Script review from /rendered

Where you lost marks — and how to fix it

Re-marked visually against Edexcel 8MA0/01 mark schemes (Summer 2019 & Oct 2020). Each item ties to your exact working, not just the topic.

2019 script

72 / 100

28 marks to recover

2020 script

70 / 100

30 marks to recover

Cross-paper patterns (even when you “almost” had it)

2019 — question-by-question (lost or partial)

Summer 2019 · 8MA0/01Full marks on Q1, Q2, Q3, Q4, Q5, Q7, Q13
Q6(b) Cosine rule · surds ~3 marks lost

Your script

Part (a) perfect: x = 2√3. Part (b): you wrote the cosine rule formula then crossed it out — no value for BC.

Mark scheme

BC² = (6√3)² + (4√3)² − 2(6√3)(4√3)cos60° = 84 → BC = 2√21 cm.

Specific struggle

Stopping after the “show that” in (a); not substituting x into side lengths before cosine rule.
Targeted drill (same error type) Triangle: AB = 3x, AC = 2x, ∠A = 60°, area = 18√3. (i) Show x = 2√3. (ii) Hence BC in simplest surd — must substitute x before cos rule.
Q8(b) Binomial · estimation 1 mark lost

Your script

(a) correct: 64 + 144x + 135x². (b) circled but blank — no explanation.

Mark scheme

Set 3x/4 = −0.075 so x = −0.1, then substitute into first three terms of expansion for 1.925⁶ (since 2 + 3x/4 = 1.925).
Targeted drill Explain (no calc) how (2 + 3x/4)⁶ estimates 1.925⁶. Then: what x makes 2 + 3x/4 = 1.925?
Q9(c) Quadratic model · interpretation 2 marks lost

Your script

You wrote T(4) = 432 for “mass mined in 2023”.

Mark scheme

Mass in 2023 = change in total: T(5) − T(4) = 93 tonnes (not T(4) alone).

Specific struggle

Reading “total by year n” as “amount in that calendar year”.
Targeted drill T = 1200 − 3(n−20)². Find tin mined (i) up to 1 Jan 2020, (ii) during 2023 only, (iii) maximum total. Box: “cumulative vs yearly”.
Q9(d) Model limitations 0–2 marks (partial)

Your script

“total amount mined can't decrease” and n ≤ 20 — direction right, explanation thin.

Mark scheme

Need n ≤ 20 and reason: total mass cannot decrease after peak at n = 20.
Q10(a)(ii) & Q11 Circles · cubics · graphs ~6–10 marks (combined)

Your script

Q10: completed square to (x−2)²+(y+4)²=28 but didn’t state centre/radius on p3; (b) later: k = 2±√28. Q11(b): discriminant on original cubic (wrong quadratic); (d) roots noted but k = 4, −3/2 not stated clearly.

Mark scheme

Q10: centre (2,−4), r = √28. Q11(b): factorise to (x−4)²(2x+3) → only two distinct roots; (d) k = 4 or −3/2.

Specific struggle

Mixing “which polynomial” when using Δ; not linking repeated root (x−4)² to “two distinct roots”.
Q12(a) Trig identity proof partial — (b) completed on next page

Your script

Used sin²θ = 1−cos²θ, reached −10cos²θ−7cosθ+12 over denominator; substitution u = cos θ but no factorisation on p4. Page 5: solved → {120.96°, 300.96°}.

Mark scheme

Factor numerator: (3+2cosθ)(4−5cosθ) to show equivalence.
Q14(b)(ii) Exponentials · logs · context ~2 marks lost

Your script

Rough work: T ≈ 8.26 years — no conversion to years and months on answer lines.

Mark scheme

T ≈ 8.24 → 8 years 3 months (nearest month). Numerical answer only without ln working → 0.
Q15 Proof · divisibility mod 8 4 marks lost

Your script

Scribbled (k+1)³+2, k³+2 — no even/odd cases, no mod 8 conclusion. Red X.

Mark scheme

Case n even: n=2m → remainder 2 mod 8. Case n odd: n=2m+1 → remainder 4 mod 8. Conclude not divisible by 8.
Targeted drill Prove: (i) n³−n divisible by 6; (ii) n³+2 not divisible by 8 for n∈ℕ. Template: “Let n be even… Let n be odd… Therefore…”
Q16 Vector geometry 5 marks lost

Your script

(i) wrote “The” only. (ii) tiny sketch, no calculation. Red X on both.

Mark scheme

(i) Vectors parallel, same direction (angle 0°). (ii) Use cosine rule / dot product with |m|=3, |m−n|=6, angle 30° between m and n.

2020 — question-by-question (lost or partial)

October 2020 · 8MA0/01Strong: Q1, Q3, Q4, Q6, Q7, Q10, Q12(a)(c)
Q2 Vectors · bearing Method OK — check coordinates

Your script

Displacement uses −7i − 3j (as if B were (−3,−5) not (−1,−5)). Bearing 246.8°, speed 2.77 — marked 6/6 but arithmetic follows wrong vector.

Mark scheme

Correct displacement −5i − 3j, bearing ≈ 239.0°, speed ≈ 2.04 km/h.

Hidden risk

Copying coordinates incorrectly while method still scores — fix before exams.
Q5(b) Sine/cosine rule · optimisation 3 marks lost
Roof structure (simplified) — your error was totalling lengths, not finding ACB

Your script

ACB = 128.9° correct. Total steel: working shows 9+7+7+16+12 = 51 m — spurious 9, AD rounded up wrong way.

Mark scheme

AD ≈ 15.09 → buy 16 m; total 12+7+7+16 = 42 m (whole metres).
Q8 Exponential models · asymptote ~5–6 marks

Your script

(a) “19” (wrong; should be t=0). (b) t = 10.72 OK-ish. (c) “It is the minimum.” (d) blank. Red X on section.

Mark scheme

(a) 83°C. (c) θ → 18 as t→∞. (d) From graph: A+B=94, A+Be⁻¹=50 → asymptote μ = (50e−94)/(e−1) ≈ 24.4°C.
Q9 Trig graphs · equations ~5 marks

Your script

(a) c=−180, d=−3 ✓. (b)(i) blank. (b)(ii) (−144,−3) ✓. (c) quadratic in sin θ OK; sin⁻¹(1/3)≈22.02° wrong; answers 562.02 & 157.98 not in interval.

Mark scheme

(b)(i) (−720,−3). (c) Only θ ≈ 520.5° in (450°, 720°).
Q11 Circle tangent · quadrant constraint ~4–6 marks

Your script

(i) Used gradient of radius 3/2 for tangent (should be −2/3). Rough work later has −3/7. (ii) Centre (4,−6), r²=52−k; wrote k<4, k>36 inconsistently — rough had 36<k<52.

Mark scheme

(i) 2x+3y−11=0. (ii) Entirely in Q4 ⇒ r < 4 and r² > 0 ⇒ 36 < k < 52.
Q12(b) Log model interpretation 1 mark lost

Your script

Part (b) scribbled out / missing — need interpretation of ab (views after day 1 ≈ 239).
Q13 AM-GM proof · counterexample 5 marks lost

Your script

Wrote odd/even number chatter — not (2a−b)²≥0 nor counterexample e.g. a=5,b=−1.

Mark scheme

(a) Square ≥ 0 → divide by ab. (b) One pair with sum < 4, e.g. 4a/b+b/a = −20−1/5 < 4.
Q14 Cubic from conditions 9 marks lost

Your script

g(x)=ax³+bx²+ax, used (2,9) → 10a+4b=9 only. No g′(2)=0. Part (b) not attempted.

Mark scheme

Also 13a+4b=0 → g(x)=−3x³+39x²/4−3x; max via g″(2)<0.

Targeted practice (mirrors your exact mistakes)

Do these before the mirror paper. Each is tagged with the paper error it fixes.

P1 2019 Q15 type

Prove n³+2 is not divisible by 8 for all n∈ℕ. Use even/odd cases only.

P2 2019 Q16 + 2020 Q2 type

|a+b|=|a|+|b| — explain geometrically. Then: |m|=3, |m−n|=6, angle between m,n is 30°. Find angle between m and m−n (1 d.p.).

P3 2019 Q9c type

T=1200−3(n−20)². Find (a) total to Jan 2020, (b) tin mined in 2023 only, (c) state valid n with reason.

P4 2019 Q8b type

Expansion (2+3x/4)⁶. Explain estimating 1.925⁶ using x = −0.1 (show why).

P5 2020 Q5b type

Structure: AB=12, BC=BD=7, ∠BAC=27°. Find ACB; then minimum whole-metre steel for AB+BC+BD+AD.

P6 2020 Q8d + Q9c

θ=18+65e^(−t/8). Initial temp; t when θ=35; why not 15°C; asymptote from (0,94) and (8,50). Separate: solve 3cosθ=8tanθ on 450°<θ<720°.

P7 2020 Q11 + Q14

Tangent at P(−5,7) to x²+y²+18x−2y+30=0. Circle x²+y²−8x+12y+k=0 entirely in Q4 — range of k. Cubic g(x)=ax³+bx²+ax through origin, stationary at (2,9) — find g and show max.

P8 2020 Q13 type

Prove 4a/b + b/a ≥ 4 for a,b>0. Counterexample when not positive.

Mirror paper — lost-mark topics only

107 marks total (100 core + 7 vectors) · Exam-style · Based only on questions where you dropped marks. Time: ~2 hours 15 min.

Show all working. Give exact values where appropriate. Calculators allowed where stated.

1

Triangle ABC has AB = 3x cm, AC = 2x cm, ∠CAB = 60°. Area = 18√3 cm².

(a) Show that x = 2√3. (b) Hence find exact length BC as a simplified surd.

[6]
AB=3x2x60°
Q1 — same structure as your 2019 Q6
2

Binomial. (a) First 3 terms of (2+3x/4)⁶. (b) Explain how to estimate 1.925⁶ using (a) without calculating.

[5]
3

Tin model T = 1200 − 3(n−20)², n years after 1 Jan 2019.

(a) Mass up to 1 Jan 2020. (b) Maximum total mass. (c) Mass mined during 2024 only. (d) Limitation on n with reasons.

[6]
4

f(x)=2x³−13x²+8x+48. (a) Show (x−4) is a factor. (b) Show f(x)=0 has only two distinct roots. (c) How many real roots for f(x)=2? (d) y=f(x+k) through origin — find both k.

[10]
5

Show (10sin²θ−7cosθ+2)/(3+2cosθ) ≡ 4−5cosθ. Hence solve (10sin²θ−7cosθ+2)/(3+2cosθ) = 4+3sinθ for 0≤x<360°.

[7]
6

V=15700e^(−0.25t)+2300. (a) Initial value. (b)(i) Show 3925e^(−0.25T)=500 when decreasing at £500/year. (ii) Age in years and months (nearest month). (c) Value of A (lower bound). (d) One limitation.

[9]
7

Prove n³+2 is not divisible by 8 for n∈ℕ.

(i) |a+b|=|a|+|b| — geometric meaning. (ii) |m|=3, |m−n|=6, angle 30° between m,n — angle between m and m−n (1 d.p.).

[9]
8

Boat: 10:00 at (4i−2j) km, 12:45 at (−1i−5j) km, constant speed. Bearing (1 d.p.) and speed (km/h, 2 d.p.).

[6]
9

Steel frame: AB=12 m, BC=BD=7 m, ∠BAC=27°. (a) ∠ACB (1 d.p.). (b) Minimum whole metres to buy for AB+BC+BD+AD.

[6]
10

Tea: θ=18+65e^(−t/8). (a) Initial °C. (b) t when θ=35 (1 d.p.). (c) Why not 15°C. Second cup: points (0,94), (8,50) on μ=A+Be^(−t/8). (d) Asymptote equation.

Graph: y=3cos x°; P minimum with smallest negative x. (a) c,d. (b)(i) y=3cos(x/4)° (ii) y=3cos(x−36)° — image points. (c) Solve 3cosθ=8tanθ, 450°<θ<720°, 1 d.p.

[17]
11

C₁: x²+y²+18x−2y+30=0, tangent at P(−5,7). (i) Equation of tangent, integer coefficients. C₂: x²+y²−8x+12y+k=0 lies entirely in quadrant 4 — range of k.

Prove 4a/b+b/a≥4 for a,b>0; counterexample otherwise. Cubic g(x)=ax³+bx²+ax, through origin, stationary at (2,9): find g(x), prove (2,9) is a maximum.

[23]
12

Figure 2 shows points O, M and N in a plane. The position vectors of M and N relative to O are m and n respectively.

Given |m| = 5 units, |n| = 7 units, and the angle between m and n is 40°,

(a) Find |m − n|, giving your answer to 3 significant figures. (b) Find the acute angle between m and m − n, giving your answer to one decimal place.

Targets your 2019 Q16(ii) gap — diagram generated with GPT Image 2.

[7]
Figure 2: vectors m and n from O, angle 40°, vector m−n from N to M
Figure 2 — vectors in the plane (gpt-image-2-2026-04-21)

Total: 107 marks (100 core + 7 vectors)

Answer key (self-mark)
  • 1: BC = 2√21
  • 2: 64+144x+135x²; x=−0.1 from 3x/4
  • 3: 117; 1200; T(6)−T(5)=93; n≤20 total can’t decrease
  • 4: (x−4)²(2x+3); 2 roots; 3 roots for f=2; k=4, −3/2
  • 5: factorise; {120.96°, 300.96°} (θ=x)
  • 6: £18000; 8 years 3 months; A=2300
  • 7: mod 8 cases; collinear same direction; ≈14.5° (check working)
  • 8: bearing ≈239.0°; speed ≈2.04
  • 9: 128.9°; 42 m
  • 10: 83°C; t≈10.7; θ→18; μ=(50e−94)/(e−1); P(−180,−3); (−720,−3); (−144,−3); θ≈520.5°
  • 11: 2x+3y−11=0; 36<k<52; AM-GM; a=5,b=−1; g(x)=−3x³+39x²/4−3x; g″(2)<0
  • 12: |m−n| ≈ 4.15 (3 s.f.); acute angle ≈ 25.3° (1 d.p.) — cosine rule + dot product