians.

Now solve for \(\theta\):
For \(X_1\):
\(2\theta - 0.6435 = 0.4115 \implies 2\theta = 1.0550 \implies \theta \approx 0.53\) (to 2 d.p.).
For \(X_2\):
\(2\theta - 0.6435 = 2.7301 \implies 2\theta = 3.3736 \implies \theta \approx 1.69\) (to 2 d.p.).

Both solutions are within the interval \(0 \le \theta < \pi\).

**(a)**
- **M1**: Attempts to find \(R\) or \(\alpha\). Award for \(R = \sqrt{4^2+3^2}\) or \(\tan\alpha = \pm\frac{3}{4}\) or \(\tan\alpha = \pm\frac{4}{3}\).
- **A1**: Correct value of \(R = 5\).
- **A1**: Correct value of \(\alpha \approx 0.644\) (accept answers which round to this, e.g., \(0.6435\)).

**(b)**
- **M1**: Sets up the equation \(\sin(2\theta - \text{their }\alpha) = \frac{2}{\text{their } R}\).
- **M1**: Finds at least one correct value for \(2\theta - \alpha\) using \(\arcsin(0.4)\) (approx. \(0.412\)).
- **M1**: Uses the symmetry of the sine graph to find a second value in the interval, e.g., \(\pi - 0.412 = 2.730\).
- **A1**: One correct value of \(\theta\) to 2 d.p. (either \(0.53\) or \(1.69\)).
- **A1**: Both correct values of \(\theta\) to 2 d.p. (\(\theta = 0.53\) and \(\theta = 1.69\)). Lose this mark for any extra solutions within the range.

(a) Using the product rule on \(y = u v\) where \(u = x^2\) and \(v = \ln(3x-2)\): \(\frac{\text{d}u}{\text{d}x} = 2x\) and \(\frac{\text{d}v}{\text{d}x} = \frac{3}{3x - 2}\). Thus, \(\frac{\text{d}y}{\text{d}x} = x^2 \left(\frac{3}{3x - 2}\right) + 2x \ln(3x - 2)\). Substituting \(x = 1\): \(\left.\frac{\text{d}y}{\text{d}x}\right|_{x=1} = (1)^2 \left(\frac{3}{3(1) - 2}\right) + 2(1) \ln(3(1) - 2) = 3 + 2\ln(1) = 3\). So the gradient of the tangent at \(P\) is 3. (b) At \(x = 1\), the \(y\)-coordinate of \(P\) is \(y = 1^2 \ln(3(1) - 2) = 0\). So \(P\) is \((1, 0)\). The gradient of the normal is \(-\frac{1}{3}\). The equation of the normal is \(y - 0 = -\frac{1}{3}(x - 1)\). Multiplying by 3 gives \(3y = -x + 1\). Rearranging into the form \(ax + by + c = 0\) gives \(x + 3y - 1 = 0\) (or any integer multiple).

(a) M1: Applies product rule to differentiate \(x^2 \ln(3x - 2)\) to obtain an expression of the form \(Ax\ln(3x-2) + \frac{Bx^2}{3x-2}\). A1: Correctly differentiates \(\ln(3x-2)\) to get \(\frac{3}{3x-2}\). A1: Correct expression for \(\frac{\text{d}y}{\text{d}x} = 2x \ln(3x - 2) + \frac{3x^2}{3x - 2}\). A1: Substitutes \(x = 1\) to find gradient is 3. (b) B1: Correctly identifies \(y = 0\) when \(x = 1\). M1: Uses the perpendicular gradient rule to find the normal gradient and attempts to form the equation of the normal using their coordinates of \(P\). A1: Correct equation in the form \(x + 3y - 1 = 0\) or equivalent integer form with all terms on one side.

(a) Using the identity \(R\cos(2\theta - \alpha) = R\cos(2\theta)\cos\alpha + R\sin(2\theta)\sin\alpha\), we compare coefficients with \(3\cos(2\theta) + 4\sin(2\theta)\): \(R\cos\alpha = 3\) and \(R\sin\alpha = 4\). Squaring and adding gives \(R^2 = 3^2 + 4^2 = 25\), so \(R = 5\). Dividing the equations gives \(\tan\alpha = \frac{4}{3}\), so \(\alpha = \arctan(4/3) \approx 0.927\) radians (to 3 decimal places). (b) Using part (a), the equation is \(5\cos(2\theta - 0.9273) = 2\), which simplifies to \(\cos(2\theta - 0.9273) = 0.4\). Since \(0 \le \theta < \pi\), the interval for \(2\theta - 0.9273\) is \(-0.9273 \le 2\theta - 0.9273 < 5.3559\). The principal value is \(2\theta - 0.9273 = \arccos(0.4) \approx 1.1593\), leading to \(2\theta = 2.0866 \implies \theta \approx 1.04\). The second value in the interval is \(2\theta - 0.9273 = 2\pi - 1.1593 \approx 5.1239\), leading to \(2\theta = 6.0512 \implies \theta \approx 3.03\).

(a) B1: Correct value \(R = 5\). M1: Attempts to find \(\alpha\) using \(\tan\alpha = \pm \frac{4}{3}\) or equivalent. A1: Correct value \(\alpha \approx 0.927\) (to 3 decimal places). (b) M1: Sets up the equation \(\cos(2\theta - \alpha) = \frac{2}{R}\) using their \(R\). M1: Finds at least one value for \(2\theta - \alpha\) using \(\arccos(0.4)\) (approx. 1.159). A1: Correctly identifies both appropriate angles for \(2\theta - \alpha\) within the interval: approx. 1.159 and 5.124. M1: Solves for \(\theta\) by adding their \(\alpha\) and dividing by 2. A1: Both answers correct: \(\theta \approx 1.04\) and \(\theta \approx 3.03\) (to 2 decimal places).

(a) Evaluating \(f(x)\) at the boundaries: \(f(1.4) = e^{0.6} - 3(1.4)^2 + 5 \approx 1.8221 - 5.88 + 5 = 0.9421\). \(f(1.5) = e^{0.5} - 3(1.5)^2 + 5 \approx 1.6487 - 6.75 + 5 = -0.1013\). Since \(f(x)\) is continuous and there is a change of sign between \(x=1.4\) and \(x=1.5\), there must be a root \(\alpha\) in the interval \([1.4, 1.5]\). (b) Setting \(f(x) = 0\) gives \(e^{2-x} - 3x^2 + 5 = 0 \implies 3x^2 = 5 + e^{2-x}\). Taking the positive square root because \(x > 0\) gives \(x = \sqrt{\frac{5 + e^{2-x}}{3}}\). (c) With \(x_0 = 1.4\): \(x_1 = \sqrt{\frac{5 + e^{0.6}}{3}} \approx 1.5080\). \(x_2 = \sqrt{\frac{5 + e^{2-1.50799...}}{3}} \approx 1.4872\). \(x_3 = \sqrt{\frac{5 + e^{2-1.48723...}}{3}} \approx 1.4911\). (d) The upper and lower bounds for \(1.490\) to 3 decimal places are \(1.4895\) and \(1.4905\). Evaluating \(f(x)\) at these points: \(f(1.4895) = e^{2-1.4895} - 3(1.4895)^2 + 5 \approx 0.0103 > 0\). \(f(1.4905) = e^{2-1.4905} - 3(1.4905)^2 + 5 \approx -0.0003 < 0\). Since \(f(x)\) is continuous and there is a change of sign in the interval \([1.4895, 1.4905]\), the root \(\alpha\) lies within this interval and must be \(1.490\) to 3 decimal places.

(a) M1: Attempts to evaluate both \(f(1.4)\) and \(f(1.5)\). A1: Correct evaluations showing a sign change with a concluding remark mentioning continuity. (b) B1: Correct algebraic derivation. (c) M1: Attempts to compute \(x_1\) using the iterative formula. A1: Correct values for \(x_1 \approx 1.5080\) and \(x_2 \approx 1.4872\) (to 4 d.p.). A1: Correct value for \(x_3 \approx 1.4911\) (to 4 d.p.). (d) M1: Evaluates \(f(1.4895)\) and \(f(1.4905)\). A1: Correctly calculates values to show a change of sign (approx. 0.010 and -0.0003) with a final conclusion.

(a) We use the quotient rule with \(u = \sin(2x)\) and \(v = 2 - \cos(2x)\).
Then:
\[ \frac{\text{d}u}{\text{d}x} = 2\cos(2x) \]
and
\[ \frac{\text{d}v}{\text{d}x} = 2\sin(2x) \]
Applying the quotient rule:
\[ \frac{\text{d}y}{\text{d}x} = \frac{v \frac{\text{d}u}{\text{d}x} - u \frac{\text{d}v}{\text{d}x}}{v^2} \]
\[ \frac{\text{d}y}{\text{d}x} = \frac{(2 - \cos(2x))(2\cos(2x)) - (\sin(2x))(2\sin(2x))}{(2 - \cos(2x))^2} \]
Expand the numerator:
\[ 4\cos(2x) - 2\cos^2(2x) - 2\sin^2(2x) \]
Using the trigonometric identity \(\sin^2(2x) + \cos^2(2x) = 1\):
\[ 4\cos(2x) - 2(\cos^2(2x) + \sin^2(2x)) = 4\cos(2x) - 2 \]
Thus,
\[ \frac{\text{d}y}{\text{d}x} = \frac{4\cos(2x) - 2}{(2 - \cos(2x))^2} \]

(b) For a stationary point, we set \(\frac{\text{d}y}{\text{d}x} = 0\):
\[ \frac{4\cos(2x) - 2}{(2 - \cos(2x))^2} = 0 \]
Since the denominator is non-zero, this requires:
\[ 4\cos(2x) - 2 = 0 \implies \cos(2x) = \frac{1}{2} \]
Since \(0 \le x \le \frac{\pi}{2}\), the range for \(2x\) is \(0 \le 2x \le \pi\).
Thus,
\[ 2x = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \implies x = \frac{\pi}{6} \]
To find the corresponding \(y\)-coordinate, substitute \(x = \frac{\pi}{6}\) into the equation for \(C\):
\[ y = \frac{\sin(2\cdot \frac{\pi}{6})}{2 - \cos(2\cdot \frac{\pi}{6})} = \frac{\sin(\frac{\pi}{3})}{2 - \cos(\frac{\pi}{3})} \]
Since \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\) and \(\cos(\frac{\pi}{3}) = \frac{1}{2}\):
\[ y = \frac{\frac{\sqrt{3}}{2}}{2 - \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3} \]
So, the exact coordinates of the stationary point are \(\left(\frac{\pi}{6}, \frac{\sqrt{3}}{3}\right)\).

(a)
M1: Attempts to use the quotient rule. Look for a form of \(\frac{(2-\cos(2x))(A\cos(2x)) - \sin(2x)(B\sin(2x))}{(2-\cos(2x))^2}\) where \(A\) and \(B\) are non-zero constants.
A1: Correct derivatives of numerator and denominator: \(\frac{\text{d}u}{\text{d}x} = 2\cos(2x)\) and \(\frac{\text{d}v}{\text{d}x} = 2\sin(2x)\).
A1: Correct unsimplified quotient rule expression: \(\frac{(2-\cos(2x))(2\cos(2x)) - (\sin(2x))(2\sin(2x))}{(2-\cos(2x))^2}\).
A1*: Fully correct simplification showing intermediate steps clearly to arrive at the given answer. Must use \(\sin^2(2x) + \cos^2(2x) = 1\).

(b)
M1: Sets the numerator \(4\cos(2x) - 2 = 0\) and attempts to solve for \(2x\) or \(x\).
A1: Obtains \(x = \frac{\pi}{6}\). (Accept \(30^\circ\) only as an intermediate step; final answer must be in radians).
M1: Substitutes their \(x = \frac{\pi}{6}\) into the equation for \(y\) to find the \(y\)-coordinate.
A1: Correct exact coordinates \(\left(\frac{\pi}{6}, \frac{\sqrt{3}}{3}\right)\) or equivalent.

(a) For \(x \in \mathbb{R}\), we have \(\text{e}^{2x} > 0\).
Therefore, \(-\text{e}^{2x} < 0\), which gives \(3 - \text{e}^{2x} < 3\).
So the range of \(\text{f}\) is \(\text{f}(x) < 3\).

(b) To find the inverse function, let \(y = 3 - \text{e}^{2x}\).
Rearranging for \(x\):
\[ \text{e}^{2x} = 3 - y \]
\[ 2x = \ln(3 - y) \]
\[ x = \frac{1}{2}\ln(3 - y) \]
Replacing \(y\) with \(x\):
\[ \text{f}^{-1}(x) = \frac{1}{2}\ln(3 - x) \]
The domain of \(\text{f}^{-1}\) is the range of \(\text{f}\), which is \(x < 3\).

(c) Find the expression for \(\text{fg}(x)\):
\[ \text{fg}(x) = \text{f}(\text{g}(x)) = 3 - \text{e}^{2\left(\frac{2}{x+1}\right)} = 3 - \text{e}^{\frac{4}{x+1}} \]
We set this equal to \(3 - \text{e}\):
\[ 3 - \text{e}^{\frac{4}{x+1}} = 3 - \text{e} \]
\[ \text{e}^{\frac{4}{x+1}} = \text{e}^1 \]
Taking the natural logarithm of both sides:
\[ \frac{4}{x+1} = 1 \]
\[ x + 1 = 4 \implies x = 3 \]
Since \(3 > 0\), this lies in the domain of \(\text{g}\).

(a)
B1: Correct range \(\text{f}(x) < 3\) or \(y < 3\).

(b)
M1: Attempts to make \(x\) the subject of the formula. Look for a correct process of rearranging and applying \(\ln\).
A1: Correct intermediate step, e.g., \(2x = \ln(3-y)\).
A1: \(\text{f}^{-1}(x) = \frac{1}{2}\ln(3 - x)\) or equivalent.
B1ft: Domain is \(x < 3\) (follow through their range from part a).

(c)
M1: Formulates the composite function \(\text{fg}(x) = 3 - \text{e}^{\frac{4}{x+1}}\).
M1: Sets their \(\text{fg}(x)\) equal to \(3 - \text{e}\) and equates exponents of \(\text{e}\) to get a linear equation in \(x\).
A1: \(x = 3\) only.

(a) Let \(\text{f}(x) = x^3 - 5x^2 + 7\).
Evaluate \(\text{f}(x)\) at the endpoints of the interval:
\[ \text{f}(4.5) = (4.5)^3 - 5(4.5)^2 + 7 = 91.125 - 101.25 + 7 = -3.125 \]
\[ \text{f}(4.8) = (4.8)^3 - 5(4.8)^2 + 7 = 110.592 - 115.2 + 7 = 2.392 \]
Since \(\text{f}(x)\) is continuous on \([4.5, 4.8]\) and there is a change of sign between \(\text{f}(4.5) < 0\) and \(\text{f}(4.8) > 0\), there is at least one root \(\alpha\) in the interval \([4.5, 4.8]\).

(b) Start with the equation:
\[ x^3 - 5x^2 + 7 = 0 \]
Rearrange to group the \(x^3\) and \(x^2\) terms:
\[ x^3 = 5x^2 - 7 \]
Divide both sides by \(x\) (since \(x \ne 0\) in the interval):
\[ x^2 = 5x - \frac{7}{x} \]
Taking the square root of both sides (since \(x > 0\)):
\[ x = \sqrt{5x - \frac{7}{x}} \]
which is the required form.

(c) Using the formula with \(x_1 = 4.7\):
\[ x_2 = \sqrt{5(4.7) - \frac{7}{4.7}} = \sqrt{22.0106383...} \approx 4.69154967... \]
To 4 decimal places, \(x_2 = 4.6915\).

Using the unrounded value of \(x_2\):
\[ x_3 = \sqrt{5(4.69154967...) - \frac{7}{4.69154967...}} \approx 4.686758... \approx 4.6868 \]
Using the rounded value \(x_2 = 4.6915\):
\[ x_3 = \sqrt{5(4.6915) - \frac{7}{4.6915}} \approx 4.686730... \approx 4.6867 \]

(a)
M1: Attempts to evaluate \(\text{f}(4.5)\) and \(\text{f}(4.8)\) with at least one correct calculation to 1 decimal place.
A1: Correct values \(\text{f}(4.5) = -3.125\) and \(\text{f}(4.8) = 2.392\) (or 2.4), and a conclusion that mentions change of sign and continuity.

(b)
M1: Rearranges the cubic equation to make \(x^3\) or \(x^2\) the subject, or divides by \(x\).
A1*: Complete and mathematically sound proof leading to the given expression with no missing steps.

(c)
M1: An attempt to substitute \(x_1 = 4.7\) into the iterative formula to find \(x_2\).
A1: \(x_2 \approx 4.6915\) (must be to 4 d.p.).
A1: \(x_3 \approx 4.6868\) (or \(4.6867\) if using the rounded \(x_2\)) (must be to 4 d.p.).

(a) Let \(u = \sec^2 x + 2\tan x\), so \(y = \ln u\).

Using the chain rule:

\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{u} \frac{\mathrm{d}u}{\mathrm{d}x}\]

Differentiating \(u\) with respect to \(x\):

\[\frac{\mathrm{d}}{\mathrm{d}x}(\sec^2 x) = 2\sec x \cdot (\sec x \tan x) = 2\sec^2 x \tan x\]

\[\frac{\mathrm{d}}{\mathrm{d}x}(2\tan x) = 2\sec^2 x\]

Therefore,

\[\frac{\mathrm{d}u}{\mathrm{d}x} = 2\sec^2 x \tan x + 2\sec^2 x = 2\sec^2 x(\tan x + 1)\]

Substituting this back into the chain rule formula:

\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\sec^2 x(\tan x + 1)}{\sec^2 x + 2\tan x}\]

which is the required result.

(b) At the point where \(x = \frac{\pi}{4}\):

\[\tan\left(\frac{\pi}{4}\right) = 1\]

\[\sec^2\left(\frac{\pi}{4}\right) = \frac{1}{\cos^2(\pi/4)} = \frac{1}{(1/\sqrt{2})^2} = 2\]

Substitute these into the expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\):

\[m = \frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{x = \pi/4} = \frac{2(2)(1 + 1)}{2 + 2(1)} = \frac{8}{4} = 2\]

Now find the \(y\)-coordinate at \(x = \frac{\pi}{4}\):

\[y = \ln\left(\sec^2\left(\frac{\pi}{4}\right) + 2\tan\left(\frac{\pi}{4}\right)\right) = \ln(2 + 2(1)) = \ln 4\]

Using the equation of a straight line \(y - y_1 = m(x - x_1)\):

\[y - \ln 4 = 2\left(x - \frac{\pi}{4}\right)\]

\[y = 2x - \frac{\pi}{2} + \ln 4\]

(or \(y = 2x - \frac{\pi}{2} + 2\ln 2\))

Part (a):
- M1: For attempting to apply the chain rule to \(\ln(f(x))\), giving \(\frac{f'(x)}{f(x)}\).
- M1: For correctly differentiating \(\sec^2 x\) using the chain rule or product rule to get \(2\sec^2 x \tan x\).
- A1: For correct derivative of the inner function, \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\sec^2 x \tan x + 2\sec^2 x\).
- A1*: For factorising and showing the given result clearly with no errors in their working.

Part (b):
- M1: Finds both the value of the gradient \(m = 2\) and the exact \(y\)-coordinate \(y = \ln 4\) (or \(2\ln 2\)) at \(x = \frac{\pi}{4}\).
- M1: Uses their values of \(m\), \(x\), and \(y\) in a correct straight-line equation method, e.g., \(y - y_1 = m(x - x_1)\).
- A1: Correct exact equation in the form \(y = mx + c\), e.g., \(y = 2x - \frac{\pi}{2} + \ln 4\) or \(y = 2x - \frac{\pi}{2} + 2\ln 2\) (or equivalent exact form).

**Part (a)**
Separate the variables in the differential equation:
\[ \int \frac{1}{3600 - V} \mathrm{d}V = \int \frac{k}{t+2} \mathrm{d}t \]

Integrate both sides:
\[ -\ln|3600 - V| = k \ln|t+2| + C \]
where \(C\) is a constant of integration.
Multiply by \(-1\) and use logarithm rules:
\[ \ln|3600 - V| = -k \ln(t+2) - C = \ln\left((t+2)^{-k}\right) + D \]
Exponentiate both sides:
\[ 3600 - V = A(t+2)^{-k} \]
where \(A = e^D\) is a constant.

Apply the initial condition \(V = 600\) when \(t = 0\):
\[ 3600 - 600 = A(2)^{-k} \implies 3000 = A \cdot 2^{-k} \implies A = 3000 \cdot 2^k \]

Substitute \(A\) back into the equation:
\[ 3600 - V = 3000 \cdot 2^k (t+2)^{-k} = 3000 \left(\frac{2}{t+2}\right)^k \]

Rearranging to make \(V\) the subject:
\[ V = 3600 - 3000\left(\frac{2}{t+2}\right)^k \]

**Part (b)**
Substitute \(V = 1100\) when \(t = 2\):
\[ 1100 = 3600 - 3000\left(\frac{2}{2+2}\right)^k \]
\[ 3000\left(\frac{1}{2}\right)^k = 2500 \]
\[ \left(\frac{1}{2}\right)^k = \frac{2500}{3000} = \frac{5}{6} \]
\[ 2^{-k} = \frac{5}{6} \implies 2^k = 1.2 \]
\[ k = \frac{\ln(1.2)}{\ln(2)} \approx 0.263034 \]
To 3 significant figures, \(k = 0.263\).

**Part (c)**
At \(t = 6\):
\[ 3600 - V = 3000 \left(\frac{2}{6+2}\right)^k = 3000 \left(\frac{1}{4}\right)^k = 3000 (2^{-k})^2 \]
Since \(2^{-k} = \frac{5}{6}\):
\[ 3600 - V = 3000 \left(\frac{5}{6}\right)^2 = 3000 \times \frac{25}{36} = \frac{6250}{3} \approx 2083.33 \]

Using the differential equation:
\[ \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{k(3600 - V)}{t+2} \]
At \(t = 6\):
\[ \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{0.263034 \times \frac{6250}{3}}{8} \approx 68.498 \]
To 1 decimal place, the rate of increase of the volume of liquid is \(68.5 \text{ cm}^3/\text{s}\).

**Part (a)**
- M1: Attempts to separate variables and integrate both sides, obtaining \(\pm\ln|3600-V|\) and \(k\ln(t+2)\) (or equivalent).
- A1: Correct integration including a constant of integration: \(-\ln|3600-V| = k\ln(t+2) + C\).
- M1: Uses \(t=0, V=600\) to find the constant of integration or the constant \(A\).
- M1: Expresses \(3600-V\) as a single power function in the form \(A(t+2)^{-k}\) or equivalent.
- A1*: Fully correct proof showing all intermediate steps leading to the given expression.

**Part (b)**
- M1: Substitutes \(t=2, V=1100\) to obtain an equation in terms of \(k\).
- A1: Obtains \(k \approx 0.263\) (or any value rounding to 0.263).

**Part (c)**
- M1: Attempts to find the value of \(3600-V\) or \(V\) at \(t = 6\), or substitutes expression for \(V(t)\) and \(k\) into the derivative.
- A1: Correctly evaluates the rate of increase to 1 decimal place: \(68.5\) (accept answers in range \([68.4, 68.6]\) depending on rounding of \(k\)).

**Part (a)**
Find \\frac{\\mathrm{d}x}{\\mathrm{d}\\theta}:
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2 - 2\cos(2\theta) \]
Using the double angle identity \\cos(2\\theta) = 1 - 2\\sin^2(\\theta), this simplifies to:
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2(1 - \cos(2\theta)) = 2\left(2\sin^2(\theta)\right) = 4\sin^2(\theta) \]

Find \\frac{\\mathrm{d}y}{\\mathrm{d}\\theta}:
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = -4\sin(\theta) \]

Use the chain rule to find \\frac{\\mathrm{d}y}{\\mathrm{d}x}:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{-4\sin(\theta)}{4\sin^2(\theta)} = -\frac{1}{\sin(\theta)} \quad (\text{or } -\csc(\theta)) \]

**Part (b)**
At \\theta = \\frac{\\pi}{3}:
1. Find the coordinates of the point:
\[ x = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \]
\[ y = 4\cos\left(\frac{\pi}{3}\right) = 4\left(\frac{1}{2}\right) = 2 \]

2. Find the gradient of the tangent:
\[ m = \left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|\\!_{\theta = \frac{\pi}{3}} = -\frac{1}{\sin\left(\frac{\pi}{3}\right)} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \]

3. Use the equation of a straight line:
\[ y - y_1 = m(x - x_1) \]
\[ y - 2 = -\frac{2\sqrt{3}}{3} \left( x - \left( \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right) \right) \]
\[ y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{2\sqrt{3}}{3}\left(\frac{2\pi}{3}\right) - \frac{2\sqrt{3}}{3}\left(\frac{\sqrt{3}}{2}\right) \]
Note that:
\[ \frac{2\sqrt{3}}{3}\left(\frac{\sqrt{3}}{2}\right) = \frac{2 \times 3}{6} = 1 \]

So:
\[ y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} - 1 \]
\[ y = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + 1 \]

**Part (a)**
- M1: Differentiates \(x\) and \(y\) with respect to \(\theta\). Award for either \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 2 - 2\cos(2\theta)\) or \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = -4\sin(\theta)\).
- A1: Both derivatives correct.
- M1: Applies the chain rule \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}\).
- A1: Correct simplified derivative in terms of \(\theta\): \(-\frac{1}{\sin(\theta)}\) or \(-\csc(\theta)\).

**Part (b)**
- B1: Correctly calculates \(x = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\) and \(y = 2\) when \(\theta = \frac{\pi}{3}\).
- M1: Substitutes \(\theta = \frac{\pi}{3}\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient of the tangent.
- M1: Applies the straight line equation \(y - y_1 = m(x - x_1)\) with their coordinates and gradient.
- A1: Obtains the correct exact equation in the required form: \(y = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + 1\) (or exact equivalent in that form).

**Part (a)**
To find the intersection of \(l_1\) and \(l_2\), equate the \(x\) and \(y\) components:
From the \(x\)-component:
\[ 1 + 2\lambda = 5 - \mu \implies 2\lambda + \mu = 4 \quad (1) \]

From the \(y\)-component:
\[ 4 - \lambda = -1 + 2\mu \implies \lambda + 2\mu = 5 \quad (2) \]

From (1), \\mu = 4 - 2\\lambda. Substitute this into (2):
\[ \lambda + 2(4 - 2\lambda) = 5 \]
\[ \lambda + 8 - 4\lambda = 5 \]
\[ -3\lambda = -3 \implies \lambda = 1 \]

Substitute \\lambda = 1 back to find \\mu:
\[ \mu = 4 - 2(1) = 2 \]

**Part (b)**
Now equate the \(z\)-components using \\lambda = 1 and \\mu = 2:
\[ -3 + 3\lambda = p + 2\mu \]
\[ -3 + 3(1) = p + 2(2) \]
\[ 0 = p + 4 \implies p = -4 \]

The coordinates of \(A\) can be found by substituting \\lambda = 1 into \(l_1\):
\[ \mathbf{r}_A = \begin{pmatrix} 1 \\ 4 \\ -3 \end{pmatrix} + 1 \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix} \]
So, the coordinates of \(A\) are \((3, 3, 0)\).

**Part (c)**
The direction vectors of the lines \(l_1\) and \(l_2\) are:
\[ \mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \quad \text{and} \quad \mathbf{d}_2 = \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix} \]

Calculate the scalar product:
\[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(-1) + (-1)(2) + (3)(2) = -2 - 2 + 6 = 2 \]

Calculate the magnitudes of each direction vector:
\[ |\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]
\[ |\mathbf{d}_2| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = 3 \]

Let \(\theta\) be the angle between the lines:
\[ \cos(\theta) = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|} = \frac{2}{3\sqrt{14}} \]
\[ \theta = \arccos\left(\frac{2}{3\sqrt{14}}\right) \approx 79.7356^{\circ} \]
To 1 decimal place, the acute angle is \(79.7^{\circ}\).

**Part (a)**
- M1: Sets up two simultaneous equations in \(\lambda\) and \(\mu\) using the \(x\) and \(y\) components.
- M1: Attempts to solve the simultaneous equations to find values for \(\lambda\) and \(\mu\).
- A1: Correct values \(\lambda = 1\) and \(\mu = 2\).

**Part (b)**
- M1: Uses their \(\lambda\) and \(\mu\) in the \(z\)-component equation to form an equation for \(p\).
- A1: Correct value of \(p = -4\) and correct coordinates of \(A\) as \((3, 3, 0)\) or equivalent position vector.

**Part (c)**
- M1: Identifies the correct direction vectors \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}\) (or scalar multiples) and attempts the scalar product.
- M1: Attempts to find the magnitudes of both direction vectors.
- M1: Uses the scalar product formula \(\cos\theta = \frac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}\).
- A1: Correct acute angle \(79.7^{\circ}\) (accept \(79.7\) or answers rounding to \(79.7\)).

\text{(a) Separating the variables in the differential equation gives:}

\[\int \frac{1}{P(8 - \ln P)} \,\mathrm{d}P = \int \frac{1}{5t^2} \,\mathrm{d}t\]

\text{Using the substitution } u = 8 - \ln P \implies \mathrm{d}u = -\frac{1}{P} \,\mathrm{d}P \text{:}

\[\int -\frac{1}{u} \,\mathrm{d}u = \int \frac{1}{5}t^{-2} \,\mathrm{d}t\]

\[-\ln|8 - \ln P| = -\frac{1}{5t} + C\]

\[\ln(8 - \ln P) = \frac{1}{5t} + D \quad \text{where } D = -C\]

\text{Using the boundary conditions } t = 1, P = \mathrm{e}^4 \text{:}

\[\ln(8 - 4) = \frac{1}{5(1)} + D \implies \ln 4 = 0.2 + D \implies D = \ln 4 - 0.2\]

\text{Substituting } D \text{ back into the equation:}

\[\ln(8 - \ln P) = \frac{1}{5t} + \ln 4 - 0.2\]

\[8 - \ln P = \mathrm{e}^{\frac{1}{5t} - 0.2 + \ln 4} = 4\mathrm{e}^{\frac{1}{5t} - 0.2}\]

\[\ln P = 8 - 4\mathrm{e}^{\frac{1}{5t} - 0.2}\]

\[P = \exp\left(8 - 4\mathrm{e}^{\frac{1}{5t} - 0.2}\right)\]

\text{(b) As } t \to \infty, \frac{1}{5t} \to 0\text{, so:}

\[P \to \exp\left(8 - 4\mathrm{e}^{-0.2}\right) = \mathrm{e}^{8 - 4\mathrm{e}^{-0.2}}\]

\text{Thus, } a = 8, b = 4, c = -0.2.

(a)
M1: Separation of variables to obtain \(\int \frac{1}{P(8 - \ln P)} \,\mathrm{d}P = \int \frac{k}{t^2} \,\mathrm{d}t\) where \(k\) is a constant.
M1: Integrates LHS to obtain \(\pm \ln(8 - \ln P)\).
A1: Correct integration: \(-\ln(8 - \ln P) = -\frac{1}{5t} + C\) or equivalent.
M1: Applies boundary conditions \(t = 1\), \(P = \mathrm{e}^4\) to find their constant of integration.
A1: Correct constant value, e.g., \(D = \ln 4 - 0.2\) if using the form \(\ln(8 - \ln P) = \frac{1}{5t} + D\).
M1: Rearranges the expression to isolate \(P\), demonstrating correct index/log rules.
A1: Fully correct final expression: \(P = \exp\left(8 - 4\mathrm{e}^{\frac{1}{5t} - 0.2}\right)\) or equivalent.

(b)
M1: Considers the limit as \(t \to \infty\) to deduce that \(\frac{1}{5t} \to 0\).
A1: Obtains the correct limiting value \(\mathrm{e}^{8 - 4\mathrm{e}^{-0.2}}\).

\text{(a) Equating the } x \text{ and } y \text{ components of } l_1 \text{ and } l_2 \text{ gives:}

\[-1 + 2\lambda = 3 + \mu \implies 2\lambda - \mu = 4 \quad \text{(1)}\]

\[5 - \lambda = 6 - 2\mu \implies \lambda - 2\mu = -1 \quad \text{(2)}\]

\text{From (1), } \mu = 2\lambda - 4. \text{ Substituting this into (2):}

\[\lambda - 2(2\lambda - 4) = -1 \implies -3\lambda + 8 = -1 \implies 3\lambda = 9 \implies \lambda = 3\]

\text{Substituting } \lambda = 3 \text{ into (1) yields } \mu = 2(3) - 4 = 2.

\text{Using } \lambda = 3 \text{ in } l_1 \text{, the position vector of the intersection point } A \text{ is:}

\[\mathbf{r}_A = \begin{pmatrix} -1 \\ 5 \\ 2 \end{pmatrix} + 3\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 11 \end{pmatrix}\]

\text{So the coordinates of } A \text{ are } (5, 2, 11).

\text{Equating the } z \text{ components at the intersection point:}

\[2 + 3\lambda = 3 + p\mu \implies 2 + 3(3) = 3 + 2p \implies 11 = 3 + 2p \implies 2p = 8 \implies p = 4 \quad (\text{as required})\]

\text{(b) The direction vector of the line } l_1 \text{ is } \mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}.

\text{The vector } \vec{AB} \text{ is:}

\[\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ 1 \\ 9 \end{pmatrix} - \begin{pmatrix} 5 \\ 2 \\ 11 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \\ -2 \end{pmatrix}\]

\text{Using the scalar product formula:}

\[\mathbf{d}_1 \cdot \vec{AB} = 2(-3) + (-1)(-1) + 3(-2) = -6 + 1 - 6 = -11\]

\[|\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}\]

\[|\vec{AB}| = \sqrt{(-3)^2 + (-1)^2 + (-2)^2} = \sqrt{14}\]

\text{Since the acute angle is required, we use the absolute value of the scalar product:}

\[\cos\theta = \frac{|\mathbf{d}_1 \cdot \vec{AB}|}{|\mathbf{d}_1| |\vec{AB}|} = \frac{|-11|}{\sqrt{14}\sqrt{14}} = \frac{11}{14}\]

(a)
M1: Sets up two simultaneous equations using the \(x\) and \(y\) components.
M1: Solves the equations to find values for \(\lambda\) and \(\mu\).
A1: Correct values: \(\lambda = 3\) and \(\mu = 2\).
A1: Obtains the correct coordinates of \(A(5, 2, 11)\).
B1: Uses the \(z\) components with their parameter values to show that \(p = 4\).

(b)
M1: Correctly calculates \(\vec{AB}\) (allow one sign error).
M1: Performs a scalar product calculation between their \(\vec{AB}\) and the direction vector \(\mathbf{d}_1\).
M1: Calculates the magnitude of both vectors.
A1: Correctly evaluates \(\cos\theta = \frac{11}{14}\) (must be positive).

\text{(a) Express the function as:}

\[f(x) = (3 - 5x)(1 - 2x)^{-1/2}\]

\text{Using the Binomial expansion on } (1 - 2x)^{-1/2} \text{:}

\[(1 - 2x)^{-1/2} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-2x)^2 + \mathcal{O}(x^3)\]

\[= 1 + x + \frac{3}{8}(4x^2) + \mathcal{O}(x^3)\]

\[= 1 + x + \frac{3}{2}x^2 + \mathcal{O}(x^3)\]

\text{Now multiply by } (3 - 5x) \text{:}

\[f(x) = (3 - 5x)\left(1 + x + \frac{3}{2}x^2\right) + \mathcal{O}(x^3)\]

\[= 3\left(1 + x + \frac{3}{2}x^2\right) - 5x(1 + x) + \mathcal{O}(x^3)\]

\[= 3 + 3x + \frac{9}{2}x^2 - 5x - 5x^2\]

\[= 3 - 2x - \frac{1}{2}x^2\]

\text{(b) Substituting } x = 0.01 \text{ into the exact definition of } f(x) \text{:}

\[f(0.01) = \frac{3 - 5(0.01)}{\sqrt{1 - 2(0.01)}} = \frac{2.95}{\sqrt{0.98}} = \frac{2.95}{\sqrt{\frac{98}{100}}} = \frac{29.5}{\sqrt{98}} = \frac{29.5}{7\sqrt{2}} = \frac{59}{14\sqrt{2}}\]

\text{Rearranging to isolate } \sqrt{2} \text{:}

\[\sqrt{2} = \frac{59}{14 f(0.01)}\]

\text{Using the binomial approximation for } f(0.01) \text{:}

\[f(0.01) \approx 3 - 2(0.01) - 0.5(0.01)^2 = 3 - 0.02 - 0.00005 = 2.97995\]

\text{Therefore:}

\[\sqrt{2} \approx \frac{59}{14 \times 2.97995} = \frac{59}{41.7193} \approx 1.41421356...\]

\text{To 6 decimal places, } \sqrt{2} \approx 1.414214.

(a)
M1: Expresses the fraction as \((3 - 5x)(1 - 2x)^{-1/2}\).
M1: Applies the binomial expansion formula to \((1 - 2x)^{-1/2}\) with at least two terms correct.
A1: Correct simplified expansion of \((1 - 2x)^{-1/2}\) as \(1 + x + \frac{3}{2}x^2\).
M1: Multiplies their binomial expansion by \((3 - 5x)\) and groups terms by powers of \(x\).
A1: Correctly finds the coefficient of the linear term to be \(-2\).
A1: Correctly finds the coefficient of the quadratic term to be \(-\frac{1}{2}\).

(b)
M1: Substitutes \(x = 0.01\) into the exact formula for \(f(x)\) and relates \(f(0.01)\) to \(\sqrt{2}\).
A1: Shows that \(\sqrt{2} = \frac{59}{14 f(0.01)}\).
M1: Evaluates their quadratic expansion at \(x = 0.01\) to find an approximation for \(f(0.01)\).
A1: Reaches the approximation \(1.414214\) correct to 6 decimal places.

(a) Rewrite \(f(x)\) as:

\[f(x) = (2+x)(1-x)^{-\frac{1}{2}}\]

Using the binomial expansion formula for \((1-x)^{-\frac{1}{2}}\):

\[(1-x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-x)^2 + \dots\]

\[(1-x)^{-\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \dots\]

Now, multiply this expansion by \((2+x)\):

\[f(x) = (2+x)\left(1 + \frac{1}{2}x + \frac{3}{8}x^2 + \dots\right)\]

\[f(x) = 2\left(1 + \frac{1}{2}x + \frac{3}{8}x^2\right) + x\left(1 + \frac{1}{2}x\right) + \dots\]

\[f(x) = 2 + x + \frac{3}{4}x^2 + x + \frac{1}{2}x^2 + \dots\]

\[f(x) = 2 + 2x + \frac{5}{4}x^2\]

(b) Substitute \(x = \frac{1}{100} = 0.01\) into the expression for \(f(x)\):

\[f(0.01) = \frac{2 + 0.01}{\sqrt{1 - 0.01}} = \frac{2.01}{\sqrt{0.99}}\]

Since \(\sqrt{0.99} = \sqrt{\frac{99}{100}} = \frac{3\sqrt{11}}{10}\):

\[f(0.01) = \frac{2.01}{\frac{3\sqrt{11}}{10}} = \frac{20.1}{3\sqrt{11}} = \frac{6.7}{\sqrt{11}}\]

Therefore, we have:

\[\sqrt{11} = \frac{6.7}{f(0.01)}\]

Substituting \(x = 0.01\) into our binomial expansion:

\[f(0.01) \approx 2 + 2(0.01) + 1.25(0.01)^2 = 2 + 0.02 + 0.000125 = 2.020125\]

Hence, the approximation for \(\sqrt{11}\) is:

\[\sqrt{11} \approx \frac{6.7}{2.020125} \approx 3.31662026\dots\]

Giving the answer to 5 decimal places:

\[\sqrt{11} \approx 3.31662\]

**Part (a):**
* **M1**: Attempts to expand \((1-x)^{-\frac{1}{2}}\) with correct binomial structure. Must see \(1 + nx + \frac{n(n-1)}{2}x^2\) with \(n = -1/2\) and \(y = -x\).
* **A1**: Correct term in \(x\), which is \(\frac{1}{2}x\).
* **A1**: Correct term in \(x^2\), which is \(\frac{3}{8}x^2\).
* **M1**: Attempts to multiply their expansion by \((2+x)\) and collect terms.
* **A1**: Correct linear terms \(2 + 2x\).
* **A1**: Correct quadratic term \(\frac{5}{4}x^2\) (or \(1.25x^2\)).

**Part (b):**
* **M1**: Shows that substituting \(x = 0.01\) into \(f(x)\) leads to an expression involving \(\sqrt{11}\), i.e., \(f(0.01) = \frac{6.7}{\sqrt{11}}\) or equivalent.
* **M1**: Substitutes \(x = 0.01\) into their expansion from part (a) to find a numerical value (e.g., \(2.020125\)).
* **A1**: Correctly calculates \(\sqrt{11} \approx 3.31662\) to 5 decimal places.

(a) Separating the variables:

\[\int \frac{3000}{P(600-P)} \, dP = \int 1 \, dt\]

Using partial fractions:

\[\frac{3000}{P(600-P)} = \frac{A}{P} + \frac{B}{600-P}\]

\[3000 = A(600-P) + BP\]

Letting \(P = 0 \implies 3000 = 600A \implies A = 5\)

Letting \(P = 600 \implies 3000 = 600B \implies B = 5\)

So, the integral becomes:

\[\int \left( \frac{5}{P} + \frac{5}{600-P} \right) \, dP = \int 1 \, dt\]

Integrating both sides:

\[5\ln(P) - 5\ln(600-P) = t + C\]

\[5\ln\left(\frac{P}{600-P}\right) = t + C\]

Using the initial condition \(t = 0\), \(P = 100\):

\[5\ln\left(\frac{100}{500}\right) = 0 + C \implies C = 5\ln\left(\frac{1}{5}\right) = -5\ln(5)\]

Thus:

\[5\ln\left(\frac{P}{600-P}\right) = t - 5\ln(5)\]

Divide by 5:

\[\ln\left(\frac{P}{600-P}\right) = \frac{t}{5} - \ln(5)\]

\[\ln\left(\frac{5P}{600-P}\right) = \frac{t}{5}\]

Exponentiate both sides:

\[\frac{5P}{600-P} = e^{\frac{t}{5}}\]

\[5P = (600-P)e^{\frac{t}{5}}\]

\[5P = 600e^{\frac{t}{5}} - Pe^{\frac{t}{5}}\]

\[P\left(5 + e^{\frac{t}{5}}\right) = 600e^{\frac{t}{5}}\]

\[P = \frac{600e^{\frac{t}{5}}}{5 + e^{\frac{t}{5}}}\]

Dividing the numerator and denominator by \(e^{\frac{t}{5}}\):

\[P = \frac{600}{1 + 5e^{-\frac{t}{5}}}\]

(b) Set \(P = 500\):

\[500 = \frac{600}{1 + 5e^{-\frac{t}{5}}}\]

\[1 + 5e^{-\frac{t}{5}} = \frac{600}{500} = 1.2\]

\[5e^{-\frac{t}{5}} = 0.2\]

\[e^{-\frac{t}{5}} = 0.04 = \frac{1}{25}\]

\[e^{\frac{t}{5}} = 25\]

\[\frac{t}{5} = \ln(25)\]

\[t = 5\ln(25) \approx 16.094\dots\]

To 1 decimal place, \(t = 16.1\) years.

**Part (a):**
* **M1**: Attempts to separate variables and sets up the integration: \(\int \frac{1}{P(600-P)} \, dP = \int \frac{1}{3000} \, dt\) (or equivalent with constants).
* **M1**: Expresses LHS as partial fractions of the form \(\frac{A}{P} + \frac{B}{600-P}\).
* **A1**: Correct partial fractions (e.g. \(\frac{5}{P} + \frac{5}{600-P}\) if 3000 kept on LHS).
* **M1**: Integrates to obtain terms in \(\ln(P)\) and \(\ln(600-P)\).
* **A1ft**: Correct integration including constant of integration: \(5\ln(P) - 5\ln(600-P) = t + C\) (follow through on their partial fractions).
* **M1**: Uses \(t = 0\) and \(P = 100\) to find \(C\) (or equivalent constant).
* **A1**: Reach a fully correct, simplified explicit formula for \(P\), such as \(P = \frac{600}{1 + 5e^{-\frac{t}{5}}}\) or \(P = \frac{600e^{\frac{t}{5}}}{5 + e^{\frac{t}{5}}}\).

**Part (b):**
* **M1**: Sets \(P = 500\) in their expression for \(P\) (or an earlier stage).
* **M1**: Rearranges to find \(e^{-\frac{t}{5}}\) or \(e^{\frac{t}{5}}\) and takes natural logarithms.
* **A1**: \(16.1\) (accept \(16.1\) only; do not accept \(16.0\) or \(16\)).

(a) Let the regions in the Venn diagram be defined as:
- All three: \(A \cap D \cap M = 6\)
- Art and Drama only: \((A \cap D) \setminus M = 15 - 6 = 9\)
- Drama and Music only: \((D \cap M) \setminus A = 18 - 6 = 12\)
- Art and Music only: \((A \cap M) \setminus D = 12 - 6 = 6\)
- Art only: \(48 - (9 + 6 + 6) = 27\)
- Drama only: \(55 - (9 + 6 + 12) = 28\)
- Music only: \(40 - (6 + 6 + 12) = 16\)
- Total attending at least one club: \(27 + 28 + 16 + 9 + 12 + 6 + 6 = 104\)
- None of the clubs: \(120 - 104 = 16\)

(b)(i) The number of students attending Drama but not Music is \(28 \text{ (only Drama)} + 9 \text{ (Art and Drama only)} = 37\).
Therefore, the probability is \(\frac{37}{120} \approx 0.308\) (to 3 s.f.).

(b)(ii) The number of students who attend Art is \(48\).
Of these, the number who attend at least two clubs is \(9 \text{ (A and D only)} + 6 \text{ (A and M only)} + 6 \text{ (all three)} = 21\).
Therefore, the conditional probability is \(\frac{21}{48} = \frac{7}{16} = 0.4375 \approx 0.438\) (to 3 s.f.).

(c) Two events \(A\) and \(M\) are independent if \(P(A \cap M) = P(A) \times P(M)\).
- \(P(A \cap M) = \frac{12}{120} = 0.1\)
- \(P(A) \times P(M) = \frac{48}{120} \times \frac{40}{120} = 0.4 \times \frac{1}{3} = \frac{2}{15} \approx 0.133\)
Since \(0.1 \neq 0.133\) (or \(\frac{1}{10} \neq \frac{2}{15}\)), the events are not statistically independent.

(a)
- M1: Draw three overlapping circles in a bounding box with labels \(A, D, M\).
- A1: 6 in the intersection of all three, and at least two of {9, 12, 6} correct in the other intersections.
- A1: Frequencies 27, 28, 16 correct in the single-club regions.
- A0.5: 16 correct in the region outside the circles.
(b)(i)
- M1: For identifying the correct region/frequencies: \(28 + 9 = 37\) or \(P(D) - P(D \cap M)\).
- A1: \(\frac{37}{120}\) or equivalent decimal to 3 s.f. (0.308).
(b)(ii)
- M1: Correct denominator of 48 or fraction form of conditional probability \(\frac{P(\text{at least 2} \cap A)}{P(A)}\).
- M1: Correct numerator of 21 (or \(9 + 6 + 6\)).
- A1: \(\frac{21}{48}\) or \(\frac{7}{16}\) or 0.438.
(c)
- M1: State a valid test for independence (e.g., comparing \(P(A \cap M)\) with \(P(A) \times P(M)\), or \(P(A|M)\) with \(P(A)\)).
- M1: Correct calculation of the two sides of their chosen comparison, e.g., \(P(A) \times P(M) = \frac{2}{15} \approx 0.133\) and \(P(A \cap M) = 0.1\).
- A1: Clear statement showing that the two values are not equal.
- A1: Correct conclusion of 'not independent' based on their calculations.

(a) From the standard normal distribution table:
- For \(P(W < 246) = 0.15\), the critical \(z\)-value is negative. Since \(\Phi(1.0364) = 0.85\), we have:
\(\frac{246 - \mu}{\sigma} = -1.0364 \implies 246 - \mu = -1.0364\sigma \implies \mu - 1.0364\sigma = 246\)
- For \(P(W > 255) = 0.05\), we have \(P(W < 255) = 0.95\). Since \(\Phi(1.6449) = 0.95\), we have:
\(\frac{255 - \mu}{\sigma} = 1.6449 \implies 255 - \mu = 1.6449\sigma \implies \mu + 1.6449\sigma = 255\)
(Accept \(z\)-values of \(-1.04\) and \(1.64\) or \(1.645\)).

(b) Subtracting the first equation from the second equation:
\(1.6449\sigma - (-1.0364\sigma) = 255 - 246\)
\(2.6813\sigma = 9 \implies \sigma = \frac{9}{2.6813} \approx 3.3565 \approx 3.36\) (to 3 s.f.)
Substituting \(\sigma = 3.3565\) into the second equation:
\(\mu + 1.6449 \times 3.3565 = 255 \implies \mu \approx 255 - 5.521 \approx 249.48 \approx 249\) (to 3 s.f.)

(c) We want to find the conditional probability \(P(W < 250 \mid W > 246)\):
\(P(W < 250 \mid W > 246) = \frac{P(246 < W < 250)}{P(W > 246)}
- The denominator is \)P(W > 246) = 1 - 0.15 = 0.85\).
- For the numerator: \(P(246 < W < 250) = P(W < 250) - P(W < 246) = P(W < 250) - 0.15\).
Standardizing for \(W = 250\):
\(Z = \frac{250 - 249.48}{3.3565} \approx 0.155\) (or \(0.15\) to \(0.16\))
\(P(W < 250) = \Phi(0.155) \approx 0.5616\)
So, \(P(246 < W < 250) = 0.5616 - 0.15 = 0.4116\).
Thus, the conditional probability is:
\(\frac{0.4116}{0.85} \approx 0.4842 \approx 0.484\) (to 3 s.f.).
(Accept any answer in the range [0.481, 0.487] due to rounding of \(Z\)-values).

(a)
- M1: Attempting to standardize with \(\mu\) and \(\sigma\) set equal to \(z\)-values.
- B1: For obtaining \(z\)-values of approximately \(-1.04\) (or -1.0364) and \(1.64\) (or 1.6449).
- A1: One correct equation, e.g. \(\mu - 1.0364\sigma = 246\) (allow equivalent equations with \(z \in [-1.04, -1.03]\)).
- A1: Second correct equation, e.g. \(\mu + 1.6449\sigma = 255\) (allow equivalent equations with \(z \in [1.64, 1.65]\)).
(b)
- M1: Attempting to solve the simultaneous equations to eliminate one variable.
- A1: For finding \(\sigma \approx 3.36\) (accept any value in range [3.30, 3.40]).
- M1: Substituting their \(\sigma\) back into one of the equations to find \(\mu\).
- A1: \(\mu \approx 249\) (accept any value in range [249, 250]).
(c)
- M1: Using the conditional probability formula \(\frac{P(246 < W < 250)}{P(W > 246)}\).
- B1: Showing that the denominator is \(0.85\).
- M1: Standardizing 250 using their \(\mu\) and \(\sigma\) and attempting to find \(P(W < 250)\).
- A1: Obtaining a numerator \(P(246 < W < 250)\) in the range [0.409, 0.414].
- A0.5: Correct final probability in the range [0.481, 0.487].

(a) First calculate the sums of squares:
- \(S_{TT} = \sum T^2 - \frac{(\sum T)^2}{n} = 2656 - \frac{144^2}{8} = 2656 - 2592 = 64\)
- \(S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 5624 - \frac{208^2}{8} = 5624 - 5408 = 216\)
- \(S_{Ty} = \sum Ty - \frac{(\sum T)(\sum y)}{n} = 3854 - \frac{144 \times 208}{8} = 3854 - 3744 = 110\)
Now, calculate the PMCC \(r\):
\(r = \frac{S_{Ty}}{\sqrt{S_{TT} S_{yy}}} = \frac{110}{\sqrt{64 \times 216}} = \frac{110}{\sqrt{13824}} \approx 0.93557 \approx 0.936\) (to 3 s.f.).

(b) The gradient \(b\) is:
\(b = \frac{S_{Ty}}{S_{TT}} = \frac{110}{64} = 1.71875 \approx 1.72\) (to 3 s.f.)
The mean values are:
\(\bar{T} = \frac{144}{8} = 18\) and \(\bar{y} = \frac{208}{8} = 26\)
The intercept \(a\) is:
\(a = \bar{y} - b\bar{T} = 26 - 1.71875 \times 18 = 26 - 30.9375 = -4.9375 \approx -4.94\) (to 3 s.f.)
So the regression equation of \(y\) on \(T\) is:
\(y = -4.94 + 1.72T\)

(c) The product moment correlation coefficient is unchanged by linear coding (as both scale factors, \(1/2\) and \(1/5\), are positive).
Therefore, the PMCC between \(u\) and \(v\) is also \(0.936\).

(d) Method 1 (Substituting coding into regression line of \(y\) on \(T\)):
We have \(y = 5v + 20\) and \(T = 2u + 15\).
Substituting these into the regression equation:
\(5v + 20 = -4.9375 + 1.71875(2u + 15)\)
\(5v + 20 = -4.9375 + 3.4375u + 25.78125\)
\(5v + 20 = 20.84375 + 3.4375u\)
\(5v = 0.84375 + 3.4375u\)
\(v = 0.16875 + 0.6875u\)
Therefore, \(c \approx 0.169\) and \(d \approx 0.688\) (to 3 s.f.).
So the regression line of \(v\) on \(u\) is:
\(v = 0.169 + 0.688u\)

Method 2 (Using summary statistics for \(u\) and \(v\)):
\(\bar{u} = \frac{\bar{T}-15}{2} = 1.5\), \(\bar{v} = \frac{\bar{y}-20}{5} = 1.2\)
\(S_{uu} = \frac{S_{TT}}{2^2} = \frac{64}{4} = 16\)
\(S_{uv} = \frac{S_{Ty}}{2 \times 5} = \frac{110}{10} = 11\)
\(d = \frac{S_{uv}}{S_{uu}} = \frac{11}{16} = 0.6875 \approx 0.688\)
\(c = \bar{v} - d\bar{u} = 1.2 - 0.6875 \times 1.5 = 0.16875 \approx 0.169\)

(a)
- M1: Attempting to calculate the sums of squares \(S_{TT}\), \(S_{yy}\), or \(S_{Ty}\) (at least one correct).
- A1: All three sums of squares correct (\(S_{TT} = 64\), \(S_{yy} = 216\), \(S_{Ty} = 110\)).
- M1: Correctly substituting their sums of squares into the formula for \(r\).
- A0.5: \(r \approx 0.936\) (accept 0.935 to 0.936).
(b)
- M1: For calculating \(b = \frac{S_{Ty}}{S_{TT}}\) using their values.
- A1: \(b = 1.71875\) or 1.72.
- M1: Attempting to find \(a\) using \(a = \bar{y} - b\bar{T}\) with their \(b\) and calculated means \(\bar{T} = 18\), \(\bar{y} = 26\).
- A0.5: \(a = -4.94\) and writing the correct equation \(y = -4.94 + 1.72T\).
(c)
- B1: Stating the value is \(0.936\) (or their answer from part a).
- B1: A correct reason stating that linear coding (with positive coefficients) does not affect the PMCC.
(d)
- M1: Substituting \(y = 5v + 20\) and \(T = 2u + 15\) into their regression equation from part (b), or calculating coded means and sums of squares.
- A1: Finding the gradient \(d = 0.6875\) or 0.688.
- M1: Solving for intercept \(c\).
- A0.5: Writing the correct equation \(v = 0.169 + 0.688u\) (accept \(v = 0.17 + 0.69u\) if 2 s.f. is used).

(a)
\(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 3624 - \frac{168^2}{8} = 3624 - 3528 = 96\)
\(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 1916.2 - \frac{168 \times 88.4}{8} = 1916.2 - 1856.4 = 59.8\)

(b)
First calculate \(S_{yy}\):
\(S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 1042.5 - \frac{88.4^2}{8} = 1042.5 - 976.82 = 65.68\)
Now calculate the PMCC, \(r\):
\(r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{59.8}{\sqrt{96 \times 65.68}} = \frac{59.8}{\sqrt{6305.28}} = \frac{59.8}{79.40579...} \approx 0.753\) (to 3 s.f.)

(c)
Yes, because \(r = 0.753\) is close to 1, indicating a reasonably strong positive linear correlation.

(d)
\(b = \frac{S_{xy}}{S_{xx}} = \frac{59.8}{96} \approx 0.622916... \approx 0.623\) (to 3 s.f.)
\(\bar{x} = \frac{168}{8} = 21\)
\(\bar{y} = \frac{88.4}{8} = 11.05\)
\(a = \bar{y} - b\bar{x} = 11.05 - 0.622916... \times 21 = 11.05 - 13.08125 = -2.03125 \approx -2.03\) (to 3 s.f.)
So the regression line equation is:
\(y = -2.03 + 0.623x\)

(e)
For \(x = 23\):
Using the exact/more accurate values:
\(y = -2.03125 + 0.622916 \times 23 = 12.29583...\) (hundreds of dollars)
Since \(y\) is in hundreds of dollars, the predicted sales are:
\(12.29583... \times 100 = 1229.58... \approx 1230\) dollars (to 3 s.f.)

(a)
M1: Correct method for calculating \(S_{xx}\) or \(S_{xy}\).
A1: Correct value for \(S_{xx} = 96\).
A1: Correct value for \(S_{xy} = 59.8\).

(b)
M1: Attempt to calculate \(S_{yy}\) and substitute into the PMCC formula.
A1: \(r \approx 0.753\) (accept 0.753 or 0.75).

(c)
B1: State 'yes' with a valid reason mentioning 'strong positive linear correlation'. Do not accept just 'positive correlation'.

(d)
M1: Correct method for calculating \(b = \frac{S_{xy}}{S_{xx}}\).
A1: \(b \approx 0.623\) (accept 0.62).
M1: Correct method for \(a\) using \(\bar{y} - b\bar{x}\).
A1: Fully correct equation: \(y = -2.03 + 0.623x\) (or equivalent 3 s.f. values. Accept \(a\) in range \([-2.04, -2.02]\) and \(b\) in range \([0.62, 0.63]\)).

(e)
M1: Substituting \(x = 23\) into their regression equation and multiplying by 100.
A1: 1230 dollars (accept \(1230\) or \(1230\) dollars. Accept answers in range \([1220, 1240]\) due to rounding).

(a) Let the regions be identified as:
- All three: \(D \cap M \cap S = 15\)
- Drama and Music only: \(22 - 15 = 7\)
- Music and Sport only: \(28 - 15 = 13\)
- Drama and Sport only: \(25 - 15 = 10\)
- Drama only: \(48 - (7 + 10 + 15) = 48 - 32 = 16\)
- Music only: \(54 - (7 + 13 + 15) = 54 - 35 = 19\)
- Sport only: \(62 - (10 + 13 + 15) = 62 - 38 = 24\)
- None of the three: \(120 - (16 + 19 + 24 + 7 + 13 + 10 + 15) = 120 - 104 = 16\).

(b) (i) The number of students participating in only one activity is \(16 + 19 + 24 = 59\).
Probability = \(\frac{59}{120}\) (or \(0.492\) to 3 s.f.)

(ii) \(\mathrm{P}(D | S) = \frac{\mathrm{P}(D \cap S)}{\mathrm{P}(S)} = \frac{25}{62}\) (or \(0.403\) to 3 s.f.)

(c) There are 54 students who participate in Music out of 120.
\(\mathrm{P}(\text{Both } M) = \frac{54}{120} \times \frac{53}{119} = \frac{9}{20} \times \frac{53}{119} = \frac{477}{2380}\) (or \(0.200\) to 3 s.f.)

(d) Out of the 48 students who participate in Drama:
- The number who also participate in Sport is \(10 + 15 = 25\).
- The number who do not participate in Sport is \(48 - 25 = 23\).
We are choosing 3 students from the 48, and we want exactly 2 of them to participate in Sport.
This can be chosen in \(\binom{25}{2} \times \binom{23}{1}\) ways.
Total ways of choosing 3 from 48 is \(\binom{48}{3}\).
\(\mathrm{P}(\text{Exactly 2}) = \frac{\binom{25}{2} \times \binom{23}{1}}{\binom{48}{3}} = \frac{300 \times 23}{17296} = \frac{6900}{17296} = \frac{1725}{4324}\) (or \(0.399\) to 3 s.f.)

(a)
M1: Three overlapping circles labeled D, M, S with '15' in the central intersection.
A1: Correct values for intersections (7, 13, 10).
A1: Correct values for single-activity regions (16, 19, 24).
A1: Correct value outside circles (16) and box labeled.

(b)(i)
M1: Summing their three single-activity regions and dividing by 120.
A1: \(\frac{59}{120}\) or 0.492 (accept 0.49).

(b)(ii)
M1: Identification of correct denominator (62) or standard conditional probability formula.
A1: \(\frac{25}{62}\) or 0.403 (accept 0.40).

(c)
M1: Attempting multiplication of fractions without replacement, i.e., \(\frac{54}{120} \times \frac{53}{119}\).
A1: \(\frac{477}{2380}\) or 0.200 (accept 0.2).

(d)
M1: Finding that 25 Drama students do Sport, and 23 do not.
M1: Standard combination method \(\frac{\binom{25}{2} \times \binom{23}{1}}{\binom{48}{3}}\) or equivalent probability method (e.g. \(3 \times \frac{25}{48} \times \frac{24}{47} \times \frac{23}{46}\)).
A1: \(\frac{1725}{4324}\) or 0.399 (accept 0.399 to 3 s.f.).

(a) Standardising the two equations using the standard normal variable \(Z\):
From \(\mathrm{P}(W < 990) = 0.15\), we get:
\(\frac{990 - \mu}{\sigma} = -1.0364\) (or \(-1.04\))
\(\mu - 1.0364\sigma = 990\) (Equation 1)

From \(\mathrm{P}(W > 1025) = 0.05\), we get:
\(\frac{1025 - \mu}{\sigma} = 1.6449\) (or \(1.64\))
\(\mu + 1.6449\sigma = 1025\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\(2.6813\sigma = 35 \Rightarrow \sigma = \frac{35}{2.6813} = 13.053...\) grams.
Rounding to 3 s.f., \(\sigma = 13.1\) grams (as shown).

Now find \(\mu\):
\(\mu = 990 + 1.0364 \times 13.053 = 1003.528... \approx 1003.5\) g (or \(1003.6\) g depending on z-value used).

(b) We want to find \(\mathrm{P}(W > 1000)\):
Standardising:
\(Z = \frac{1000 - 1003.528}{13.053} = -0.2703...\)
Using tables for \(Z = -0.27\):
\(\mathrm{P}(Z > -0.27) = \Phi(0.27) = 0.6064\) (accept values in range \([0.605, 0.608]\)).
So the probability is \(0.606\) (to 3 s.f.).

(c) Let \(p = \mathrm{P}(W > 1000) \approx 0.6064\).
The number of ways to choose exactly two out of three packs is \(3\).
\(\mathrm{P}(\text{Exactly 2}) = 3 \times p^2 \times (1 - p) = 3 \times (0.6064)^2 \times (1 - 0.6064) = 3 \times 0.36772 \times 0.3936 = 0.434\) (to 3 s.f., accept \(0.432 - 0.436\)).

(a)
M1: Attempting to standardise either 990 or 1025 with \(\mu\) and \(\sigma\).
B1: Either \(z = -1.0364\) (accept range \([-1.04, -1.03]\)) or \(z = 1.6449\) (accept range \([1.64, 1.65]\)) correct.
M1: Setting up two linear equations in \(\mu\) and \(\sigma\) and attempting to solve simultaneously to find \(\sigma\).
A1: Correctly showing \(\sigma = 13.1\) (from a value in \([13.0, 13.1]\)).
M1: Substituting their \(\sigma\) back into one of the equations to find \(\mu\).
A1: \(\mu = 1003.5\) (accept \(1003.5\) or \(1003.6\) or \(1004\)).

(b)
M1: Standardising 1000 with their \(\mu\) and \(\sigma\).
A1: Correct \(z\)-value in range \([-0.28, -0.26]\).
A1: Probability in range \([0.605, 0.608]\).

(c)
M1: Using \(3 \times p^2 \times (1 - p)\) with their probability from part (b).
M1: Substituting their value of \(p\) into the binomial formula-like expression.
A1: Answer in range \([0.432, 0.436]\) (to 3 s.f.).

(a) Let \(\lambda\) represent the mean daily sales of organic honey.
\(H_0: \lambda = 4\)
\(H_1: \lambda > 4\)

(b) Let \(Y\) be the number of jars sold in a 3-day period.
Under \(H_0\), \(Y \sim \text{Po}(12)\) because the daily rate is multiplied by 3.
We want to test if \(Y \ge 18\).
Using the cumulative Poisson tables for \(\lambda = 12\):
\(\text{P}(Y \ge 18) = 1 - \text{P}(Y \le 17) = 1 - 0.9370 = 0.0630\).
Since \(0.0630 > 0.05\), the result is not statistically significant.
We do not reject \(H_0\). There is insufficient evidence at the 5% level of significance to suggest that the mean daily sales of organic honey have increased.

(c) Let \(W\) be the number of jars sold in a 10-day period.
Under \(H_0\), \(W \sim \text{Po}(40)\).
Since \(\lambda = 40 > 10\), we can approximate \(W\) using a Normal distribution:
\(W \approx \text{N}(40, 40)\).
For a two-tailed test at the 5% significance level, we want the probability in each tail to be as close as possible to 0.025.

For the lower tail:
\(\text{P}(W \le c_1) \approx 0.025\)
Using a continuity correction:
\(\text{P}\left(Z \le \frac{c_1 + 0.5 - 40}{\sqrt{40}}\right) \approx 0.025\)
\(\frac{c_1 - 39.5}{\sqrt{40}} \approx -1.96\)
\(c_1 - 39.5 \approx -1.96 \times 6.3246 \approx -12.396\)
\(c_1 \approx 27.104 \implies c_1 = 27\).
Checking the probabilities:
If \(c_1 = 27\): \(\text{P}(W \le 27) = \text{P}\left(Z \le \frac{27.5 - 40}{\sqrt{40}}\right) = \text{P}(Z \le -1.976) = 0.0241\).
If \(c_1 = 28\): \(\text{P}(W \le 28) = \text{P}\left(Z \le \frac{28.5 - 40}{\sqrt{40}}\right) = \text{P}(Z \le -1.818) = 0.0345\).
Since \(0.0241\) is closer to \(0.025\) than \(0.0345\), the lower critical region is \(W \le 27\).

For the upper tail:
\(\text{P}(W \ge c_2) \approx 0.025\)
Using a continuity correction:
\(\text{P}\left(Z \ge \frac{c_2 - 0.5 - 40}{\sqrt{40}}\right) \approx 0.025\)
\(\frac{c_2 - 40.5}{\sqrt{40}} \approx 1.96\)
\(c_2 - 40.5 \approx 12.396\)
\(c_2 \approx 52.896 \implies c_2 = 53\).
Checking the probabilities:
If \(c_2 = 53\): \(\text{P}(W \ge 53) = \text{P}\left(Z \ge \frac{52.5 - 40}{\sqrt{40}}\right) = \text{P}(Z \ge 1.976) = 0.0241\).
If \(c_2 = 52\): \(\text{P}(W \ge 52) = \text{P}\left(Z \ge \frac{51.5 - 40}{\sqrt{40}}\right) = \text{P}(Z \ge 1.818) = 0.0345\).
Since \(0.0241\) is closer to \(0.025\) than \(0.0345\), the upper critical region is \(W \ge 53\).

Therefore, the critical region is \(W \le 27\) or \(W \ge 53\).

(a) B1: For writing both hypotheses correctly with the parameter \(\lambda\) defined.
(b) M1: For identifying the correct distribution \(Y \sim \text{Po}(12)\) under the null hypothesis.
M1: For attempting the calculation \(1 - \text{P}(Y \le 17)\).
A1: For obtaining the probability of \(0.0630\).
M1: For comparing the probability with \(0.05\).
A1: For a fully correct conclusion in context, stating that there is insufficient evidence to suggest that the mean daily sales have increased.
(c) B1: For defining the approximating Normal distribution parameters: \(W \approx \text{N}(40, 40)\).
M1: For utilizing the continuity correction in at least one tail (e.g., using \(c_1 + 0.5\) or \(c_2 - 0.5\)).
M1: For equating the standardized score to \(\pm 1.96\) (or using a suitable critical value from standard normal tables).
A1: For proving that either \(c_1 = 27\) or \(c_2 = 53\) is closer to the tail probability limit.
A1: For establishing the complete critical region as \(W \le 27\) or \(W \ge 53\) (allow union notation).

(a) Since \(\text{F}(x)\) is a continuous distribution function, \(\text{F}(4) = 1\).
\(k(4^2 - 4) = 1\)
\(12k = 1 \implies k = \frac{1}{12}\) (as required).

(b) The probability density function \(\text{f}(x)\) is the derivative of \(\text{F}(x)\).
For \(1 \le x \le 4\):
\(\text{f}(x) = \frac{\text{d}}{\text{d}x}\left[\frac{1}{12}(x^2 - x)\right] = \frac{1}{12}(2x - 1)\).
For all other values of \(x\), \(\text{f}(x) = 0\).
Therefore:
\[\text{f}(x) = \begin{cases} \frac{1}{12}(2x - 1) & 1 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}\]

(c) The mean of \(X\), \(\text{E}(X)\), is given by:
\(\text{E}(X) = \int_{1}^{4} x \cdot \text{f}(x) \, \text{d}x\)
\(\text{E}(X) = \int_{1}^{4} \frac{1}{12}x(2x - 1) \, \text{d}x = \frac{1}{12} \int_{1}^{4} (2x^2 - x) \, \text{d}x\)
\(\text{E}(X) = \frac{1}{12} \left[ \frac{2}{3}x^3 - \frac{1}{2}x^2 \right]_{1}^{4}\)
Evaluating the integration limits:
At \(x = 4\): \(\frac{2}{3}(64) - \frac{1}{2}(16) = \frac{128}{3} - 8 = \frac{104}{3}\)
At \(x = 1\): \(\frac{2}{3}(1) - \frac{1}{2}(1) = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}\)
Difference: \(\frac{104}{3} - \frac{1}{6} = \frac{208 - 1}{6} = \frac{207}{6} = \frac{69}{2}\)
So: \(\text{E}(X) = \frac{1}{12} \times \frac{69}{2} = \frac{23}{8} = 2.875\).

(d) Let \(m\) be the median of \(X\), so \(\text{F}(m) = 0.5\).
\(\frac{1}{12}(m^2 - m) = 0.5\)
\(m^2 - m = 6 \implies m^2 - m - 6 = 0\)
\((m - 3)(m + 2) = 0\)
Since the domain of \(X\) is \(1 \le x \le 4\), we must choose the root in this interval.
Therefore, \(m = 3\).

(a) M1: For setting \(\text{F}(4) = 1\) using the given cumulative distribution function.
A1: For showing \(12k = 1\) leading to \(k = \frac{1}{12}\) with clear working.
(b) M1: For attempting to differentiate the cumulative distribution function.
A1: For getting \(\frac{1}{12}(2x - 1)\) in the range \([1, 4]\).
A1: For writing down the complete piecewise function, including "0 otherwise" or equivalent.
(c) M1: For setting up the integral \(\int x \cdot \text{f}(x) \, \text{d}x\).
M1: For successfully integrating to obtain \(\left[ \frac{2}{3}x^3 - \frac{1}{2}x^2 \right]\) (ignoring constants).
A1: For correctly evaluating and simplifying to get the exact value \(2.875\) (or \(\frac{23}{8}\)).
(d) M1: For setting \(\text{F}(m) = 0.5\).
M1: For establishing and solving the quadratic equation \(m^2 - m - 6 = 0\).
A1: For identifying \(m = 3\) and rejecting the negative solution \(m = -2\).

(a) Let \(X\) be the number of germinated seeds in a sample of 150. Under the exact model, \(X \sim \text{B}(150, 0.08)\).
(i) Using the Binomial distribution:
\(\text{P}(X = 10) = \binom{150}{10} (0.08)^{10} (0.92)^{140}\)
Using a calculator: \(\text{P}(X = 10) \approx 0.1061\).

(ii) Using the Poisson approximation:
\(\lambda = np = 150 \times 0.08 = 12\).
We approximate \(X\) as \(Y \sim \text{Po}(12)\).
\(\text{P}(Y = 10) = \frac{e^{-12} \times 12^{10}}{10!} \approx 0.1048\)
(Alternatively, using cumulative Poisson tables: \(\text{P}(Y \le 10) - \text{P}(Y \le 9) = 0.3472 - 0.2424 = 0.1048\)).

(b) The Poisson approximation is justified because the number of trials \(n\) is large (\(n = 150 > 50\)) and the probability of success \(p\) is small (\(p = 0.08 < 0.1\)).

(c) Let \(p_1\) be the probability that a tray of 150 seeds contains fewer than 10 germinated seeds.
Using the Poisson distribution \(Y \sim \text{Po}(12)\):
\(p_1 = \text{P}(Y < 10) = \text{P}(Y \le 9) = 0.2424\).

Let \(T\) be the number of trays out of 120 that contain fewer than 10 germinated seeds.
\(T \sim \text{B}(120, 0.2424)\).
We want to find \(\text{P}(T \ge 25)\).
Since \(n = 120\) is large and \(p_1\) is not near 0 or 1, we can use a Normal approximation to the Binomial:
Mean: \(\mu = n p_1 = 120 \times 0.2424 = 29.088\)
Variance: \(\sigma^2 = n p_1 (1 - p_1) = 29.088 \times (1 - 0.2424) = 22.037\)
Standard deviation: \(\sigma = \sqrt{22.037} \approx 4.694\)

Applying a continuity correction:
\(\text{P}(T \ge 25) \approx \text{P}(T \ge 24.5)\)
Standardizing:
\(Z = \frac{24.5 - 29.088}{4.694} = \frac{-4.588}{4.694} \approx -0.9774\)
We require \(\text{P}(Z \ge -0.9774) = \text{P}(Z \le 0.9774) = \Phi(0.9774)\).
Using normal tables:
If using \(z = 0.98\): \(\Phi(0.98) = 0.8365\).
If using a more precise value of \(z = 0.977\): \(\Phi(0.977) = 0.8357\).
Both values (and any in the range \([0.835, 0.837]\)) are acceptable.

(a) M1: For writing down the binomial formula or expressing \(\text{P}(X=10)\) with correct parameters.
A1: For obtaining \(0.1061\).
B1: For finding \(\lambda = 12\).
M1: For using \(\text{P}(Y \le 10) - \text{P}(Y \le 9)\) or the Poisson probability formula for \(Y=10\).
A1: For obtaining \(0.1048\).
(b) B1: For stating both reasons (large \(n\) and small \(p\)) with numerical support.
(c) B1: For establishing the probability \(p_1 = 0.2424\) using the Poisson model.
M1: For identifying the parameters of the approximating Normal distribution: \(\mu = 29.088\) and \(\sigma^2 = 22.037\).
M1: For applying the continuity correction to get \(24.5\).
M1: For standardizing correctly: \(z = \frac{24.5 - 29.088}{\sqrt{22.037}}\).
A1: For obtaining a final probability in the range \([0.835, 0.837]\).

(a) Let \(X\) be the number of faults in a 4-week period.
Under the null hypothesis \(H_0\), \(X \sim \text{Po}(6)\) since the rate is \(1.5 \times 4 = 6\) faults per 4 weeks.
The hypotheses are:
\(H_0: \lambda = 6\) (or \(1.5\) per week)
\(H_1: \lambda < 6\) (or \(<1.5\) per week)

We require \(P(X \le c) \le 0.05\).
From the Poisson cumulative distribution tables for \(\mu = 6\):
\(P(X \le 1) = 0.0174\)
\(P(X \le 2) = 0.0620\)
Since \(0.0174 \le 0.05\) and \(0.0620 > 0.05\), the critical value is \(1\).
Therefore, the critical region is \(X \le 1\) (or \(X = 0, 1\)).

(b) The actual significance level of the test is the probability of rejecting \(H_0\) when it is true, which is:
\(P(X \le 1) = 0.0174\) or \(1.74\%\).

(c) Let \(Y\) be the number of faults in a 10-week period.
Under \(H_0\), the rate for 10 weeks is \(\lambda' = 1.5 \times 10 = 15\).
So, \(Y \sim \text{Po}(15)\).
The hypotheses are:
\(H_0: \lambda' = 15\)
\(H_1: \lambda' < 15\)

We observe \(y = 8\).
The p-value is:
\(P(Y \le 8)\)
From Poisson tables with \(\mu = 15\):
\(P(Y \le 8) = 0.0374\)
Since \(0.0374 < 0.05\) (or since 8 lies in the critical region, as the critical value for \(\mu = 15\) at 5% is 8 since \(P(Y \le 8) = 0.0374 \le 0.05\)), we reject \(H_0\).
There is sufficient evidence at the 5% level of significance to suggest that the rate of server faults has decreased.

(a)
B1: Both hypotheses correct with parameter \(\lambda\) or \(\mu\) defined. Accept \(H_0: \lambda = 1.5\), \(H_1: \lambda < 1.5\) or \(H_0: \lambda = 6\), \(H_1: \lambda < 6\).
M1: Realising that the distribution under \(H_0\) is \(\text{Po}(6)\).
M1: Attempt to find \(P(X \le c) \le 0.05\) by checking values from Poisson cumulative tables for \(\mu = 6\).
A1: Correct probabilities shown: \(P(X \le 1) = 0.0174\) and \(P(X \le 2) = 0.0620\).
A1: Correct critical region: \(X \le 1\) or \(X = 0, 1\).

(b)
B1: Correct actual significance level of \(0.0174\) (or \(1.74\%\)).

(c)
B1: Both hypotheses correct in terms of \(\lambda\) or \(\mu\). Accept \(H_0: \lambda = 15\), \(H_1: \lambda < 15\).
M1: Use of \(\text{Po}(15)\) as the test distribution.
M1: Finding the probability \(P(Y \le 8) = 0.0374\) from tables.
A1: Correct comparison of their probability with \(0.05\) (e.g., \(0.0374 < 0.05\)) and a decision to reject \(H_0\).
A1ft: Correct conclusion in context, e.g., 'There is sufficient evidence to support the claim that the rate of server faults has decreased.'

(a) Since \(f(x)\) is a probability density function, the total area under the curve is 1:
\[\int_{0}^{2} k x^2 \, dx + \int_{2}^{4} \frac{3}{10}(4 - x) \, dx = 1\]
Evaluating the first integral:
\[\int_{0}^{2} k x^2 \, dx = \left[ \frac{k x^3}{3} \right]_0^2 = \frac{8}{3}k\]
Evaluating the second integral:
\[\int_{2}^{4} \frac{3}{10}(4 - x) \, dx = \frac{3}{10} \left[ 4x - \frac{x^2}{2} \right]_2^4 = \frac{3}{10} \left[ (16 - 8) - (8 - 2) \right] = \frac{3}{10} (8 - 6) = \frac{6}{10} = \frac{3}{5}\]
Thus,
\[\frac{8}{3}k + \frac{3}{5} = 1 \implies \frac{8}{3}k = \frac{2}{5} \implies k = \frac{6}{40} = \frac{3}{20} \quad \text{(as required)}\]

(b) The cumulative distribution function \(F(x)\) is given by:
For \(x < 0\), \(F(x) = 0\).
For \(0 \le x \le 2\):
\[F(x) = \int_{0}^{x} \frac{3}{20} t^2 \, dt = \left[ \frac{t^3}{20} \right]_0^x = \frac{x^3}{20}\]
Note that \(F(2) = \frac{2^3}{20} = \frac{8}{20} = \frac{2}{5}\).

For \(2 < x \le 4\):
\[F(x) = F(2) + \int_{2}^{x} \frac{3}{10}(4 - t) \, dt = \frac{2}{5} + \frac{3}{10} \left[ 4t - \frac{t^2}{2} \right]_2^x\]
\[= \frac{2}{5} + \frac{3}{10} \left( 4x - \frac{x^2}{2} - 6 \right) = \frac{4}{10} + \frac{12x - 1.5x^2 - 18}{10} = \frac{12x - 1.5x^2 - 14}{10} = \frac{24x - 3x^2 - 28}{20}\]

For \(x > 4\), \(F(x) = 1\).

Thus,
\[F(x) = \begin{cases} 0 & x < 0 \\ \frac{x^3}{20} & 0 \le x \le 2 \\ \frac{24x - 3x^2 - 28}{20} & 2 < x \le 4 \\ 1 & x > 4 \end{cases}\]

(c) Using the cumulative distribution function:
\[P(1.5 < X < 3) = F(3) - F(1.5)\]
Using \(F(x)\) for \(2 < x \le 4\):
\[F(3) = \frac{24(3) - 3(3^2) - 28}{20} = \frac{72 - 27 - 28}{20} = \frac{17}{20} = 0.85\]
Using \(F(x)\) for \(0 \le x \le 2\):
\[F(1.5) = \frac{1.5^3}{20} = \frac{3.375}{20} = 0.16875\]
Therefore,
\[P(1.5 < X < 3) = 0.85 - 0.16875 = 0.68125 \text{ (or } \frac{109}{160}\text{)}\]

(a)
M1: Setting the sum of the two integrals equal to 1.
A1: Correctly evaluating the second integral as \(\frac{3}{5}\) (or \(0.6\)) and the first as \(\frac{8}{3}k\).
A1*: Completing the algebra to show \(k = \frac{3}{20}\) with no errors seen.

(b)
M1: Correct method for \(F(x)\) in the interval \(0 \le x \le 2\), integrating their \(k x^2\).
A1: \(F(x) = \frac{x^3}{20}\) (or equivalent) for \(0 \le x \le 2\).
M1: Correct method for \(F(x)\) in the interval \(2 < x \le 4\), involving \(F(2) + \int_{2}^{x} f(t) \, dt\).
A1: \(F(x) = \frac{24x - 3x^2 - 28}{20}\) (or equivalent) for \(2 < x \le 4\).
A1: Full definition of \(F(x)\) including the boundaries (0 for \(x < 0\) and 1 for \(x > 4\)).

(c)
M1: Using \(P(1.5 < X < 3) = F(3) - F(1.5)\) or integrating \(f(x)\) over the two separate intervals.
A1: Correct values for both components, e.g., \(F(3) = 0.85\) and \(F(1.5) = 0.16875\).
A1: Correct final answer of \(0.68125\) (or \(\frac{109}{160}\) or \(0.681\) to 3 s.f.).

(a) Let \(X\) be the number of defective ornaments in a box of 150.
\(X \sim \text{B}(150, 0.018)\).
Since \(n = 150\) is large and \(p = 0.018\) is small, we can approximate \(X\) using a Poisson distribution with mean:
\[\lambda = np = 150 \times 0.018 = 2.7\]
So \(X \approx Y \sim \text{Po}(2.7)\).

(i) For exactly 3 defective ornaments:
\[P(Y = 3) = \frac{e^{-2.7} \times 2.7^3}{3!} \approx 0.220 \quad \text{(3 s.f.)}\]

(ii) For more than 4 defective ornaments:
\[P(Y > 4) = 1 - P(Y \le 4)\]
Using the Poisson cumulative probability:
\[P(Y \le 4) = e^{-2.7} \left( 1 + 2.7 + \frac{2.7^2}{2} + \frac{2.7^3}{6} + \frac{2.7^4}{24} \right)\]
\[= e^{-2.7} \left( 1 + 2.7 + 3.645 + 3.2805 + 2.2143 \right) = e^{-2.7} \times 12.8398 \approx 0.8629\]
Thus,
\[P(Y > 4) = 1 - 0.8629 \approx 0.1371 \approx 0.137 \quad \text{(3 s.f.)}\]

(b) Let \(W\) be the number of boxes containing more than 4 defective ornaments out of 80.
We have \(W \sim \text{B}(80, 0.137)\).
Since \(n = 80\) is large, we can approximate \(W\) using a normal distribution:
Mean:
\[\mu = np = 80 \times 0.137 = 10.96\]
Variance:
\[\sigma^2 = np(1-p) = 10.96 \times (1 - 0.137) = 10.96 \times 0.863 = 9.4585\]
So, \(W \approx V \sim \text{N}(10.96, 9.4585)\).

We want to find \(P(W \ge 15)\). Applying a continuity correction:
\[P(W \ge 15) \approx P(V \ge 14.5)\]
Standardizing:
\[z = \frac{14.5 - 10.96}{\sqrt{9.4585}} = \frac{3.54}{3.0755} \approx 1.151\]
Using normal distribution tables:
\[P(Z \ge 1.15) = 1 - \Phi(1.15) = 1 - 0.8749 = 0.1251 \approx 0.125\]

(a)
B1: State the Poisson approximation to be used with \(\lambda = 2.7\).
M1: Using the Poisson formula to find \(P(Y = 3)\).
A1: \(0.220\) (accept answers in range \(0.220 - 0.221\)).
M1: Expressing \(P(Y > 4)\) as \(1 - P(Y \le 4)\) and attempting to sum at least 4 terms.
A1: \(0.137\) (accept answers in range \(0.137 - 0.138\)).

(b)
B1: Identifying the binomial model \(W \sim \text{B}(80, 0.137)\) and stating the mean \(\mu = 10.96\) and variance \(\sigma^2 \approx 9.46\).
M1: Correct use of continuity correction, transforming \(W \ge 15\) to \(V \ge 14.5\).
M1: Standardizing with their mean and standard deviation (must have square root of variance).
A1: Correct \(z\)-value of \(1.15\) (or \(1.151\)).
M1: Calculating \(1 - \Phi(z)\) for their \(z\)-value.
A1: Awrt \(0.125\) (accept \(0.124 - 0.126\)).

(a) Let \(\lambda\) represent the mean number of laptops sold per week. The hypotheses are \(H_0: \lambda = 2.5\) and \(H_1: \lambda > 2.5\). (b) Let \(X\) represent the number of laptops sold in a 4-week period. Under \(H_0\), the weekly rate is 2.5, so for a 4-week period, the mean is \(2.5 \times 4 = 10\). Thus, \(X \sim \text{Po}(10)\). We observe \(x = 15\). We want to find \(\text{P}(X \ge 15) = 1 - \text{P}(X \le 14)\). From the cumulative Poisson table with \(\lambda = 10\), we have \(\text{P}(X \le 14) = 0.9165\). Thus, \(\text{P}(X \ge 15) = 1 - 0.9165 = 0.0835\). Since \(0.0835 > 0.05\), this result is not significant at the 5% level. There is insufficient evidence to support the shopkeeper's belief that the mean weekly sales of laptops has increased. (Alternatively, the critical region for this test is \(X \ge 16\) since \(\text{P}(X \ge 16) = 1 - 0.9513 = 0.0487 < 0.05\) and \(\text{P}(X \ge 15) = 0.0835 > 0.05\). Since 15 does not lie within the critical region, we fail to reject \(H_0\)). (c) Let \(Y\) represent the number of desktop computers sold in a 10-week period. Under \(H_0\), the weekly mean is 0.8, so the 10-week mean is \(0.8 \times 10 = 8\). Thus, \(Y \sim \text{Po}(8)\). We require a two-tailed test with 10% significance, meaning we aim for approximately 5% (0.05) in each tail. For the lower tail: From cumulative Poisson tables with \(\lambda = 8\), \(\text{P}(Y \le 3) = 0.0424\) (difference from 0.05 is \(|0.0424 - 0.05| = 0.0076\)) and \(\text{P}(Y \le 4) = 0.0996\) (difference from 0.05 is \(|0.0996 - 0.05| = 0.0496\)). Since 0.0424 is closer to 0.05, the lower boundary of the critical region is \(Y \le 3\). For the upper tail: \(\text{P}(Y \ge 13) = 1 - \text{P}(Y \le 12) = 1 - 0.9362 = 0.0638\) (difference from 0.05 is \(|0.0638 - 0.05| = 0.0138\)) and \(\text{P}(Y \ge 14) = 1 - \text{P}(Y \le 13) = 1 - 0.9658 = 0.0342\) (difference from 0.05 is \(|0.0342 - 0.05| = 0.0158\)). Since 0.0638 is closer to 0.05, the upper boundary of the critical region is \(Y \ge 13\). Therefore, the critical region is \(Y \le 3\) or \(Y \ge 13\). (d) The actual significance level of this test is the sum of the probabilities of the two tails: \(\text{P}(Y \le 3) + \text{P}(Y \ge 13) = 0.0424 + 0.0638 = 0.1062\) (or 10.62%).

(a) B1: For both hypotheses correct with \(\lambda\) defined. Allow hypotheses defined in terms of the 4-week mean: \(H_0: \lambda = 10\), \(H_1: \lambda > 10\). (b) M1: For identifying the correct distribution \(\text{Po}(10)\). M1: For attempting to find \(\text{P}(X \ge 15)\) or establishing the critical value 16. A1: For \(\text{P}(X \ge 15) = 0.0835\) (or for finding critical region \(X \ge 16\)). M1: For comparing their probability with 0.05 (or comparing 15 with their critical value 16). A1: For a correct conclusion in context, stating that there is insufficient evidence to suggest that the mean number of weekly sales of laptops has increased. (c) M1: For identifying the correct distribution \(Y \sim \text{Po}(8)\). M1: For comparing \(\text{P}(Y \le 3)\) and \(\text{P}(Y \le 4)\) to 0.05 and selecting lower limit of 3. M1: For comparing \(\text{P}(Y \ge 13)\) and \(\text{P}(Y \ge 14)\) to 0.05 and selecting upper limit of 13. A1: For a correct critical region of \(Y \le 3\) or \(Y \ge 13\) (do not accept \(3 \ge Y \ge 13\)). (d) B1: For 0.1062 or 10.62%.

Let \(T\) be the total weight of a packed box.

\(T = B_1 + B_2 + \dots + B_{12} + C_1 + C_2 + \dots + C_{12} + W\)

(a)
\(\text{E}(T) = 12\text{E}(B) + 12\text{E}(C) + \text{E}(W) = 12(18) + 12(5) + 80 = 216 + 60 + 80 = 356\)

Since the individual variables are independent and normally distributed:
\(\text{Var}(T) = 12\text{Var}(B) + 12\text{Var}(C) + \text{Var}(W)\)
\(\text{Var}(T) = 12(0.8) + 12(0.2) + 4 = 9.6 + 2.4 + 4 = 16\)

Thus, \(T \sim \text{N}(356, 16)\).

We require \(\text{P}(T > 362)\):
\(\text{P}(T > 362) = \text{P}\left(Z > \frac{362 - 356}{\sqrt{16}}\right) = \text{P}(Z > 1.5)\)
From the normal distribution tables, \(\Phi(1.5) = 0.9332\).
So, \(\text{P}(Z > 1.5) = 1 - 0.9332 = 0.0668\).

(b)
Let \(Y = C_1 + C_2 + C_3 + C_4 - B\).

\(\text{E}(Y) = 4\text{E}(C) - \text{E}(B) = 4(5) - 18 = 2\)
\(\text{Var}(Y) = 4\text{Var}(C) + \text{Var}(B) = 4(0.2) + 0.8 = 0.8 + 0.8 = 1.6\)

Thus, \(Y \sim \text{N}(2, 1.6)\).

We require \(\text{P}(Y > 0)\):
\(\text{P}(Y > 0) = \text{P}\left(Z > \frac{0 - 2}{\sqrt{1.6}}\right) = \text{P}(Z > -1.5811)\)

Using \(Z = -1.58\):
\(\text{P}(Z > -1.58) = \Phi(1.58) = 0.9429\) (or \(0.9431\) if using calculator/interpolation with \(Z = -1.5811\)).

(c)
We must assume that the weights of all individual bottles, caps, and the box are independent of one another.

(a)
- M1: For setting up the correct expectation equation: \(\text{E}(T) = 12(18) + 12(5) + 80 = 356\).
- M1: For setting up the correct variance equation: \(\text{Var}(T) = 12(0.8) + 12(0.2) + 4 = 16\).
- M1: For standardising with their mean and standard deviation: \(\frac{362 - 356}{\sqrt{16}}\).
- A1: For obtaining 0.0668 (accept 0.0668 to 0.067).

(b)
- M1: For finding the mean of the linear combination: \(4(5) - 18 = 2\).
- M1: For finding the variance of the linear combination: \(4(0.2) + 0.8 = 1.6\).
- M1: For standardising: \(\frac{0 - 2}{\sqrt{1.6}}\).
- A1: For obtaining 0.9429 (accept 0.943).

(c)
- B1: For stating "independent" or "independence".
- B1: Contextualised to the weights of the bottles, caps, and boxes.

(a)
\(H_0\): There is no association between the type of sport played and the preferred time of day to exercise (the two variables are independent).
\(H_1\): There is an association between the type of sport played and the preferred time of day to exercise (the two variables are not independent).

(b)
First, we calculate the row, column, and grand totals:
- Row Totals: Football = \(18 + 32 + 30 = 80\), Tennis = \(22 + 18 + 20 = 60\), Swimming = \(20 + 10 + 30 = 60\).
- Column Totals: Morning = \(18 + 22 + 20 = 60\), Afternoon = \(32 + 18 + 10 = 60\), Evening = \(30 + 20 + 30 = 80\).
- Grand Total \(N = 200\).

Expected frequency for Football and Morning:
\(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{80 \times 60}{200} = 24\).

The remaining expected frequencies are:
- Football & Afternoon: \(\frac{80 \times 60}{200} = 24\)
- Football & Evening: \(\frac{80 \times 80}{200} = 32\)
- Tennis & Morning: \(\frac{60 \times 60}{200} = 18\)
- Tennis & Afternoon: \(\frac{60 \times 60}{200} = 18\)
- Tennis & Evening: \(\frac{60 \times 80}{200} = 24\)
- Swimming & Morning: \(\frac{60 \times 60}{200} = 18\)
- Swimming & Afternoon: \(\frac{60 \times 60}{200} = 18\)
- Swimming & Evening: \(\frac{60 \times 80}{200} = 24\)

(c)
We calculate \(\sum \frac{(O-E)^2}{E}\):
- Football & Morning: \(\frac{(18-24)^2}{24} = \frac{36}{24} = 1.5\)
- Football & Afternoon: \(\frac{(32-24)^2}{24} = \frac{64}{24} \approx 2.667\)
- Football & Evening: \(\frac{(30-32)^2}{32} = \frac{4}{32} = 0.125\)
- Tennis & Morning: \(\frac{(22-18)^2}{18} = \frac{16}{18} \approx 0.889\)
- Tennis & Afternoon: \(\frac{(18-18)^2}{18} = 0\)
- Tennis & Evening: \(\frac{(20-24)^2}{24} = \frac{16}{24} \approx 0.667\)
- Swimming & Morning: \(\frac{(20-18)^2}{18} = \frac{4}{18} \approx 0.222\)
- Swimming & Afternoon: \(\frac{(10-18)^2}{18} = \frac{64}{18} \approx 3.556\)
- Swimming & Evening: \(\frac{(30-24)^2}{24} = \frac{36}{24} = 1.5\)

Test Statistic:
\(X^2 = 1.5 + 2.667 + 0.125 + 0.889 + 0 + 0.667 + 0.222 + 3.556 + 1.5 = 11.125\)

Degrees of Freedom:
\(df = (r-1)(c-1) = (3-1)(3-1) = 4\)

At the 5% level of significance with 4 degrees of freedom, the critical value is:
\(\chi^2_4(0.05) = 9.488\)

Since our test statistic \(11.125 > 9.488\), we reject the null hypothesis \(H_0\).
There is significant evidence at the 5% level of significance to suggest that there is an association between the type of sport played by students and their preferred time of day to exercise.

(a)
- B1: Both hypotheses correctly stated in terms of "association" or "independence".
- B1: Hypotheses fully contextualised with both variables (sport played and preferred exercise time).

(b)
- M1: Showing correct method for at least one cell: e.g., \(\frac{80 \times 60}{200} = 24\).
- A1: Correctly completing all expected frequencies for Football: (24, 24, 32).
- A1: Correctly completing all remaining expected frequencies: Tennis (18, 18, 24) and Swimming (18, 18, 24).

(c)
- M1: Attempting to calculate \(\sum \frac{(O-E)^2}{E}\) with at least 4 terms correct.
- A1: Test statistic value of 11.125 (or \(11.1\) or \(11.13\)).
- B1: Correct degrees of freedom (\(df = 4\)) and critical value of 9.488.
- M1: Comparing their test statistic with their critical value.
- A1: Correct conclusion in context, rejecting \(H_0\).

(a)
Total weight of 100 bags = 100800 grams.
Unbiased estimate of the mean \(\mu\) is the sample mean, \(\bar{x}\):
\(\bar{x} = \frac{100800}{100} = 1008\) grams.

(b)
For a 95% confidence interval, the critical value \(z = 1.9600\).
Using the formula \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\):
\(1008 \pm 1.9600 \times \frac{15}{\sqrt{100}}\)
\(= 1008 \pm 1.9600 \times 1.5\)
\(= 1008 \pm 2.94\)

The confidence interval is \((1005.06, 1010.94)\) grams.

(c)
For a 99% confidence interval, the critical value \(z = 2.5758\) (or \(2.58\)).
The width of the interval is given by \(2 \times z \frac{\sigma}{\sqrt{N}}\).
We want:
\(2 \times 2.5758 \times \frac{15}{\sqrt{N}} \le 5\)
\(\frac{77.274}{\sqrt{N}} \le 5\)
\(\sqrt{N} \ge 15.4548\)
\(N \ge 238.85\)

Since \(N\) must be an integer, the minimum value of \(N\) is 239.
(If using \(z = 2.58\), \(\sqrt{N} \ge 15.48 \implies N \ge 239.63\), so \(N = 240\)).

(d)
The distribution of the weight of flour in a bag is not known to be normal. However, because the sample size is large (\(n = 100 > 30\)), the Central Limit Theorem ensures that the sample mean \(\bar{X}\) is approximately normally distributed, allowing us to use standard normal critical values.

(a)
- B1: Correctly calculating \(\bar{x} = 1008\) (must have correct units if stated, or just the number).

(b)
- B1: Using critical value \(z = 1.9600\) (accept 1.96).
- M1: For substituting their \(\bar{x}\), \(\sigma = 15\), and \(n=100\) into the confidence interval formula \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\).
- A1: For the lower bound of 1005.06 (accept 1005).
- A1: For the upper bound of 1010.94 (accept 1011).

(c)
- M1: For formulating the width expression: \(2 \times z \frac{15}{\sqrt{N}}\).
- B1: For using \(z = 2.5758\) (or 2.58).
- M1: Setting their width expression \(\le 5\) and solving for \(N\).
- A1: Correctly identifying the integer minimum value of \(N = 239\) (or 240 if \(z=2.58\) was used).

(d)
- B1: For explaining that since the parent distribution is unknown/not normal, the CLT is required to assume the sample mean is approximately normal for large \(n\).

(a) The sample mean is given by:
\(\bar{x} = \frac{\sum fx}{\sum f}\)
\(\sum fx = (0 \times 46) + (1 \times 72) + (2 \times 50) + (3 \times 22) + (4 \times 8) + (5 \times 2)\)
\(\sum fx = 0 + 72 + 100 + 66 + 32 + 10 = 280\)
\(\sum f = 200\)
\(\bar{x} = \frac{280}{200} = 1.4\)

(b) Since the sum of all expected frequencies must equal the total frequency of 200:
Expected frequency for \(x \ge 4\) is:
\(200 - (49.32 + 69.05 + 48.33 + 22.56) = 200 - 189.26 = 10.74\)
Alternatively, using the exact Poisson probability:
\(P(X \ge 4) = 1 - e^{-1.4}\left(1 + 1.4 + \frac{1.4^2}{2} + \frac{1.4^3}{6}\right) \approx 0.053725\)
Expected frequency \(= 200 \times 0.053725 = 10.75\)

(c)
\(H_0\): A Poisson distribution is a suitable model for the number of system alerts per hour.
\(H_1\): A Poisson distribution is not a suitable model for the number of system alerts per hour.

Since the expected frequency for \(x \ge 4\) is 10.74 (or 10.75), which is \(\ge 5\), no further pooling is needed, but we must group the observed frequencies for \(x = 4\) and \(x = 5\) into \(x \ge 4\).
Observed frequency for \(x \ge 4\) is \(8 + 2 = 10\).

We set up the test statistic calculation:
- \(x = 0\): \(O = 46\), \(E = 49.32\), \(\frac{(O-E)^2}{E} = \frac{(46 - 49.32)^2}{49.32} = 0.2235\)
- \(x = 1\): \(O = 72\), \(E = 69.05\), \(\frac{(O-E)^2}{E} = \frac{(72 - 69.05)^2}{69.05} = 0.1260\)
- \(x = 2\): \(O = 50\), \(E = 48.33\), \(\frac{(O-E)^2}{E} = \frac{(50 - 48.33)^2}{48.33} = 0.0577\)
- \(x = 3\): \(O = 22\), \(E = 22.56\), \(\frac{(O-E)^2}{E} = \frac{(22 - 22.56)^2}{22.56} = 0.0139\)
- \(x \ge 4\): \(O = 10\), \(E = 10.74\), \(\frac{(O-E)^2}{E} = \frac{(10 - 10.74)^2}{10.74} = 0.0510\) (or 0.0523 using 10.75)

Test statistic:
\(\chi^2 = \sum \frac{(O-E)^2}{E} \approx 0.2235 + 0.1260 + 0.0577 + 0.0139 + 0.0510 = 0.4721\) (or 0.4734 using 10.75)

Degrees of freedom:
Number of groups after pooling is 5. We estimated 1 parameter (\(\lambda = 1.4\)) from the data.
\(\nu = 5 - 1 - 1 = 3\).

Critical Value:
At the 5% level of significance with \(\nu = 3\), the critical value is \(\chi^2_3(0.05) = 7.815\).

Conclusion:
Since the test statistic \(0.472 < 7.815\), we fail to reject \(H_0\). There is insufficient evidence to suggest that the Poisson distribution is not a suitable model. We conclude that the Poisson distribution with mean 1.4 is a suitable model for these data.

(a)
- M1: Standard formula for sample mean, \(\frac{\sum fx}{\sum f}\), seen with numbers.
- A1: Correctly shows 1.4 (must show numerator 280 and denominator 200 or equivalent).

(b)
- M1: For \(200 - (\text{sum of other expected frequencies})\) or using the exact Poisson probability.
- A1: 10.74 or 10.75.

(c)
- B1: Both hypotheses correct with reference to Poisson and context.
- M1: Realising that \(x = 4\) and \(x = 5\) need to be pooled (observed frequency = 10).
- M1: Calculating \(\frac{(O-E)^2}{E}\) or \(\frac{O^2}{E}\) with at least 3 correct terms.
- A1: Test statistic value in range \([0.47, 0.48]\).
- B1: Degrees of freedom \(\nu = 3\).
- B1: Critical value of 7.815.
- M1: Comparing their test statistic with their critical value.
- A1: Conclusion in context, stating that there is no evidence that the Poisson distribution is not a suitable model (do not accept definitive statements like 'the data is Poisson').

(a) Let \(T\) be the total weight of a packed box:
\(T = X_1 + X_2 + X_3 + X_4 + Y_1 + Y_2 + W\)

Since the variables are independent:
\(\text{E}(T) = 4\text{E}(X) + 2\text{E}(Y) + \text{E}(W)\)
\(\text{E}(T) = 4(150) + 2(240) + 80 = 600 + 480 + 80 = 1160\text{ g}\)

\(\text{Var}(T) = 4\text{Var}(X) + 2\text{Var}(Y) + \text{Var}(W)\)
\(\text{Var}(T) = 4(16) + 2(36) + 4 = 64 + 72 + 4 = 140\text{ g}^2\)

Thus, \(T \sim \text{N}(1160, 140)\).

We want to find \(P(T < 1150)\):
\(Z = \frac{1150 - 1160}{\sqrt{140}} = \frac{-10}{11.832} \approx -0.8452\)

Using standard normal tables:
\(P(Z < -0.85) = 1 - 0.8023 = 0.1977\)
Using linear interpolation or precise values:
\(P(Z < -0.8452) \approx 0.1990\) (or 0.199 to 3 s.f.)

(b) Let \(\bar{T}\) be the sample mean of 10 independent packed boxes.
\(\bar{T} \sim \text{N}\left(\mu_T, \frac{\sigma_T^2}{10}\right)\)
\(\bar{T} \sim \text{N}\left(1160, \frac{140}{10}\right) \implies \bar{T} \sim \text{N}(1160, 14)\)

We want to find \(P(\bar{T} > 1162)\):
\(Z = \frac{1162 - 1160}{\sqrt{14}} = \frac{2}{3.7417} \approx 0.5345\)

Using standard tables:
\(P(Z > 0.53) = 1 - 0.7019 = 0.2981\)
Using more precise standard normal tables or linear interpolation for \(z = 0.5345\):
\(P(Z > 0.5345) \approx 0.2965\) (or 0.296 to 3 s.f.)

(a)
- M1: Setting up the linear combination for \(T\) and stating its mean \(\text{E}(T) = 1160\).
- M1: Attempting to calculate \(\text{Var}(T) = 4\text{Var}(X) + 2\text{Var}(Y) + \text{Var}(W)\). Must not use \(4^2\text{Var}(X)\).
- A1: Correct variance of 140.
- M1: Standardising with 1150, their mean, and their standard deviation.
- A1: Correct probability in range \([0.197, 0.200]\).

(b)
- M1: Stating or using the distribution of the sample mean \(\bar{T}\) with mean \(\mu_T\).
- M1: Stating or using the variance of \(\bar{T}\) as \(\frac{\text{Var}(T)}{10}\).
- A1: Variance of 14 (or standard deviation of \(\sqrt{14}\)).
- M1: Standardising 1162 with their mean and \(\sqrt{14}\).
- A1: Finding \(1 - \Phi(z)\) for their \(z\) value.
- A1: Correct probability in range \([0.295, 0.299]\).

(a) The lifetimes of the batteries are not stated to be normally distributed. Since both sample sizes are large (\(n_P = 80 > 30\) and \(n_Q = 100 > 30\)), the Central Limit Theorem allows us to assume that the sample means \(\bar{X}_P\) and \(\bar{X}_Q\) are approximately normally distributed, which justifies the use of the standard normal (\(z\)) test.

(b) Hypotheses:
\(H_0: \mu_P = \mu_Q\)
\(H_1: \mu_P \neq \mu_Q\)

We use a two-tailed \(z\)-test.
The standard error of the difference between the means is:
\(\text{SE} = \sqrt{\frac{s_P^2}{n_P} + \frac{s_Q^2}{n_Q}} = \sqrt{\frac{4.8^2}{80} + \frac{5.5^2}{100}} = \sqrt{\frac{23.04}{80} + \frac{30.25}{100}}\)
\(\text{SE} = \sqrt{0.288 + 0.3025} = \sqrt{0.5905} \approx 0.7684\)

The test statistic is:
\(z = \frac{\bar{x}_P - \bar{x}_Q}{\text{SE}} = \frac{45.2 - 43.5}{0.7684} = \frac{1.7}{0.7684} \approx 2.212\)

Critical value:
At the 5% level of significance for a two-tailed test, the critical values are \(\pm 1.96\).

Conclusion:
Since \(2.212 > 1.96\), we reject \(H_0\). There is significant evidence at the 5% level of significance to suggest a difference between the mean lifetimes of Brand \(P\) and Brand \(Q\) batteries.

(c) Any one of the following:
- The two samples are independent of each other.
- The sample standard deviations are unbiased estimates of the population standard deviations.

(a)
- B1: Explaining that the lifetimes are not normally distributed and that large samples allow normal approximation of sample means.

(b)
- B1: Correct hypotheses, using \(\mu_P\) and \(\mu_Q\) with clear definitions.
- M1: For attempting to calculate the standard error \(\sqrt{\frac{s_P^2}{n_P} + \frac{s_Q^2}{n_Q}}\).
- A1: Standard error of \(\approx 0.768\) (accept 0.77).
- M1: Standardising the difference between the means, i.e., \(\frac{1.7}{\text{SE}}\).
- A1: Test statistic \(z \approx 2.21\) (accept range \([2.20, 2.22]\)).
- B1: Critical values \(\pm 1.96\).
- M1: Comparing their test statistic with their critical value.
- A1: Rejection of \(H_0\) and a fully correct conclusion in context.

(c)
- B2: Clearly stating an appropriate assumption (e.g., independence of samples, or sample variances are good estimates of population variances). Accept 1 mark for partially complete explanation.

(a) Let \(T = C + B\) be the total weight of a packed box.
Since \(C\) and \(B\) are independent normal random variables,
\[\text{E}(T) = \text{E}(C) + \text{E}(B) = 375 + 40 = 415\]
\[\text{Var}(T) = \text{Var}(C) + \text{Var}(B) = 15.75 + 0.25 = 16\]
So, \(T \sim \text{N}(415, 16)\).
We want to find \(\text{P}(T < 410)\).
Standardising:
\[Z = \frac{410 - 415}{\sqrt{16}} = \frac{-5}{4} = -1.25\]
\[\text{P}(T < 410) = \text{P}(Z < -1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056\]

(b) Let \(S = T_1 + T_2 + \dots + T_9\) be the total weight of 9 packed boxes.
\[\text{E}(S) = 9 \times \text{E}(T) = 9 \times 415 = 3735\]
\[\text{Var}(S) = 9 \times \text{Var}(T) = 9 \times 16 = 144\]
So, \(S \sim \text{N}(3735, 144)\).
We want to find \(\text{P}(S > 3753)\).
Standardising:
\[Z = \frac{3753 - 3735}{\sqrt{144}} = \frac{18}{12} = 1.5\]
\[\text{P}(S > 3753) = \text{P}(Z > 1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668\]

(c) We want to find \(\text{P}(C > 9B)\), which is equivalent to \(\text{P}(C - 9B > 0)\).
Let \(W = C - 9B\).
\[\text{E}(W) = \text{E}(C) - 9\text{E}(B) = 375 - 9(40) = 15\]
\[\text{Var}(W) = \text{Var}(C) + 9^2\text{Var}(B) = 15.75 + 81(0.25) = 15.75 + 20.25 = 36\]
So, \(W \sim \text{N}(15, 36)\).
We want to find \(\text{P}(W > 0)\).
Standardising:
\[Z = \frac{0 - 15}{\sqrt{36}} = \frac{-15}{6} = -2.5\]
\[\text{P}(W > 0) = \text{P}(Z > -2.5) = \text{P}(Z < 2.5) = \Phi(2.5) = 0.9938\]

(a)
- M1: For finding \(\text{E}(T) = 375 + 40 = 415\).
- A1: For finding \(\text{Var}(T) = 15.75 + 0.25 = 16\).
- A1: For standardising correctly and finding the probability \(0.1056\) (accept awrt 0.106).

(b)
- B1: For finding \(\text{E}(S) = 3735\).
- M1: For attempting to find \(\text{Var}(S) = 9 \times 16 = 144\) (do not accept \(9^2 \times 16\)).
- M1: For standardising with their mean and standard deviation: \(\frac{3753 - \text{mean}}{\sqrt{\text{variance}}}\).
- A1: For standardising to \(1.5\) and finding the probability \(0.0668\) (accept awrt 0.067).

(c)
- M1: For attempting to find \(\text{E}(C - 9B) = 15\).
- M1: For attempting to find \(\text{Var}(C - 9B)\) using the correct formula \(\text{Var}(C) + 9^2\text{Var}(B)\).
- A1: For finding \(\text{Var}(C - 9B) = 36\).
- A1: For standardising to \(-2.5\) and finding the probability \(0.9938\) (accept awrt 0.994).