2023 Edexcel IAL Mathematics (YMA01) Practice Paper with Answers

Let the initial list be: 11.2, 5.4, 8.5, 3.2, 9.1, 7.6, 4.3, 6.7. (a) Bubble Sort descending: Pass 1: Compare adjacent elements and swap if the left is smaller than the right. 11.2 vs 5.4 (no swap), 5.4 vs 8.5 (swap), 5.4 vs 3.2 (no swap), 3.2 vs 9.1 (swap), 3.2 vs 7.6 (swap), 3.2 vs 4.3 (swap), 3.2 vs 6.7 (swap). State at end of Pass 1: 11.2, 8.5, 5.4, 9.1, 7.6, 4.3, 6.7, 3.2. Pass 2: 11.2 vs 8.5 (no swap), 8.5 vs 5.4 (no swap), 5.4 vs 9.1 (swap), 5.4 vs 7.6 (swap), 5.4 vs 4.3 (no swap), 4.3 vs 6.7 (swap). State at end of Pass 2: 11.2, 8.5, 9.1, 7.6, 5.4, 6.7, 4.3, 3.2. Pass 3: 11.2 vs 8.5 (no swap), 8.5 vs 9.1 (swap), 8.5 vs 7.6 (no swap), 7.6 vs 5.4 (no swap), 5.4 vs 6.7 (swap). State at end of Pass 3: 11.2, 9.1, 8.5, 7.6, 6.7, 5.4, 4.3, 3.2. Pass 4: No swaps occur. List is sorted. (b) First-Fit Decreasing: Sorted list is 11.2, 9.1, 8.5, 7.6, 6.7, 5.4, 4.3, 3.2. Capacity = 20. Item 11.2 goes into Box 1 (space left 8.8). Item 9.1 goes into Box 2 (space left 10.9). Item 8.5 goes into Box 3 (space left 11.5). Item 7.6 goes into Box 2 (space left 3.3). Item 6.7 goes into Box 1 (space left 2.1). Item 5.4 goes into Box 3 (space left 6.1). Item 4.3 goes into Box 3 (space left 1.8). Item 3.2 goes into Box 2 (space left 0.1). Final packing: Box 1: [11.2, 6.7], Box 2: [9.1, 7.6, 3.2], Box 3: [8.5, 5.4, 4.3]. (c) Lower bound calculation: Sum of weights = 11.2 + 9.1 + 8.5 + 7.6 + 6.7 + 5.4 + 4.3 + 3.2 = 56.0. Lower bound = 56.0 / 20 = 2.8. Since the lower bound is 2.8, the minimum integer number of boxes needed is 3. Since we achieved packing in 3 boxes, it is optimal.

Part (a): M1 for a correct first pass of bubble sort. A1 for correct end of Pass 1 state. M1 for Pass 2 and Pass 3. A1 for the correct sorted list. A1 for stating that the algorithm stops after Pass 4 with no swaps. Part (b): M1 for ordering the list first. M1 for placing at least 5 items correctly. A1 for Box 1 and Box 2 correct. A1 for Box 3 correct. Part (c): M1 for calculating the sum and dividing by 20. A1 for correct lower bound of 3. A1 for a clear conclusion linking the lower bound to the number of boxes used.

Dijkstra's Algorithm details: Node A: Final Label = 0, Order of labelling = 1. Temporary labels at B (12), C (8), D (15). Node C: Final Label = 8, Order = 2. Temporary labels updated: D becomes 14, E becomes 19, F becomes 22. Node B: Final Label = 12, Order = 3. Temporary label at E remains 19. Node D: Final Label = 14, Order = 4. Temporary labels updated: F becomes 21, G becomes 32. Node E: Final Label = 19, Order = 5. Temporary labels updated: H becomes 32. Node F: Final Label = 21, Order = 6. Temporary labels updated: G becomes 30, H remains 32. Node G: Final Label = 30, Order = 7. Temporary label at H remains 32. Node H: Final Label = 32, Order = 8. (b) The shortest paths are A - C - E - H (length 8 + 11 + 13 = 32) and A - C - D - F - H (length 8 + 6 + 7 + 11 = 32). (c) The computer works backwards from H. At each vertex V, it checks if d(V) - d(U) = weight(UV) for any connected vertex U. If so, UV is on a shortest path. For example, for H (32), it checks E (32 - 19 = 13, yes) and F (32 - 21 = 11, yes).

Part (a): M1 for a systematic method starting at A. A1 for correct final labels and order for A, C, B. A1 for correct final labels and order for D, E, F. A1 for correct final labels and order for G, H. A1 for showing all correct temporary labels at all vertices. Part (b): B1 for length 32. B1 for path A - C - E - H. B1 for path A - C - D - F - H. Part (c): M1 for explaining backtracking from the destination to the start. A1 for stating the formula d(V) - d(U) = weight(UV) and showing a correct sample step.

(a) A network can only be traversed without repeating any edges if all vertices have even degrees (an Eulerian graph). Since this network contains odd vertices, some edges must be repeated. (b) Let's list the degrees: P: 2, Q: 2, R: 4, S: 3 (odd), T: 3 (odd), U: 3 (odd), V: 1 (odd). Odd vertices are S, T, U, and V. (c) Possible pairings of the odd vertices: Pair 1: ST and UV. Shortest path ST = 80, UV = 70. Total = 80 + 70 = 150. Pair 2: SU and TV. Shortest path SU = 140, TV = T-U-V = 170 + 70 = 240. Total = 140 + 240 = 380. Pair 3: SV and TU. Shortest path SV = S-U-V = 140 + 70 = 210, TU = 170. Total = 210 + 170 = 380. The minimum sum is 150, which is achieved by repeating edges ST and UV. Total length of all edges = 120 + 150 + 90 + 100 + 110 + 80 + 140 + 170 + 70 = 1030m. Minimum route length = 1030 + 150 = 1180m. (d) If the inspector can start and finish at different vertices, these two vertices will be the only odd vertices in the modified route, meaning the other two odd vertices must be paired and repeated. To minimize the route, we choose the start/finish pair that leaves the smallest pairing for the other two. Possible pairs to repeat: Repeat UV (70), leaving S and T as start/finish. Total length = 1030 + 70 = 1100m. Repeat ST (80), leaving U and V as start/finish. Total length = 1030 + 80 = 1110m. Repeat SU (140), leaving T and V as start/finish. Total length = 1030 + 140 = 1170m. Repeat SV (210), leaving T and U as start/finish. Total length = 1030 + 210 = 1240m. Repeat TU (170), leaving S and V as start/finish. Total length = 1030 + 170 = 1200m. Repeat TV (240), leaving S and U as start/finish. Total length = 1030 + 240 = 1270m. The minimum route length is 1100m, starting at S and finishing at T (or vice versa).

Part (a): B1 for stating that odd vertices exist or the graph is non-Eulerian. Part (b): B2 for listing all four odd vertices S, T, U, V (award B1 for two or three correct). Part (c): M1 for listing the three pairings of odd vertices. A1 for correct shortest paths for ST and UV, SU and TV, SV and TU. A1 for correct totals of 150, 380, and 380. A1 for identifying ST and UV as the edges to repeat. B1 for total weight 1030. A1 for final route length 1180. Part (d): M1 for explaining that we choose start/finish to leave the smallest remaining pairing to repeat. A1 for identifying UV of length 70 as the repeated path. A1 for 1100m and start/finish at S and T.

(a) Let x be the number of Classic boxes and y be the number of Premium boxes. Maximize Profit P = 15x + 25y, subject to: Assembly time: 2x + y <= 120. Polishing time: x + 3y <= 150. Lower demand limit: y >= 0.5x (or 2y >= x). Upper demand limit: y <= 3x. Minimum Classic boxes: x >= 10. Non-negativity: x, y >= 0. (b) Graphing the boundary lines: Line 1: 2x + y = 120 passes through (60, 0) and (0, 120). Line 2: x + 3y = 150 passes through (150, 0) and (0, 50). Line 3: y = 0.5x passes through (0, 0) and (40, 20). Line 4: y = 3x passes through (0, 0) and (20, 60). Line 5: x = 10 is a vertical line. Shading the unwanted regions leaves the feasible region R. (c) The vertices of R are: 1) Intersection of x = 10 and y = 0.5x: (10, 5). P = 15(10) + 25(5) = 150 + 125 = £275. 2) Intersection of x = 10 and y = 3x: (10, 30). P = 15(10) + 25(30) = 150 + 750 = £900. 3) Intersection of y = 3x and x + 3y = 150: (15, 45). P = 15(15) + 25(45) = 225 + 1125 = £1350. 4) Intersection of 2x + y = 120 and x + 3y = 150: (42, 36). P = 15(42) + 25(36) = 630 + 900 = £1530. 5) Intersection of 2x + y = 120 and y = 0.5x: (48, 24). P = 15(48) + 25(24) = 720 + 600 = £1320. Comparing these, the maximum profit is £1530, achieved with 42 Classic and 36 Premium boxes.

Part (a): B1 for defining the variables and writing the correct profit function. B1 for 2x + y <= 120 and x + 3y <= 150. B1 for y >= 0.5x and y <= 3x. B1 for x >= 10 and x, y >= 0. Part (b): M1 for drawing at least three boundary lines correctly. A1 for drawing all five boundary lines correctly. A1 for correctly shading the unwanted regions. A1 for clearly labelling the feasible region R. Part (c): M1 for testing at least two vertices of the feasible region, or drawing a correct objective line. A1 for finding the coordinates of the key vertex (42, 36). A1 for 42 Classic and 36 Premium. A1 for maximum profit £1530.

(a) Constructing the activity network: Activities A and B start at Node 1. Node 2 is the end of A. Node 4 is the end of B. C starts at Node 2, E starts at Node 4. D depends on both A and B, so we use dummy 1 from Node 2 to Node 3, and dummy 2 from Node 4 to Node 3. D then starts at Node 3 and ends at Node 5. F depends on C and D, and G depends on D and E. Thus, we use dummy 3 from Node 5 to Node 6 (where C ends and F starts), and dummy 4 from Node 5 to Node 7 (where E ends and G starts). F ends at Node 8, and G ends at Node 8. H starts at Node 8 and ends at Node 9. (b) Earliest event times: Node 1 = 0. Node 2 (end of A) = 4. Node 4 (end of B) = 6. Node 3 (start of D) = max(4, 6) = 6. Node 5 (end of D) = 6 + 3 = 9. Node 6 (start of F) = max(4 + 5, 9) = 9. Node 7 (start of G) = max(6 + 7, 9) = 13. Node 8 (start of H) = max(9 + 4, 13 + 8) = 21. Node 9 (end) = 21 + 5 = 26. Latest event times: Node 9 = 26. Node 8 = 26 - 5 = 21. Node 6 = 21 - 4 = 17. Node 7 = 21 - 8 = 13. Node 5 = min(17, 13) = 13. Node 3 = 13 - 3 = 10. Node 2 = min(17 - 5, 10) = 10. Node 4 = min(13 - 7, 10) = 6. Node 1 = min(10 - 4, 6 - 6) = 0. (c) Critical activities have zero float: B (6 - 0 - 6 = 0), E (13 - 6 - 7 = 0), G (21 - 13 - 8 = 0), H (26 - 21 - 5 = 0). So critical activities are B, E, G, H. Project duration = 26 days. (d) Total float for D = Latest finish of D - Earliest start of D - Duration of D = Latest event time of Node 5 - Earliest event time of Node 3 - Duration of D = 13 - 6 - 3 = 4 days.

Part (a): M1 for starting A and B at a single node. A1 for correct placement of C and E. M1 for using dummies correctly to feed A and B into D. A1 for using dummies correctly to feed D into F and G. A1 for H ending at a single terminal node. Part (b): M1 for a forward pass. A1 for correct earliest event times at all nodes. M1 for a backward pass. A1 for correct latest event times at all nodes. Part (c): B1 for correct minimum completion time of 26 days. B1 for listing all critical activities: B, E, G, H. Part (d): M1 for a correct float formula using their values. A1 for total float of 4 days.

(a) A bipartite graph has two sets of vertices. Left set: {A, B, C, D, E}. Right set: {F, M, O, T, H}. Edges connect preferences: A to {F, T}, B to {O, T}, C to {F, M, H}, D to {M, O}, E to {O, H}. (b) The second graph shows only the matched edges: A - F, C - M, D - O, E - H. (c) An alternating path must start at an unmatched student (B) and alternate between unmatched and matched edges, ending at an unmatched internship (T). Starting at B (unmatched): B - O (unmatched edge), O - D (matched edge), D - M (unmatched edge), M - C (matched edge), C - F (unmatched edge), F - A (matched edge), A - T (unmatched edge). Since T is unmatched, this is a complete alternating path: B - O = D - M = C - F = A - T. Changing the state of the edges in the matching (unmatched becomes matched and vice versa): B - O, D - M, C - F, A - T are now matched. The untouched match E - H remains. The complete matching is: A - T, B - O, C - F, D - M, E - H. (d) If Alice prefers only Finance, Emily prefers only Finance, and Chloe only prefers Finance and Marketing, then the subset of students S = {Alice, Chloe, Emily} has a combined preference set of only {Finance, Marketing}. Since the size of the student subset (|S| = 3) is greater than the size of their combined preference set (2), by Hall's Marriage Theorem, it is impossible to find a complete matching because we cannot assign three distinct internships to these three students.

Part (a): M1 for two sets of vertices. A1 for correct edges for Alice, Ben, and Chloe. A1 for correct edges for Daniel and Emily. Part (b): B2 for drawing the initial matching correctly (award B1 if one error). Part (c): M1 for a path starting at B and ending at T. A1 for the correct alternating path: B - O = D - M = C - F = A - T (or equivalent notation). M1 for demonstrating the change in matching status. A1 for listing the correct final matching. Part (d): M1 for identifying the subset of students {Alice, Chloe, Emily}. A1 for explaining that three students only prefer two internships, making a complete matching impossible.

(a) The position vector of \( B \) at time \( t \) hours is given by:
\(\mathbf{r}_B = (3\mathbf{i} + 12\mathbf{j}) + t(2\mathbf{i} - 5\mathbf{j}) = (3 + 2t)\mathbf{i} + (12 - 5t)\mathbf{j}\)

(b) (i) The position vector of \( C \) at time \( t \) hours is:
\(\mathbf{r}_C = (15\mathbf{i} - 4\mathbf{j}) + t(-\mathbf{i} + u\mathbf{j}) = (15 - t)\mathbf{i} + (-4 + ut)\mathbf{j}\)

For a collision to occur, there must be a value of \( t > 0 \) such that \(\mathbf{r}_B = \mathbf{r}_C\).
Equating the \(\mathbf{i}\) components:
\(3 + 2t = 15 - t \implies 3t = 12 \implies t = 4\) hours.

Equating the \(\mathbf{j}\) components at \( t = 4 \):
\(12 - 5(4) = -4 + u(4)\)
\(-8 = -4 + 4u \implies 4u = -4 \implies u = -1\).

(ii) The collision occurs \( 4 \) hours after 12:00, which is at 16:00.
Substituting \( t = 4 \) into \(\mathbf{r}_B\):
\(\mathbf{r}_B = (3 + 2(4))\mathbf{i} + (12 - 5(4))\mathbf{j} = 11\mathbf{i} - 8\mathbf{j}\).
Thus, the position vector of the point of collision is \(11\mathbf{i} - 8\mathbf{j}\) (in km).

(a) M1: Attempts \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) for boat \( B \). A1: Correct expression: \((3 + 2t)\mathbf{i} + (12 - 5t)\mathbf{j}\) (or equivalent).

(b)(i) M1: Writes down an expression for \(\mathbf{r}_C\) in terms of \( t \) and \( u \). M1: Equates \(\mathbf{i}\) components to find the collision time. A1: Obtains \( t = 4 \). M1: Equates \(\mathbf{j}\) components using their \( t \) to form an equation in \( u \). A1: Correct value \( u = -1 \).

(b)(ii) B1: Correct time of 4 hours (or 16:00). M1: Substitutes their \( t \) into their \(\mathbf{r}_B\) or \(\mathbf{r}_C\). A1: Correct vector \(11\mathbf{i} - 8\mathbf{j}\).

(a) Consider the system as a whole (truck + trailer):
Total mass, \( M = 1200 + 800 = 2000 \text{ kg} \).
Equation of motion:
\( D - 400 - 200 = M a \)
\( D - 600 = 2000 \times 1.5 \)
\( D - 600 = 3000 \implies D = 3600 \text{ N} \).

(b) Consider the motion of the trailer only:
Let the tension in the towbar be \( T \) N.
Equation of motion:
\( T - 200 = 800 \times 1.5 \)
\( T - 200 = 1200 \implies T = 1400 \text{ N} \).

(c) When the towbar breaks, the only horizontal force acting on the trailer is the resistance force of 200 N.
Let the new acceleration (deceleration) of the trailer be \( a_t \).
\( -200 = 800 a_t \implies a_t = -0.25 \text{ m s}^{-2} \).
Using \( v^2 = u^2 + 2as \) with \( u = 15 \), \( v = 0 \), and \( a_t = -0.25 \):
\( 0^2 = 15^2 + 2(-0.25)s \)
\( 0 = 225 - 0.5s \implies 0.5s = 225 \implies s = 450 \text{ m} \).

(a) M1: Attempts equation of motion for the whole system with correct number of terms. A1: Correctly substituted equation: \( D - 600 = 2000 \times 1.5 \). A1: \( D = 3600 \).

(b) M1: Attempts equation of motion for either the truck or the trailer. A1: Correct equation for trailer: \( T - 200 = 800 \times 1.5 \) (or for truck: \( 3600 - T - 400 = 1200 \times 1.5 \)). A1: \( T = 1400 \text{ N} \).

(c) M1: Attempts equation of motion for the trailer after the towbar breaks. A1: Finds acceleration \( a = -0.25 \text{ m s}^{-2} \). M1: Uses a constant acceleration formula to find distance \( s \) with their \( a \) and \( u = 15 \), \( v = 0 \). A1: Correct substitution: \( 0 = 225 - 0.5s \). A1: \( s = 450 \text{ m} \).

(a) The forces acting on the particle are:
1. Weight \( mg \) acting vertically downwards.
2. Normal reaction \( R \) acting perpendicular to and out of the plane.
3. Force of magnitude \( 12 \text{ N} \) acting up the plane.
4. Friction force \( F_{\max} = \mu R \) acting up the plane (opposing the tendency to slide down).

(b) (i) Since \( \tan \alpha = 0.75 \), we have \( \sin \alpha = 0.6 \) and \( \cos \alpha = 0.8 \).
Resolving perpendicular to the plane:
\( R = mg \cos \alpha = 0.8 mg \).
Thus, maximum friction is \( F_{\max} = \mu R = 0.8 \mu mg \).

Case 1: On the point of sliding down the plane (friction acts up the plane):
\( 12 + \mu R = mg \sin \alpha \)
\( 12 + 0.8 \mu mg = 0.6 mg \) (Equation 1)

Case 2: On the point of sliding up the plane (friction acts down the plane):
\( 24 = mg \sin \alpha + \mu R \)
\( 24 = 0.6 mg + 0.8 \mu mg \) (Equation 2)

Adding Equation 1 and Equation 2:
\( 12 + 24 = (0.6 mg - 0.8 \mu mg) + (0.6 mg + 0.8 \mu mg) \)
\( 36 = 1.2 mg \implies mg = 30 \).
Using \( g = 9.8 \text{ m s}^{-2} \):
\( m = \frac{30}{9.8} \approx 3.06 \text{ kg} \) (or \( 3.1 \text{ kg} \) to 2 s.f.).

(ii) Substitute \( mg = 30 \) into Equation 1:
\( 12 + 0.8 \mu (30) = 0.6 (30) \)
\( 12 + 24 \mu = 18 \)
\( 24 \mu = 6 \implies \mu = 0.25 \).

(a) B1: Correctly shows weight, normal reaction, and 12 N force in correct directions. B1: Friction shown pointing up the inclined plane with correct labelling.

(b)(i) B1: Finds \( \sin \alpha = 0.6 \) and \( \cos \alpha = 0.8 \). M1: Resolves perpendicular to the plane to find \( R = 0.8 mg \). M1: Sets up the equation of equilibrium parallel to the plane when sliding down (Eq 1). M1: Sets up the equation of equilibrium parallel to the plane when sliding up (Eq 2). M1: Solves the simultaneous equations to find \( mg \) or \( m \). A1: \( m \approx 3.06 \text{ kg} \) (accept \( 3.1 \text{ kg} \) or exact fraction \(\frac{150}{49}\)).

(b)(ii) M1: Substitutes their \( mg \) value back into one of the equilibrium equations to solve for \( \mu \). A1: Correct working shown. A1: \( \mu = 0.25 \).

(a) The graph starts at \( (0, 4) \). It rises linearly with a gradient of 1.5 up to the point \( (T, 4 + 1.5T) \). It then decreases linearly with a gradient of -0.5 for a duration of \( 2T \) seconds (reaching \( t = 3T \)) to the point \( (3T, V) \).

(b) During the first stage (constant acceleration \( 1.5 \) for \( T \) seconds):
\( v(T) = u + a_1 T = 4 + 1.5T \).
During the second stage (constant deceleration \( 0.5 \) for \( 2T \) seconds):
\( V = v(T) - 0.5(2T) = 4 + 1.5T - T = 4 + 0.5T \). (As shown).

(c) The distance traveled is the area under the speed-time graph. This is the sum of the areas of the two trapeziums representing the two phases of motion.

Area of the first trapezium (from \( t = 0 \) to \( t = T \)):
\( A_1 = \frac{1}{2}(4 + (4 + 1.5T)) \times T = \frac{1}{2}(8 + 1.5T)T = 4T + 0.75T^2 \).

Area of the second trapezium (from \( t = T \) to \( t = 3T \)):
\( A_2 = \frac{1}{2}((4 + 1.5T) + (4 + 0.5T)) \times 2T = \frac{1}{2}(8 + 2T) \times 2T = (8 + 2T)T = 8T + 2T^2 \).

Total Area:
\( S = A_1 + A_2 = 4T + 0.75T^2 + 8T + 2T^2 = 12T + 2.75T^2 \).

We are given \( S = 171 \):
\( 2.75T^2 + 12T - 171 = 0 \).
Multiply by 4 to clear decimals:
\( 11T^2 + 48T - 684 = 0 \).

Factoring the quadratic:
\( (11T + 114)(T - 6) = 0 \).
Since \( T \) must be positive, \( T = 6 \).

(a) B1: Correct general shape of graph (two straight segments, positive then negative gradient). B1: Correct labels for initial speed of 4, peak speed at time \( T \), and final speed \( V \) at time \( 3T \). B1: Correctly indicates gradients or coordinates of key points.

(b) M1: Attempts to write expression for speed at \( t = T \). M1: Applies deceleration formula over the second interval of duration \( 2T \). A1: Correctly shows that \( V = 4 + 0.5T \) with intermediate steps.

(c) M1: Expresses the area under the graph in terms of \( T \) (either as two trapeziums or integration/kinematic equations). A1: Correct expression for area in terms of \( T \), e.g., \( 12T + 2.75T^2 \). M1: Equates area to 171 and forms a three-term quadratic equation. A1: Correctly simplifies quadratic to \( 11T^2 + 48T - 684 = 0 \) (or equivalent). A1: Solves to obtain \( T = 6 \) (rejects negative root).

(a) Let the weight of the beam be \( 15g \) acting at the centre of mass \( G \), where \( AG = x \).

Case 1: Particle of mass \( M \) at \( A \).
The beam is on the point of tilting about \( C \), so the reaction force at \( D \) is 0.
Taking moments about \( C \):
\( M g \times AC = 15g \times CG \)
Since \( AC = 1 \text{ m} \) and \( CG = x - 1 \text{ m} \):
\( M g (1) = 15g (x - 1) \implies M = 15(x - 1) \). (As shown).

Case 2: Particle of mass \( M \) at \( B \).
The beam is on the point of tilting about \( D \), so the reaction force at \( C \) is 0.
Taking moments about \( D \):
\( 15g \times GD = M g \times DB \)
Since \( DB = 1.5 \text{ m} \) and \( AD = 6 - 1.5 = 4.5 \text{ m} \), we have \( GD = 4.5 - x \):
\( 15g (4.5 - x) = M g (1.5) \)
\( 1.5 M = 15(4.5 - x) \implies M = 10(4.5 - x) \).

(b) (i) Equating the two expressions for \( M \):
\( 15(x - 1) = 10(4.5 - x) \)
Divide both sides by 5:
\( 3(x - 1) = 2(4.5 - x) \)
\( 3x - 3 = 9 - 2x \)
\( 5x = 12 \implies x = 2.4 \).

(ii) Substitute \( x = 2.4 \) into the expression for \( M \):
\( M = 15(2.4 - 1) = 15(1.4) = 21 \).

(a) M1: Identifies that tilting about \( C \) implies reaction at \( D \) is 0. M1: Sets up correct moment equation about \( C \) with correct terms. A1: Correctly derives \( M = 15(x-1) \). M1: Identifies that tilting about \( D \) implies reaction at \( C \) is 0. M1: Sets up correct moment equation about \( D \) with correct terms. A1: Correctly obtains \( M = 10(4.5 - x) \) or equivalent.

(b)(i) M1: Equates their two expressions for \( M \). A1: Correct algebraic working. A1: \( x = 2.4 \).

(b)(ii) M1: Substitutes their \( x \) back to find \( M \). A1: \( M = 21 \).

(a) (i) Let the acceleration of the system be \( a \text{ m s}^{-2} \) and the tension in the string be \( T \) N.
For block \( A \), the normal reaction is \( R = 4g \).
Maximum frictional force is \( F = \mu R = 0.4 \times 4g = 1.6g \).
Equation of motion for block \( A \):
\( T - 1.6g = 4a \) (Equation 1)

Equation of motion for sphere \( B \):
\( 3g - T = 3a \) (Equation 2)

Adding Equation 1 and Equation 2:
\( 3g - 1.6g = 7a \)
\( 1.4g = 7a \implies a = 0.2g = 1.96 \text{ m s}^{-2} \) (or \( 2.0 \text{ m s}^{-2} \) to 2 s.f.).

(ii) Using Equation 2 to find tension:
\( T = 3(g - a) = 3(g - 0.2g) = 2.4g = 2.4 \times 9.8 = 23.52 \text{ N} \) (or \( 24 \text{ N} \) to 2 s.f.).

(b) The speed \( v \) of block \( A \) after 1.2 seconds, just before the string breaks:
\( v = u + at = 0 + (1.96)(1.2) = 2.352 \text{ m s}^{-1} \).

After the string breaks, the only horizontal force acting on block \( A \) is friction \( 1.6g \) acting in the opposite direction of motion.
Let the new acceleration (deceleration) of \( A \) be \( a' \).
\( -1.6g = 4a' \implies a' = -0.4g = -3.92 \text{ m s}^{-2} \).

Using \( v_f^2 = v^2 + 2a's \) with \( v_f = 0 \):
\( 0^2 = 2.352^2 + 2(-3.92)s \)
\( 0 = 5.531904 - 7.84s \)
\( s = \frac{5.531904}{7.84} = 0.7056 \text{ m} \) (or \( 0.71 \text{ m} \) to 2 s.f.).

(a)(i) M1: Resolves vertically for block \( A \) to find normal reaction and attempts friction force \( F = \mu R \). A1: Correct friction force \( 1.6g \) (or 15.7 N). M1: Sets up equations of motion for both block \( A \) and sphere \( B \). A1: Correct system of equations. A1: Correct acceleration \( a = 1.96 \text{ m s}^{-2} \) (accept \( 2.0 \text{ m s}^{-2} \)).

(a)(ii) M1: Uses one of their equations to find \( T \). A1: Correct tension \( T = 23.5 \text{ N} \) or \( 24 \text{ N} \).

(b) M1: Calculates speed of \( A \) at 1.2 s using their acceleration. A1: \( v = 2.352 \text{ m s}^{-1} \). M1: Formulates new equation of motion for \( A \) to find new deceleration. A1: Correct further distance \( s \approx 0.706 \text{ m} \) (accept \( 0.71 \text{ m} \)).

(a) Using the equation \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) for particle \( P \):
\(\mathbf{r}_P = (4\mathbf{i} - 2\mathbf{j}) + (3\mathbf{i} + \mathbf{j})t + \frac{1}{2}(\mathbf{i} - \mathbf{j})t^2\)
\(\mathbf{r}_P = (4 + 3t + 0.5t^2)\mathbf{i} + (-2 + t - 0.5t^2)\mathbf{j}\).

(b) Similarly, for particle \( Q \):
\(\mathbf{r}_Q = (-2\mathbf{i} + 6\mathbf{j}) + (5\mathbf{i} - 2\mathbf{j})t + \frac{1}{2}(0.5\mathbf{i} + a\mathbf{j})t^2\)
\(\mathbf{r}_Q = (-2 + 5t + 0.25t^2)\mathbf{i} + (6 - 2t + 0.5at^2)\mathbf{j}\).

(c) First, evaluate the positions at \( t = 4 \) seconds:
\(\mathbf{r}_P(4) = (4 + 3(4) + 0.5(4^2))\mathbf{i} + (-2 + 4 - 0.5(4^2))\mathbf{j} = (4 + 12 + 8)\mathbf{i} + (-2 + 4 - 8)\mathbf{j} = 24\mathbf{i} - 6\mathbf{j}\).

\(\mathbf{r}_Q(4) = (-2 + 5(4) + 0.25(4^2))\mathbf{i} + (6 - 2(4) + 0.5a(4^2))\mathbf{j} = (-2 + 20 + 4)\mathbf{i} + (6 - 8 + 8a)\mathbf{j} = 22\mathbf{i} + (-2 + 8a)\mathbf{j}\).

The vector \( \vec{PQ} = \mathbf{r}_Q - \mathbf{r}_P \):

\( \vec{PQ} = (22\mathbf{i} + (-2 + 8a)\mathbf{j}) - (24\mathbf{i} - 6\mathbf{j}) = -2\mathbf{i} + (4 + 8a)\mathbf{j} \).

Since the vector \( \vec{PQ} \) is parallel to \( \mathbf{i} \), its \( \mathbf{j} \)-component must be equal to 0:
\( 4 + 8a = 0 \implies 8a = -4 \implies a = -0.5 \).

(a) M1: Attempts to use \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) for \( P \). A1: Correct substitution. A1: Correct simplified vector form.

(b) M1: Attempts to use the position vector formula for \( Q \). A1: Correct expression containing \( a \). A1: Correct simplified vector form.

(c) M1: Substitutes \( t = 4 \) into their expression for \(\mathbf{r}_P\). M1: Substitutes \( t = 4 \) into their expression for \(\mathbf{r}_Q\). M1: Finds \( \vec{PQ} = \mathbf{r}_Q - \mathbf{r}_P \). A1: Sets the \(\mathbf{j}\) component of \( \vec{PQ} \) to 0. A1: Correct value of \( a = -0.5 \).

(a) When the car travels up the hill at its maximum speed, its acceleration is zero. The driving force is \(F = \frac{P}{v} = \frac{40000}{25} = 1600\text{ N}\). Resolving forces parallel to the incline: \(F - R - Mg\sin\theta = 0 \implies 1600 - R - 1200(9.8)\left(\frac{1}{14}\right) = 0 \implies 1600 - R - 840 = 0 \implies R = 760\). (b) When traveling down the hill at speed \(20\text{ m s}^{-1}\), the driving force is \(F_{\text{down}} = \frac{32000}{20} = 1600\text{ N}\). The weight component acts down the slope, and resistance acts up the slope. Resolving parallel to the incline down the hill: \(F_{\text{down}} + Mg\sin\theta - R = Ma \implies 1600 + 840 - 760 = 1200a \implies 1680 = 1200a \implies a = 1.4\text{ m s}^{-2}\).

(a) M1: Use \(P = Fv\) to find the driving force. A1: Correct driving force of \(1600\text{ N}\). M1: Set up the equation of motion along the plane with \(a=0\) and substitute known values. A1: Obtain \(R = 760\). (b) M1: Find the new driving force \(F_{\text{down}} = 1600\text{ N}\). M1: Formulate the equation of motion down the plane incorporating the driving force, gravity component, and resistance. A1: Correct equation of motion: \(1600 + 840 - 760 = 1200a\). M1: Solve for \(a\). A1: Obtain \(a = 1.4\text{ m s}^{-2}\).

(a) Let the mass per unit area be \(\sigma\). The area of the large square is \(36a^2\) with centre of mass at \((3a, 3a)\). The area of the removed square is \(4a^2\) with centre of mass at \((2a, 4a)\). The remaining area of lamina \(L\) is \(32a^2\). (i) Taking moments about the y-axis (the side \(AD\)): \(32a^2 \bar{x} = 36a^2(3a) - 4a^2(2a) = 108a^3 - 8a^3 = 100a^3 \implies \bar{x} = 3.125a\). (ii) Taking moments about the x-axis (the side \(AB\)): \(32a^2 \bar{y} = 36a^2(3a) - 4a^2(4a) = 108a^3 - 16a^3 = 92a^3 \implies \bar{y} = 2.875a\). (b) When suspended from \(A\), the downward vertical passes through \(A\) and the centre of mass \(G(\bar{x}, \bar{y})\). The angle \(\theta\) that the side \(AB\) makes with the vertical \(AG\) is given by \(\tan \theta = \frac{\bar{y}}{\bar{x}} = \frac{2.875a}{3.125a} = \frac{23}{25} = 0.92\). Thus, \(\theta = \arctan(0.92) \approx 42.61^{\circ} \approx 43^{\circ}\) to the nearest degree.

(a) M1: Set up a moments table or equation for areas and positions. A1: Correct coordinates and areas for both the large square and the removed square. M1: Calculate \(\bar{x}\) using moments about the y-axis. A1: Correct \(\bar{x} = 3.125a\). A1: Correct \(\bar{y} = 2.875a\) using moments about the x-axis. (b) M1: Identify that the angle satisfies \(\tan \theta = \frac{\bar{y}}{\bar{x}}\) or its reciprocal depending on the defined angle. A1: Correct trigonometric ratio expression. M1: Calculate the numerical value of the angle. A1: Round correctly to the nearest degree to obtain \(43^{\circ}\).

(a) Differentiating velocity with respect to time to find the acceleration: \(\mathbf{a} = \frac{\text{d}\mathbf{v}}{\text{d}t} = (6t - 4)\mathbf{i} + (k - 6t)\mathbf{j}\). At \(t = 2\), \(\mathbf{a} = (6(2) - 4)\mathbf{i} + (k - 6(2))\mathbf{j} = 8\mathbf{i} + (k - 12)\mathbf{j}\). Since \(\mathbf{a}\) is perpendicular to \(\mathbf{i} - \mathbf{j}\), their dot product is zero: \((8\mathbf{i} + (k - 12)\mathbf{j}) \cdot (\mathbf{i} - \mathbf{j}) = 0 \implies 8(1) + (k - 12)(-1) = 0 \implies 8 - k + 12 = 0 \implies k = 20\). (b) Substituting \(k = 20\) and \(t = 3\) into the velocity vector: \(\mathbf{v} = (3(3)^2 - 4(3))\mathbf{i} + (20(3) - 3(3)^2)\mathbf{j} = (27 - 12)\mathbf{i} + (60 - 27)\mathbf{j} = 15\mathbf{i} + 33\mathbf{j}\). The square of the speed of \(P\) is \(v^2 = 15^2 + 33^2 = 225 + 1089 = 1314\text{ m}^2\text{ s}^{-2}\). The kinetic energy of \(P\) is \(KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.5)(1314) = 328.5\text{ J}\).

(a) M1: Differentiate velocity vector with respect to \(t\) to find acceleration. A1: Correct acceleration vector expression. M1: Substitute \(t = 2\) into the acceleration vector. M1: Apply the dot product condition for perpendicular vectors. A1: Solve the equation to show \(k = 20\). (b) M1: Substitute \(k = 20\) and \(t = 3\) into the velocity expression. A1: Correct velocity vector \(15\mathbf{i} + 33\mathbf{j}\). M1: Use Pythagoras to find \(v^2\). M1: Apply \(KE = \frac{1}{2}mv^2\). A1: Obtain \(328.5\text{ J}\).

(a) Let the initial direction of motion of \(A\) be positive. Therefore, the initial velocities are \(u_A = 3u\) and \(u_B = -u\). Let the velocities after the collision be \(v_A\) and \(v_B\). By conservation of linear momentum: \(2m(3u) + 3m(-u) = 2mv_A + 3mv_B \implies 3u = 2v_A + 3v_B\) (Equation 1). By Newton's law of restitution: \(v_B - v_A = e(3u - (-u)) = 4eu \implies v_A = v_B - 4eu\) (Equation 2). Substitute Equation 2 into Equation 1: \(3u = 2(v_B - 4eu) + 3v_B \implies 3u = 5v_B - 8eu \implies 5v_B = 8eu + 3u \implies v_B = \frac{u}{5}(8e + 3)\) as required. (b) Find \(v_A\) using Equation 2: \(v_A = \frac{u}{5}(8e + 3) - 4eu = \frac{u}{5}(3 - 12e)\). Since the initial direction of \(A\) was positive, its direction is reversed if its final velocity is negative, so \(v_A < 0\). Since \(u > 0\), this requires \(3 - 12e < 0 \implies 12e > 3 \implies e > \frac{1}{4}\). Since the coefficient of restitution satisfies \(e \le 1\), the range of values is \(\frac{1}{4} < e \le 1\).

(a) M1: Set up the conservation of momentum equation. A1: Correct simplified momentum equation. M1: Set up Newton's law of restitution. A1: Correct restitution equation. M1: Solve the simultaneous equations to eliminate \(v_A\). A1: Show the given expression for \(v_B\) clearly. (b) M1: Obtain an expression for \(v_A\) in terms of \(u\) and \(e\). A1: Correct expression for \(v_A\). M1: Set up the inequality \(v_A < 0\) and solve for \(e\). A1: State the final range \(\frac{1}{4} < e \le 1\) (accept equivalent notation, must include upper limit of 1).

(a) Resolving forces vertically for the system in equilibrium: \(R = mg + 3mg = 4mg\), where \(R\) is the normal reaction at the ground. (b) When the ladder is on the point of slipping, the friction force \(F\) at \(A\) is at its maximum: \(F = \mu R = 0.5(4mg) = 2mg\). Resolving horizontally: \(S = F = 2mg\), where \(S\) is the normal reaction at the wall. Let the man climb a distance \(x\) along the ladder from \(A\). Taking moments about \(A\): \(S(2l\sin\theta) = mg(l\cos\theta) + 3mg(x\cos\theta)\). Since \(\tan\theta = 4/3\), we have \(\sin\theta = 4/5\) and \(\cos\theta = 3/5\). Substituting the known values into the moments equation: \(2mg\left(2l \times \frac{4}{5}\right) = mg\left(l \times \frac{3}{5}\right) + 3mg\left(x \times \frac{3}{5}\right) \implies \frac{16}{5}l = \frac{3}{5}l + \frac{9}{5}x \implies 16l = 3l + 9x \implies 13l = 9x \implies x = \frac{13}{9}l\).

(a) M1: Resolve vertically for the ladder. A1: Show clearly that \(R = 4mg\). (b) M1: Use \(F = \mu R\) to calculate the maximum friction. A1: Correct friction force \(2mg\). M1: Resolve horizontally to find the normal reaction at the wall \(S\). A1: Correct reaction \(S = 2mg\). M1: Take moments about a point, e.g., about \(A\). A1: Correct moments equation. M1: Substitute trigonometric ratios and values of \(S\) and solve for \(x\). A1: Obtain \(x = \frac{13}{9}l\).

(a) From \(\tan\alpha = 3/4\), we have \(\sin\alpha = 0.6\) and \(\cos\alpha = 0.8\). The normal reaction is \(R = mg\cos\alpha = 4(9.8)(0.8) = 31.36\text{ N}\). The friction force is \(F = \mu R = 0.5(31.36) = 15.68\text{ N}\). Let the distance travelled up the plane be \(d\). The initial kinetic energy is \(KE = \frac{1}{2}mu^2 = \frac{1}{2}(4)(14^2) = 392\text{ J}\). The potential energy gained is \(PE = mgh = mgd\sin\alpha = 4(9.8)(0.6)d = 23.52d\). The work done against friction is \(W = Fd = 15.68d\). By the work-energy principle: \(KE = PE + W \implies 392 = 23.52d + 15.68d \implies 392 = 39.2d \implies d = 10\text{ m}\). (b) When the package returns to \(A\) from the top, let its final speed be \(v\). The loss in potential energy is \(235.2\text{ J}\), and the work done against friction is still \(156.8\text{ J}\). By the work-energy principle: \(\frac{1}{2}mv^2 = PE_{\text{loss}} - W \implies 2v^2 = 235.2 - 156.8 = 78.4 \implies v^2 = 39.2 \implies v = \sqrt{39.2} \approx 6.26\text{ m s}^{-1}\).

(a) M1: Calculate normal reaction \(R\). A1: Correct friction force \(F = 15.68\text{ N}\). M1: Set up the work-energy equation with KE, PE, and work done against friction. A1: Correct terms: \(KE = 392\), \(PE = 23.52d\), \(W = 15.68d\). M1: Solve for \(d\). A1: Obtain \(10\text{ m}\). (b) M1: Set up the work-energy equation for the downward journey. A1: Correct equation \(2v^2 = 235.2 - 156.8\). M1: Solve for \(v\). A1: Obtain \(6.26\text{ m s}^{-1}\) (or \(\sqrt{39.2}\)).

(a) Differentiating the displacement to get velocity: \(v = \frac{\text{d}s}{\text{d}t} = 3t^2 - 18t + 15\). Setting \(v = 0\): \(3(t^2 - 6t + 5) = 0 \implies 3(t - 1)(t - 5) = 0 \implies t = 1\) and \(t = 5\). (b) Set \(v = 96\text{ m s}^{-1}\): \(3t^2 - 18t + 15 = 96 \implies 3t^2 - 18t - 81 = 0 \implies t^2 - 6t - 27 = 0 \implies (t - 9)(t + 3) = 0\). Since \(t \ge 0\), \(t = 9\). The acceleration is \(a = \frac{\text{d}v}{\text{d}t} = 6t - 18\). At \(t = 9\), \(a = 6(9) - 18 = 36\text{ m s}^{-2}\). (c) The particle changes direction of motion at \(t = 1\) and \(t = 5\). We compute displacement at key times: \(s(0) = 10\), \(s(1) = 1^3 - 9(1)^2 + 15(1) + 10 = 17\), \(s(5) = 5^3 - 9(5)^2 + 15(5) + 10 = -15\), \(s(6) = 6^3 - 9(6)^2 + 15(6) + 10 = -8\). The distance travelled in each interval is: \(d_1 = |17 - 10| = 7\), \(d_2 = |-15 - 17| = 32\), \(d_3 = |-8 - (-15)| = 7\). Total distance \(= 7 + 32 + 7 = 46\text{ m}\).

(a) M1: Differentiate displacement to find velocity. A1: Correct velocity expression. M1: Equate velocity to zero and solve the quadratic. A1: Obtain \(t = 1\) and \(t = 5\). (b) M1: Equate velocity expression to \(96\) and solve for positive \(t\). A1: Correct time \(t = 9\). A1: Correct acceleration of \(36\text{ m s}^{-2}\). (c) M1: Identify the need to calculate displacement at turning points \(t=1\) and \(t=5\). A1: Correct displacements at \(t = 0, 1, 5, 6\). M1: Calculate sum of individual absolute displacement differences. A1: Obtain total distance of \(46\text{ m}\).

(a) Using Newton's second law, \(F = ma\). Here \(a = v\frac{\mathrm{d}v}{\mathrm{d}x}\), so \(\frac{3}{2x + 1} = 0.5 v \frac{\mathrm{d}v}{\mathrm{d}x}\). Rearranging gives: \(v \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{6}{2x + 1}\). Separating variables and integrating both sides with respect to \(x\): \(\int v \mathrm{d}v = \int \frac{6}{2x + 1} \mathrm{d}x \implies \frac{1}{2}v^2 = 3\ln(2x + 1) + C\). When \(x = 0\), \(v = u = 4\): \(\frac{1}{2}(4)^2 = 3\ln(1) + C \implies 8 = C\). Thus, \(\frac{1}{2}v^2 = 3\ln(2x + 1) + 8 \implies v^2 = 16 + 6\ln(2x + 1)\). (b) When \(x = 6\): \(v^2 = 16 + 6\ln(13)\). \(v^2 \approx 16 + 6(2.5649) = 31.3897\). \(v = \sqrt{31.3897} \approx 5.6026\text{ ms}^{-1}\). To 3 significant figures, the speed of \(P\) is \(5.60\text{ ms}^{-1}\).

(a) M1: Set up the equation of motion using \(a = v\frac{\mathrm{d}v}{\mathrm{d}x}\). A1: Correct differential equation \(0.5v\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{3}{2x+1}\). M1: Integrate both sides to obtain terms in \(v^2\) and \(\ln(2x+1)\). A1: Correct integration: \(\frac{1}{2}v^2 = 3\ln(2x+1) + C\). M1: Use initial conditions \(x = 0, v = 4\) to find \(C\). A1: Find \(C = 8\) and correctly establish the given formula for \(v^2\). (b) M1: Substitute \(x = 6\) into the formula for \(v^2\). A1: Calculate \(v^2 = 16 + 6\ln(13)\) (approx. 31.4). A1.7: State final speed as \(5.60\text{ ms}^{-1}\) (allow 5.6).

(a) The extension of the string when \(P\) is at \(B\) is \(x = 2.5 - 1.5 = 1.0\text{ m}\). The elastic potential energy (EPE) stored is: \(\text{EPE} = \frac{\lambda x^2}{2l} = \frac{24 \times (1.0)^2}{2 \times 1.5} = 8\text{ J}\). (b) When the string first becomes slack, the extension is \(0\), so the EPE is completely converted into kinetic energy (KE). By the principle of conservation of mechanical energy: \(\text{KE}_i + \text{EPE}_i = \text{KE}_f + \text{EPE}_f \implies 0 + 8 = \frac{1}{2}mv^2 + 0 \implies 8 = \frac{1}{2}(0.8)v^2 \implies 0.4v^2 = 8 \implies v^2 = 20 \implies v = \sqrt{20} = 2\sqrt{5} \approx 4.47\text{ ms}^{-1}\).

(a) M1: Use \(x = AB - l\) to find the extension. M1: Apply the formula for EPE, \(\text{EPE} = \frac{\lambda x^2}{2l}\). A2: Correct value of \(8\text{ J}\) (award 2 marks, or 1 mark if a minor calculation slip occurs). (b) M1: Set up the conservation of energy equation: \(\text{EPE}_i = \text{KE}_f\). A1: Correct equation \(8 = 0.4v^2\). M1: Solve for \(v^2\) to get \(v^2 = 20\). A2.7: Show final speed is \(4.47\text{ ms}^{-1}\) to 3 s.f. (allow \(2\sqrt{5}\)).

(a) Let \(v\) be the speed of \(P\) at the instant the string becomes slack. The radial equation of motion towards the centre \(O\) is: \(T + mg\cos\theta = \frac{mv^2}{L}\). When the string becomes slack, \(T = 0\), so \(mg\cos\theta = \frac{mv^2}{L} \implies v^2 = gL\cos\theta\). Since \(\cos\theta = \frac{1}{3}\), \(v^2 = \frac{1}{3}gL\). By conservation of mechanical energy from the lowest point (initial position, where GPE = 0) to the point where the string becomes slack (height \(h = L + L\cos\theta = \frac{4}{3}L\)): \(\frac{1}{2}mU^2 = \frac{1}{2}mv^2 + mgh \implies U^2 = v^2 + 2gh \implies U^2 = \frac{1}{3}gL + 2g\left(\frac{4}{3}L\right) = \frac{1}{3}gL + \frac{8}{3}gL = 3gL \implies U = \sqrt{3gL}\). (b) After the string becomes slack, \(P\) moves as a free projectile under gravity. The velocity \(v\) is perpendicular to \(OP\) and makes an angle \(\theta\) with the horizontal. The horizontal component of velocity is \(v_x = v\cos\theta\). At the maximum height, the vertical component of velocity is \(0\), so the kinetic energy is \(\frac{1}{2}mv_x^2\). By conservation of energy from the release point (height \(y_0 = \frac{4}{3}L\)) to the maximum height \(H\): \(\frac{1}{2}mv^2 + mgy_0 = \frac{1}{2}mv_x^2 + mgH \implies H = y_0 + \frac{v^2 - v_x^2}{2g} = y_0 + \frac{v^2(1 - \cos^2\theta)}{2g} = \frac{4}{3}L + \frac{\frac{1}{3}gL(1 - \frac{1}{9})}{2g} = \frac{4}{3}L + \frac{4}{27}L = \frac{40}{27}L\).

(a) M1: Set up radial equation of motion for circular path: \(T + mg\cos\theta = \frac{mv^2}{L}\). A1: Identify \(T = 0\) to get \(v^2 = gL\cos\theta\). M1: Set up energy equation between projection and slack point. A1: Correctly express the height above the lowest point as \(L(1 + \cos\theta)\). A1: Substitute \(\cos\theta = \frac{1}{3}\) to get \(v^2 = \frac{1}{3}gL\) and height \(\frac{4}{3}L\). A2: Combine equations to show \(U = \sqrt{3gL}\). (b) M1: State that \(P\) behaves as a projectile after string becomes slack. M1: Find horizontal component of velocity or vertical component \(v_y = v\sin\theta\). A1: Correctly identify vertical component or set up energy equation for maximum height. A0.7: Find the maximum height is \(\frac{40}{27}L\).

(a) Using the simple harmonic motion formula \(v^2 = \omega^2(a^2 - x^2)\): For \(x = 0.6\), \(4^2 = \omega^2(a^2 - 0.6^2) \implies 16 = \omega^2(a^2 - 0.36)\). For \(x = 0.8\), \(3^2 = \omega^2(a^2 - 0.8^2) \implies 9 = \omega^2(a^2 - 0.64)\). Dividing the first equation by the second: \(\frac{16}{9} = \frac{a^2 - 0.36}{a^2 - 0.64} \implies 16a^2 - 10.24 = 9a^2 - 3.24 \implies 7a^2 = 7 \implies a^2 = 1 \implies a = 1\text{ m}\). (b) From the first equation, \(16 = \omega^2(1 - 0.36) = 0.64\omega^2 \implies \omega^2 = 25 \implies \omega = 5\text{ rad s}^{-1}\). The period \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5}\text{ s} \approx 1.26\text{ s}\). (c) The displacement is given by \(x = a\cos(\omega t)\) when starting from maximum displacement. Here, \(a = 1\) and \(\omega = 5\). To travel a total distance of \(1.5\text{ m}\) from \(x = 1\), the particle must reach a displacement of \(x = 1 - 1.5 = -0.5\text{ m}\). Substituting \(x = -0.5\) into the equation: \(-0.5 = 1\cos(5t) \implies 5t = \arccos(-0.5) = \frac{2\pi}{3}\text{ radians}\). \(t = \frac{2\pi}{15}\text{ s} \approx 0.419\text{ s}\).

(a) M1: State the equation \(v^2 = \omega^2(a^2 - x^2)\) and set up equations for both cases. A1: Correct equations: \(16 = \omega^2(a^2 - 0.36)\) and \(9 = \omega^2(a^2 - 0.64)\). M1: Divide the equations to eliminate \(\omega^2\) and solve for \(a^2\). A1: Correctly show \(a = 1\text{ m}\). (b) M1: Substitute \(a = 1\) to find \(\omega = 5\). A2: Calculate \(T = \frac{2\pi}{5}\text{ s}\) or \(1.26\text{ s}\). (c) M1: Identify that travelling \(1.5\text{ m}\) from \(x = 1\) leads to \(x = -0.5\). M1: Set up the displacement equation \(-0.5 = \cos(5t)\). A1.7: Correctly calculate \(t = \frac{2\pi}{15} \approx 0.419\text{ s}\).

(a) The curve is \(y = 4 - x^2 \implies x^2 = 4 - y\). The limits of integration for \(y\) are from \(y = 0\) to \(y = 4\) (since the region is in the first quadrant and bounded by \(y = 0\)). The volume \(V\) is: \(V = \pi \int_{0}^{4} x^2 \mathrm{d}y = \pi \int_{0}^{4} (4 - y) \mathrm{d}y = \pi \left[ 4y - \frac{1}{2}y^2 \right]_0^4 = \pi \left(16 - 8\right) = 8\pi\). (b) The \(y\)-coordinate of the centre of mass, \(\bar{y}\), of a solid of revolution about the \(y\)-axis is given by: \(\bar{y} V = \pi \int_{0}^{4} y x^2 \mathrm{d}y \implies 8\pi \bar{y} = \pi \int_{0}^{4} y(4 - y) \mathrm{d}y\). Integrating: \(\int_{0}^{4} (4y - y^2) \mathrm{d}y = \left[ 2y^2 - \frac{1}{3}y^3 \right]_0^4 = 32 - \frac{64}{3} = \frac{32}{3}\). Thus: \(8\pi \bar{y} = \frac{32}{3}\pi \implies \bar{y} = \frac{32}{24} = \frac{4}{3}\).

(a) M1: Recall volume of revolution formula about the \(y\)-axis, \(V = \pi \int x^2 \mathrm{d}y\). A1: Correct limits from \(0\) to \(4\). M1: Correctly substitute \(x^2 = 4-y\) and integrate. A1: Show that the volume is \(8\pi\). (b) M1: Recall formula for centre of mass \(\bar{y} V = \pi \int y x^2 \mathrm{d}y\). A1: Set up correct integral \(\int_{0}^{4} (4y - y^2) \mathrm{d}y\). M1: Perform integration correctly to obtain \(\left[ 2y^2 - \frac{1}{3}y^3 \right]_0^4\). A1: Obtain value \(\frac{32}{3}\pi\) for the integral. A2.7: Find \(\bar{y} = \frac{4}{3}\) (accept \(1.33\)).

(a) Let \(\theta\) be the angle that the radius from \(O\) to \(P\) makes with the downward vertical. The radius of the horizontal circle is \(r = R\sin\theta\), and the depth is \(d = R\cos\theta\). The normal reaction force \(R_N\) acts along the radius towards \(O\). Resolving vertically: \(R_N\cos\theta = mg\). Resolving horizontally towards the centre of the circle: \(R_N\sin\theta = m r \omega^2 = m R \sin\theta \omega^2\). Since \(\sin\theta \neq 0\), we have \(R_N = m R \omega^2\). Substituting this into the vertical equation: \((m R \omega^2)\cos\theta = mg \implies R\cos\theta\omega^2 = g\). Since \(d = R\cos\theta\), this gives \(d\omega^2 = g \implies d = \frac{g}{\omega^2}\). (b) Using \(g = 9.8\text{ ms}^{-2}\) and \(\omega = 7\text{ rad s}^{-1}\): \(d = \frac{9.8}{7^2} = \frac{9.8}{49} = 0.2\text{ m}\). (c) The magnitude of the normal reaction force is: \(R_N = m R \omega^2 = 0.3 \times 0.5 \times 7^2 = 0.15 \times 49 = 7.35\text{ N}\).

(a) M1: Identify forces acting on \(P\) and resolve vertically to get \(R_N\cos\theta = mg\). M1: Resolve horizontally to get \(R_N\sin\theta = mr\omega^2\). A1: Correctly write \(r = R\sin\theta\). M1: Eliminate \(R_N\) or \(\sin\theta\) to relate \(\cos\theta\) to \(g, R, \omega\). A2: Establish the given formula \(d = \frac{g}{\omega^2}\). (b) M2: Substitute \(g=9.8\) and \(\omega=7\) to get \(d = 0.2\text{ m}\). (c) M1: Use formula \(R_N = mR\omega^2\) or \(R_N = \frac{mg}{\cos\theta}\). A1.7: Calculate \(R_N = 7.35\text{ N}\).

(a) Let \(x\) be the height of \(P\) above \(O\). Initially, \(x_i = 0.5\text{ m}\). The extension of the spring is \(e_i = 0.5 - 0.4 = 0.1\text{ m}\). The total initial energy is: \(E = \text{EPE}_i + \text{GPE}_i = \frac{\lambda e_i^2}{2l} + mgx_i = \frac{19.6 \times 0.1^2}{2 \times 0.4} + 0.5 \times 9.8 \times 0.5 = 0.245 + 2.45 = 2.695\text{ J}\). At the turning point \(x_f\), \(\text{KE} = 0\), so: \(\frac{19.6(x_f - 0.4)^2}{0.8} + 4.9x_f = 2.695 \implies 24.5(x_f - 0.4)^2 + 4.9x_f - 2.695 = 0\). Expanding and simplifying: \(24.5(x_f^2 - 0.8x_f + 0.16) + 4.9x_f - 2.695 = 0 \implies 24.5x_f^2 - 14.7x_f + 1.225 = 0\). Dividing by 24.5: \(x_f^2 - 0.6x_f + 0.05 = 0 \implies (x_f - 0.5)(x_f - 0.1) = 0\). The roots are \(x_f = 0.5\text{ m}\) (the initial release point) and \(x_f = 0.1\text{ m}\). Thus, the particle first comes to instantaneous rest at a height of \(0.1\text{ m}\) above \(O\). (b) The maximum speed is attained at the equilibrium position where the net force is zero. Let the compression at equilibrium be \(e_0\): \(\frac{\lambda e_0}{l} = mg \implies \frac{19.6 e_0}{0.4} = 0.5 \times 9.8 \implies 49 e_0 = 4.9 \implies e_0 = 0.1\text{ m}\). So the equilibrium height is \(x_0 = 0.4 - 0.1 = 0.3\text{ m}\). Using conservation of energy at \(x = 0.3\): \(\text{EPE}_0 + \text{GPE}_0 + \text{KE}_0 = 2.695 \implies \frac{19.6 \times 0.1^2}{0.8} + 0.5 \times 9.8 \times 0.3 + \frac{1}{2}(0.5)v_{\max}^2 = 2.695\). \(0.245 + 1.47 + 0.25v_{\max}^2 = 2.695 \implies 1.715 + 0.25v_{\max}^2 = 2.695 \implies 0.25v_{\max}^2 = 0.98 \implies v_{\max}^2 = 3.92\). Thus, \(v_{\max} = \sqrt{3.92} \approx 1.98\text{ ms}^{-1}\).

(a) M1: Calculate the total energy at release. A1: Correct initial energy \(E = 2.695\text{ J}\). M1: Set up the energy equation at the turning point: \(\text{EPE} + \text{GPE} = 2.695\). A1: Correct quadratic equation \(24.5(x - 0.4)^2 + 4.9x = 2.695\). M1: Solve the quadratic equation. A2: Show that \(x_f = 0.1\text{ m}\) is a valid root. (b) M1: Find the equilibrium position where forces balance. A1: Correct equilibrium height \(x = 0.3\text{ m}\) (or compression \(0.1\text{ m}\)). M1: Apply energy conservation at \(x = 0.3\) to find kinetic energy. A0.7: Calculate maximum speed \(1.98\text{ ms}^{-1}\) (3 s.f.).

For (a), a quadratic equation \( ax^2 + bx + c = 0 \) has no real roots if the discriminant \( b^2 - 4ac < 0 \). Here, \( a = 1 \), \( b = k + 5 \), and \( c = 2k + 9 \). Therefore, \( (k + 5)^2 - 4(1)(2k + 9) < 0 \). Expanding the brackets gives \( k^2 + 10k + 25 - 8k - 36 < 0 \), which simplifies to \( k^2 + 2k - 11 < 0 \) as required.

For (b), to find the critical values, we solve \( k^2 + 2k - 11 = 0 \) using the quadratic formula: \( k = \frac{-2 \pm \sqrt{2^2 - 4(1)(-11)}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3} \). Since we require \( k^2 + 2k - 11 < 0 \), the set of possible values of \( k \) is \( -1 - 2\sqrt{3} < k < -1 + 2\sqrt{3} \).

(a) M1: Attempts to use the discriminant \( b^2 - 4ac < 0 \) with correct substitution. A1: Correctly expands and simplifies to the given inequality with no errors shown.
(b) M1: Solves the quadratic equation \( k^2 + 2k - 11 = 0 \) using a valid method to find surd roots. A1: Obtains \( -1 \pm 2\sqrt{3} \) or equivalent. M1: Chooses the inside region for the inequality. A1: Fully correct inequality \( -1 - 2\sqrt{3} < k < -1 + 2\sqrt{3} \).

For (a), the gradient of \( l_1 \) is \( m = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3} \). The equation of \( l_1 \) is \( y - 5 = -\frac{4}{3}(x + 2) \). Multiplying by 3: \( 3y - 15 = -4x - 8 \), which simplifies to \( 4x + 3y - 7 = 0 \).

For (b), the midpoint of \( AB \) is \( M = \left(\frac{-2 + 4}{2}, \frac{5 - 3}{2}\right) = (1, 1) \). The gradient of the perpendicular line \( l_2 \) is \( m_2 = -\frac{1}{m} = \frac{3}{4} \). The equation of \( l_2 \) is \( y - 1 = \frac{3}{4}(x - 1) \), which simplifies to \( y = \frac{3}{4}x + \frac{1}{4} \) (or \( y = 0.75x + 0.25 \)).

(a) M1: Correct method to find the gradient of \( l_1 \). M1: Uses their gradient and one of the points to write down a linear equation. A1: Correct equation in the form \( ax + by + c = 0 \) with integer coefficients.
(b) B1: Correct coordinates for the midpoint \( (1, 1) \). M1: Finds the perpendicular gradient by taking the negative reciprocal. M1: Uses their perpendicular gradient and midpoint to find the equation. A1: Correct equation in the form \( y = mx + c \).

For (a), \( 5\sin(2x - 40^\circ) = 2 \implies \sin(2x - 40^\circ) = 0.4 \).

For (b), let \( \theta = 2x - 40^\circ \). Since \( 0 \le x \le 180^\circ \), the interval for \( \theta \) is \( -40^\circ \le \theta \le 320^\circ \). Finding the principal value: \( \theta = \arcsin(0.4) \approx 23.578^\circ \). The second value in the interval is \( 180^\circ - 23.578^\circ = 156.422^\circ \). Setting up the equations: \( 2x - 40 = 23.578 \implies 2x = 63.578 \implies x \approx 31.8^\circ \). \( 2x - 40 = 156.422 \implies 2x = 196.422 \implies x \approx 98.2^\circ \). Both solutions lie within the range \( 0 \le x \le 180^\circ \).

(a) B1: States \( \sin(2x - 40^\circ) = 0.4 \).
(b) M1: Finds the principal angle \( \approx 23.6^\circ \). M1: Finds the second angle \( 180^\circ - \text{PV} \) (approx \( 156.4^\circ \)). M1: Sets up correct equations of the form \( 2x - 40 = \theta \) to solve for \( x \). A1: One correct value of \( x \) (either \( 31.8^\circ \) or \( 98.2^\circ \)). A1: Both values correct and no extra values in the range.

For (a), write \( y = 2x^2 - 5x - 4x^{-1} \). Differentiating term by term: \( \frac{dy}{dx} = 4x - 5 + 4x^{-2} = 4x - 5 + \frac{4}{x^2} \).

For (b), at \( x = 2 \), \( y = 2(2)^2 - 5(2) - \frac{4}{2} = 8 - 10 - 2 = -4 \). So \( P \) is \( (2, -4) \). The gradient of the tangent at \( P \) is \( m_t = 4(2) - 5 + \frac{4}{4} = 8 - 5 + 1 = 4 \). The gradient of the normal is \( m_n = -\frac{1}{4} \). The equation of the normal is \( y - (-4) = -\frac{1}{4}(x - 2) \). Multiplying by 4: \( 4(y + 4) = -(x - 2) \implies 4y + 16 = -x + 2 \implies x + 4y + 14 = 0 \).

(a) M1: Differentiates \( 2x^2 - 5x \) correctly to obtain \( 4x - 5 \). M1: Expresses \( -\frac{4}{x} \) as \( -4x^{-1} \) and differentiates to obtain \( 4x^{-2} \). A1: Fully correct derivative.
(b) B1: Correctly evaluates \( y = -4 \) at \( x = 2 \). M1: Substitutes \( x = 2 \) into their derivative to find the tangent gradient. M1: Finds the negative reciprocal of their tangent gradient and attempts to find the equation of the normal. A1: Correct equation in the form \( ax + by + c = 0 \).

For (a), divide both terms in the numerator by the denominator: \( f'(x) = \frac{3\sqrt{x}}{\sqrt{x}} - \frac{4}{\sqrt{x}} = 3 - 4x^{-\frac{1}{2}} \) as required.

For (b), integrate \( f'(x) \): \( f(x) = \int (3 - 4x^{-\frac{1}{2}}) dx = 3x - \frac{4x^{\frac{1}{2}}}{\frac{1}{2}} + C = 3x - 8x^{\frac{1}{2}} + C \). Since the curve passes through \( (4, 11) \): \( 11 = 3(4) - 8(4)^{\frac{1}{2}} + C \implies 11 = 12 - 16 + C \implies 11 = -4 + C \implies C = 15 \). Thus, \( f(x) = 3x - 8\sqrt{x} + 15 \).

(a) B1: Correct algebraic manipulation to show the given form.
(b) M1: Attempts to integrate the expression. M1: At least one term integrated correctly (e.g. \( 3x \) or \( -8x^{1/2} \)). A1: Correct integrated expression including the constant of integration \( + C \). M1: Uses the point \( (4, 11) \) to find the value of \( C \). A1: Correct value of \( C = 15 \). A1: Fully correct expression for \( f(x) \).

For (a), the curve crosses the \( x \)-axis when \( y = 0 \). So \( \frac{2}{x} + 1 = 0 \implies \frac{2}{x} = -1 \implies x = -2 \). The coordinates of \( A \) are \( (-2, 0) \).

For (b)(i), \( y = f(x - 3) \) represents a translation of the curve by 3 units to the right. The vertical asymptote becomes \( x = 3 \) and the horizontal asymptote remains \( y = 1 \). The point \( A \) is translated to \( (1, 0) \).

For (b)(ii), \( y = 2f(x) \) represents a vertical stretch of scale factor 2. The vertical asymptote remains \( x = 0 \) and the horizontal asymptote becomes \( y = 2 \). The point \( A \) remains at \( (-2, 0) \).

(a) M1: Sets \( f(x) = 0 \) and solves for \( x \). A1: Correct coordinates \( (-2, 0) \).
(b) (i) M1: Sketches a graph of the correct shape with a vertical translation of asymptotes. A1: Correctly labels the asymptotes as \( x = 3 \) and \( y = 1 \), and the intercept as \( (1, 0) \).
(ii) M1: Sketches a graph with correct shape and altered horizontal asymptote. A1: Correctly labels the asymptotes as \( x = 0 \) and \( y = 2 \), and the intercept as \( (-2, 0) \).

For (a), substitute \( y = 2x - 3 \) into the equation for curve \( C \): \( (2x - 3)^2 = 4x + 9 \). Expanding the left-hand side: \( 4x^2 - 12x + 9 = 4x + 9 \). Subtracting \( 4x + 9 \) from both sides: \( 4x^2 - 16x = 0 \) as required.

For (b), solve \( 4x^2 - 16x = 0 \implies 4x(x - 4) = 0 \), which gives \( x = 0 \) and \( x = 4 \). Find the corresponding \( y \)-coordinates: for \( x = 0 \), \( y = 2(0) - 3 = -3 \); for \( x = 4 \), \( y = 2(4) - 3 = 5 \). The points are \( P(0, -3) \) and \( Q(4, 5) \). The distance \( PQ = \sqrt{(4 - 0)^2 + (5 - (-3))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \).

(a) M1: Substitutes the linear equation into the quadratic equation. M1: Expands \( (2x - 3)^2 \) correctly. A1: Simplifies to the given equation with no errors shown.
(b) B1: Correctly solves to find \( x = 0, 4 \). M1: Evaluates the corresponding \( y \)-coordinates of both points. M1: Uses the distance formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) on their points. A1: Obtains the correct exact distance of \( 4\sqrt{5} \).

For (a), the area is the integral between the intersection points \( x = -3 \) and \( x = 2 \) of the upper curve minus the lower line: \( \int_{-3}^{2} ((9 - x^2) - (x + 3)) dx = \int_{-3}^{2} (6 - x - x^2) dx \).

For (b), integrating the expression: \( \int (6 - x - x^2) dx = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{2} \). Evaluating at the upper limit \( x = 2 \): \( 6(2) - \frac{2^2}{2} - \frac{2^3}{3} = 12 - 2 - \frac{8}{3} = \frac{22}{3} \). Evaluating at the lower limit \( x = -3 \): \( 6(-3) - \frac{(-3)^2}{2} - \frac{(-3)^3}{3} = -18 - \frac{9}{2} + 9 = -\frac{27}{2} \). Calculating the area: \( \frac{22}{3} - \left(-\frac{27}{2}\right) = \frac{44 + 81}{6} = \frac{125}{6} \).

(a) B1: Correct integral expression including correct limits.
(b) M1: Attempts to integrate the expression term-by-term. A1: Correct integration. M1: Substitutes the limits \( 2 \) and \( -3 \) into their integrated expression. A1: Correct values for both limits substituted. A1: Correct exact area of \( \frac{125}{6} \).

**Part (a)**

The line \(l_1\) can be rearranged as \(4y = 3x + 12 \implies y = \frac{3}{4}x + 3\).

Therefore, the gradient of \(l_1\) is \(m_1 = \frac{3}{4}\).

Since \(l_2\) is perpendicular to \(l_1\), its gradient is:
\(m_2 = -\frac{1}{m_1} = -\frac{4}{3}\).

Using the point-gradient formula with \(A(4, k)\):
\(y - k = -\frac{4}{3}(x - 4)\)

Multiply by 3:
\(3(y - k) = -4(x - 4) \implies 3y - 3k = -4x + 16\)

Rearranging into the required form:
\(3y + 4x = 3k + 16\)

**Part (b)**

(i) Since \(B\) lies on the \(x\)-axis, its \(y\)-coordinate is \(0\).

Substituting \(y = 0\) into the equation of \(l_1\):
\(3x - 4(0) + 12 = 0 \implies 3x = -12 \implies x = -4\).

So, the coordinates of \(B\) are \((-4, 0)\).

(ii) Since \(B(-4, 0)\) is the intersection point, it must also lie on \(l_2\).

Substituting \(x = -4\) and \(y = 0\) into our equation from part (a):
\(3(0) + 4(-4) = 3k + 16\)

\(-16 = 3k + 16 \implies 3k = -32 \implies k = -\frac{32}{3}\).

**Part (a) [3 Marks]**
* **M1**: Identifies the gradient of \(l_1\) is \(\frac{3}{4}\) and finds the perpendicular gradient is \(-\frac{4}{3}\).
* **M1**: Applies the line equation formula using their perpendicular gradient and the point \((4, k)\).
* **A1**: Correct equation in the specified form, e.g., \(3y + 4x = 3k + 16\) (accept any equivalent integer coefficients).

**Part (b) [4 Marks]**
* **M1**: Sets \(y = 0\) in the equation of \(l_1\) to find the \(x\)-coordinate of \(B\).
* **A1**: Correct coordinates for \(B\): \((-4, 0)\) or \(x = -4, y = 0\).
* **M1**: Substitutes their coordinates of \(B\) into their equation for \(l_2\).
* **A1**: Correct value of \(k = -\frac{32}{3}\) (or exact equivalent fraction/decimal).

**Part (a)**

To find \(y\), integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\(y = \int \left( 3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}} \right) \mathrm{d}x\)

Integrating term by term:
\(y = 3 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - 4 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C\)
\(y = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + C\)

Since the point \(P(4, 5)\) lies on \(C\), substitute \(x = 4\) and \(y = 5\):
\(5 = 2(4)^{\frac{3}{2}} - 8(4)^{\frac{1}{2}} + C\)
\(5 = 2(8) - 8(2) + C\)
\(5 = 16 - 16 + C \implies C = 5\)

Thus, the equation of the curve is:
\(y = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + 5\)

**Part (b)**

To find the gradient of the tangent at \(P(4, 5)\), substitute \(x = 4\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\(m = 3(4)^{\frac{1}{2}} - \frac{4}{\sqrt{4}} = 3(2) - \frac{4}{2} = 6 - 2 = 4\)

Using the equation of a straight line with gradient \(m = 4\) and point \((4, 5)\):
\(y - 5 = 4(x - 4)\)
\(y - 5 = 4x - 16 \implies y = 4x - 11\)

**Part (a) [5 Marks]**
* **M1**: Attempts to integrate by rewriting \(\frac{4}{\sqrt{x}}\) as \(4x^{-\frac{1}{2}}\) and increasing at least one power by 1.
* **A1**: At least one term correctly integrated (e.g., \(2x^{\frac{3}{2}}\) or \(-8x^{\frac{1}{2}}\)).
* **A1**: Fully correct integrated expression with constant of integration, \(2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + C\).
* **M1**: Substitutes \(x = 4\) and \(y = 5\) to find \(C\).
* **A1**: Correct final equation: \(y = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + 5\).

**Part (b) [2 Marks]**
* **M1**: Substitutes \(x = 4\) into the given \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient of the tangent.
* **A1**: Correct equation of the tangent in the required form: \(y = 4x - 11\).

**Part (a)**

For the quadratic-type equation \(3kx^2 + (k + 4)x + k = 0\) to have real roots, the discriminant must be greater than or equal to zero (\(b^2 - 4ac \ge 0\)), or if \(k=0\), the equation is \(4x = 0\), which has a real root \(x=0\).

Identifying the coefficients:
\(a = 3k\)
\(b = k + 4\)
\(c = k\)

Setting the discriminant \(\Delta \ge 0\):
\((k + 4)^2 - 4(3k)(k) \ge 0\)

Expanding and simplifying:
\(k^2 + 8k + 16 - 12k^2 \ge 0\)

\(-11k^2 + 8k + 16 \ge 0\)

Multiplying by \(-1\) (which reverses the inequality sign):
\(11k^2 - 8k - 16 \le 0\) (as required).

**Part (b)**

To solve \(11k^2 - 8k - 16 \le 0\), first find the critical values by solving the quadratic equation \(11k^2 - 8k - 16 = 0\):

Using the quadratic formula:
\(k = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(11)(-16)}}{2(11)}\)
\(k = \frac{8 \pm \sqrt{64 + 704}}{22}\)
\(k = \frac{8 \pm \sqrt{768}}{22}\)

Simplify the surd:
\(\sqrt{768} = \sqrt{256 \times 3} = 16\sqrt{3}\)

So:
\(k = \frac{8 \pm 16\sqrt{3}}{22} = \frac{4 \pm 8\sqrt{3}}{11}\)

Since we want \(11k^2 - 8k - 16 \le 0\), we look for the region between the critical values:
\(\frac{4 - 8\sqrt{3}}{11} \le k \le \frac{4 + 8\sqrt{3}}{11}\)

**Part (a) [3 Marks]**
* **M1**: Uses the discriminant \(b^2 - 4ac\) with \(a=3k, b=k+4, c=k\).
* **M1**: Sets their discriminant \(\ge 0\) and attempts to expand and simplify.
* **A1***: Correctly shows that \(11k^2 - 8k - 16 \le 0\) with no errors in algebra or inequality signs.

**Part (b) [4 Marks]**
* **M1**: Attempts to solve the quadratic equation \(11k^2 - 8k - 16 = 0\) using a correct method (e.g. quadratic formula).
* **A1**: Finds the correct unsimplified or partially simplified roots, e.g., \(\frac{8 \pm \sqrt{768}}{22}\).
* **A1**: Simplifies the roots to \(\frac{4 \pm 8\sqrt{3}}{11}\).
* **A1**: Expresses the correct inequality range: \(\frac{4 - 8\sqrt{3}}{11} \le k \le \frac{4 + 8\sqrt{3}}{11}\) (accept equivalent interval notation).

(a) Let \(a = \sqrt{2}\) and \(b = -\sqrt{2}\). Since both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum \(a + b = \sqrt{2} + (-\sqrt{2}) = 0\) is a rational number, the statement is false. (b) For any positive real numbers \(x\) and \(y\), their square roots \(\sqrt{x}\) and \(\sqrt{y}\) are real. Since the square of any real number is non-negative: \((\sqrt{x} - \sqrt{y})^2 \ge 0 \implies x - 2\sqrt{xy} + y \ge 0 \implies x + y \ge 2\sqrt{xy}\). (c) Applying the inequality from part (b) to the pairs \((a, b)\), \((b, c)\), and \((c, a)\), we get: \(a + b \ge 2\sqrt{ab}\), \(b + c \ge 2\sqrt{bc}\), and \(c + a \ge 2\sqrt{ca}\). Since \(a\), \(b\), and \(c\) are positive, all terms are positive. Multiplying these three inequalities gives: \((a+b)(b+c)(c+a) \ge (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca}) = 8\sqrt{a^2 b^2 c^2} = 8abc\).

(a)
B1: States two correct irrational numbers whose sum is rational.
B1: Shows their sum is rational and concludes the statement is false.
(b)
M1: Starts with a valid inequality such as \((\sqrt{x} - \sqrt{y})^2 \ge 0\).
A1: Expands to \(x - 2\sqrt{xy} + y \ge 0\).
A1*: Fully correct proof with no errors, concluding with \(x + y \ge 2\sqrt{xy}\).
(c)
M1: Applies the result of part (b) to \(a+b\), \(b+c\), and \(c+a\).
A1: Multiplies the inequalities together and simplifies to get \((a+b)(b+c)(c+a) \ge 8abc\).

(a) Using the remainder theorem, \(f(1) = -3 \implies 2(1)^3 + a(1)^2 + b(1) + 4 = -3 \implies a + b + 6 = -3 \implies a + b = -9\). Also, \(f(-1) = 9 \implies 2(-1)^3 + a(-1)^2 + b(-1) + 4 = 9 \implies a - b + 2 = 9 \implies a - b = 7\). Adding the two equations: \(2a = -2 \implies a = -1\). Substituting back gives \(-1 + b = -9 \implies b = -8\). (b) Since \(f(x) = 2x^3 - x^2 - 8x + 4\), we test \(x = 1/2\): \(f(1/2) = 2(1/8) - (1/4) - 8(1/2) + 4 = 1/4 - 1/4 - 4 + 4 = 0\). Since \(f(1/2) = 0\), by the factor theorem, \(2x - 1\) is a factor of \(f(x)\). (c) Dividing \(f(x)\) by \(2x - 1\): \(f(x) = (2x - 1)(x^2 - 4)\). Factorising the quadratic difference of two squares: \(f(x) = (2x - 1)(x - 2)(x + 2)\).

(a)
M1: Attempts to use the remainder theorem with \(f(1) = -3\).
A1: Correct equation, e.g., \(a + b = -9\).
M1: Attempts to use the remainder theorem with \(f(-1) = 9\).
A1: Correct equation, e.g., \(a - b = 7\).
M1: Solves the simultaneous equations to find \(a\) and \(b\).
A1: \(a = -1\) and \(b = -8\).
(b)
B1: Evaluates \(f(1/2)\) and shows it equals \(0\), concluding that \(2x-1\) is a factor.
(c)
M1: Attempts to divide or factorise \(f(x)\) by \((2x - 1)\) to find a quadratic factor.
A1: Correct complete factorisation \((2x - 1)(x - 2)(x + 2)\) (any order).

(a) The sum of the first two terms is \(a + ar = a(1+r) = 15\). The sum to infinity is \(\frac{a}{1-r} = 27 \implies a = 27(1-r)\). Substituting this into the first equation: \(27(1-r)(1+r) = 15 \implies 27(1-r^2) = 15 \implies 1-r^2 = \frac{15}{27} = \frac{5}{9} \implies r^2 = 1 - \frac{5}{9} = \frac{4}{9}\). (b)(i) Since all terms are positive, we must have \(r > 0\), so \(r = \sqrt{4/9} = 2/3\). Then \(a = 27(1 - 2/3) = 9\). (ii) The sum of the first 5 terms is \(S_5 = \frac{9(1 - (2/3)^5)}{1 - 2/3} = 27\left(1 - \frac{32}{243}\right) = 27\left(\frac{211}{243}\right) = \frac{211}{9}\).

(a)
M1: Writes a correct equation for the sum of the first two terms, e.g., \(a + ar = 15\) or \(a(1+r) = 15\).
M1: Writes a correct equation for the sum to infinity, e.g., \(\frac{a}{1-r} = 27\).
M1: Eliminates \(a\) from their equations to obtain an equation in terms of \(r\) only.
A1*: Correctly shows \(r^2 = \frac{4}{9}\) with no errors.
(b)
B1: Deduces that \(r = \frac{2}{3}\) (since terms are positive) and finds \(a = 9\).
M1: Attempts to use the sum formula with \(n = 5\), \(a = 9\), and \(r = 2/3\).
A1: Correct simplified fraction \(\frac{211}{9}\) (or \(23\frac{4}{9}\)).

(a) The radius \(r\) is the distance from \(P\) to \(Q\): \(r^2 = (7-3)^2 + (1 - (-2))^2 = 4^2 + 3^2 = 25\). Thus the equation of \(C\) is \((x-3)^2 + (y+2)^2 = 25\). (b) The gradient of the radius \(PQ\) is \(m_{PQ} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}\). The gradient of the tangent is perpendicular to this: \(m = -\frac{4}{3}\). The equation of the tangent through \(Q(7, 1)\) is \(y - 1 = -\frac{4}{3}(x - 7) \implies 3(y - 1) = -4(x - 7) \implies 3y - 3 = -4x + 28 \implies 4x + 3y - 31 = 0\). (c) Since \(R(k, 5)\) lies on the tangent, we substitute \(x = k\) and \(y = 5\) into the equation of the tangent: \(4k + 3(5) - 31 = 0 \implies 4k + 15 - 31 = 0 \implies 4k = 16 \implies k = 4\).

(a)
M1: Attempts to find the radius of the circle using the distance formula between \((3, -2)\) and \((7, 1)\).
A1: Correct radius \(r = 5\) (or \(r^2 = 25\)).
A1: Correct equation for \(C\): \((x-3)^2 + (y+2)^2 = 25\) (or equivalent).
(b)
M1: Attempts to find the gradient of the radius \(PQ\).
A1: Gradient is \(\frac{3}{4}\).
M1: Uses the perpendicular gradient rule to find the gradient of the tangent (\(-\frac{4}{3}\)) and attempts to find the equation of the line through \(Q(7, 1)\).
A1: \(4x + 3y - 31 = 0\) (or any integer multiple thereof).
(c)
B1: Substitutes \(y = 5\) into their tangent equation to find \(k = 4\).

(a) Using the identity \(\cos^2 x = 1 - \sin^2 x\), we substitute this into the equation: \(6(1 - \sin^2 x) - \sin x - 5 = 0 \implies 6 - 6\sin^2 x - \sin x - 5 = 0 \implies -6\sin^2 x - \sin x + 1 = 0\). Multiplying the entire equation by \(-1\) yields: \(6\sin^2 x + \sin x - 1 = 0\). (b) Factorising the quadratic equation in \(\sin x\): \((3\sin x - 1)(2\sin x + 1) = 0\). This gives two possible cases: 1) \(\sin x = \frac{1}{3}\): \(x = \arcsin(1/3) \approx 19.47^{\circ}\). The second solution in the interval is \(180^{\circ} - 19.47^{\circ} = 160.53^{\circ}\). To 1 dp, these are \(19.5^{\circ}\) and \(160.5^{\circ}\). 2) \(\sin x = -\frac{1}{2}\): \(x = 180^{\circ} - (-30^{\circ}) = 210^{\circ}\) and \(x = 360^{\circ} - 30^{\circ} = 330^{\circ}\). Thus, the complete set of solutions in the range is \(x = 19.5^{\circ}, 160.5^{\circ}, 210^{\circ}, 330^{\circ}\).

(a)
M1: Uses the identity \(\cos^2 x = 1 - \sin^2 x\).
A1*: Obtains the given quadratic equation \(6\sin^2 x + \sin x - 1 = 0\) with no algebraic errors.
(b)
M1: Attempts to factorise or solve the quadratic in \(\sin x\).
A1: Correct values \(\sin x = \frac{1}{3}\) and \(\sin x = -\frac{1}{2}\).
M1: Finds at least one correct angle for either value (e.g., \(19.5^{\circ}\) or \(210^{\circ}\)).
A1: For \(x = 19.5^{\circ}\) and \(x = 160.5^{\circ}\) (accept \(19.47^{\circ}\) rounded to 1 dp).
A1: For \(x = 210^{\circ}\) and \(x = 330^{\circ}\).
A1: Deducts 1 mark if there are any extra solutions in the range, or if the answers are not rounded to 1 decimal place where appropriate.

(a) Applying the subtraction law of logarithms: \(\log_3\left(\frac{y+10}{y-2}\right) = 2\). Converting from logarithmic to exponential form: \(\frac{y+10}{y-2} = 3^2 = 9\). Rearranging the equation: \(y + 10 = 9(y - 2) \implies y + 10 = 9y - 18 \implies 28 = 8y \implies y = 3.5\). (b) Let \(u = 3^x\). The equation becomes: \(u^2 - 5u - 14 = 0\). Factorising: \((u - 7)(u + 2) = 0 \implies u = 7\) or \(u = -2\). Since \(3^x > 0\) for all real \(x\), we reject \(u = -2\). Solving \(3^x = 7\) by taking natural logs: \(x \ln 3 = \ln 7 \implies x = \frac{\ln 7}{\ln 3} \approx 1.7712\). To 2 decimal places, \(x = 1.77\).

(a)
M1: Uses the subtraction law of logs: \(\log_3\left(\frac{y+10}{y-2}\right) = 2\).
M1: Removes logs correctly: \(\frac{y+10}{y-2} = 3^2 = 9\).
A1: Solves to find \(y = 3.5\) (or \(\frac{7}{2}\)).
(b)
M1: Recognises the quadratic structure and substitutes \(u = 3^x\) to get \(u^2 - 5u - 14 = 0\).
A1: Solves to get \(3^x = 7\) (and rejects \(3^x = -2\)).
M1: Uses logs to solve \(3^x = 7\), e.g. \(x = \frac{\ln 7}{\ln 3}\).
A1: \(x \approx 1.77\) (accept only 1.77, reject other decimal places unless 1.77 is shown).

(a) The total surface area \(A\) of a closed box with a square base of side \(x\) and height \(h\) consists of two square ends and four rectangular sides: \(A = 2x^2 + 4xh\). Since \(A = 150\), we have \(2x^2 + 4xh = 150 \implies 4xh = 150 - 2x^2 \implies h = \frac{150 - 2x^2}{4x} = \frac{37.5}{x} - 0.5x\). The volume \(V\) of the box is given by \(V = x^2 h = x^2\left(\frac{150 - 2x^2}{4x}\right) = \frac{x(150 - 2x^2)}{4} = 37.5x - 0.5x^3\). (b) To find stationary points, we find \(\frac{dV}{dx}\) and set it to zero: \(\frac{dV}{dx} = 37.5 - 1.5x^2 = 0 \implies 1.5x^2 = 37.5 \implies x^2 = 25\). Since length must be positive, \(x = 5\). To justify that it is a maximum, we find the second derivative: \(\frac{d^2V}{dx^2} = -3x\). When \(x = 5\), \(\frac{d^2V}{dx^2} = -15 < 0\), which confirms a maximum. The maximum volume is \(V = 37.5(5) - 0.5(5^3) = 187.5 - 62.5 = 125\text{ cm}^3\).

(a)
M1: Writes an expression for the surface area: \(2x^2 + 4xh\).
A1: Equates to 150: \(2x^2 + 4xh = 150\) and makes \(h\) the subject: \(h = \frac{150 - 2x^2}{4x}\) (or equivalent).
M1: Writes the volume formula \(V = x^2 h\) and substitutes the expression for \(h\).
A1*: Simplifies correctly to show \(V = 37.5x - 0.5x^3\) with no errors.
(b)
M1: Differentiates \(V\) with respect to \(x\) to get \(\frac{dV}{dx} = 37.5 - 1.5x^2\) (or equivalent).
A1: Sets \(\frac{dV}{dx} = 0\) and solves to find \(x = 5\) (rejects \(x = -5\)).
M1: Finds \(\frac{d^2V}{dx^2} = -3x\) and substitutes \(x = 5\) to show it is negative (confirming a maximum), or uses another valid method.
A1: Calculates the maximum volume as \(125\text{ cm}^3\).

(a) The curve crosses the \(x\)-axis where \(y = 0\): \(9x - x^3 = 0 \implies x(9 - x^2) = 0 \implies x(3-x)(3+x) = 0\). The roots are \(x = -3\), \(x = 0\), and \(x = 3\). Thus, the coordinates are \((-3, 0)\), \((0, 0)\), and \((3, 0)\). (b) Since the region \(R\) is bounded by the curve and the positive \(x\)-axis, the limits of integration are \(x = 0\) and \(x = 3\). The area is: \(\int_{0}^{3} (9x - x^3) \, dx = \left[ \frac{9}{2}x^2 - \frac{1}{4}x^4 \right]_{0}^{3} = \left( \frac{9}{2}(3)^2 - \frac{1}{4}(3)^4 \right) - (0) = \frac{81}{2} - \frac{81}{4} = \frac{81}{4} = 20.25\).

(a)
M1: Sets \(y = 0\) and factorises the cubic equation to find the values of \(x\).
A1: Correct coordinates: \((0, 0)\), \((3, 0)\), and \((-3, 0)\).
(b)
M1: Identifies the limits of integration as \(0\) and \(3\).
M1: Integrates \(9x - x^3\) to get \(\frac{9}{2}x^2 - \frac{1}{4}x^4\) (constant of integration not required).
A1: Correct integrated expression.
M1: Substitutes the limits \(3\) and \(0\) into their integrated expression.
A1: Correct exact area of \(\frac{81}{4}\) (or \(20.25\)).

(a) Since \(P(2, 5)\) lies on \(C\), substitute \(x = 2\) and \(y = 5\) into the equation of \(C\):
\[(2)^2 + (5)^2 - 10(2) + 4(5) + k = 0\]
\[4 + 25 - 20 + 20 + k = 0\]
\[29 + k = 0 \implies k = -29\]

(b) Complete the square for the equation of \(C\):
\[x^2 - 10x + y^2 + 4y - 29 = 0\]
\[(x - 5)^2 - 25 + (y + 2)^2 - 4 - 29 = 0\]
\[(x - 5)^2 + (y + 2)^2 = 58\]
Thus, the centre of \(C\) is \((5, -2)\).

(c) The gradient of the radius from the centre \((5, -2)\) to \(P(2, 5)\) is:
\[m_{\text{radius}} = \frac{5 - (-2)}{2 - 5} = \frac{7}{-3} = -\frac{7}{3}\]
The tangent at \(P\) is perpendicular to the radius, so its gradient is:
\[m_{\text{tangent}} = -\frac{1}{-\frac{7}{3}} = \frac{3}{7}\]
The equation of the tangent is:
\[y - 5 = \frac{3}{7}(x - 2)\]
\[7(y - 5) = 3(x - 2)\]
\[7y - 35 = 3x - 6\]
\[3x - 7y + 29 = 0\]

**(a)**
* **M1**: Attempts to substitute \(x = 2\) and \(y = 5\) into the equation of the circle.
* **A1**: Correctly shows that \(k = -29\) with no errors.

**(b)**
* **M1**: Attempts to complete the square for both \(x\) and \(y\) terms. Condone sign errors.
* **A1**: Correct coordinates \((5, -2)\) or \(x = 5, y = -2\).

**(c)**
* **M1**: Attempts to find the gradient of the radius \(QP\) where \(Q\) is their centre from (b).
* **M1**: Uses the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent.
* **M1**: Attempts to find the equation of the straight line through \(P(2, 5)\) using their perpendicular gradient.
* **A1**: Correct equation in the required form, e.g. \(3x - 7y + 29 = 0\) (or any non-zero integer multiple of this).

Using the laws of logarithms:
First apply the power law to the first term:
\[2\log_3(x-2) = \log_3(x-2)^2\]
Now write the left-hand side as a single logarithm using the division law:
\[\log_3(x-2)^2 - \log_3(x+4) = \log_3\left(\frac{(x-2)^2}{x+4}\right)\]
So the equation becomes:
\[\log_3\left(\frac{(x-2)^2}{x+4}\right) = 1\]
Convert from logarithmic to exponential form:
\[\frac{(x-2)^2}{x+4} = 3^1 = 3\]
Multiply both sides by \(x+4\):
\[(x-2)^2 = 3(x+4)\]
\[x^2 - 4x + 4 = 3x + 12\]
Rearrange into a quadratic equation:
\[x^2 - 7x - 8 = 0\]
Factorise the quadratic:
\[(x - 8)(x + 1) = 0\]
This gives possible solutions:
\[x = 8 \quad \text{or} \quad x = -1\]
We must check the validity of these solutions in the original equation:
For \(\log_3(x-2)\) to be defined, we require \(x-2 > 0 \implies x > 2\).
- If \(x = 8\), then \(x-2 = 6 > 0\) and \(x+4 = 12 > 0\), which is valid.
- If \(x = -1\), then \(x-2 = -3 < 0\), which is undefined.
Therefore, the only valid solution is \(x = 8\).

* **M1**: Correct use of the power law of logarithms to write \(2\log_3(x-2)\) as \(\log_3(x-2)^2\).
* **M1**: Correct use of the subtraction/division law of logarithms to express the LHS as a single logarithm: \(\log_3\left(\frac{(x-2)^2}{x+4}\right)\).
* **M1**: Removes logarithms correctly to obtain \(\frac{(x-2)^2}{x+4} = 3\).
* **M1**: Expands and rearranges to form a 3-term quadratic equation.
* **A1**: Correct quadratic equation \(x^2 - 7x - 8 = 0\).
* **M1**: Solves their 3-term quadratic equation to obtain two values of \(x\) (e.g., \(x = 8\) and \(x = -1\)).
* **A1**: Identifies \(x = 8\) as the only solution and provides a valid reason for rejecting \(x = -1\) (e.g., stating that \(x-2 > 0\) is required, or that \(\log_3(-3)\) is undefined).

(a) To sketch the graph of \( f(x) = |2x - 5| - 3 \):
- The vertex occurs where the expression inside the modulus is zero: \( 2x - 5 = 0 \implies x = 2.5 \). At this point, \( y = -3 \). So, the vertex is \( (2.5, -3) \).
- The y-intercept is found by setting \( x = 0 \): \( f(0) = |-5| - 3 = 2 \). So, the y-intercept is \( (0, 2) \).
- The x-intercepts are found by setting \( f(x) = 0 \):
\( |2x - 5| - 3 = 0 \implies |2x - 5| = 3 \)
This gives two cases:
1) \( 2x - 5 = 3 \implies 2x = 8 \implies x = 4 \)
2) \( 2x - 5 = -3 \implies 2x = 2 \implies x = 1 \)
So, the x-intercepts are \( (1, 0) \) and \( (4, 0) \).

(b) To solve \( |2x - 5| - 3 = x + 2 \):
\( |2x - 5| = x + 5 \)
Case 1 (for \( x \ge 2.5 \)):
\( 2x - 5 = x + 5 \implies x = 10 \)
Since \( 10 \ge 2.5 \), this is a valid solution.

Case 2 (for \( x < 2.5 \)):
\( -(2x - 5) = x + 5 \implies -2x + 5 = x + 5 \implies 3x = 0 \implies x = 0 \)
Since \( 0 < 2.5 \), this is also a valid solution.

So the solutions are \( x = 0 \) and \( x = 10 \).

Part (a):
- M1: For drawing a V-shaped graph with a minimum point in the 4th quadrant.
- A1: For the correct coordinates of the vertex: \( (2.5, -3) \) or \( \left(\frac{5}{2}, -3\right) \).
- A1.5: For all three correct intercepts: \( (0, 2) \), \( (1, 0) \), and \( (4, 0) \) clearly indicated on the sketch.

Part (b):
- M1: For forming and solving a linear equation for the positive branch: \( 2x - 5 = x + 5 \).
- A1: For the solution \( x = 10 \).
- M1: For forming and solving a linear equation for the negative branch: \( -(2x - 5) = x + 5 \).
- A1: For the solution \( x = 0 \) and demonstrating both solutions are valid.

(a) We use the quotient rule: \( y = \frac{u}{v} \) where \( u = e^{2x-2} \) and \( v = 3x - 1 \).
- \( \frac{\mathrm{d}u}{\mathrm{d}x} = 2e^{2x-2} \)
- \( \frac{\mathrm{d}v}{\mathrm{d}x} = 3 \)

Using the quotient rule formula:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{(3x - 1)(2e^{2x-2}) - (e^{2x-2})(3)}{(3x - 1)^2} \]

Factorizing the numerator:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{2x-2}[2(3x - 1) - 3]}{(3x - 1)^2} = \frac{e^{2x-2}(6x - 5)}{(3x - 1)^2} \]

(b) At \( x = 1 \):
- \( y = \frac{e^0}{3(1) - 1} = \frac{1}{2} \)
- The gradient of the tangent, \( m \), is given by substituting \( x = 1 \) into \( \frac{\mathrm{d}y}{\mathrm{d}x} \):
\[ m = \frac{e^0(6(1) - 5)}{(3(1) - 1)^2} = \frac{1}{4} \]

Using the equation of a straight line:
\[ y - y_1 = m(x - x_1) \implies y - \frac{1}{2} = \frac{1}{4}(x - 1) \]

Multiply by 4 to clear fractions:
\[ 4y - 2 = x - 1 \implies x - 4y + 1 = 0 \]

Thus, the equation in the required form is \( x - 4y + 1 = 0 \).

Part (a):
- M1: Applies the quotient rule correctly with \( u = e^{2x-2} \) and \( v = 3x - 1 \).
- A1: For obtaining correct derivatives of both terms: \( u' = 2e^{2x-2} \) and \( v' = 3 \).
- A1.5: Correct fully simplified expression for the derivative: \( \frac{e^{2x-2}(6x - 5)}{(3x - 1)^2} \).

Part (b):
- B1: Finds the correct y-coordinate: \( y = \frac{1}{2} \).
- M1: Evaluates the gradient of the tangent at \( x = 1 \) using their derivative.
- A1: Correct gradient \( m = \frac{1}{4} \).
- A1: Correct equation of the tangent in the required form: \( x - 4y + 1 = 0 \) (or any integer multiple thereof, such as \( -x + 4y - 1 = 0 \)).

(a) Using the identity \( 1 + \tan^2\theta = \sec^2\theta \), we have:
\[ \text{LHS} = \frac{2\tan\theta}{\sec^2\theta} \]

Since \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) and \( \sec^2\theta = \frac{1}{\cos^2\theta} \):
\[ \text{LHS} = \frac{2\left(\frac{\sin\theta}{\cos\theta}\right)}{\frac{1}{\cos^2\theta}} = 2\left(\frac{\sin\theta}{\cos\theta}\right) \times \cos^2\theta = 2\sin\theta\cos\theta \]

Using the double-angle identity \( 2\sin\theta\cos\theta = \sin 2\theta \):
\[ \text{LHS} = \sin 2\theta = \text{RHS} \]

(b) The given equation can be written as:
\[ 2 \left( \frac{2\tan x}{1 + \tan^2 x} \right) = \sqrt{3} \]

Using the identity from part (a), this is:
\[ 2\sin 2x = \sqrt{3} \implies \sin 2x = \frac{\sqrt{3}}{2} \]

Given \( 0 \le x < \pi \), we have \( 0 \le 2x < 2\pi \).

The solutions for \( 2x \) are:
\[ 2x = \frac{\pi}{3}, \quad 2x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

Solving for \( x \):
\[ x = \frac{\pi}{6}, \quad x = \frac{\pi}{3} \]

Part (a):
- M1: Uses the identity \( 1 + \tan^2\theta = \sec^2\theta \) or writes \( \tan\theta \) and \( \tan^2\theta \) in terms of sine and cosine.
- M1: Simplifies the fraction to a product of sine and cosine terms, reaching \( 2\sin\theta\cos\theta \).
- A1.5*: Fully correct proof with all steps shown clearly and no notation errors.

Part (b):
- M1: Realises the connection to part (a) to write the equation as \( 2\sin 2x = \sqrt{3} \).
- M1: Correctly finds at least one principal value for \( 2x \) (e.g., \( 2x = \frac{\pi}{3} \)).
- A1: Obtains \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{3} \).
- A1: Gives no extra solutions within the range.

(a) Let \( f(x) = 3x^2 - e^x - 2 \).
- \( f(1) = 3(1)^2 - e^1 - 2 = 1 - e \approx -1.718 < 0 \)
- \( f(2) = 3(2)^2 - e^2 - 2 = 10 - e^2 \approx 2.611 > 0 \)
Since there is a change of sign in the interval \( [1, 2] \) and \( f(x) \) is a continuous function, there exists a root \( \alpha \) in \( [1, 2] \).

(b) \( 3x^2 - e^x - 2 = 0 \implies 3x^2 = e^x + 2 \implies x^2 = \frac{e^x + 2}{3} \).
Taking the positive square root (since the root lies in \( [1, 2] \) where \( x > 0 \)):
\[ x = \sqrt{\frac{e^x + 2}{3}} \]

(c) Using the iterative formula with \( x_0 = 1.2 \):
- \( x_1 = \sqrt{\frac{e^{1.2} + 2}{3}} \approx 1.3317 \)
- \( x_2 = \sqrt{\frac{e^{1.331680} + 2}{3}} \approx 1.3889 \)
- \( x_3 = \sqrt{\frac{e^{1.388942} + 2}{3}} \approx 1.4155 \)

(d) Choose the interval \( [1.4395, 1.4405] \):
- \( f(1.4395) = 3(1.4395)^2 - e^{1.4395} - 2 \approx -0.0021 < 0 \)
- \( f(1.4405) = 3(1.4405)^2 - e^{1.4405} - 2 \approx 0.0023 > 0 \)
Since there is a sign change and \( f(x) \) is continuous, the root \( \alpha \) lies in \( [1.4395, 1.4405] \), so \( \alpha = 1.440 \) correct to 3 decimal places.

Part (a):
- M1: Attempts to evaluate \( f(1) \) and \( f(2) \) with at least one correct calculation.
- A0.5: Concludes with correct values and mentions sign change and continuity.

Part (b):
- M1: Isolates the \( x^2 \) term on one side of the equation.
- A1: Shows clear algebraic steps to arrive at the given iterative form.

Part (c):
- M1: Correctly substitutes \( x_0 = 1.2 \) to find \( x_1 \).
- A1: Obtains \( x_1 \approx 1.3317 \) and \( x_2 \approx 1.3889 \).
- A1: Obtains \( x_3 \approx 1.4155 \).

Part (d):
- B1: Evaluates \( f(1.4395) \) and \( f(1.4405) \) to show a change of sign and concludes \( \alpha = 1.440 \).

(a) Using the reverse chain rule:
\[ \int (2x + 1)^4 \, \mathrm{d}x = \frac{(2x + 1)^5}{5 \times 2} + C = \frac{(2x + 1)^5}{10} + C \]

(b) Let \( u = 2x - 3 \).
Then \( \mathrm{d}u = 2\,\mathrm{d}x \implies \mathrm{d}x = \frac{1}{2}\,\mathrm{d}u \).
Also, \( x = \frac{u+3}{2} \).

Now, change the limits:
- When \( x = 2 \), \( u = 2(2) - 3 = 1 \).
- When \( x = 4 \), \( u = 2(4) - 3 = 5 \).

Substitute into the integral:
\[ \int_{2}^{4} \frac{x}{(2x - 3)^2} \, \mathrm{d}x = \int_{1}^{5} \frac{\frac{u+3}{2}}{u^2} \left(\frac{1}{2}\,\mathrm{d}u\right) = \frac{1}{4} \int_{1}^{5} \frac{u+3}{u^2} \, \mathrm{d}u \]
\[ = \frac{1}{4} \int_{1}^{5} \left(\frac{1}{u} + 3u^{-2}\right) \mathrm{d}u \]

Integrating term by term:
\[ = \frac{1}{4} \left[ \ln|u| - \frac{3}{u} \right]_{1}^{5} \]

Apply the limits:
\[ = \frac{1}{4} \left[ \left(\ln 5 - \frac{3}{5}\right) - \left(\ln 1 - 3\right) \right] \]
\[ = \frac{1}{4} \left[ \ln 5 - \frac{3}{5} + 3 \right] \]
\[ = \frac{1}{4} \left[ \ln 5 + \frac{12}{5} \right] = \frac{3}{5} + \frac{1}{4}\ln 5 \]

Thus, \( p = \frac{3}{5} \), \( q = \frac{1}{4} \), and \( r = 5 \).

Part (a):
- M1: Integrates to get the form \( k(2x+1)^5 \).
- A1.5: Correct answer: \( \frac{(2x+1)^5}{10} + C \) (must include \( +C \)).

Part (b):
- M1: For attempting the substitution with correct derivative relation, i.e., \( \mathrm{d}x = \frac{1}{2}\,\mathrm{d}u \).
- A1: For correctly changing the limits to \( u = 1 \) and \( u = 5 \).
- M1: For substituting \( x = \frac{u+3}{2} \) and obtaining the integral in terms of \( u \).
- A1: For correct integration of the expression in terms of \( u \), yielding \( \ln|u| - \frac{3}{u} \).
- A1: For substituting limits and correctly simplifying to get \( \frac{3}{5} + \frac{1}{4}\ln 5 \) or equivalent.

(a) At \( t = 0 \), \( M = 150 \). Substituting these values into the model:
\[ 150 = A e^0 \implies A = 150 \]

(b) At \( t = 10 \), \( M = 90 \). Using \( A = 150 \):
\[ 90 = 150 e^{-10k} \]
\[ e^{-10k} = \frac{90}{150} = \frac{3}{5} \]
\[ -10k = \ln\left(\frac{3}{5}\right) \implies 10k = -\ln\left(\frac{3}{5}\right) = \ln\left(\frac{5}{3}\right) \]
\[ k = \frac{1}{10}\ln\left(\frac{5}{3}\right) \]

(c) The rate of change of the mass is given by \( \frac{\mathrm{d}M}{\mathrm{d}t} \):
\[ \frac{\mathrm{d}M}{\mathrm{d}t} = -k A e^{-kt} = -k M \]

When \( t = 20 \):
\[ M = 150 e^{-20k} = 150 \left( e^{-10k} \right)^2 = 150 \left(\frac{3}{5}\right)^2 = 150 \times \frac{9}{25} = 54 \text{ grams} \]

So,
\[ \frac{\mathrm{d}M}{\mathrm{d}t} = -k \times 54 = -\left[ \frac{1}{10}\ln\left(\frac{5}{3}\right) \right] \times 54 = -5.4\ln\left(\frac{5}{3}\right) \approx -2.76 \text{ g/year} \]

Therefore, the rate of decrease when \( t = 20 \) is \( 2.76 \) grams per year.

Part (a):
- B1: States \( A = 150 \).

Part (b):
- M1: Sets up the equation \( 90 = 150 e^{-10k} \) and attempts to solve for \( e^{-10k} \).
- A1.5: Correctly applies laws of logarithms to show the negative sign inversion: \( -\ln(3/5) = \ln(5/3) \) and reaches the given result.

Part (c):
- M1: Differentiates the equation with respect to \( t \) to get \( \frac{\mathrm{d}M}{\mathrm{d}t} = -kAe^{-kt} \).
- M1: Correctly evaluates \( M \) at \( t = 20 \) as \( 54 \) (either exact or decimal approximation).
- A1: Evaluates the rate of change at \( t = 20 \) as \( -2.76 \) or \( 2.76 \).
- A1: Final answer given as a positive rate of decrease: \( 2.76 \).

(a) To find \( f^{-1}(x) \), let \( y = f(x) \):
\[ y = \frac{2x + 5}{x - 3} \]
\[ y(x - 3) = 2x + 5 \]
\[ yx - 3y = 2x + 5 \]
\[ yx - 2x = 3y + 5 \]
\[ x(y - 2) = 3y + 5 \]
\[ x = \frac{3y + 5}{y - 2} \]

So,
\[ f^{-1}(x) = \frac{3x + 5}{x - 2} \]

The domain of \( f^{-1}(x) \) is the range of \( f(x) \).
We can write \( f(x) = \frac{2(x-3) + 11}{x-3} = 2 + \frac{11}{x-3} \).
Since \( x \ne 3 \), \( \frac{11}{x-3} \ne 0 \), so \( f(x) \ne 2 \).
Thus, the range of \( f(x) \) is \( \{ y \in \mathbb{R} : y \ne 2 \} \).
Therefore, the domain of \( f^{-1}(x) \) is \( x \in \mathbb{R}, \ x \ne 2 \).

(b) To find \( \operatorname{fg}(5) \):
First, calculate \( g(5) \):
\[ g(5) = 5^2 - 4 = 21 \]

Next, substitute this into \( f(x) \):
\[ \operatorname{fg}(5) = f(21) = \frac{2(21) + 5}{21 - 3} = \frac{42 + 5}{18} = \frac{47}{18} \]

Part (a):
- M1: Sets \( y = \frac{2x+5}{x-3} \) and multiplies both sides by \( x-3 \).
- M1: Gathers terms in \( x \) on one side and factorizes.
- A1: Correct expression: \( f^{-1}(x) = \frac{3x+5}{x-2} \) (or equivalent).
- B1.5: Correct domain: \( x \in \mathbb{R}, \ x \ne 2 \).

Part (b):
- M1: Correctly evaluates \( g(5) = 21 \).
- M1: Substitutes their value for \( g(5) \) into the expression for \( f \).
- A1: Correct exact value of \( \frac{47}{18} \).

(a) We apply the quotient rule with \( u = \sin x \) and \( v = 2 + \cos x \).
- \( \frac{\mathrm{d}u}{\mathrm{d}x} = \cos x \)
- \( \frac{\mathrm{d}v}{\mathrm{d}x} = -\sin x \)

Using the quotient rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} \]
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2 + \cos x)(\cos x) - (\sin x)(-\sin x)}{(2 + \cos x)^2} \]
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\cos x + \cos^2 x + \sin^2 x}{(2 + \cos x)^2} \]

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\cos x + 1}{(2 + \cos x)^2} \]

(b) Stationary points occur where \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).
\[ \frac{2\cos x + 1}{(2 + \cos x)^2} = 0 \implies 2\cos x + 1 = 0 \implies \cos x = -\frac{1}{2} \]

Since \( 0 \le x \le \pi \), the only solution is:
\[ x = \frac{2\pi}{3} \]

Now, find the y-coordinate by substituting \( x = \frac{2\pi}{3} \) into the original equation:
\[ y = \frac{\sin\left(\frac{2\pi}{3}\right)}{2 + \cos\left(\frac{2\pi}{3}\right)} = \frac{\frac{\sqrt{3}}{2}}{2 - \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3} \]

So the exact coordinates of the stationary point are \( \left(\frac{2\pi}{3}, \frac{\sqrt{3}}{3}\right) \).

Part (a):
- M1: Applies the quotient rule with \( u = \sin x \) and \( v = 2 + \cos x \).
- A1: For the correct unsimplified numerator: \( (2 + \cos x)\cos x - \sin x(-\sin x) \).
- M1: Uses the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify numerator.
- A0.5*: Correctly proves the given result with no errors and correct notation.

Part (b):
- M1: Sets the numerator of \( \frac{\mathrm{d}y}{\mathrm{d}x} \) to 0 and attempts to solve for \( x \).
- A1: Obtains \( x = \frac{2\pi}{3} \).
- M1: Substitutes their \( x \) value back into the original curve equation.
- A1: Obtains the exact coordinate \( \left(\frac{2\pi}{3}, \frac{\sqrt{3}}{3}\right) \) or equivalent, e.g. \( \left(\frac{2\pi}{3}, \frac{1}{\sqrt{3}}\right) \).

(a) We use the compound angle formula:
\[R\sin(2\theta - \alpha) = R\sin(2\theta)\cos(\alpha) - R\cos(2\theta)\sin(\alpha)\]

Comparing this with \(3\sin(2\theta) - 4\cos(2\theta)\), we get:
\[R\cos(\alpha) = 3\]
\[R\sin(\alpha) = 4\]

Squaring and adding these equations:
\[R^2\cos^2(\alpha) + R^2\sin^2(\alpha) = 3^2 + 4^2\]
\[R^2 = 9 + 16 = 25 \implies R = 5\]

To find \(\alpha\):
\[\tan(\alpha) = \frac{R\sin(\alpha)}{R\cos(\alpha)} = \frac{4}{3}\]
\[\alpha = \arctan\left(\frac{4}{3}\right) \approx 0.927295\dots \approx 0.927 \text{ radians}\]

So,
\[3\sin(2\theta) - 4\cos(2\theta) = 5\sin(2\theta - 0.927)\]

(b) Using the result from part (a), the equation
\[3\sin(2x) - 4\cos(2x) = 2.5\]
can be written as:
\[5\sin(2x - 0.927295) = 2.5\]
\[\sin(2x - 0.927295) = 0.5\]

Let \(\phi = 2x - 0.927295\).
Since \(0 \le x \le \pi\), we have \(0 \le 2x \le 2\pi\),
so \(-0.927295 \le \phi \le 2\pi - 0.927295 \approx 5.35589\).

In this interval, the solutions for \(\sin(\phi) = 0.5\) are:
\[\phi = \frac{\pi}{6} \approx 0.523599\]
\[\phi = \frac{5\pi}{6} \approx 2.617994\]

Now, solve for \(x\):
1) \(2x - 0.927295 = 0.523599 \implies 2x = 1.450894 \implies x \approx 0.7254 \approx 0.73\)
2) \(2x - 0.927295 = 2.617994 \implies 2x = 3.545289 \implies x \approx 1.7726 \approx 1.77\)

Thus, the solutions in the interval are \(x \approx 0.73\) and \(x \approx 1.77\).

**Part (a)**
* **B1**: Identifies or states that \(R = 5\).
* **M1**: Uses a correct trigonometric identity or division to find \(\alpha\), e.g., \(\tan(\alpha) = \pm \frac{4}{3}\) or \(\tan(\alpha) = \pm \frac{3}{4}\).
* **A1**: \(\alpha \approx 0.927\) (must be 3 decimal places).

**Part (b)**
* **M1**: Sets their \(R\sin(2x - \alpha) = 2.5\) and attempts to make \(\sin(2x - \alpha)\) the subject, yielding \(\sin(2x - \alpha) = 0.5\).
* **M1**: Finds at least one value for \(2x - \alpha\) using \(\arcsin(0.5)\), e.g. \(\frac{\pi}{6}\) or \(\frac{5\pi}{6}\) (accept in degrees or decimals for this mark, e.g. \(30^\circ\) or \(0.524\) radians).
* **A1**: One correct value of \(x\) to 2 decimal places (\(x \approx 0.73\) or \(x \approx 1.77\)).
* **A1**: Both correct values of \(x\) to 2 decimal places (\(x \approx 0.73\) and \(x \approx 1.77\)) and no extra values in the interval.

(a) We differentiate \(y = \mathrm{e}^{-2x} \sin(2x)\) using the product rule:
\[\frac{\mathrm{d}(uv)}{\mathrm{dx}} = u\frac{\mathrm{d}v}{\mathrm{dx}} + v\frac{\mathrm{d}u}{\mathrm{dx}}\]
Let \(u = \mathrm{e}^{-2x}\) and \(v = \sin(2x)\).
Then
\[\frac{\mathrm{d}u}{\mathrm{dx}} = -2\mathrm{e}^{-2x}\]
\[\frac{\mathrm{d}v}{\mathrm{dx}} = 2\cos(2x)\]

Substituting these into the product rule formula:
\[\frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{e}^{-2x}(2\cos(2x)) + \sin(2x)(-2\mathrm{e}^{-2x})\]
\[\frac{\mathrm{dy}}{\mathrm{dx}} = 2\mathrm{e}^{-2x}\cos(2x) - 2\mathrm{e}^{-2x}\sin(2x)\]
Factorising out \(2\mathrm{e}^{-2x}\):
\[\frac{\mathrm{dy}}{\mathrm{dx}} = 2\mathrm{e}^{-2x}(\cos(2x) - \sin(2x))\]
Thus, \(A = 2\).

(b) At stationary points of \(C\), we have
\[\frac{\mathrm{dy}}{\mathrm{dx}} = 0\]
\[2\mathrm{e}^{-2x}(\cos(2x) - \sin(2x)) = 0\]
Since \(2\mathrm{e}^{-2x} \ne 0\) for all real \(x\), we must have:
\[\cos(2x) - \sin(2x) = 0\]
\[\sin(2x) = \cos(2x)\]
Dividing both sides by \(\cos(2x)\) (which is valid as \(\cos(2x) \neq 0\) when \(\sin(2x) = \cos(2x)\)):
\[\tan(2x) = 1\]

Since \(0 < x < \pi\), we have \(0 < 2x < 2\pi\).
In this interval, the solutions for \(\tan(2x) = 1\) are:
\[2x = \frac{\pi}{4} \implies x = \frac{\pi}{8}\]
\[2x = \frac{5\pi}{4} \implies x = \frac{5\pi}{8}\]

Now find the corresponding exact \(y\)-coordinates:
1) For \(x = \frac{\pi}{8}\):
\[y = \mathrm{e}^{-2\left(\frac{\pi}{8}\right)} \sin\left(2 \cdot \frac{\pi}{8}\right) = \mathrm{e}^{-\frac{\pi}{4}} \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\mathrm{e}^{-\frac{\pi}{4}}\]
So the first stationary point is
\[\left(\frac{\pi}{8}, \frac{\sqrt{2}}{2}\mathrm{e}^{-\frac{\pi}{4}}\right)\]

2) For \(x = \frac{5\pi}{8}\):
\[y = \mathrm{e}^{-2\left(\frac{5\pi}{8}\right)} \sin\left(2 \cdot \frac{5\pi}{8}\right) = \mathrm{e}^{-\frac{5\pi}{4}} \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\mathrm{e}^{-\frac{5\pi}{4}}\]
So the second stationary point is
\[\left(\frac{5\pi}{8}, -\frac{\sqrt{2}}{2}\mathrm{e}^{-\frac{5\pi}{4}}\right)\]

**Part (a)**
* **M1**: Attempts to apply the product rule to \(y = \mathrm{e}^{-2x} \sin(2x)\). Must see an expression of the form \(\pm k_1 \mathrm{e}^{-2x} \sin(2x) \pm k_2 \mathrm{e}^{-2x} \cos(2x)\) where \(k_1, k_2 > 0\).
* **A1**: For finding correct derivatives of individual parts, i.e., \(\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{e}^{-2x}) = -2\mathrm{e}^{-2x}\) and \(\frac{\mathrm{d}}{\mathrm{dx}}(\sin(2x)) = 2\cos(2x)\).
* **A1**: Fully correct derivative factorised into the required form: \(2\mathrm{e}^{-2x}(\cos(2x) - \sin(2x))\). (Accept \(A = 2\)).

**Part (b)**
* **M1**: Sets their \(\frac{\mathrm{dy}}{\mathrm{dx}} = 0\) and deduces \(\tan(2x) = 1\) (or equivalent equation in sine and cosine).
* **A1**: Finds at least one correct value for \(x\) in the interval, i.e., \(x = \frac{\pi}{8}\) or \(x = \frac{5\pi}{8}\).
* **A1**: Finds both correct values for \(x\): \(x = \frac{\pi}{8}\) and \(x = \frac{5\pi}{8}\) (and no others in the interval).
* **M1**: Substitutes at least one of their \(x\) values back into the original curve equation to find a \(y\)-value, showing a correct method for evaluating the exact trigonometric part (e.g., \(\sin(\pi/4) = \frac{\sqrt{2}}{2}\)).
* **A1**: Correct exact coordinates of both stationary points: \(\left(\frac{\pi}{8}, \frac{\sqrt{2}}{2}\mathrm{e}^{-\frac{\pi}{4}}\right)\) and \(\left(\frac{5\pi}{8}, -\frac{\sqrt{2}}{2}\mathrm{e}^{-\frac{5\pi}{4}}\right)\) (or equivalent simplified exact forms, e.g., \(\frac{1}{\sqrt{2}}\)).

Rewrite the function using negative fractional indices:
\[ f(x) = 4(9 - 6x)^{-\frac{1}{2}} \]

Factor out 9 to obtain a binomial in the form \( (1 + y)^n \):
\[ f(x) = 4 \times 9^{-\frac{1}{2}} \left(1 - \frac{6}{9}x\right)^{-\frac{1}{2}} = \frac{4}{3} \left(1 - \frac{2}{3}x\right)^{-\frac{1}{2}} \]

Apply the general binomial expansion formula \( (1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots \) with \( n = -\frac{1}{2} \) and \( y = -\frac{2}{3}x \):
\[ \left(1 - \frac{2}{3}x\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{2}{3}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{2}{3}x\right)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}\left(-\frac{2}{3}x\right)^3 + \dots \]

Simplify the terms inside the bracket:
\[ = 1 + \frac{1}{3}x + \frac{3}{8}\left(\frac{4}{9}x^2\right) - \frac{5}{16}\left(-\frac{8}{27}x^3\right) \]
\[ = 1 + \frac{1}{3}x + \frac{1}{6}x^2 + \frac{5}{54}x^3 \]

Multiply by the constant factor of \( \frac{4}{3} \):
\[ f(x) = \frac{4}{3}\left(1 + \frac{1}{3}x + \frac{1}{6}x^2 + \frac{5}{54}x^3\right) = \frac{4}{3} + \frac{4}{9}x + \frac{2}{9}x^2 + \frac{10}{81}x^3 \]

For the expansion to be valid, we require:
\[ \left|-\frac{2}{3}x\right| < 1 \implies |x| < \frac{3}{2} \]

M1: Attempts to write \( f(x) \) in the form \( k(1 - ax)^{-1/2} \), finding \( k = \frac{4}{3} \) and \( a = \frac{2}{3} \).
A1: Correct simplified factor outside the bracket \( \frac{4}{3} \) (or equivalent).
M1: Applies the binomial expansion formula correctly to \( (1 - ax)^{-1/2} \) up to the \( x^3 \) term with their \( a \).
A1: Correct unsimplified expansion inside the bracket.
A1: Correct simplified expansion: \( \frac{4}{3} + \frac{4}{9}x + \frac{2}{9}x^2 + \frac{10}{81}x^3 \) (or equivalent).
B1: States correct validity of \( |x| < \frac{3}{2} \) (or equivalent interval).

(a) Differentiate \( x \) and \( y \) with respect to \( t \):
\[ \frac{dx}{dt} = 3t^2 - 3 \]
\[ \frac{dy}{dt} = 2t + 2 = 2(t+1) \]

Using the chain rule for parametric differentiation:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2(t+1)}{3t^2 - 3} \]

Factorise the denominator:
\[ \frac{dy}{dx} = \frac{2(t+1)}{3(t-1)(t+1)} = \frac{2}{3(t-1)} \quad (\text{for } t \neq -1) \]

(b) Find the coordinates of the point on the curve at \( t = 2 \):
\[ x = 2^3 - 3(2) = 8 - 6 = 2 \]
\[ y = 2^2 + 2(2) = 4 + 4 = 8 \]

Find the gradient of the tangent at \( t = 2 \):
\[ m = \left. \frac{dy}{dx} \right|_{t=2} = \frac{2}{3(2-1)} = \frac{2}{3} \]

Using the equation of a straight line:
\[ y - y_1 = m(x - x_1) \implies y - 8 = \frac{2}{3}(x - 2) \]
Multiply by 3 to clear the fraction:
\[ 3y - 24 = 2x - 4 \implies 2x - 3y + 20 = 0 \]

(a)
M1: Differentiates \( x \) and \( y \) with respect to \( t \) to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
A1: Correct derivatives: \( \frac{dx}{dt} = 3t^2 - 3 \) and \( \frac{dy}{dt} = 2t + 2 \).
M1: Applies \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) and attempts to factorise and simplify.
A1: Correct simplified derivative \( \frac{2}{3(t-1)} \).

(b)
B1: Finds correct coordinates \( (2, 8) \).
M1: Substitutes \( t = 2 \) into their \( \frac{dy}{dx} \) to find the gradient.
M1: Uses their coordinates and gradient to form a linear equation.
A1: Obtains \( 2x - 3y + 20 = 0 \) (or any integer multiple thereof).

Let \( u = 1 + e^x \).
Then \( \frac{du}{dx} = e^x \implies dx = \frac{du}{e^x} = \frac{du}{u-1} \).

Rewrite \( e^{2x} \) in terms of \( u \):
\[ e^{2x} = (e^x)^2 = (u-1)^2 \]

Now find the limits of integration in terms of \( u \):
When \( x = 0 \), \( u = 1 + e^0 = 1 + 1 = 2 \).
When \( x = \ln 3 \), \( u = 1 + e^{\ln 3} = 1 + 3 = 4 \).

Substitute all parts into the integral:
\[ \int_{2}^{4} \frac{(u-1)^2}{u} \cdot \frac{du}{u-1} = \int_{2}^{4} \frac{u-1}{u} \, du \]

Separate the fraction:
\[ \int_{2}^{4} \left(1 - \frac{1}{u}\right) \, du \]

Integrate with respect to \( u \):
\[ \left[ u - \ln|u| \right]_{2}^{4} \]

Apply the limits of integration:
\[ = (4 - \ln 4) - (2 - \ln 2) \]
\[ = 4 - 2 - \ln 4 + \ln 2 \]
\[ = 2 - \ln\left(\frac{4}{2}\right) = 2 - \ln 2 \]

This can be written in the form \( a + \ln b \) as \( 2 + \ln\left(\frac{1}{2}\right) \), so \( a = 2 \) and \( b = 0.5 \).

M1: Finds the differential relation \( du = e^x \, dx \) or equivalent, and substitutes to get \( dx \) in terms of \( du \) and \( u \).
M1: Changes limits correctly from \( x \)-values of \( 0 \) and \( \ln 3 \) to \( u \)-values of \( 2 \) and \( 4 \).
M1: Expresses the entire integrand in terms of \( u \) resulting in a simplified integrand \( \frac{u-1}{u} \) or equivalent.
A1: Correct integral in \( u \): \( \int (1 - \frac{1}{u}) \, du \) with correct limits.
M1: Integrates correctly to obtain \( u - \ln|u| \).
A1: Substitutes limits and applies logarithm laws to achieve the final simplified exact answer \( 2 - \ln 2 \) (or \( 2 + \ln 0.5 \)).

(a) Equate the position vectors of \( l_1 \) and \( l_2 \) to find the intersection point:
\[ \begin{pmatrix} 1 + 2\lambda \\ 5 - \lambda \\ -1 + 3\lambda \end{pmatrix} = \begin{pmatrix} -2 + \mu \\ a + 2\mu \\ 7 - \mu \end{pmatrix} \]

This gives three simultaneous equations:
1) \( 1 + 2\lambda = -2 + \mu \implies \mu - 2\lambda = 3 \)
2) \( 5 - \lambda = a + 2\mu \)
3) \( -1 + 3\lambda = 7 - \mu \implies \mu + 3\lambda = 8 \)

Subtract equation (1) from equation (3) to eliminate \( \mu \):
\[ (\mu + 3\lambda) - (\mu - 2\lambda) = 8 - 3 \]
\[ 5\lambda = 5 \implies \lambda = 1 \]

Substitute \( \lambda = 1 \) back into equation (1):
\[ \mu - 2(1) = 3 \implies \mu = 5 \]

Check these parameters in the z-coordinate equation (3):
LHS: \( -1 + 3(1) = 2 \)
RHS: \( 7 - 5 = 2 \) (Consistent!)

Now use equation (2) to find the constant \( a \):
\[ 5 - 1 = a + 2(5) \]
\[ 4 = a + 10 \implies a = -6 \]

(b) To find the coordinates of \( P \), substitute \( \lambda = 1 \) into the equation for \( l_1 \):
\[ \mathbf{r} = \begin{pmatrix} 1 + 2(1) \\ 5 - 1 \\ -1 + 3(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix} \]

Hence, the coordinates of \( P \) are \( (3, 4, 2) \).

(a)
M1: Writes down simultaneous equations by equating components of \( l_1 \) and \( l_2 \).
M1: Solves a system of two equations to find values for \( \lambda \) and \( \mu \).
A1: Correct values \( \lambda = 1 \) and \( \mu = 5 \).
M1: Substitutes \( \lambda \) and \( \mu \) into the equation involving \( a \) to solve for \( a \).
A1: Correct value \( a = -6 \).

(b)
M1: Substitutes their parameter \( \lambda \) (or \( \mu \)) back into the line equation to find the position vector of \( P \).
A1: Coordinates of \( P \) given as \( (3, 4, 2) \) or vector form \( 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \).

Separate the variables to prepare for integration:
\[ \int \frac{1}{y - 3} \, dy = \int \frac{2x}{x^2 + 1} \, dx \]

Integrate both sides:
\[ \ln|y - 3| = \ln(x^2 + 1) + C \]
where \( C \) is the constant of integration. (Note: \( x^2 + 1 > 0 \), so absolute value signs are not needed there).

Use the initial condition \( y = 5 \) when \( x = 0 \):
\[ \ln|5 - 3| = \ln(0 + 1) + C \]
\[ \ln 2 = \ln 1 + C \implies C = \ln 2 \]

Substitute the value of \( C \) back into the integrated equation:
\[ \ln|y - 3| = \ln(x^2 + 1) + \ln 2 \]

Use logarithm laws to combine the terms on the right-hand side:
\[ \ln|y - 3| = \ln\left[2(x^2 + 1)\right] \]

Exponentiate both sides:
\[ |y - 3| = 2(x^2 + 1) \]

Since \( y = 5 \) is greater than 3, we can remove the absolute value signs:
\[ y - 3 = 2x^2 + 2 \]
\[ y = 2x^2 + 5 \]

M1: Separates variables to get \( \int \frac{1}{y-3} \, dy = \int \frac{2x}{x^2+1} \, dx \) (allow missing integral signs).
A1: Integrates LHS to \( \ln(y-3) \) or \( \ln|y-3| \).
M1: Integrates RHS using substitution or recognition to get \( \ln(x^2+1) \).
A1: Correct integrated equation: \( \ln|y-3| = \ln(x^2+1) + C \) (or equivalent with constant).
M1: Uses the boundary condition \( x=0, y=5 \) to find a value for \( C \) (or equivalent constant in product form).
A1: Simplifies to final explicit form \( y = 2x^2 + 5 \).

Assume for contradiction that \( \log_3 5 \) is a rational number.

Then, by definition of a rational number, we can write:
\[ \log_3 5 = \frac{p}{q} \]
where \( p \) and \( q \) are positive integers (since \( \log_3 5 > 0 \)).

Using the definition of a logarithm, rewrite the equation in exponential form:
\[ 3^{\frac{p}{q}} = 5 \]

Raise both sides to the power of \( q \):
\[ 3^p = 5^q \]

Analyze both sides of this equation:
- The left-hand side \( 3^p \) is a power of 3, meaning it is an odd integer whose only prime factor is 3.
- The right-hand side \( 5^q \) is a power of 5, meaning it is an odd integer whose only prime factor is 5.

By the Fundamental Theorem of Arithmetic (unique factorization of integers into prime factors), an integer cannot have two different prime factorizations. A power of 3 cannot equal a power of 5 for positive integers \( p, q \ge 1 \).

This is a contradiction. Therefore, the assumption that \( \log_3 5 \) is rational must be false.
Hence, \( \log_3 5 \) is an irrational number.

B1: Sets up the proof by contradiction by assuming \( \log_3 5 \) is rational.
M1: Expresses \( \log_3 5 \) as \( \frac{p}{q} \) where \( p \) and \( q \) are integers (specify \( p, q \ge 1 \)).
M1: Converts the logarithmic equation into exponential form, \( 3^{p/q} = 5 \).
A1: Correctly derives the equation \( 3^p = 5^q \).
M1: Explains why \( 3^p = 5^q \) is impossible (e.g., uniqueness of prime factorisation, powers of 3 are not divisible by 5, etc.).
A1: Completes the proof with a clear concluding statement that the contradiction proves \( \log_3 5 \) is irrational.

(a) Let
\[ \frac{5x + 7}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3} \]

Multiply by the common denominator:
\[ 5x + 7 = A(x+3) + B(x-1) \]

Let \( x = 1 \):
\[ 5(1) + 7 = A(1+3) \implies 12 = 4A \implies A = 3 \]

Let \( x = -3 \):
\[ 5(-3) + 7 = B(-3-1) \implies -8 = -4B \implies B = 2 \]

So,
\[ \frac{5x + 7}{(x-1)(x+3)} = \frac{3}{x-1} + \frac{2}{x+3} \]

(b) Integrate using the partial fractions from part (a):
\[ \int_{2}^{5} \left( \frac{3}{x-1} + \frac{2}{x+3} \right) \, dx = \left[ 3\ln|x-1| + 2\ln|x+3| \right]_{2}^{5} \]

Evaluate at the upper limit \( x = 5 \):
\[ 3\ln(4) + 2\ln(8) = \ln(4^3) + \ln(8^2) = \ln(64) + \ln(64) = \ln(4096) \]

Evaluate at the lower limit \( x = 2 \):
\[ 3\ln(1) + 2\ln(5) = 0 + \ln(25) = \ln(25) \]

Subtract the lower limit evaluation from the upper limit evaluation:
\[ \ln(4096) - \ln(25) = \ln\left(\frac{4096}{25}\right) \]

(a)
M1: Sets up correct partial fraction form \( \frac{A}{x-1} + \frac{B}{x+3} \).
A1: Correct value for \( A = 3 \).
A1: Correct value for \( B = 2 \).

(b)
M1: Integrates their partial fractions to obtain terms in \( \ln \).
A1: Correct integration: \( [3\ln|x-1| + 2\ln|x+3|] \) (or equivalent).
M1: Substitutes the limits \( 5 \) and \( 2 \) and applies logarithmic properties to simplify.
A1: Obtains \( \ln\left(\frac{4096}{25}\right) \) (accept exact equivalents like \( \ln 163.84 \)).

The formula for the volume of revolution about the \( x \)-axis is:
\[ V = \pi \int_{0}^{2} y^2 \, dx \]

Substitute the equation of the curve:
\[ V = \pi \int_{0}^{2} (3x e^{-x})^2 \, dx = 9\pi \int_{0}^{2} x^2 e^{-2x} \, dx \]

Use integration by parts twice to evaluate \( \int x^2 e^{-2x} \, dx \).

**First Integration by Parts:**
Let \( u = x^2 \implies du = 2x \, dx \)
Let \( dv = e^{-2x} \, dx \implies v = -\frac{1}{2}e^{-2x} \)

\[ \int x^2 e^{-2x} \, dx = -\frac{1}{2}x^2 e^{-2x} - \int \left(-\frac{1}{2}e^{-2x}\right)(2x) \, dx \]
\[ = -\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} \, dx \]

**Second Integration by Parts:**
For \( \int x e^{-2x} \, dx \):
Let \( u = x \implies du = dx \)
Let \( dv = e^{-2x} \, dx \implies v = -\frac{1}{2}e^{-2x} \)

\[ \int x e^{-2x} \, dx = -\frac{1}{2}x e^{-2x} - \int \left(-\frac{1}{2}e^{-2x}\right) \, dx \]
\[ = -\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} \]

Combine the results:
\[ \int x^2 e^{-2x} \, dx = \left[ -e^{-2x}\left(\frac{1}{2}x^2 + \frac{1}{2}x + \frac{1}{4}\right) \right] \]

Now evaluate from \( 0 \) to \( 2 \):
At \( x = 2 \):
\[ -e^{-4}\left(\frac{1}{2}(4) + \frac{1}{2}(2) + \frac{1}{4}\right) = -e^{-4}\left(2 + 1 + \frac{1}{4}\right) = -\frac{13}{4}e^{-4} \]

At \( x = 0 \):
\[ -e^{0}\left(0 + 0 + \frac{1}{4}\right) = -\frac{1}{4} \]

Subtract the limit values:
\[ \int_{0}^{2} x^2 e^{-2x} \, dx = -\frac{13}{4}e^{-4} - \left(-\frac{1}{4}\right) = \frac{1}{4} - \frac{13}{4}e^{-4} = \frac{1}{4}(1 - 13e^{-4}) \]

Multiply by \( 9\pi \) to find the volume:
\[ V = 9\pi \times \frac{1}{4}(1 - 13e^{-4}) = \frac{9}{4}\pi(1 - 13e^{-4}) \]

This is in the required form with \( a = \frac{9}{4} \), \( b = 1 \), and \( c = 13 \).

B1: Sets up the correct volume formula including limits, \( V = \pi \int_0^2 (3xe^{-x})^2 \, dx \).
M1: Squares the integrand correctly to obtain \( 9\pi \int x^2 e^{-2x} \, dx \).
M1: Applies integration by parts for the first time on \( x^2 e^{-2x} \).
A1: Correct first stage of integration by parts.
M1: Applies integration by parts for the second time on \( x e^{-2x} \).
A1: Correct completed integration: \( -e^{-2x}(\frac{1}{2}x^2 + \frac{1}{2}x + \frac{1}{4}) \).
M1: Substitutes the limits \( 2 \) and \( 0 \) correctly.
A1: Simplifies to the final exact answer \( \frac{9}{4}\pi(1 - 13e^{-4}) \).

(a)
Cumulative frequencies:
- $10-19$: $8$
- $20-24$: $23$
- $25-29$: $45$
Median lies in the class $25-29$ (boundaries $24.5$ to $29.5$, class width $5$).
$$\text{Median} = 24.5 + \frac{40 - 23}{22} \times 5 = 24.5 + \frac{17}{22} \times 5 \approx 28.36 \approx 28.4\text{ minutes}$$

(b)
Midpoints ($x$) and calculations:
- $10-19$ (midpoint $14.5$): $f = 8$, $fx = 116$, $fx^2 = 1682$
- $20-24$ (midpoint $22$): $f = 15$, $fx = 330$, $fx^2 = 7260$
- $25-29$ (midpoint $27$): $f = 22$, $fx = 594$, $fx^2 = 16038$
- $30-39$ (midpoint $34.5$): $f = 20$, $fx = 690$, $fx^2 = 23805$
- $40-59$ (midpoint $49.5$): $f = 15$, $fx = 742.5$, $fx^2 = 36753.75$

$$\sum f = 80$$
$$\sum fx = 2472.5$$
$$\sum fx^2 = 85538.75$$

$$\text{Mean } \bar{x} = \frac{2472.5}{80} \approx 30.906 \approx 30.9\text{ minutes}$$
$$\text{Variance } \sigma^2 = \frac{85538.75}{80} - \bar{x}^2 = 1069.234 - 955.196 = 114.038$$
$$\text{Standard Deviation } \sigma = \sqrt{114.038} \approx 10.679 \approx 10.7\text{ minutes}$$

(c)
Class $20-24$ has a class width (CW) of $24.5 - 19.5 = 5$.
This is represented by a width of $1\text{ cm}$. Therefore, $1\text{ cm}$ represents $5$ units of class width.
Class $40-59$ has boundaries $39.5$ to $59.5$, so CW $= 20$.
$$\text{Width of } 40-59 \text{ bar} = \frac{20}{5} = 4\text{ cm}$$

Frequency density (FD) of $20-24 = \frac{15}{5} = 3$.
Height of this bar $= 6\text{ cm}$, so scaling factor $k$ is given by:
$$\text{Height} = k \times \text{FD} \implies 6 = k \times 3 \implies k = 2$$

FD of $40-59 = \frac{15}{20} = 0.75$.
$$\text{Height of } 40-59 \text{ bar} = 2 \times 0.75 = 1.5\text{ cm}$$

(d)
$$\text{Mean} \approx 30.9$$
$$\text{Median} \approx 28.4$$
Since $\text{Mean} > \text{Median}$ ($30.9 > 28.4$), the distribution is positively skewed.

(a)
- M1: For identifying the correct class boundaries $24.5$ and $29.5$, and setting up a correct interpolation expression.
- A1: For $\frac{40-23}{22} \times 5$ or equivalent.
- A1: For 28.4 (awarded for 28.4 or 28.36).

(b)
- M1: For $\sum fx$ attempt using midpoints.
- A1: For Mean $\approx 30.9$.
- M1: For correct standard deviation formula substituting $\sum fx^2$ and their mean.
- A1: For SD $\approx 10.7$.

(c)
- M1: For correct width scale calculation to find 4 cm.
- A1: Width $= 4\text{ cm}$.
- M1: For correct frequency density relationship to find height.
- A1: Height $= 1.5\text{ cm}$ (or equivalent fractional forms).

(d)
- M1: Comparing their mean and median.
- A1: Concluding positive skewness with numerical evidence.

(a)
$$S_{dd} = 154200 - \frac{1120^2}{10} = 154200 - 125440 = 28760$$
$$S_{tt} = 72.85 - \frac{24.5^2}{10} = 72.85 - 60.025 = 12.825$$
$$S_{dt} = 3120 - \frac{1120 \times 24.5}{10} = 3120 - 2744 = 376$$

(b)
$$r = \frac{S_{dt}}{\sqrt{S_{dd} \times S_{tt}}} = \frac{376}{\sqrt{28760 \times 12.825}} = \frac{376}{\sqrt{368847}} \approx 0.6191 \approx 0.619$$

(c)
$$b = \frac{S_{dt}}{S_{dd}} = \frac{376}{28760} \approx 0.0130737 \approx 0.0131$$
$$\bar{d} = 112, \quad \bar{t} = 2.45$$
$$a = \bar{t} - b\bar{d} = 2.45 - (0.0130737 \times 112) = 2.45 - 1.46425 = 0.98575 \approx 0.986$$
So the regression equation is:
$$t = 0.986 + 0.0131d$$

(d)
$b = 0.0131$: For each additional mile travelled on a business trip, the duration of the trip increases by approximately 0.0131 hours (or about $0.78$ minutes).

(e)
$$t = 0.98575 + 0.0130737 \times 180 \approx 3.339 \approx 3.34\text{ hours}$$
This estimate is reliable because $d = 180$ lies within the range of the observed data ($[40, 220]$ miles), which means we are interpolating. Furthermore, there is a moderate positive correlation ($r = 0.619$) supporting the linear model.

(a)
- M1: Attempting to calculate at least one of $S_{dd}, S_{tt}, S_{dt}$ with correct formula.
- A1: Two correct values.
- A1: All three correct values ($S_{dd} = 28760$, $S_{tt} = 12.825$, $S_{dt} = 376$).

(b)
- M1: For correct substitution into PMCC formula.
- A1: For $r \approx 0.619$.

(c)
- M1: For a correct attempt at $b$.
- A1: $b \approx 0.0131$ (or $0.01307$ or better).
- M1: For a correct attempt at $a$ using their $b$.
- A1: $t = 0.986 + 0.0131d$ (allow $0.986$ and $0.0131$ or better).

(d)
- B1.5: Complete interpretation mentioning both distance increase and time increase with units.

(e)
- M1: For substituting $d = 180$ into their regression line.
- A1: $3.34$ hours (or 3.33) and stating it is reliable because it is an interpolation / within range.

(a)
Since $A$ and $B$ are independent:
$$P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.5 = 0.2$$

(b)
Since $B$ and $C$ are mutually exclusive, $P(B \cap C) = 0$, so their circles do not overlap. This also means $P(A \cap B \cap C) = 0$.
Let's find the other regions:
- $P(A \cap B \cap C') = P(A \cap B) - 0 = 0.2$
- $P(A \cap C \cap B') = P(A \cap C) - 0 = 0.15$
- $P(A \text{ only}) = P(A) - P(A \cap B) - P(A \cap C) = 0.4 - 0.2 - 0.15 = 0.05$
- $P(B \text{ only}) = P(B) - P(A \cap B) = 0.5 - 0.2 = 0.3$
- $P(C \text{ only}) = P(C) - P(A \cap C) = 0.35 - 0.15 = 0.2$
- $P(\text{Outside all circles}) = 1 - (0.05 + 0.2 + 0.3 + 0.15 + 0.2) = 1 - 0.9 = 0.1$

The Venn diagram should show three circles, with $B$ and $C$ non-overlapping, and the calculated probabilities correctly placed in each region.

(c)
$$P(A' \cup B') = 1 - P(A \cap B) = 1 - 0.2 = 0.8$$

(d)
$$P((A \cup C)' | B') = \frac{P((A \cup C)' \cap B')}{P(B')}$$
First:
$$P(B') = 1 - P(B) = 1 - 0.5 = 0.5$$
Now, $(A \cup C)' \cap B'$ is the region that is outside $A$, outside $C$, and outside $B$. This is exactly the region outside all three circles:
$$P((A \cup C)' \cap B') = 0.1$$
Therefore:
$$P((A \cup C)' | B') = \frac{0.1}{0.5} = 0.2$$

(a)
- B1: 0.2

(b)
- M1: Three circles drawn with $B$ and $C$ non-overlapping.
- A1: Correctly placing $P(A \cap B) = 0.2$ and $P(A \cap C) = 0.15$.
- A1: Correctly placing $P(A \text{ only}) = 0.05$, $P(B \text{ only}) = 0.3$, $P(C \text{ only}) = 0.2$.
- A1: Correctly placing $0.1$ outside the circles.
- A1: Correctly specifying/drawing the entire sample space box and labeling $A$, $B$, $C$.

(c)
- M1: For $1 - P(A \cap B)$ or correct sum of regions from Venn diagram.
- A1: 0.8

(d)
- M1: For writing or implying a correct conditional probability expression $\frac{P((A \cup C)' \cap B')}{P(B')}$.
- A1.5: For identifying numerator is 0.1.
- A1: For denominator is 0.5.
- A1: For 0.2.

(a)
Sum of probabilities must equal 1:
$$5k + 4k + 3k + 2k + k = 1 \implies 15k = 1 \implies k = \frac{1}{15} \quad \text{(shown)}$$

(b)
Using $k = \frac{1}{15}$, the probabilities are $\frac{5}{15}$, $\frac{4}{15}$, $\frac{3}{15}$, $\frac{2}{15}$, and $\frac{1}{15}$.
$$E(X) = \sum x P(X=x) = (-2)\left(\frac{5}{15}\right) + (-1)\left(\frac{4}{15}\right) + 0\left(\frac{3}{15}\right) + 1\left(\frac{2}{15}\right) + 2\left(\frac{1}{15}\right)$$
$$E(X) = \frac{-10 - 4 + 0 + 2 + 2}{15} = -\frac{10}{15} = -\frac{2}{3}$$

Now, find $E(X^2)$:
$$E(X^2) = \sum x^2 P(X=x) = 4\left(\frac{5}{15}\right) + 1\left(\frac{4}{15}\right) + 0\left(\frac{3}{15}\right) + 1\left(\frac{2}{15}\right) + 4\left(\frac{1}{15}\right)$$
$$E(X^2) = \frac{20 + 4 + 0 + 2 + 4}{15} = \frac{30}{15} = 2$$

Thus:
$$Var(X) = E(X^2) - [E(X)]^2 = 2 - \left(-\frac{2}{3}\right)^2 = 2 - \frac{4}{9} = \frac{14}{9}$$

(c)
(i)
$$E(Y) = E(3X - 1) = 3E(X) - 1 = 3\left(-\frac{2}{3}\right) - 1 = -2 - 1 = -3$$

(ii)
$$Var(Y) = Var(3X - 1) = 9Var(X) = 9\left(\frac{14}{9}\right) = 14$$

(iii)
$$P(Y > X) = P(3X - 1 > X) = P(2X > 1) = P(X > 0.5)$$
Since $X$ is discrete, $X > 0.5$ corresponds to $X = 1$ or $X = 2$.
$$P(X > 0.5) = P(X=1) + P(X=2) = 2k + k = 3k = \frac{3}{15} = 0.2$$

(a)
- M1: For $5k + 4k + 3k + 2k + k = 1$.
- A1: For obtaining $k = 1/15$ clearly.

(b)
- M1: Correct method for $E(X)$.
- A1: $E(X) = -2/3$ (or exact decimal equivalent).
- M1: Correct method for $E(X^2)$.
- A1: $Var(X) = 14/9$ (or exact decimal equivalent $1.5\dot{5}$).

(c)
- M1: Using $E(aX+b) = aE(X)+b$ for (i).
- A1: $E(Y) = -3$.
- M1: Using $Var(aX+b) = a^2 Var(X)$ for (ii).
- A1: $Var(Y) = 14$.
- M1: Re-arranging inequality $Y > X$ to $X > 0.5$.
- A1: $0.2$ (or $1/5$).

(a)
For $P(M < 110) = 0.0808$:
$$P\left(Z < \frac{110 - \mu}{\sigma}\right) = 0.0808$$
From the standard normal table, the z-score corresponding to a left-tail probability of $0.0808$ is $-1.40$.
$$\frac{110 - \mu}{\sigma} = -1.40 \implies 110 - \mu = -1.40\sigma \quad \text{(shown)}$$

For $P(M > 175) = 0.1151$:
$$P\left(Z > \frac{175 - \mu}{\sigma}\right) = 0.1151 \implies P\left(Z < \frac{175 - \mu}{\sigma}\right) = 1 - 0.1151 = 0.8849$$
From the standard normal table, the z-score corresponding to a cumulative probability of $0.8849$ is $1.20$.
$$\frac{175 - \mu}{\sigma} = 1.20 \implies 175 - \mu = 1.20\sigma \quad \text{(shown)}$$

(b)
Subtracting equation (i) from equation (ii):
$$(175 - \mu) - (110 - \mu) = 1.20\sigma - (-1.40\sigma)$$
$$65 = 2.60\sigma \implies \sigma = \frac{65}{2.60} = 25$$
Substituting $\sigma = 25$ back into (ii):
$$175 - \mu = 1.20(25) \implies 175 - \mu = 30 \implies \mu = 145$$

(c)
$$P(M < 130) = P\left(Z < \frac{130 - 145}{25}\right) = P(Z < -0.6)$$
$$P(Z < -0.6) = 1 - \Phi(0.6) = 1 - 0.7257 = 0.2743 \approx 0.274$$

(d)
Let $X$ be the number of pears in a box of 3 that have a mass of less than 130g.
$$X \sim \text{B}(3, 0.2743)$$
$$P(X = 1) = \binom{3}{1} (0.2743)^1 (1 - 0.2743)^2 = 3 \times 0.2743 \times (0.7257)^2$$
$$P(X = 1) = 3 \times 0.2743 \times 0.52664 \approx 0.4334 \approx 0.433$$

(a)
- M1: For attempting to standardise with $110$ or $175$.
- A1: For finding z-value $-1.40$.
- A1: For finding z-value $1.20$.
- A1: Both equations correctly shown with signs verified.

(b)
- M1: Attempting to solve simultaneous equations to eliminate $\mu$.
- A1: $\sigma = 25$.
- M1: Attempting to find $\mu$ using their $\sigma$.
- A1: $\mu = 145$.

(c)
- M1: Standardising $130$ with their $\mu$ and $\sigma$.
- A1: $0.274$ (or $0.2743$).

(d)
- M1: Recognising a Binomial scenario with $n=3$ and their $p$ from part (c).
- A1: $\binom{3}{1} p (1-p)^2$ substituted correctly.
- A1: $0.433$ (or $0.433$ to $0.434$ depending on rounding of $p$).

(a)
The tree diagram should have:
- First branch: $A$ (probability $0.6$) and $B$ (probability $0.4$).
- Second branch from $A$: Defective, $D$ ($0.04$) and Not Defective, $D'$ ($0.96$).
- Second branch from $B$: Defective, $D$ ($0.06$) and Not Defective, $D'$ ($0.94$).

(b)
$$P(D) = P(A \cap D) + P(B \cap D) = (0.6 \times 0.04) + (0.4 \times 0.06)$$
$$P(D) = 0.024 + 0.024 = 0.048$$

(c)
$$P(A | D) = \frac{P(A \cap D)}{P(D)} = \frac{0.6 \times 0.04}{0.048} = \frac{0.024}{0.048} = 0.5$$

(d)
Let $W$ be the profit/loss per case.
- Defective ($D$): $W = -15$, with probability $0.048$
- Non-defective ($D'$): $W = 5$, with probability $1 - 0.048 = 0.952$

$$\text{Expected Profit } E(W) = \sum w P(W=w) = 5(0.952) + (-15)(0.048)$$
$$E(W) = 4.76 - 0.72 = 4.04\text{ dollars}$$
So the expected profit is $$4.04$ per case.

(a)
- M1: Two stage tree diagram drawn.
- A1: Correct branch probabilities for $A$ and $B$.
- A1: Correct conditional probabilities for $D$ and $D'$ on both branches.

(b)
- M1: For $(0.6 \times 0.04) + (0.4 \times 0.06)$ or equivalent.
- A1.5: 0.048 (or equivalent fraction $6/125$).

(c)
- M1: For conditional probability expression $\frac{P(A \cap D)}{P(D)}$ using their values.
- A1: For numerator $0.024$.
- A1: $0.5$ (or $1/2$).

(d)
- M1: Identifying the two possible profit outcomes ($5$ and $-15$).
- A1: Finding the probability of a non-defective case is $0.952$.
- M1: Setting up expectation sum $5 \times P(D') - 15 \times P(D)$.
- A1: $4.04$ (or $$4.04$).

(a) To find \(k\), we use the property that the total area under the probability density function is equal to 1:

\[\int_{1}^{3} k(x^2 - x) \, dx = 1\]

\[k \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{1}^{3} = 1\]

Substituting the limits:

\[k \left\{ \left( \frac{27}{3} - \frac{9}{2} \right) - \left( \frac{1}{3} - \frac{1}{2} \right) \right\} = 1\]

\[k \left\{ \left( 9 - 4.5 \right) - \left( -\frac{1}{6} \right) \right\} = 1\]

\[k \left\{ 4.5 + \frac{1}{6} \right\} = 1\]

\[k \left\{ \frac{9}{2} + \frac{1}{6} \right\} = 1\]

\[k \left( \frac{27 + 1}{6} \right) = 1 \implies k \left( \frac{28}{6} \right) = 1 \implies \frac{14}{3}k = 1\]

\[k = \frac{3}{14} \quad \text{(as required)}\]

(b) The cumulative distribution function \(F(x)\) is given by \(\int_{-\infty}^{x} f(t) \, dt\).

For \(x < 1\), \(F(x) = 0\).

For \(1 \le x \le 3\):

\[F(x) = \int_{1}^{x} \frac{3}{14}(t^2 - t) \, dt = \frac{3}{14} \left[ \frac{t^3}{3} - \frac{t^2}{2} \right]_{1}^{x}\]

\[F(x) = \frac{3}{14} \left[ \left(\frac{x^3}{3} - \frac{x^2}{2}\right) - \left(\frac{1}{3} - \frac{1}{2}\right) \right]\]

\[F(x) = \frac{3}{14} \left( \frac{x^3}{3} - \frac{x^2}{2} + \frac{1}{6} \right)\]

\[F(x) = \frac{3}{14} \left( \frac{2x^3 - 3x^2 + 1}{6} \right) = \frac{2x^3 - 3x^2 + 1}{28}\]

For \(x > 3\), \(F(x) = 1\).

Hence, the cumulative distribution function is:

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

(c) The expectation \(\text{E}(X)\) is given by:

\[\text{E}(X) = \int_{1}^{3} x \cdot f(x) \, dx = \int_{1}^{3} \frac{3}{14}x(x^2 - x) \, dx = \frac{3}{14} \int_{1}^{3} (x^3 - x^2) \, dx\]

\[\text{E}(X) = \frac{3}{14} \left[ \frac{x^4}{4} - \frac{x^3}{3} \right]_{1}^{3}\]

\[\text{E}(X) = \frac{3}{14} \left\{ \left( \frac{81}{4} - \frac{27}{3} \right) - \left( \frac{1}{4} - \frac{1}{3} \right) \right\}\]

\[\text{E}(X) = \frac{3}{14} \left\{ \left( 20.25 - 9 \right) - \left( -\frac{1}{12} \right) \right\}\]

\[\text{E}(X) = \frac{3}{14} \left\{ 11.25 + \frac{1}{12} \right\} = \frac{3}{14} \left\{ \frac{45}{4} + \frac{1}{12} \right\}\]

\[\text{E}(X) = \frac{3}{14} \left( \frac{135 + 1}{12} \right) = \frac{3}{14} \left( \frac{136}{12} \right) = \frac{3}{14} \times \frac{34}{3} = \frac{34}{14} = \frac{17}{7}\]

(a)
M1: Integrates \(k(x^2-x)\) with correct power terms.
A1: Obtains \(\frac{14}{3}k\) after substituting limits \(3\) and \(1\).
A1*: Equates to 1 and shows clearly that \(k = \frac{3}{14}\).

(b)
M1: Writes down integral for \(F(x)\) with limit \(x\) and integrates correctly.
A1: Correct integration with correct substitution of lower limit \(1\).
A1: Simplifies to \(\frac{2x^3 - 3x^2 + 1}{28}\) for \(1 \le x \le 3\).
B1: Correctly states the other parts of the piecewise function (0 for \(x < 1\) and 1 for \(x > 3\)).

(c)
M1: Attempts to evaluate \(\int x f(x) \, dx\) with correct limit values.
A1: Correct integration to get \(\frac{x^4}{4} - \frac{x^3}{3}\).
M1: Correctly substitutes limits 1 and 3 into their integrated expression.
A1: Evaluates to \(\frac{17}{7}\) or equivalent fraction.

(a) Let \(X\) be the number of seeds that germinate. Under the null hypothesis, \(X \sim \text{B}(20, 0.3)\).

State the hypotheses:
\[H_0: p = 0.3\]
\[H_1: p > 0.3\]

We want to find the probability of observing 10 or more germinated seeds under \(H_0\):
\[P(X \ge 10) = 1 - P(X \le 9)\]

From the binomial cumulative distribution tables for \(n = 20\) and \(p = 0.3\):
\[P(X \le 9) = 0.9520\]

Therefore:
\[P(X \ge 10) = 1 - 0.9520 = 0.0480\]

Since \(0.0480 < 0.05\), the result is significant at the 5% level.
We reject \(H_0\). There is sufficient evidence at the 5% level of significance to support the gardener's belief that the germination rate is higher than 30%.

(b) For a two-tailed test at the 5% level of significance, the significance level in each tail should be as close to 2.5% (0.025) as possible, without exceeding it.

Let's check the lower tail:
\[P(X \le 1) = 0.0076 \quad (< 0.025)\]
\[P(X \le 2) = 0.0355 \quad (> 0.025)\]
So the lower critical value is 1, and the lower critical region is \(X \le 1\).

Let's check the upper tail:
We require \(P(X \ge c_2) \le 0.025\), which is equivalent to \(1 - P(X \le c_2 - 1) \le 0.025 \implies P(X \le c_2 - 1) \ge 0.975\).

From binomial tables:
\[P(X \le 9) = 0.9520 \quad (< 0.975)\]
\[P(X \le 10) = 0.9829 \quad (\ge 0.975)\]

Thus, \(c_2 - 1 = 10 \implies c_2 = 11\).
So the upper critical region is \(X \ge 11\).

The critical region is \(X \le 1\) or \(X \ge 11\).

The actual significance level is:
\[P(X \le 1) + P(X \ge 11) = 0.0076 + (1 - 0.9829) = 0.0076 + 0.0171 = 0.0247 \quad \text{or} \quad 2.47\%\]

(a)
B1: Correctly states both hypotheses: \(H_0: p = 0.3\) and \(H_1: p > 0.3\) with parameter \(p\) defined.
M1: Attempts to find \(P(X \ge 10)\) using binomial tables with \(n=20\), \(p=0.3\).
A1: Correctly calculates \(0.0480\) (or identifies the critical region boundary as \(X \ge 10\) since \(P(X \ge 10) = 0.0480\) and \(P(X \ge 9) = 0.1133\)).
M1: Compares \(0.0480\) with \(0.05\) and makes a decision to reject \(H_0\).
A1: Concludes in context, stating there is sufficient evidence that the germination rate is higher.

(b)
M1: Recognizes two-tailed test requires finding critical boundaries for both ends, attempting to find probabilities close to \(0.025\).
A1: Correctly identifies lower boundary as 1 (with \(P(X \le 1) = 0.0076\)).
A1: Correctly identifies upper boundary as 11 (with \(P(X \ge 11) = 0.0171\)).
A1: States the complete critical region as \(X \le 1\) or \(X \ge 11\).
M1: Attempts to add tail probabilities to find actual significance level.
A1: Correct actual significance level of \(0.0247\) or \(2.47\%\).

(a) Let \(Y\) be the number of support requests in a 2-hour period.
Since the hourly rate is 2.5, the mean for a 2-hour period is \(\lambda = 2.5 \times 2 = 5\).
Thus, \(Y \sim \text{Po}(5)\).

We need to find:
\[P(Y = 4) = \frac{e^{-5} \cdot 5^4}{4!} = \frac{625 e^{-5}}{24} \approx 0.175467...\]

To 4 decimal places, this is \(0.1755\).

(b) Let \(X\) be the number of support requests in a 1-hour period. Thus, \(X \sim \text{Po}(2.5)\).

We need to find:
\[P(X > 3) = 1 - P(X \le 3)\]

Using the cumulative Poisson table or formula for \(\lambda = 2.5\):
\[P(X \le 3) = e^{-2.5} \left( 1 + 2.5 + \frac{2.5^2}{2!} + \frac{2.5^3}{3!} \right)\]
\[P(X \le 3) = e^{-2.5} \left( 1 + 2.5 + 3.125 + 2.60417 \right) = e^{-2.5} \times 9.22917 \approx 0.7576\]

Therefore:
\[P(X > 3) = 1 - 0.7576 = 0.2424\]

(c) Let \(W\) be the number of hours in an 8-hour workday where the requests in that hour are more than 3.
Each hour is independent, and the probability of success in any given hour is \(p = 0.2424\) (from part b).
Thus, \(W \sim \text{B}(8, 0.2424)\).

We need to find:
\[P(W = 2) = \binom{8}{2} p^2 (1-p)^6\]
\[P(W = 2) = 28 \times (0.2424)^2 \times (0.7576)^6\]

Calculate the components:
\[(0.2424)^2 \approx 0.058758\]
\[(0.7576)^6 \approx 0.191371\]

Therefore:
\[P(W = 2) = 28 \times 0.058758 \times 0.191371 \approx 0.31485...\]

To 4 decimal places, this is \(0.3149\).

(a)
M1: Identifies the correct distribution \(Y \sim \text{Po}(5)\).
M1: Applies the Poisson probability formula for \(P(Y = 4)\).
A1: Correct probability of \(0.1755\) (accept 0.175).

(b)
M1: Identifies the correct distribution \(X \sim \text{Po}(2.5)\) and writes the expression \(1 - P(X \le 3)\).
M1: Attempts to evaluate \(P(X \le 3)\) using formula or tables.
A1: Correct probability of \(0.2424\) (accept 0.242).

(c)
M1: Identifies the need for a Binomial distribution \(W \sim \text{B}(8, p)\) where \(p\) is their answer to part (b).
M1: Applies the Binomial probability formula for \(P(W = 2)\) with \(n=8\).
A1: Correct structure of the expression: \(\binom{8}{2} p^2 (1-p)^6\) using their \(p\).
M1: Evaluation of the binomial term using a decimal value of \(p\).
A1: Final correct answer of \(0.3149\) (allow rounding errors from early rounding of \(p\), e.g., accept \(0.314\) to \(0.316\)).

(a) When the wire is cut at point \(X\), the two pieces have lengths \(X\) and \(20-X\).

The length of the shorter piece is \(\min(X, 20-X)\), and the length of the longer piece is \(\max(X, 20-X)\).

By symmetry, we can assume \(X\) is in the range \([0, 10]\) for half of the outcomes, in which case the shorter piece is \(X\) and the longer piece is \(20-X\).

We want to find \(P(R > 3)\):
\[R = \frac{\max(X, 20-X)}{\min(X, 20-X)} > 3\]

If \(X < 10\):
\[\frac{20-X}{X} > 3 \implies 20-X > 3X \implies 4X < 20 \implies X < 5\]

If \(X > 10\), the pieces are reversed, so:
\[\frac{X}{20-X} > 3 \implies X > 3(20-X) \implies 4X > 60 \implies X > 15\]

So \(R > 3\) corresponds to the event \(\{X < 5\} \cup \{X > 15\}\).

Since \(X \sim \text{U}[0, 20]\):
\[P(R > 3) = P(X < 5) + P(X > 15) = \frac{5}{20} + \frac{5}{20} = \frac{10}{20} = 0.5 \quad \text{(as required)}\]

(b) We want to find \(F_R(r) = P(R \le r)\) for \(r \ge 1\).
First, find \(P(R > r)\) for any \(r \ge 1\):
\[R > r \iff X < \frac{20}{r+1} \quad \text{or} \quad X > 20 - \frac{20}{r+1}\]

Since \(X\) is uniformly distributed on \([0, 20]\):
\[P\left(X < \frac{20}{r+1}\right) = \frac{\frac{20}{r+1}}{20} = \frac{1}{r+1}\]
\[P\left(X > 20 - \frac{20}{r+1}\right) = \frac{\frac{20}{r+1}}{20} = \frac{1}{r+1}\]

So:
\[P(R > r) = \frac{1}{r+1} + \frac{1}{r+1} = \frac{2}{r+1}\]

Thus, the cumulative distribution function is:
\[F_R(r) = P(R \le r) = 1 - P(R > r) = 1 - \frac{2}{r+1} = \frac{r-1}{r+1} \quad \text{for} \quad r \ge 1\]

For \(r < 1\), \(F_R(r) = 0\).

(c) The probability density function \(f_R(r)\) is the derivative of \(F_R(r)\):
\[f_R(r) = \frac{d}{dr} F_R(r) = \frac{d}{dr} \left( 1 - 2(r+1)^{-1} \right) = 2(r+1)^{-2} = \frac{2}{(r+1)^2} \quad \text{for} \quad r \ge 1\]

So:
\[f_R(r) = \begin{cases} \frac{2}{(r+1)^2} & r \ge 1 \\ 0 & \text{otherwise} \end{cases}\]

(a)
M1: Formulates the ratio \(R\) in terms of \(X\), setting up the inequality \(\frac{20-X}{X} > 3\) or equivalent.
A1: Correctly solves the inequality to find \(X < 5\) (or \(X > 15\)).
M1: Uses the uniform distribution probability formula to find \(P(X < 5) + P(X > 15)\).
A1*: Shows clearly that the probability is \(0.5\).

(b)
M1: Generalizes the method in part (a) to set up the inequality for general \(r\), e.g., \(\frac{20-X}{X} > r\).
A1: Solves the inequality to obtain the boundary condition \(X < \frac{20}{r+1}\).
M1: Writes down the probability expression \(P(R > r) = \frac{2}{r+1}\).
M1: Relates CDF to survival probability: \(F_R(r) = 1 - P(R > r)\).
A1: Obtains \(F_R(r) = \frac{r-1}{r+1}\) (must specify the range \(r \ge 1\)).

(c)
M1: Differentiates \(F_R(r)\) with respect to \(r\).
A1: Obtains \(f_R(r) = \frac{2}{(r+1)^2}\) for \(r \ge 1\) (and 0 otherwise).

(a) Let \(\lambda\) be the mean number of complaints per 10-day period.
Under the null hypothesis, the rate is 1.2 per day, so over 10 days, \(\lambda = 1.2 \times 10 = 12\).
Let \(Y\) be the number of complaints in a 10-day period. Under \(H_0\), \(Y \sim \text{Po}(12)\).

State the hypotheses:
\[H_0: \lambda = 12\]
\[H_1: \lambda < 12\]

We observe \(y = 6\) complaints. We calculate the probability of observing 6 or fewer complaints:
\[P(Y \le 6) \quad \text{under} \quad Y \sim \text{Po}(12)\]

From the Poisson cumulative distribution table for \(\lambda = 12\):
\[P(Y \le 6) = 0.0458\]

Since \(0.0458 < 0.05\), the result is significant at the 5% level.
We reject \(H_0\). There is sufficient evidence at the 5% level of significance to support the manager's claim that the rate of complaints has decreased.

(b) A Type I error occurs when the null hypothesis is rejected when it is actually true (i.e. concluding the rate has decreased when it remains at 1.2 per day).

This occurs if the observed number of complaints falls inside the critical region. Let's find the critical region for this 5% test:
We want \(P(Y \le c) \le 0.05\).
From tables for \(\lambda = 12\):
\[P(Y \le 5) = 0.0191\]
\[P(Y \le 6) = 0.0458\]
\[P(Y \le 7) = 0.0895\]

So the critical region is \(Y \le 6\).

Therefore, a Type I error occurs if 6 or fewer complaints are received in the 10-day period when the true rate is indeed 1.2 complaints per day.
The probability of this Type I error is:
\[P(Y \le 6) = 0.0458\]

(c) The null hypothesis is not rejected if the observed number of complaints lies outside the critical region \(Y \le 6\).
Therefore, \(H_0\) is not rejected if \(Y \ge 7\).
The minimum number of complaints for which the null hypothesis would not be rejected is 7.

(a)
B1: States correct hypotheses: \(H_0: \lambda = 12\) and \(H_1: \lambda < 12\) (or in terms of rate per day, e.g. \(H_0: r = 1.2, H_1: r < 1.2\)).
M1: Identifies the correct distribution \(Y \sim \text{Po}(12)\) and attempts to find \(P(Y \le 6)\).
A1: Evaluates the probability as \(0.0458\).
M1: Compares \(0.0458\) with \(0.05\) and makes a decision to reject \(H_0\).
A1: Concludes in context, stating there is sufficient evidence that the complaint rate has decreased.

(b)
B1: Clear explanation of Type I error in this context (concluding rate has decreased when it has not, or rejecting \(H_0\) when it is true).
M1: Realizes that the probability of Type I error is equal to the actual significance level (probability of critical region \(Y \le 6\)).
A1: Correct probability of \(0.0458\).

(c)
M1: Identifies that the critical region consists of values \(Y \le 6\), so non-rejection region is \(Y \ge 7\).
A1: Concludes that the minimum number of complaints is 7.

(a) Since 40% of the coins are 10p and 60% are 20p, the probability distribution of \(X\) is:

\[\begin{array}{c|cc} x & 10 & 20 \\ \hline P(X = x) & 0.4 & 0.6 \end{array}\]

(b) The mean \(\mu = \text{E}(X)\) is:
\[\mu = 10(0.4) + 20(0.6) = 4 + 12 = 16\text{p}\]

To find the variance \(\sigma^2 = \text{Var}(X)\), we first find \(\text{E}(X^2)\):
\[\text{E}(X^2) = 10^2(0.4) + 20^2(0.6) = 100(0.4) + 400(0.6) = 40 + 240 = 280\]

Then:
\[\sigma^2 = \text{E}(X^2) - \mu^2 = 280 - 16^2 = 280 - 256 = 24\text{p}^2\]

(c) We select a sample of size 3 from the bag. Let the values of the three selected coins be \((C_1, C_2, C_3)\). Since the bag is large, sampling is effectively with replacement.

The possible ordered samples, their sample mean \(\bar{x} = \frac{C_1 + C_2 + C_3}{3}\), and their probabilities are:

1. \((10, 10, 10)\):
\[\bar{x} = 10\text{, } \text{Probability} = 0.4^3 = 0.064\]

2. \((10, 10, 20)\), \((10, 20, 10)\), \((20, 10, 10)\):
\[\bar{x} = \frac{40}{3}\text{, } \text{Probability} = 3 \times (0.4^2 \times 0.6) = 3 \times 0.16 \times 0.6 = 0.288\]

3. \((10, 20, 20)\), \((20, 10, 20)\), \((20, 20, 10)\):
\[\bar{x} = \frac{50}{3}\text{, } \text{Probability} = 3 \times (0.4 \times 0.6^2) = 3 \times 0.4 \times 0.36 = 0.432\]

4. \((20, 20, 20)\):
\[\bar{x} = 20\text{, } \text{Probability} = 0.6^3 = 0.216\]

We compile these into the sampling distribution table of \(\bar{X}\):

\[\begin{array}{c|cccc} \bar{x} & 10 & \frac{40}{3} & \frac{50}{3} & 20 \\ \hline P(\bar{X} = \bar{x}) & 0.064 & 0.288 & 0.432 & 0.216 \end{array}\]

(Note: we check that \(0.064 + 0.288 + 0.432 + 0.216 = 1.000\).)

(a)
B1: Correctly lists values 10 and 20.
B1: Correctly assigns probabilities 0.4 and 0.6.

(b)
M1: Applies formula for \(\text{E}(X)\).
A1: Correct mean of 16.
M1: Applies formula for \(\text{Var}(X)\) using \(\text{E}(X^2)\).
A1: Correct variance of 24.

(c)
M1: Identifies the four possible combinations/means (10, 40/3, 50/3, 20) or lists all 8 possible outcomes.
M1: Shows method to calculate probability for at least one of the mixed combinations, including the multiplier 3.
A1: Correct probability of \(0.064\) for \(\bar{x} = 10\) and \(0.216\) for \(\bar{x} = 20\).
A1: Correct probability of \(0.288\) for \(\bar{x} = 40/3\).
A1: Correct probability of \(0.432\) for \(\bar{x} = 50/3\).
B1: Presents the final sampling distribution in a clear table or list format.

(a) The probability density function \(f(y)\) is the derivative of \(F(y)\) with respect to \(y\).

For \(2 \le y \le 4\):
\[f(y) = \frac{d}{dy} \left[ \frac{1}{12}(y^2 - 4) \right] = \frac{1}{12}(2y) = \frac{y}{6}\]

For \(y < 2\) and \(y > 4\), \(f(y) = 0\).

Therefore, the probability density function is:
\[f(y) = \begin{cases} \frac{y}{6} & 2 \le y \le 4 \\ 0 & \text{otherwise}
\end{cases}\]

(b) The median \(m\) of \(Y\) satisfies the equation \(F(m) = 0.5\).
Since the median must lie in the interval \([2, 4]\):
\[\frac{1}{12}(m^2 - 4) = 0.5\]
\[m^2 - 4 = 6\]
\[m^2 = 10\]
\[m = \sqrt{10} \approx 3.162277...\]

To 3 significant figures, the median is \(3.16\).

(c) The mode is the value of \(y\) for which the probability density function \(f(y)\) is maximized.
On the interval \([2, 4]\), the function \(f(y) = \frac{y}{6}\) is a linear function with a positive gradient of \(\frac{1}{6}\), which means it is strictly increasing.

Thus, the maximum value of \(f(y)\) occurs at the upper boundary of the domain, which is \(y = 4\).
Hence, the mode of \(Y\) is 4.

(d) We find the probability using the cumulative distribution function:
\[P(2.5 < Y \le 3.5) = F(3.5) - F(2.5)\]

Calculate \(F(3.5)\):
\[F(3.5) = \frac{1}{12}(3.5^2 - 4) = \frac{1}{12}(12.25 - 4) = \frac{8.25}{12} = \frac{33}{48} = \frac{11}{16} = 0.6875\]

Calculate \(F(2.5)\):
\[F(2.5) = \frac{1}{12}(2.5^2 - 4) = \frac{1}{12}(6.25 - 4) = \frac{2.25}{12} = \frac{9}{48} = \frac{3}{16} = 0.1875\]

Subtracting these two values:
\[P(2.5 < Y \le 3.5) = 0.6875 - 0.1875 = 0.5\]

(a)
M1: Differentiates \(F(y)\) to find \(f(y)\).
A1: Correctly obtains \(\frac{y}{6}\).
B1: Correctly states \(f(y) = 0\) elsewhere with domain limits included.

(b)
M1: Sets \(F(m) = 0.5\).
A1: Obtains the equation \(m^2 = 10\).
A1: Evaluates to \(3.16\) (must be to 3 sig figs).

(c)
M1: Explains that \(f(y)\) is an increasing function on \([2, 4]\).
A1*: Concludes clearly that the maximum is at \(y=4\), so the mode is 4.

(d)
M1: Uses \(F(3.5) - F(2.5)\) (or sets up the equivalent integral \(\int_{2.5}^{3.5} \frac{y}{6} \, dy\)).
A1: Correct evaluation of both terms: \(F(3.5) = \frac{11}{16}\) and \(F(2.5) = \frac{3}{16}\).
A1: Obtains final correct probability of \(0.5\).

Let \( T \sim N(80, 4^2) \) and \( W \sim N(250, 8^2) \). (a) The weight of a filled tin is \( F = T + W \). Since \( T \) and \( W \) are independent, \( F \sim N(80+250, 4^2+8^2) = N(330, 80) \). We want \( P(F < 320) = P\left(Z < \frac{320-330}{\sqrt{80}}\right) = P(Z < -1.118) = 1 - \Phi(1.12) = 1 - 0.8686 = 0.1314 \) (or \( 0.1318 \) using a calculator). (b) Let the total weight be \( X = B + F_1 + F_2 + F_3 + F_4 + F_5 + F_6 \), where \( B \sim N(150, 10^2) \). Then \( E(X) = E(B) + 6E(F) = 150 + 6(330) = 2130 \). Since the variables are independent, \( Var(X) = Var(B) + 6Var(F) = 100 + 6(80) = 580 \). So \( X \sim N(2130, 580) \). We want \( P(X > 2150) = P\left(Z > \frac{2150-2130}{\sqrt{580}}\right) = P(Z > 0.8305) = 1 - \Phi(0.83) = 1 - 0.7967 = 0.2033 \) (or \( 0.2031 \) using a calculator). (c) We want \( P(W > 3.2T) \implies P(W - 3.2T > 0) \). Let \( Y = W - 3.2T \). Then \( E(Y) = E(W) - 3.2E(T) = 250 - 3.2(80) = -6 \). Since \( W \) and \( T \) are independent, \( Var(Y) = Var(W) + 3.2^2 Var(T) = 64 + 10.24(16) = 227.84 \). We want \( P(Y > 0) = P\left(Z > \frac{0 - (-6)}{\sqrt{227.84}}\right) = P(Z > 0.3975) = P(Z < 0.3975) \approx \Phi(0.40) = 0.6554 \) (or \( 0.6545 \) using a calculator).

(a) M1: For attempting to find the distribution of \( T+W \) (mean = 330 and variance = 80). A1: Correct standardization of 320. A1: Correct probability in the range [0.1314, 0.1318]. (b) M1: Attempting to find the mean of the total weight (2130). M1: Attempting to find the variance of the total weight (580). A1: Correct distribution \( X \sim N(2130, 580) \). M1: Correct standardization of 2150. A1: Correct probability in the range [0.2031, 0.2033]. (c) M1: For identifying the required linear combination \( Y = W - 3.2T \) and finding \( E(Y) = -6 \). M1: Attempting to find \( Var(Y) = 227.84 \). A1: Correct probability in the range [0.6545, 0.6554].

(a) Unbiased estimate of population mean: \( \bar{x} = \frac{\sum x}{n} = \frac{122.4}{8} = 15.3 \). Unbiased estimate of population variance: \( s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) = \frac{1}{7} \left( 1874.5 - \frac{122.4^2}{8} \right) = \frac{1}{7} (1874.5 - 1872.72) = \frac{1.78}{7} \approx 0.2543 \approx 0.254 \). (b) For the second sample, \( n_2 = 12 \), \( s_2^2 = 0.182 \). The pooled variance is \( s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = \frac{1.78 + (11 \times 0.182)}{8 + 12 - 2} = \frac{1.78 + 2.002}{18} = \frac{3.782}{18} \approx 0.2101 \approx 0.210 \). (c) The Central Limit Theorem states that if a random sample of size \(n\) is taken from any population with mean \(\mu\) and finite variance \(\sigma^2\), then as \(n \to \infty\) (or is sufficiently large, usually \(n \ge 30\)), the sampling distribution of the sample mean \(\bar{X}\) is approximately normally distributed with mean \(\mu\) and variance \(\frac{\sigma^2}{n}\).

(a) M1: Correct formula or method for unbiased mean. A1: Mean = 15.3. M1: Correct formula or method for unbiased variance. A1: Variance = 0.254 (accept 1.78/7). (b) M1: For identifying that \( (n_1-1)s_1^2 = 1.78 \). M1: For attempting to find \( (n_2-1)s_2^2 = 2.002 \). M1: Correct pooled variance formula. A1: Pooled variance = 0.210 (accept 0.21). (c) B1: States that sample size \(n\) must be large (or \(n \to \infty\)). B1: Mentions that the sample mean \(\bar{X}\) is approximately normally distributed. B1: Specifies the mean is \(\mu\) and variance is \(\frac{\sigma^2}{n}\).

(a) We have \( n = 150 \), \( \bar{x} = 48.2 \), and \( s = 5.4 \). Since \( n \) is large, we can estimate \( \sigma \) with \( s \). For a 98% confidence interval, the critical value \( z \) is \( 2.3263 \) (or \( 2.33 \) from tables). The confidence interval is \( \bar{x} \pm z \frac{s}{\sqrt{n}} = 48.2 \pm 2.3263 \times \frac{5.4}{\sqrt{150}} = 48.2 \pm 2.3263 \times 0.4409 = 48.2 \pm 1.0257 = (47.174, 49.226) \approx (47.2, 49.2) \text{ cm (3 sf)}. (b) The width of the confidence interval is \) 2 z \frac{s}{\sqrt{n}} \). We require \( 2 \times 2.3263 \times \frac{5.4}{\sqrt{n}} \le 1.5 \implies \frac{25.124}{\sqrt{n}} \le 1.5 \implies \sqrt{n} \ge \frac{25.124}{1.5} \approx 16.7493 \implies n \ge 280.54 \). Thus, the minimum sample size required is 281.

(a) B1: Correct critical value \( z = 2.3263 \) (or \( 2.33 \)). M1: Correct formula for confidence interval. A1: Correct standard error term \( \frac{5.4}{\sqrt{150}} \approx 0.441 \). M1: Attempting to calculate the limits. A1: Correct interval (47.2, 49.2) (accept 47.17 to 49.23). (b) M1: Expression for width \( 2 z \frac{s}{\sqrt{n}} \). A1: Correct inequality \( 2 \times 2.3263 \times \frac{5.4}{\sqrt{n}} \le 1.5 \). M1: Rearranging for \( \sqrt{n} \). A1: Obtaining \( \sqrt{n} \ge 16.75 \) (or \( 16.78 \) if using \( z=2.33 \)). M1: Squaring to find \( n \ge 280.5 \) (or \( n \ge 281.5 \) if using \( z=2.33 \)). A1: Concluding with the next integer, 281 (or 282 if using \( z=2.33 \)).

(a) Let \( \mu_1 \) be the mean reaction time of tennis players and \( \mu_2 \) be the mean reaction time of squash players. We test: \( H_0: \mu_1 = \mu_2 \) against \( H_1: \mu_1 \ne \mu_2 \). The sample statistics are: \( n_1 = 60 \), \( \bar{x}_1 = 0.185 \), \( s_1 = 0.024 \); and \( n_2 = 80 \), \( \bar{x}_2 = 0.178 \), \( s_2 = 0.030 \). Since the samples are large, we can use the test statistic: \( z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{0.185 - 0.178}{\sqrt{\frac{0.024^2}{60} + \frac{0.030^2}{80}}} = \frac{0.007}{\sqrt{0.0000096 + 0.00001125}} = \frac{0.007}{\sqrt{0.00002085}} = \frac{0.007}{0.004566} \approx 1.533 \). At the 5% level of significance for a two-tailed test, the critical values are \( \pm 1.96 \). Since \( |1.533| < 1.96 \), we fail to reject \( H_0 \). There is insufficient evidence at the 5% significance level to suggest that the mean reaction time of tennis players is different from that of squash players. (b) Since the distributions of reaction times of individual players are not known to be normal, we must rely on the Central Limit Theorem. The theorem states that because the sample sizes (\( n_1 = 60 \) and \( n_2 = 80 \)) are large (\( \ge 30 \)), the sample means \( \bar{X}_1 \) and \( \bar{X}_2 \) will be approximately normally distributed, justifying the use of the \( z \)-test.

(a) B1: State \( H_0: \mu_1 = \mu_2 \) and \( H_1: \mu_1 \ne \mu_2 \) correctly. M1: Identify the correct test statistic formula. M1: Calculate the variance of the difference of means \( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \approx 0.00002085 \). A1: Correct standard error \( \approx 0.00457 \). A1: Correct test statistic value \( z \approx 1.53 \). B1: Correct critical values of \( \pm 1.96 \). M1: Make a valid comparison between the test statistic and critical value. A1: Conclude in context, stating there is no evidence of a difference. (b) B1: State that individual distributions of reaction times are unknown. B1: Mention that sample sizes are large (\( \ge 30 \)). B1: State that this means the sample means are approximately normally distributed.

(a) Mean \( \lambda = \frac{(18 \times 0) + (32 \times 1) + (27 \times 2) + (15 \times 3) + (8 \times 4) + (0 \times 5)}{100} = \frac{32 + 54 + 45 + 32}{100} = \frac{163}{100} = 1.63 \). (b) For a Poisson distribution with \( \lambda = 1.63 \), the probability of '4 or more' weeds is: \( P(X \ge 4) = 1 - P(X \le 3) = 1 - e^{-1.63}\left(1 + 1.63 + \frac{1.63^2}{2} + \frac{1.63^3}{6}\right) = 1 - e^{-1.63}(1 + 1.63 + 1.32845 + 0.72179) = 1 - 0.19593 \times 4.68024 = 1 - 0.91699 = 0.08301 \). Expected frequency \( E = 100 \times 0.08301 = 8.30 \) (shown). (c) We test: \( H_0 \): A Poisson distribution is a suitable model. \( H_1 \): A Poisson distribution is not a suitable model. Combining '4' and '5 or more', the observed frequency for '4 or more' is \( 8 + 0 = 8 \). Observed: 18, 32, 27, 15, 8. Expected: 19.59, 31.94, 26.03, 14.14, 8.30. Since all expected frequencies are \( \ge 5 \), no pooling is required. Test statistic \( \chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(18-19.59)^2}{19.59} + \frac{(32-31.94)^2}{31.94} + \frac{(27-26.03)^2}{26.03} + \frac{(15-14.14)^2}{14.14} + \frac{(8-8.30)^2}{8.30} = 0.1291 + 0.0001 + 0.0361 + 0.0523 + 0.0108 = 0.2284 \). Number of classes after pooling \( k = 5 \). We estimated 1 parameter (\( \lambda \)) from the sample, so degrees of freedom \( df = k - 1 - 1 = 3 \). Critical value \( \chi^2_3(0.10) = 6.251 \). Since \( 0.2284 < 6.251 \), we fail to reject \( H_0 \). There is no significant evidence at the 10% level that a Poisson distribution is not a suitable model.

(a) M1: For calculating total weeds (163) / 100. A1: Correctly showing 1.63. (b) M1: Attempting to find \( P(X \le 3) \) using Poisson formula with \( \lambda = 1.63 \). A1: Correct calculation of \( P(X \le 3) \approx 0.917 \). A1: Correctly showing expected frequency \( 100 \times P(X \ge 4) = 8.30 \). (c) B1: Correct hypotheses stated. M1: Attempting to calculate \( \sum \frac{(O-E)^2}{E} \). A1: Correct test statistic value of 0.228 (accept 0.22 to 0.24). B1: Identifying correct degrees of freedom (3) and critical value (6.251). M1: Correct comparison of test statistic with critical value. A1: Concluding in context that Poisson is a suitable model.

(a) \( H_0 \): There is no association between age group and preferred holiday destination (they are independent). \( H_1 \): There is an association between age group and preferred holiday destination (they are not independent). (b) Row totals: Under 30 = 80; 30 to 50 = 80; Over 50 = 40. Total \( N = 200 \). Column totals: Beach = 90; City Break = 70; Countryside = 40. Expected frequencies are \( E = \frac{\text{Row Total} \times \text{Col Total}}{200} \). Expected frequencies table: For Under 30: Beach \( = \frac{80 \times 90}{200} = 36 \), City Break \( = \frac{80 \times 70}{200} = 28 \), Countryside \( = \frac{80 \times 40}{200} = 16 \). For 30 to 50: Beach \( = 36 \), City Break \( = 28 \), Countryside \( = 16 \). For Over 50: Beach \( = \frac{40 \times 90}{200} = 18 \), City Break \( = \frac{40 \times 70}{200} = 14 \), Countryside \( = \frac{40 \times 40}{200} = 8 \). (c) We calculate \( \chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(45-36)^2}{36} + \frac{(25-28)^2}{28} + \frac{(10-16)^2}{16} + \frac{(35-36)^2}{36} + \frac{(30-28)^2}{28} + \frac{(15-16)^2}{16} + \frac{(10-18)^2}{18} + \frac{(15-14)^2}{14} + \frac{(15-8)^2}{8} = 2.25 + 0.3214 + 2.25 + 0.0278 + 0.1429 + 0.0625 + 3.5556 + 0.0714 + 6.125 = 14.8066 \approx 14.81 \). Degrees of freedom \( df = (3-1)(3-1) = 4 \). Critical value for \( \chi^2_4(0.05) = 9.488 \). Since \( 14.81 > 9.488 \), we reject \( H_0 \). There is significant evidence at the 5% level of significance to suggest an association between age group and preferred holiday destination.

(a) B1: State \( H_0 \) correctly. B1: State \( H_1 \) correctly. (b) M1: For calculating row and column totals. M1: Correct method for calculating expected frequencies (at least one cell shown). A1: Correct row 1 expected values (36, 28, 16). A1: Correct remaining expected values. (c) M1: Correct method to calculate \( \chi^2 = \sum \frac{(O-E)^2}{E} \). A1: Correctly calculated \( \chi^2 \approx 14.81 \). B1: Correct degrees of freedom (4) and critical value (9.488). M1: Correct comparison of test statistic with critical value. A1: Correctly conclude in context (reject \( H_0 \)).

(a) Let us find the differences \( d \) between the rankings of Judge A and Judge B for each singer: Singer P: \( d = 1 - 3 = -2 \implies d^2 = 4 \); Singer Q: \( d = 2 - 1 = 1 \implies d^2 = 1 \); Singer R: \( d = 3 - 5 = -2 \implies d^2 = 4 \); Singer S: \( d = 4 - 2 = 2 \implies d^2 = 4 \); Singer T: \( d = 5 - 8 = -3 \implies d^2 = 9 \); Singer U: \( d = 6 - 4 = 2 \implies d^2 = 4 \); Singer V: \( d = 7 - 6 = 1 \implies d^2 = 1 \); Singer W: \( d = 8 - 7 = 1 \implies d^2 = 1 \). Sum of \( d^2 \) is \( \sum d^2 = 4 + 1 + 4 + 4 + 9 + 4 + 1 + 1 = 28 \). Spearman’s rank correlation coefficient is \( r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} \). With \( n = 8 \), \( r_s = 1 - \frac{6 \times 28}{8(64-1)} = 1 - \frac{168}{504} = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.667 \). (b) Hypotheses: \( H_0: \rho_s = 0 \) (no correlation between the judges' rankings). \( H_1: \rho_s > 0 \) (there is a positive correlation between the judges' rankings). Critical value for a one-tailed test at the 5% level of significance with \( n = 8 \) is \( 0.6429 \). Since \( 0.667 > 0.6429 \), we reject \( H_0 \). There is significant evidence at the 5% level of significance to suggest that there is a positive correlation between the rankings of the two judges.

(a) M1: Attempting to find the differences \( d \). M1: Attempting to calculate \( d^2 \) for each singer. A1: Correct sum of \( \sum d^2 = 28 \). M1: Attempt to use the Spearman formula. A1: Correct calculation of fraction \( \frac{168}{504} \). A1: Correct final answer \( r_s = 0.667 \) (accept 2/3). (b) B1: Correct hypotheses stated. B1: Correct critical value of 0.6429 from tables. M1: Correct comparison of \( r_s \) with critical value. A1: Correctly reject \( H_0 \). A1: Conclude in context that there is a positive correlation.