Tab 1
Edexcel A Level Mathematics 2025 Paper 1 Unofficial Markscheme.
Question | Part | Answer | Notes | Marks |
1 | (a) | (4, −4) | Transformation: (x,y) → (x−2, y) | 1 |
1 | (b) | (−4, 6) | Inverse: swap x, y | 1 |
1 | (c) | (6, 5) | (x,y) → (x, 2 | 2 |
2 | (a)(i) | Centre: (−3, 4) | From (x−h)²+(y−k)²=r² | 1 |
2 | (a)(ii) | Radius: 2√6 |
| 2 |
2 | (b) in | No, origin is outside. Distance to centre = 5 > 2√6 ≈ 4.899 | Calculate √[(−3)²+4²] and compare to radius | 2 |
3 | (a) | k = 5/2 | Arithmetic sequence: 10−6k = 2k−10 | 2 |
3 | (b) | S₅₀ = −5375 | Sₙ = n/2[2a+(n−1)d], a=15, d=−5 | 3 |
4 | (a) | f(x) = (x+4)(2x²−5x+4) | Factor theorem at x=−4, then divide | 3 |
4 | (b) | Discriminant = 25−32 = −7 ⇒ no real roots ⇒ only x=−4 | Check b²−4ac | 2 |
5 | (a) | ∫₀.₄₄².⁸₉ (2/√x) dx | Definite integral from limit-of-sum | 1 |
5 | (b) | k = 2 | Direct evaluation with limits 1.44 and 2.89 for ∫(2/√x) | 2 |
6 | (a) | h = 20 − 0.3 t¹·⁵ | Solve 17.6 = A−B·4¹·⁵ and 11.9 = A−B·9¹·⁵ ⇒ A=20, B=0.3 | 4 |
6 | (b) | 20 m | h(0)=A | 1 |
6 | (c) | T ≤ (200/3)²ᐟ³ ≈ 16.44 | 20−0.3 t¹·⁵ ≥ 0 | 2 |
7 | (a) | {x: x≤ −1} ∪ {x: 2 ≤ x ≤ 5} | f′(x)≥0 on those intervals | 2 |
7 | (b) | f(x)=−3(x+1)²(x−5)² | Roots at x=−1, 5 double; use y-int to find leading coefficient | 3 |
7 | (c) | 0 < k < 243 | f(2)=−243 so shift up by k to get four real roots | 2 |
8 | (a) | Correct (5k+2)²=25k²+20k+4 ⇒ 5(5k²+4k+1)−1 | Student had 10k term error | 1 |
8 | (b) | m=5k+3 → form 5n−1; m=5k+4 → form 5n+1 OR m=5k-1 m=5k-2 | Square and factor each to show they fit 5n±1 | 4 |
9 | a) | 13.5 | 15t - te^0.2t t(15-e^0.2t), solve 15 - e^0.2t = 0 Value when y axis (think it was V?) was 0 | 2 |
9 | b) | dv/dt then show that it looks like 5ln(75/t+5 | 4 | |
9 | c) i) | t3 = | 3 | |
9 | c) ii) | 8.554 seconds | Repeated integration to find the time in seconds when the car reaches max velocity | |
10 | (a) | PR = 8i + 8j + 4k | PR = PQ + QR | 2 |
10 | (b) | ∥PQ∥=∥QR∥=6√2, ∥PR∥=12 ⇒ isosceles & right at Q | Check magnitudes & dot(PQ,QR)=0 (FM) but u can also do pythag or cosine | 4 |
11 | (a) | V=1500+18500 e^(−kt), k≈0.227 | Use V(0)=20000 → A=18500; V(2.5)=12000 → solve for k | 4 |
11 | (b) | dV/dt=−kAe^(−kt) ⇒ −k(V-500) | Differentiate & substitute A e^(−kt)=V−1500 | 3 |
11 | (c) | V→1500 as t→∞ | e^(−kt)→0 | 1 |
12 | (a) | y=15x/[(2x+3)(x−3)] at x=6 → y=2; line: y=2x−10 at x=6 → y=2 | Shows C and ℓ meet at same y | 2 |
12 | (b) | 4x³−26x²−3x+90=0 | Equate and rearrange | 2 |
12 | (c) | x=(1−√61)/4 | Divide cubic by (x−6) then solve quadratic | 4 |
13 | — | 2/25 | Likely ∫₀² [x/(2x+1)³] dx by parts (typo in original). Substitution u=2x+1 worked too | 5 |
14 | (a) | 2 cos x | Expand sin(x+30)+√3 cos(x+30) via compound-angle | 3 |
14 | (b) | θ=19.5°, 90°, 160.5° | Solve 2 cosθ(1−3 sinθ)=0 for 0≤θ<180 | 4 |
15 | (a) | P=2r+480/r | θ=480/r²−1/5; P=2r+r/5+rθ | 4 |
15 | (b) | r=4√15 | dP/dr=0 ⇒ r²=240 | 3 |
15 | (c) | d²P/dr²=960/r³>0 at r=4√15 ⇒ minimum (0.258) | 2 | |
16 | (a) | ∫_{π/6}^{π/3} (1/4)cosec²u du | x=2 sin u ⇒ dx and √(4−x²)=2 cos u; change of variables | 4 |
16 | (b) | √3/6 | (1/4)(−cot u) | 3 |
Q1: 4, Q2: 5, Q3: 5, Q4: 5, Q5: 3, Q6: 7, Q7: 7, Q8: 5, Q9: 9, Q10: 6, Q11: 8, Q12: 8, Q13: 5, Q14: 7, Q15: 9, Q16: 7. Total = 4+5+5+5+3+7+7+5+9+6+8+8+5+7+9+7 = 100.
Tab 2
PC Build Cost Tracker
Here's a detailed breakdown of the costs for the PC build:
Item | Cost (£) |
1 TB SSD | 33 |
32 GB RAM | 67 |
Ryzen 7 7800X3D | 225 |
Case (pre-owned) | 0 (60 when bought 9 years prior) |
Subtotal | 325 |
Item | Cost (£) | Notes |
Thermal paste | 5 | AliExpress, PTM7950 40×80mm |
CPU cooler | 32 | Phantom spirit 120SE |
PSU | 60 | CCL - £55 for a 750W MSI A750GN Gold, some users report loud fans with AGL version. |
Motherboard | Under 100 | This part is difficult, Mobos are more important now, than before. |
GPU | 500 | I want to do AI stuff +gaming Rumoured 5070TI super looks really juicy with 24 GB but most likely over 1K |
Subtotal | around 700 |