Cambridge IGCSE Additional Mathematics 0606/12 is the May/June 2026 Paper 1 (Non-calculator), variant 2. It lasts 2 hours, is worth 80 marks and is sat by every 0606 candidate. It had thirteen compulsory questions, and the biggest, Question 12, asked for the exact area between a curve and a straight line for 11 marks. This guide goes through the paper question by question, with the final answers, brief working and how the official mark scheme split the marks. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of both 0606 papers from this series, see IGCSE Additional Maths 0606 May/June 2026 past papers: what came up and how it was marked.

0606/12 at a glance

Question	Topic	Marks
1	Graph of y = 4cos x − 2: sketch, amplitude, period, values of k	5
2	Perpendicular bisector, lengths in surd form	7
3	Cubic polynomial: remainder, linear factors, a related equation in u	7
4	Arithmetic progression: 81st and 100th terms, sum of a block of terms	5
5	Exact equation of a tangent to a ln curve	4
6	Circular measure: exact angle and exact radius	8
7	Show that a trig expression equals cos x ÷ tan x	3
8	Small changes using the quotient rule	5
9	Logarithms: a single log to base e, a log equation in base 5	7
10	Geometric progression: common ratio, first term, sum to infinity	8
11	Vectors: a point dividing a line in a ratio, a scalar multiple	6
12	Exact area between a curve and a straight line	11
13	Proof with combinations	4
	Total	80

Question-by-question walkthrough

The mark scheme uses four kinds of mark. M (method) marks are for a valid method, even if the arithmetic goes wrong. A (accuracy) marks are for a correct answer or step, and can only be given if the M mark they rely on has been earned. B marks are for a correct result on its own, independent of method. dep means a mark depends on an earlier one. FT (follow through) lets you keep marks after an earlier error. Other notes you will see: oe (or equivalent), isw (ignore subsequent working), soi (seen or implied), nfww (not from wrong working) and SC (special case, a fixed number of marks for an incomplete route).

Question 1: the graph of y = 4cos x − 2 (5 marks)

• Sketch for −360° ≤ x ≤ 360° (B2): two full cycles of a smooth cosine curve, with maximum 2 and minimum −6, crossing y = −2 at x = −270°, −90°, 90° and 270°. One mark was for the right shape with two cycles and either the right amplitude or those four points. Full marks also needed correct curvature at the ends and a continuous line.
• Amplitude (B1): 4.
• Period (B1): 360° (2π was also allowed).
• Values of k for at least two solutions (B1): −6 ≤ k ≤ 2. The line y = k must meet the curve at least twice, which happens anywhere from the minimum to the maximum, including the end values. Writing “k ≥ −6 or k ≤ 2” was rejected, because “or” describes every number. The mark could follow through from your own graph if you had at least one mark in part (a).

Question 2: a perpendicular bisector and surd lengths (7 marks)

A is (−2, 3) and B is (8, −3).

• Find c, where C(0, c) is on the perpendicular bisector of AB (3 marks, all B): the gradient of AB is −6/10 = −3/5, so the perpendicular gradient is 5/3; the midpoint is (3, 0); the line is y = (5/3)(x − 3), so at x = 0, c = −5.
• Find n, where BD = (√n ÷ 2) × AD and D is (2, −1) (4 marks): AD² = 4² + 4² = 32 (B1) and BD² = 6² + 2² = 40 (B1). Then √40 = (√n ÷ 2) × √32 (M1). Since √40 = 2√10 and √32 = 4√2, the ratio is √10 ÷ (2√2) = √5 ÷ 2, so n = 5 (A1). Without a calculator, simplifying the surds is the whole skill here.

Question 3: a cubic polynomial (7 marks)

p(x) = 6x³ − x² − 5x + 2.

• Remainder on division by x + 2 (B1): p(−2) = −48 − 4 + 10 + 2 = −40.
• Product of linear factors (4 marks): p(−1) = 0, so x + 1 is a factor (M1). Dividing gives the quadratic 6x² − 7x + 2 (M1 dep for two terms right, A1 for it all right), which factorises, so p(x) = (x + 1)(3x − 2)(2x − 1) (A1). Other first factors, such as 2x − 1 or 3x − 2, were equally fine. Special cases covered students who spotted the factors by trial: SC B4 if all three were found and written as a product, SC B3 if all three were found but not written as a product, SC B2 for two.
• Solve 6u⁶ − u⁴ − 5u² + 2 = 0 (2 marks): this is p(u²) = 0, so u² = −1, 2/3 or 1/2. u² = −1 has no real solutions, which leaves u = ±√(2/3) and u = ±1/√2. One mark (FT from your factors) was for any two correct solutions, and extra solutions from u² + 1 = 0 were condoned.

Question 4: an arithmetic progression (5 marks)

First term 400, common difference −2.

• The 81st and 100th terms (B1 each): u₈₁ = 400 + 80(−2) = 240 and u₁₀₀ = 400 + 99(−2) = 202.
• Sum of the 81st to 100th terms (3 marks): these 20 terms form their own AP, so the sum is (20/2)(240 + 202) = 4420. M2 was for this method, or for S₁₀₀ − S₈₀ = 30 100 − 25 680. An off-by-one subtraction, such as S₁₀₀ − S₈₁ or S₁₀₀ − S₇₉, still earned M1 if the values from part (a) were used correctly. Listing and adding all twenty terms to get 4420 earned SC B3.

Question 5: tangent to a ln curve (4 marks)

The curve is y = ln((2x − 4)⁵) and the tangent is at x = 3.

• Differentiate (B1): using the log law first, y = 5ln(2x − 4), so dy/dx = 10 ÷ (2x − 4) = 5 ÷ (x − 2).
• Gradient at x = 3 (M1): 5.
• y at x = 3 (B1): ln(2⁵) = ln 32 (or 5ln 2).
• Tangent (A1): y − ln 32 = 5(x − 3), or any equivalent exact form. A decimal for ln 32 would not be exact.

Question 6: circular measure (8 marks)

A circle with centre O and radius r; the shaded sector AOB has angle α radians.

• Exact α when the sector perimeter equals the major arc ACB (4 marks): the sector perimeter is 2r + rα (B1) and the major arc is r(2π − α) (B1). Setting them equal and cancelling r (M1) gives 2 + α = 2π − α, so α = π − 1 (A1). Both expressions had to be in radians, and the M mark needed an equation in arc lengths, not areas.
• Exact r when the sector area is 18(π² − 1) (4 marks): ½r²(π − 1) = 18(π² − 1) (M1), so r² = 36(π + 1) (A1), using π² − 1 = (π − 1)(π + 1). Taking the square root (M1 dep) gives r = 6√(π + 1) (A1). The question said “simplify”, and the final mark was only for this form.

Question 7: a trig “show that” (3 marks)

Show that cosec x − sin x can be written as cos x ÷ tan x.

• M1: write cosec x as 1 ÷ sin x, so the expression is (1 − sin² x) ÷ sin x.
• A1: use the identity to get cos² x ÷ sin x.
• A1: a fully justified final step, for example cos x ÷ (sin x ÷ cos x) = cos x ÷ tan x. The scheme wanted one more line after cos² x ÷ sin x before the given answer. Working from right to left was allowed. Leaving out the x throughout lost the last mark.

Question 8: small changes (5 marks)

y = (5x + 1) ÷ (x − 2), and y increases from 4 by a small amount h.

• Differentiate (2 marks): by the quotient rule, dy/dx = [5(x − 2) − (5x + 1)] ÷ (x − 2)² = −11 ÷ (x − 2)². One M mark was for the correct structure of the rule; a product-rule route was also accepted.
• x when y = 4 (B1): 5x + 1 = 4(x − 2), so x = −9.
• Approximate change (M1 dep, A1): δx ≈ h ÷ (dy/dx). At x = −9, dy/dx = −11 ÷ 121 = −1/11, so δx ≈ −11h. Substituting x = 4 (the y value) into dy/dx scored nothing for this step.
• An alternative was to make x the subject, x = (2y + 1) ÷ (y − 5), find dx/dy and use δx ≈ (dx/dy) × h at y = 4, which gave the same −11h.

Question 9: logarithms (7 marks)

• Write 1 ÷ logₓ e − ln(x − 1) as a single log to base e (2 marks): 1 ÷ logₓ e = ln x by the change-of-base rule (B1), so the answer is ln(x ÷ (x − 1)) (B2).
• Solve 2log₅ 2x = 1 + log₅(9x − 10) (5 marks): write 2log₅ 2x as log₅ 4x², or 1 as log₅ 5 (B1). Combine to get 4x² = 5(9x − 10) (M1, A1). This gives 4x² − 45x + 50 = 0, which factorises as (4x − 5)(x − 10) = 0 (M1 for solving a 3-term quadratic). So x = 5/4 or x = 10 (A1). Both are valid, and the A mark was lost if either was rejected. Turning 1 + log₅(9x − 10) into a product of logs earned no method mark.

Question 10: a geometric progression (8 marks)

The 2nd, 3rd and 4th terms are 11x − 1, x + 5 and x − 1, with x > 0.

• Common ratio and first term (6 marks): equal ratios give (x + 5) ÷ (11x − 1) = (x − 1) ÷ (x + 5) (M1). Expanding (M1 dep) gives 10x² − 22x − 24 = 0, or 5x² − 11x − 12 = 0 (A1). Solving (M1) gives (5x + 4)(x − 3) = 0, and x > 0 means x = 3 only (A1). The terms are 32, 8 and 2, so r = 1/4 and a = 128 (A1 for both). Setting up an equation in r instead was an equally valid route.
• k when the sum to infinity is k/3 (2 marks): S∞ = 128 ÷ (1 − 1/4) = 512/3, so k = 512. The M1 needed |r| < 1.

Question 11: vectors (6 marks)

OA = (4, −2) and OB = (−4, 6) as column vectors; C is on AB with AC : CB = 3 : 1; D is on OA with OD = m × OA and CD = (5, −5.5).

• Find OD (5 marks): AB = (−8, 8) (B1). OC = OA + (3/4)AB (M1) = (−2, 4) (A1). Then OD = OC + CD (M1 FT) = (3, −1.5) (A1). The final answer had to be written as a vector.
• Find m (B1): (3, −1.5) = (3/4)(4, −2), so m = 3/4. This mark was only available if part (a) had full marks.

Question 12: exact area between a curve and a line (11 marks)

The curve is y = 10 ÷ (4x + 1) and the line L passes through (0, 6) and (1.5, 0). The shaded region lies between them, with the line above the curve.

• Equation of L (B1): y = 6 − 4x.
• Intersection points (4 marks): set 10 ÷ (4x + 1) = 6 − 4x (M1 FT) to get 4x² − 5x + 1 = 0 (A1). Solve (M1 dep) to get x = 1/4 and x = 1 (A1).
• y values (B1): y = 5 and y = 2, so P is (1/4, 5) and Q is (1, 2).
• Area under the line (B1): a trapezium, ½ × (5 + 2) × 3/4 = 21/8. Integrating 6x − 2x² between the limits was also accepted.
• Area under the curve (4 marks): ∫ 10 ÷ (4x + 1) dx = (10/4)ln(4x + 1) (M1 for the form k ln(4x + 1), A1 for k = 10/4). Substituting the limits (M1 dep on all previous M marks) gives (5/2)(ln 5 − ln 2).
• Final answer (A1): 21/8 − (5/2)ln(5/2), or 21/8 − (5/2)ln 5 + (5/2)ln 2. The last mark was lost if the subtraction was the wrong way round.

Ten of the 11 marks come before the final answer, so a slip near the end still leaves most of the marks intact.

Question 13: a proof with combinations (4 marks)

Show that, for n ≥ 3, ⁿ⁺²C₃ − ⁿC₃ = n².

• M1: write both terms with factorials: (n + 2)! ÷ [(n − 1)! 3!] − n! ÷ [(n − 3)! 3!].
• A1: simplify to (n + 2)(n + 1)n ÷ 6 − n(n − 1)(n − 2) ÷ 6.
• A1: expand: n[(n² + 3n + 2) − (n² − 3n + 2)] ÷ 6.
• A1: n × 6n ÷ 6 = n². You had to work from the left-hand side to the right-hand side.

Where the marks are on 0606/12

By topic, the 80 marks split like this:

Topic area	Questions	Marks
Calculus (tangent, small changes, area)	5, 8, 12	20
Series (arithmetic and geometric progressions)	4, 10	13
Trigonometry (graph and identity)	1, 7	8
Circular measure	6	8
Coordinate geometry	2	7
Polynomials	3	7
Logarithms	9	7
Vectors	11	6
Permutations and combinations	13	4
Total		80

• Question 12 is worth 11 marks on its own, and it pulls together a straight line, a quadratic, a trapezium and a ln integral.
• Questions 6 and 10 are worth 8 marks each. Both need exact algebra: π − 1 and 6√(π + 1) in Question 6, a quadratic and a fraction in Question 10.
• The two proofs, Questions 7 and 13, carry 7 marks between them. In these, the working is the answer.

How to self-mark 0606/12 with the mark scheme

1. Mark the method first. For each part, find the M marks in the scheme and check whether your method matches, even if your final number is wrong. In Question 10, three M marks come before x = 3.
2. Award A marks only with their M mark. An A mark for a correct value needs the method behind it. A “dep” mark, such as the second M in Question 6(b), needs the earlier mark too.
3. Use follow through properly. If you made an early slip, carry your wrong value forward and check the FT marks, such as Question 11(a), where OD follows through from your OC. But note where FT is blocked: Question 11(b) needs full marks in part (a).
4. Check exact forms. On this paper, a decimal for π − 1 or for ln 32 is not the answer the scheme asks for. Simplified surds matter in Questions 2 and 6.
5. Check “show that” steps line by line. In Question 7 you need a line after cos² x ÷ sin x, and in Question 13 you must start from the left-hand side.
6. Look up the SC and “no other solutions” notes. Question 3 has special cases for trial-and-error factors; Question 9(b) loses the A mark if a valid root is rejected. The scheme also uses cao (correct answer only) and awrt (answers which round to) in its list of abbreviations.
7. Log each lost mark as missing knowledge, a method error, an exact-form slip or a careless arithmetic error, and revise in that order.

Sit 0606/12 as a timed mock

The paper allows 120 minutes for 80 marks, a minute and a half per mark. A workable plan:

Block	Marks	Time
Question 1	5	7 minutes
Question 2	7	10 minutes
Question 3	7	10 minutes
Question 4	5	7 minutes
Question 5	4	6 minutes
Question 6	8	12 minutes
Question 7	3	5 minutes
Question 8	5	7 minutes
Question 9	7	10 minutes
Question 10	8	12 minutes
Question 11	6	9 minutes
Question 12	11	16 minutes
Question 13	4	6 minutes
Check exact forms and signs	–	3 minutes

Sit it with no calculator at all, not even to check. Question 12 has two pages of working space; set out the line, the intersections, the trapezium and the integral as separate steps so each mark is easy to find.

Other May/June 2026 0606 papers

• 0606 Paper 22 (Calculator)

Practise and download

• Download 0606/12 and its mark scheme on the Learning Portal →
• All 0606 May/June 2026 papers: overview and marking patterns
• Try a free IGCSE Additional Mathematics quiz

Need help with a 0606 topic? Book a free trial lesson with an IGCSE Additional Mathematics tutor →

Frequently asked questions

What topics were on 0606 Paper 12 May/June 2026? +

Thirteen questions: a cosine graph, a perpendicular bisector and surd lengths, a cubic polynomial, an arithmetic progression, a tangent to a ln curve, circular measure, a trig identity, small changes, logarithms, a geometric progression, vectors, the exact area between a curve and a line, and a proof with combinations.

How many marks is 0606/12 and how long is it? +

It is out of 80 marks and lasts 2 hours. No calculator is allowed. Question 12, the exact area between a curve and a line, is the biggest at 11 marks.

What was the answer to the area question in 0606/12 Question 12? +

The line is y = 6 − 4x and it meets the curve at (1/4, 5) and (1, 2). The trapezium under the line has area 21/8 and the area under the curve is (5/2)(ln 5 − ln 2), so the shaded area is 21/8 − (5/2)ln(5/2).

What was the sum to infinity in 0606/12 Question 10? +

With x = 3 the terms are 32, 8 and 2, so the common ratio is 1/4 and the first term is 128. The sum to infinity is 128 ÷ (3/4) = 512/3, so k = 512.

Do you need a calculator for 0606 Paper 1? +

No. Calculators must not be used on Paper 1, so answers such as the angle π − 1, the radius 6√(π + 1) and the area with ln terms have to be left in exact form. Paper 2 (0606/22) is the calculator paper.

Where can I download 0606/12 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge IGCSE Additional Mathematics 0606 past papers page.

Ready to Excel in Your Studies?

Get personalised help from Tutopiya's expert tutors. Whether it's IGCSE, IB, A-Levels, or any other curriculum — we match you with the perfect tutor and your first session is free.

Book Your Free Trial