Cambridge IGCSE Art and Design 0400 — 2024 Feb/March Paper 2 · Variant 0

This document has 12 pages. Blank pages are indicated. © UCLES 2017 [Turn over Cambridge O Level ADDITIONAL MATHEMATICS 4037/02 Paper 2 For examination from 2020 MARK SCHEME Maximum Mark: 80 Specimen

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 2 of 12 © UCLES 2017 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: •• the specific content of the mark scheme or the generic level descriptors for the question •• the specific skills defined in the mark scheme or in the generic level descriptors for the question •• the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: •• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate •• marks are awarded when candidates clearly demonstrate what they know and can do •• marks are not deducted for errors •• marks are not deducted for omissions •• answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen).

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 3 of 12 © UCLES 2017 GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind. MARK SCHEME NOTES The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, given for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations AG answer given awrt answer which rounds to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC special case soi seen or implied

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 4 of 12 © UCLES 2017 Question Answer Marks Partial Marks 1 (xy)4y = 486 oe 81y = 486 M1 For elimination of one variable and attempt to solve y = 6, x = 2 1 A2 A1 for each Question Answer Marks Partial Marks 2(a) –5 0 2 4 10 y x 2 B1 for shape B1 for intercepts on coordinate axes 2(b) Valid explanation, e.g. multiplying throughout by 5 does not change x values because the x-axis is invariant. 1 2(c) x ⩽ –5 2 ⩽ x ⩽ 4 1 FT their graph

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 5 of 12 © UCLES 2017 Question Answer Marks Partial Marks 3(a) y = 2 + 4 ln x ln x = y 4 2 − M1 Complete method to find the inverse g–1(x) = e x 4 2 − A1 Must be in a correct form Domain x ∈ ℝ B1 Must be in a correct form Range y > 0 B1 Must be in a correct form 3(b) g(x2 + 4) = 10 M1 For correct order 2 + 4 ln (x2 + 4) = 10 M1 dep For attempt to solve Leading to x = 1.84 or e 4 2 − only A1 One solution only 3(b) alternative h(x) = x2 + 4 = g–1 (10) M1 For correct order g–1 (10) = e2, so x2 + 4 = e2 M1 dep For attempt to solve Leading to x = 1.84 only A1 One solution only 3(c) x 4 = 2x B1 For given equation, allow in this form x2 = 2 M1 For attempt to solve, must be using derivatives x = 2 A1 One solution only, allow 1.41 or better

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 6 of 12 © UCLES 2017 Question Answer Marks Partial Marks 4 0 –1 20° 100° y x 4 B1 for shape B1 for 20° B1 for 100° B1 for (0, –1) Question Answer Marks Partial Marks 5(a)(i) 9P6 = 60 480 B1 Must be evaluated 5(a)(ii) 4P2 × 3P2 × 2 M1 For attempt at a product of 3 permutations 144 A1 Must be evaluated 5(a)(iii) 840 × 2 B1 For either 840 or for realising that there are 2 possible positions for the symbols 1680 B1 5(b)(i) 10C6 × 5C3 M1 Unsimplified form 2100 A1 5(b)(ii) 8C4 × 4C2 M1 Unsimplified form 420 A1

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 7 of 12 © UCLES 2017 Question Answer Marks Partial Marks 6(a) 27 1 6(b) t2 = 8 ln 4 M1 Correct attempt to solve e t 8 2 , from correct working t = 3.33 or better A1 6(c) Acceleration = e e t 3 8 2 4 t t 8 8 2 2 2 # − a k 2 M1 for a correct attempt to differentiate using the chain rule A1 all correct When t = 1, a = 6.98 2 M1 for use of t = 1 in their acceleration Question Answer Marks Partial Marks 7(a) lg y = x2 lg b + lg A lg b = ± 0.21 B1 lg b = ±0.21 may be implied b = 0.617 allow 0.62, 10–0.21 B1 lg A = 0.94 allow 0.93 to 0.95 B1 A = 8.71 allow awrt 8.5 to 8.9 B1 7(a) alternative 5.37 or 100.73 = Ab 1.259 or 100.1 = Ab4 B2 For both equations, allow correct to 2 SF b3 = 10–0.63 b = 0.617 allow 0.62, 10–0.21 B1 A = 8.71 allow awrt 8.5 to 8.9 B1 7(b) x = 1.5, x2 = 2.25 M1 y = their A × their b .1 52 or lg y = lg their A + (1.52 lg their b) y = 2.93 allow awrt 2.9 or 3.0 A1 7(c) lg y = 0.301 or 2 = 8.71(0.617)x2 M1 2 = their A(their b)x2 or lg 2 = (lg their b)x2 + lg(their A) x = 1.74 Allow 3 or awrt 1.7, 1.8 A1

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 8 of 12 © UCLES 2017 Question Answer Marks Partial Marks 8(a) 2r sin θ B1 their (2r sin θ) + 2rθ = 20 M1 r = sin 10 i i + A1 AG 8(b) ( ) ( ) sin cos d dr 10 1 2 i i i i + + =- When θ = 6 r, d dr i = –17.8 4 M1 for a correct attempt to differentiate A2, 1, 0, –1 each error A1, allow awrt – 17.8 Answer only scores 0/4. Question Answer Marks Partial Marks 9(a) DB = 7a – b 1 Mark final answer; allow unsimplified 9(b) DA = 3a – b 1 Mark final answer; allow unsimplified 9(c) AX = λ(4a + b) 1 Mark final answer; allow unsimplified 9(d) Valid method of finding DX M1 FT their DA + AX DX = 3a – b + λ(4a + b) A1 Allow unsimplified 9(e) 3a – b + λ(4a + b) = μ(7a – b) M1 Equating their (3a – b + λ(4a + b)) and μ × their (7a – b) Attempts to solve their (3 + 4λ = 7μ) and their (–1 + λ = – μ) M1 dep on first M1 λ = 11 4 , μ = 11 7 A2 A1 for each

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 9 of 12 © UCLES 2017 Question Answer Marks Partial Marks 10(a)(i) 0 – 4 ln 5 3 B1 correct shape B1 through (0, –4) B1 through (ln5, 0) 10(a)(ii) k ⩽ –5 1 10(b) 2 1 loga 2 + 3 loga 2 – loga 2 or loga (2 2 1 × 23 × 2–1) oe M1 Condone one error 22 1 loga 2 oe A1 Answer only scores 0/2. 10(c) log9 4x = log log x 9 4 3 3 or log3 x = log log x 3 9 9 B1 soi log3 x – log x 2 4 3 = 1 or log x 2 1 9 ` j – log9 4x = 1 M1 log3 ( )x x 4 2 1 = log3 3 or log9 x x 4 2 = log9 9 oe M1 x = 36 A1

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 10 of 12 © UCLES 2017 Question Answer Marks Partial Marks 11(a)(i) cosec sin cosec sin sin sin 1 1 − − i i i i i i = M1 For using cosec θ = sin 1 i and either attempt to multiply top and bottom by sin θ or an attempt to combine terms in the denominator = sin 1 1 2 − i or sin sin sin 1 1 2 − i i i ^ h M1 dep For correct use of 1 – sin2 θ = cos2 θ = cos 1 2i = sec2 θ A1 Completion of proof 11(a)(i) alternative cosec sin cosec cosec cosec cosec 1 − − i i i i i i = = cosec cosec 1 2 2 − i i M1 For using sin θ = cosec 1 i and an attempt to combine terms in the denominator = cot cot 1 2 2 i i + M1 dep For use of 1 + cot2 θ = cosec2 θ = tan2 θ + 1 = sec2 θ A1 Completion of proof 11(a)(ii) cos2 φ = 4 1 , cos φ = ±2 1 or tan2 φ = 3, tan φ = ± 3 or sin2 φ = 4 3 , sin φ = ± 2 3 M1 For using (i) to obtain a value for either cos2 φ, tan2 φ or sin2 φ and then taking the square root φ = 60°, 120°, 240°, 300° A2 A1 for two correct values A1 for two further correct values and no extras in the range

4037/02 Cambridge O Level – Mark Scheme For examination SPECIMEN from 2020 Page 11 of 12 © UCLES 2017 Question Answer Marks Partial Marks 11(b) tan x 4 3 1 r + = c m , , x 6 4 6 7 4 6 13 4 − − − r r r r r r = M1 For correct order of operations, can be implied by x = 12 −r , , x 12 12 11 12 23 −r r r = c m A2 A1 for x = 11 12 r A1 for x = 23 12 r If there are extra solutions in the range in addition to the two correct ones then A1A0

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