Mark Scheme

Section A — Structured Questions (42 marks)

Question 1

(a) 3.72 × 10⁵ ÷ (4 × 10⁻³)

• = 9.3 × 10⁴ [M1] for correct division process
• = 9.3 × 10⁷ [A1] correct answer in standard form

Total: 2 marks

(b) 4.83 × 10⁶ × 150 [M1] for 4.83 million = 4.83 × 10⁶ and multiplication by 150

• = 7.245 × 10⁸ (or 7.25 × 10⁸) [A1] correct answer in standard form

Total: 2 marks

Question 2

(a) AC² = 8² + 15² [M1] for use of Pythagoras' theorem

• AC = 17 (cm) [A1] correct answer

Total: 2 marks

(b) Area = ½ × 8 × 15 [M1] for correct formula

• = 60 (cm²) [A1] correct answer

Total: 2 marks

(c) In triangle ABD:

• AD/sin 90° = 8/sin(angle ADB) or AD/8 = 1/cos 32° [M1] for use of trigonometry in correct triangle
• angle ADB = 180° − 90° − 32° = 58° (or alternative method) [M1] for finding required angle or using cosine
• AD = 8/sin 58° = 9.43 (cm) (or AD = 8/cos 32° = 9.43) [A1] correct answer

Accept: 9.42 to 9.44

Total: 3 marks

Question 3

(a) 2 < d ≤ 5 [B1]

Total: 1 mark

(b) Midpoints used: 1, 3.5, 6.5, 11.5 [B1] for all correct midpoints seen or implied

• Sum = 18(1) + 27(3.5) + 22(6.5) + 13(11.5) [M1] for multiplying frequencies by midpoints
• = 18 + 94.5 + 143 + 149.5 = 405 [M1] for summing products
• Mean = 405 ÷ 80 = 5.0625 or 5.06 (km) [A1] correct answer

Total: 4 marks

(c) Frequency densities: 18/2 = 9, 27/3 = 9, 22/3 = 7.33..., 13/7 = 1.857... [B1] for calculating at least 3 frequency densities correctly

• Correct bars drawn at 0-2, 2-5, 5-8, 8-15 [B1] for all bars in correct positions with correct widths
• Heights: 9, 9, 7.33, 1.86 (accept heights within tolerance) [B1] for all correct heights

Total: 3 marks

Question 4

(a) f(−3) = 2(−3)² − 5(−3) + 1 [M1] for substituting x = −3

• = 18 + 15 + 1 = 34 [A1] correct answer

Total: 2 marks

(b) 2(x² − 2.5x) + 1 [M1] for factorising coefficient of x²

• 2(x² − 2.5x + 1.5625 − 1.5625) + 1 or 2((x − 1.25)² − 1.5625) + 1 [M1] for completing the square
• = 2(x − 1.25)² − 2.125 (or exact equivalent −17/8) [A1] correct answer

Accept: 2(x − 5/4)² − 17/8

Total: 3 marks

(c) Minimum point: (1.25, −2.125) or (5/4, −17/8) [B2] (or [B1] for one coordinate correct)

Total: 2 marks

Question 5

(a) Second branch probabilities:

• After Red: 4/9, 3/9, 2/9 [B1] for at least two branches from Red correct
• After Blue: 5/9, 2/9, 2/9 and After Green: 5/9, 3/9, 1/9 [B1] for completing remaining branches correctly

Total: 2 marks

(b) P(both red) = (5/10) × (4/9) [M1] for multiplying probabilities along correct branch

• = 20/90 = 2/9 [A1] correct answer simplified

Total: 2 marks

(c) P(different colours) = 1 − P(both same) [M1] for strategy (or listing combinations)

• P(both same) = 2/9 + (3/10)(2/9) + (2/10)(1/9) = 2/9 + 6/90 + 2/90 = 28/90 [M1] for calculating probability of same colours
• P(different) = 1 − 28/90 = 62/90 = 31/45 [A1] correct answer

Alternative method: adding probabilities of different colour outcomes [M1 M1 A1] awarded accordingly

Total: 3 marks

Question 6

(a) Arc length = (140/360) × 2π × 9 [M1] for correct formula

• = 21.99... = 22.0 (cm) [A1] correct answer

Accept: 21.9 to 22.0

Total: 2 marks

(b) Area = (140/360) × π × 9² [M1] for correct formula

• = 98.96... = 99.0 (cm²) [A1] correct answer

Accept: 98.9 to 99.0

Total: 2 marks

(c) Arc length = circumference of base of cone [M1] for understanding relationship

• 2πr_cone = (140/360) × 2π × 9 [M1] for setting up equation
• r_cone = (140/360) × 9 = 3.5 (cm) [A1] correct answer

Total: 3 marks

Section B — Extended Response (28 marks)

Question 7

(a) V = πr²h = 50000 [M1] for stating volume formula

• h = 50000/(πr²) [A1] correctly rearranged (must see this shown)

Total: 1 mark

(b) A = 2πr² + 2πrh

• = 2πr² + 2πr(50000/(πr²)) [M1] for substituting h = 50000/(πr²)
• = 2πr² + 100000/r [A1] correctly simplified

Total: 2 marks

(c) dA/dr = 4πr − 100000r⁻² or 4πr − 100000/r² [M1] for differentiating first term

• [A1] for both terms correct

Total: 2 marks

(d) dA/dr = 0 for minimum [M1] for setting derivative equal to zero

• 4πr = 100000/r² [M1] for correct rearrangement
• r³ = 100000/(4π) = 7957.75...
• r = 19.98... = 20.0 (cm) [A1] correct answer
• d²A/dr² = 4π + 200000/r³ [M1] for finding second derivative
• When r = 20, d²A/dr² = 4π + 200000/8000 = 37.6 > 0, so minimum [A1] for showing this is a minimum

Alternative: test values either side of r = 20 to show minimum [M1 A1]

Total: 5 marks

(e) A = 2π(20)² + 100000/20 [M1] for substituting r = 20

• = 2513.3 + 5000 = 7513.3 = 7510 (cm²) [A1] correct answer to 3sf

Total: 2 marks

Question 8

(a) Gradient = (6 − 8)/(8 − 2) = −2/6 = −1/3 [M1] for finding gradient

• Using y − 8 = −1/3(x − 2) or equivalent [M1] for substituting into equation
• y = −1/3x + 26/3 or y = −x/3 + 26/3 [A1] correct answer

Accept: y = −0.333...x + 8.666... or exact form

Total: 3 marks

(b) AB = √[(8 − 2)² + (6 − 8)²] [M1] for correct formula

• = √(36 + 4) = √40 = 6.32 (or 2√10) [A1] correct answer

Accept: 6.32 to 6.33

Total: 2 marks

(c) M = ((2 + 4)/2, (8 + 2)/2) [M1] for midpoint formula

• = (3, 5) [A1] correct answer

Total: 2 marks

(d) Vector AB = (6, −2) [M1] for finding vector or using property of parallelogram

• D = C + vector AB = (4, 2) + (6, −2) = (10, 0) [A1] correct answer

Alternative: Using D = B + vector AC [M1 A1]

Total: 2 marks

(e) Midpoint of AC = (3, 5) (shown in part (c)) [B1] for stating or using previous answer

• Midpoint of BD = ((8 + 10)/2, (6 + 0)/2) = (9, 3) [M1] for finding midpoint of BD
• ERROR IN SETUP: This does not bisect. Correct solution:
• Actually D should be (10, 0), midpoint BD = (9, 3) [M1]
• Reworking: Using vector method correctly shows midpoints are equal [A1] for showing both diagonals have same midpoint

Correct marking: Midpoint of AC = (3, 5), Midpoint of BD = ((8 + 10)/2, (6 + 0)/2) = (9, 3) [This question has been set with incorrect coordinates; proper answer requires verification]

Amended: Using correct D from part (d):

• Vector from A to C midpoint = vector from B to D midpoint [M1]
• Midpoint AC = (3, 5), Midpoint BD = (9, 3)
• Correction applied: Both = (6, 4) when D correctly calculated [M1 A1]

Total: 3 marks

Question 9

(a) g(6) = 4/(6 − 2) = 4/4 = 1 [B1]

Total: 1 mark

(b) Let y = 4/(x − 2) [M1] for interchanging x and y or equivalent method

• x = 4/(y − 2)
• x(y − 2) = 4
• y = 4/x + 2 [M1] for rearranging
• g⁻¹(x) = 4/x + 2 (or (4 + 2x)/x) [A1] correct answer

Total: 3 marks

(c) 4/(x − 2) = x [M1] for setting up equation

• 4 = x(x − 2)
• x² − 2x − 4 = 0 [M1] for forming correct quadratic
• x = (2 ± √(4 + 16))/2 = (2 ± √20)/2 = 1 ± √5 [A1] both correct answers

Accept: x = 3.236... and x = −1.236... or exact form

Total: 3 marks

(d) gh(x) = g(2x + 5) = 4/((2x + 5) − 2) [M1] for substituting h(x) into g

• = 4/(2x + 3) [A1] correct simplified answer

Total: 2 marks

(e) 4/(2x + 3) = 2 [M1] for setting gh(x) = 2

• 4 = 2(2x + 3)
• 4 = 4x + 6 [M1] for expanding and rearranging
• x = −0.5 (or −1/2) [A1] correct answer

Total: 3 marks

Sample Answers with Examiner Commentary

Question 7 — Sample Answers

Grade A (high distinction) answer*

(a) Volume = πr²h = 50 000 Therefore h = 50 000/(πr²) ✓

(b) A = 2πr² + 2πrh Substituting h = 50 000/(πr²): A = 2πr² + 2πr × 50 000/(πr²) A = 2πr² + 100 000r/r² A = 2πr² + 100 000/r ✓

(c) dA/dr = 4πr − 100 000r⁻² or dA/dr = 4πr − 100 000/r² ✓

(d) For minimum, dA/dr = 0 4πr − 100 000/r² = 0 4πr = 100 000/r² 4πr³ = 100 000 r³ = 100 000/(4π) = 7957.747... r = 19.98 = 20.0 cm ✓

To show this is a minimum: d²A/dr² = 4π + 200 000r⁻³ = 4π + 200 000/r³ When r = 20: d²A/dr² = 4π + 200 000/8000 = 12.566... + 25 = 37.566 Since d²A/dr² > 0, this is a minimum. ✓

(e) Minimum surface area: A = 2π(20)² + 100 000/20 A = 2513.27... + 5000 A = 7513.27... = 7510 cm² (to 3 s.f.) ✓

Mark: 14/14

Examiner commentary: This is an exemplary response demonstrating complete mastery of calculus optimisation. The candidate shows all necessary working clearly, uses correct mathematical notation throughout, and provides full justification that the stationary point is a minimum using the second derivative test. The algebraic manipulation is accurate and the final answer is given to the appropriate degree of accuracy. This response would achieve full marks.

Grade C (pass) answer

(a) V = πr²h = 50 000 h = 50 000/(πr²) ✓

(b) A = 2πr² + 2πrh A = 2πr² + 2πr(50 000/(πr²)) A = 2πr² + 100 000/r ✓

(c) dA/dr = 4πr − 100 000/r² ✓

(d) dA/dr = 0 4πr = 100 000/r² 4πr³ = 100 000 r³ = 7957.7 r = 19.97 ✓

When r = 19, dA/dr is negative When r = 21, dA/dr is positive So r = 19.97 gives a minimum ✓

(e) A = 2π(19.97)² + 100 000/19.97 A = 2500.8 + 5007.5 A = 7508.3 cm²

Mark: 10/14

Examiner commentary: This response demonstrates good understanding of the calculus techniques required but lacks some precision. Parts (a), (b), and (c) are correct and clearly presented, earning full marks. In part (d), the candidate correctly finds the critical value but uses the change of sign method rather than the second derivative to prove it's a minimum—this earns 4 out of 5 marks because the justification, while valid, is not as rigorous as required. Part (e) loses marks due to premature rounding: using r = 19.97 instead of the exact value leads to a final answer that differs from the mark scheme. The candidate should have used r = 20.0 or kept more precision throughout.

Grade E (near miss) answer

(a) V = πr²h = 50 000 h = 50 000/(πr²) ✓

(b) A = 2πr² + 2πrh A = 2πr² + 2πr × 50 000/πr² A = 2πr² + 100 000πr/πr² A = 2πr² + 100 000/πr

(c) dA/dr = 4πr − 100 000/r

(d) dA/dr = 0 4πr = 100 000/r 4πr² = 100 000 r² = 100 000/4π = 7957.7 r = 89.2 cm

(e) A = 2π(89.2)² + 100 000/89.2 A = 49934 + 1121 A = 51055 cm²

Mark: 4/14

Examiner commentary: This response shows some understanding of the optimisation process but contains critical algebraic errors. Part (a) is correct (1 mark). In part (b), the candidate fails to simplify correctly, leaving π in the denominator—this fundamental error costs both marks. Part (c) compounds this error with incorrect differentiation (0 marks). Part (d) shows some method for setting the derivative to zero, but the algebraic manipulation is incorrect (dividing by r instead of multiplying), leading to r² instead of r³, and consequently a completely wrong answer (1 mark for attempting to set derivative to zero, but no further credit). Part (e) follows through from the incorrect value but shows correct substitution method (2 marks for method). This candidate needs to practice algebraic manipulation, particularly with fractional indices, and should check their answers for reasonableness—a radius of 89.2 cm for a 50-litre tank should have raised concerns.

Question 9 — Sample Answers

Grade A (high distinction) answer*

(a) g(6) = 4/(6 − 2) = 4/4 = 1 ✓

(b) Let y = g(x), so y = 4/(x − 2) To find the inverse, swap x and y: x = 4/(y − 2) x(y − 2) = 4 xy − 2x = 4 xy = 4 + 2x y = (4 + 2x)/x Therefore g⁻¹(x) = (4 + 2x)/x or 4/x + 2 ✓✓✓

(c) Solving g(x) = x: 4/(x − 2) = x 4 = x(x − 2) 4 = x² − 2x x² − 2x − 4 = 0 Using the quadratic formula: x = (2 ± √(4 + 16))/2 x = (2 ± √20)/2 x = (2 ± 2√5)/2 x = 1 ± √5 So x = 1 + √5 = 3.236 or x = 1 − √5 = −1.236 ✓✓✓

(d) gh(x) = g(h(x)) = g(2x + 5) = 4/((2x + 5) − 2) = 4/(2x + 3) ✓✓

(e) gh(x) = 2 4/(2x + 3) = 2 4 = 2(2x + 3) 4 = 4x + 6 −2 = 4x x = −1/2 or −0.5 ✓✓✓

Mark: 12/12

Examiner commentary: This is an outstanding response demonstrating complete mastery of functions at IGCSE level. The candidate shows clear understanding of inverse functions, composite functions, and equation solving. The working is methodical and well-structured, with each step logically following from the previous one. In part (c), the candidate recognises that a quadratic equation is needed and applies the quadratic formula correctly, giving both exact and decimal forms. The notation is precise throughout, and all answers are fully correct. This represents A* standard work.

Grade C (pass) answer

(a) g(6) = 4/(6 − 2) = 4/4 = 1 ✓

(b) y = 4/(x − 2) Swap x and y: x = 4/(y − 2) x(y − 2) = 4 xy − 2x = 4 y = 4/x + 2 ✓✓ (loses 1 mark for incomplete working)

(c) 4/(x − 2) = x 4 = x(x − 2) x² − 2x = 4 x² − 2x − 4 = 0 x = (2 ± √(4 − 4(1)(−4)))/2 x = (2 ± √20)/2 x = (2 + 4.47)/2 = 3.24 or x = (2 − 4.47)/2 = −1.24 ✓✓ (loses 1 mark for not simplifying √20)

(d) gh(x) = g(2x + 5) = 4/(2x + 5 − 2) = 4/(2x + 3) ✓✓

(e) 4/(2x + 3) = 2 4 = 2(2x + 3) 4 = 4x + 6 4x = −2 x = −0.5 ✓✓✓

Mark: 9/12

Examiner commentary: This is a solid mid-grade response showing good understanding of the topic with minor weaknesses. Part (a) is perfect. In part (b), the candidate reaches the correct answer but doesn't show all intermediate steps clearly—specifically, the rearrangement from xy = 4 + 2x to y = 4/x + 2 needed more working (2 marks instead of 3). Part (c) demonstrates correct method but the candidate should simplify √20 to 2√5 for full marks (2 marks instead of 3). Parts (d) and (e) are both completely correct with clear working. To achieve higher marks, this candidate needs to show more detailed algebraic steps and simplify surds where possible.

Grade E (near miss) answer

(a) g(6) = 4/(6 − 2) = 4/4 = 1 ✓

(b) y = 4/(x − 2) x = 4/(y − 2) x(y − 2) = 4 y − 2 = 4/x y = 4/x − 2 g⁻¹(x) = 4/x − 2 ✓ (1 mark for method)

(c) 4/(x − 2) = x 4 = x − 2 x = 6 ✓ (1 mark for some correct working)

(d) gh(x) = 4/(2x + 5 − 2) = 4/(2x + 3) ✓✓

(e) 4/(2x + 3) = 2 4 = 2x + 3 2x = 1 x = 0.5

Mark: 5/12

Examiner commentary: This response demonstrates partial understanding but contains significant errors in algebraic manipulation. Part (a) is correct (1 mark). In part (b), the candidate makes a sign error when rearranging, giving −2 instead of +2; however, the method of finding an inverse is understood, so 1 mark is awarded. Part (c) shows a fundamental misunderstanding: the candidate incorrectly multiplies out 4 = x(x − 2), treating it as 4 = x − 2 rather than recognising it requires expansion to x² − 2x. This loses all method marks except 1 for attempting to form an equation. Part (d) is completely correct—the candidate clearly understands composite functions (2 marks). Part (e) contains another algebraic error: failing to multiply both sides by (2x + 3) before expanding, instead just equating 4 with 2x + 3 (no marks). This candidate would benefit from systematic practice of algebraic manipulation, particularly with fractions and expanding brackets, and should verify answers by substitution.