Cambridge IGCSE Additional Mathematics 0606/22 is the May/June 2026 Paper 2, variant 2, the paper on which a calculator is allowed. It lasts 2 hours, is worth 80 marks and is sat by every 0606 candidate. It had twelve compulsory questions, and the biggest, Question 8 on an exponential function and its log inverse, carried 12 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.

0606/22 at a glance

Question	Topic	Marks
1	Completing the square for a minimum point, a parallel line	4
2	Discriminant: showing a quadratic never has equal roots	3
3	Differentiating a root and a cosine, integrating sec²	6
4	Sketch of y = |x² − 5|, finding f(x) from a modulus equation	7
5	Permutations of 6-digit even numbers, choosing a committee	6
6	Binomial expansion of a product: values of a, b and c	8
7	Two boats with velocity vectors: positions and distance apart	8
8	Functions: range, log inverse, sketches, a composite function	12
9	Kinematics with exponentials: stationary velocity, acceleration	7
10	Trig equation in radians with tan and sin of (3x + 1)	6
11	Two circles and a common chord: radius of the second circle	9
12	Connected rates of change for a sphere	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), awrt (answers which round to) and SC (special case, a fixed number of marks for a partly correct answer).

Question 1: completing the square (4 marks)

The curve is y = x² − 6x + 10, with minimum point A.

• Coordinates of A (M1, A1): y = (x − 3)² + 1, so A is (3, 1). The answer had to come from completing the square; a minor slip in the completed square with (3, 1) still earned SC1.
• Line through B(6, 7) parallel to AC (M1, A1): C, where the curve meets the y-axis, is (0, 10). The gradient of AC is (10 − 1) ÷ (0 − 3) = −3, so the line is y − 7 = −3(x − 6), or y = 25 − 3x. Both marks could follow through from your A. Finding the perpendicular gradient scored nothing, and writing only “y = −3x + c with c = 25” lost the A mark because it is not a complete equation.

Question 2: no equal roots (3 marks)

Show that (k − 2)x² + (2k + 3)x + 3 = 0 never has two equal roots for real k.

• M1: the discriminant: (2k + 3)² − 4(k − 2)(3).
• A1: simplify to 4k² + 33.
• A1: a clear argument: equal roots need 4k² + 33 = 0, which has no real solution because k² cannot be negative. The ”= 0” had to be stated or clearly implied, and the conclusion must not be contradicted.

Question 3: differentiation and integration (6 marks)

• Differentiate y = √(x + 1) + cos(5x + 3) (3 marks): dy/dx = ½(x + 1)^(−½) − 5sin(5x + 3). One mark was for the first term and two for the second (one for k sin(5x + 3) with a negative k). For all three marks the two terms had to be combined in one expression. Adding ”+ c” to a derivative capped the answer at 2 marks, and there was no follow through from a wrong expression.
• Integrate sec²(x/3) (3 marks): 3tan(x/3) + c. M1 for k tan(x/3), A1 for the 3, and the final mark needed the + c.

Question 4: modulus graphs (7 marks)

• Sketch y = |x² − 5| with intercepts (B3): a W shape: the parts of the parabola below the x-axis are reflected above it, giving two cusps on the x-axis at x = ±√5 and a local maximum on the y-axis at (0, 5). B2 was for the shape (two cusps roughly equidistant from O, the maximum on the y-axis, correct curvature in the middle and on the outer branches) and B1 for the intercepts.
• Find f(x) = ax + b when |f(x)| = 4 at x = 2.5 and x = −1.5 (4 marks): f(x) is 4 at one value and −4 at the other, so for example 2.5a + b = 4 and −1.5a + b = −4 (B1). Eliminating (M1) gives a = 2, b = −1 or a = −2, b = 1 (A1). The final mark needed both, and no others: f(x) = 2x − 1 or f(x) = −2x + 1 (A1). Another route used the vertex of y = |f(x)| at (0.5, 0) and the gradient from there to (2.5, 4).

Question 5: permutations and combinations (6 marks)

• 6-digit even numbers with six different digits, not starting with 0 (3 marks): split by the last digit. Ending in 0: 9 × 8 × 7 × 6 × 5 = 15 120. Ending in 2, 4, 6 or 8: 4 choices for the last digit, 8 for the first (not 0, not the last digit), then 8 × 7 × 6 × 5, giving 4 × 8 × 8 × 7 × 6 × 5 = 53 760. Total 68 880 (M2, A1). The scheme accepted several other correct splits, and gave M1 for any one correct part of the count.
• A committee of 10 from 9 students and 7 teachers, with at least 4 of each (3 marks): the options are 4 students and 6 teachers, 5 and 5, or 6 and 4. That is ⁹C₄ × ⁷C₆ + ⁹C₅ × ⁷C₅ + ⁹C₆ × ⁷C₄ = 882 + 2646 + 2940 = 6468 (M2, A1). Taking the cases that break the rule away from ¹⁶C₁₀ = 8008 was also accepted. M1 was for any two correct products.

Question 6: binomial expansion of a product (8 marks)

The first three terms of (a + x)⁴(1 + ax)⁵ are 16 + bx + cx², with a, b and c integers.

• The two expansions (B1 each): (a + x)⁴ starts a⁴ + 4a³x + 6a²x², and (1 + ax)⁵ starts 1 + 5ax + 10a²x². Binomial coefficients had to be evaluated.
• a (M1, A1): a⁴ = 16, so a = 2 or a = −2.
• b (M1 FT, A1): b = 5a⁵ + 4a³, so b = 192 when a = 2 and b = −192 when a = −2.
• c (M1 FT, A1): c = 10a⁶ + 20a⁴ + 6a², which is 984 for both values of a.
• The question asked for all possible values, and using only a = 2 limited the answer to 6 marks.

Question 7: boats and velocity vectors (8 marks)

i is due east, j is due north; distances in km, time in hours. Boat A leaves P (2i + 5j) at 13 00 with velocity 8i + 8j.

• Position of A after 2 hours (B1): 2i + 5j + 2(8i + 8j) = 18i + 21j.
• Position after t hours (B1): (2 + 8t)i + (5 + 8t)j.
• Position of B at 17 00 (B3): B leaves Q (20i + 8j) at 15 00 on a bearing of 300° at speed 5√3. Its velocity is 5√3(sin 300°)i + 5√3(cos 300°)j = −7.5i + (5√3 ÷ 2)j, about −7.5i + 4.33j (B2 for both components, B1 for one). After 2 hours B is at 5i + (8 + 5√3)j. Swapping the components earned SC1.
• Distance apart at 17 00 (3 marks): A has sailed for 4 hours, so it is at 34i + 37j. The difference is 29i + (29 − 5√3)j, so the distance is √(29² + (29 − 5√3)²) = 35.4 km (M2, A1). The M mark could follow through from your own positions, as long as they were stated.

Question 8: functions (12 marks)

f(x) = 1 + 2ˣ for x ≥ −1.

• Range of f (B1): f(x) ≥ 1.5, since f(−1) = 1.5 and f is increasing. Writing x ≥ 1.5 was not accepted.
• Domain of f⁻¹ (B1 FT): x ≥ 1.5. This one had to be in terms of x.
• f⁻¹(x) (M1, A1): swap the variables and make y the subject: f⁻¹(x) = log₂(x − 1), or lg(x − 1) ÷ lg 2. M1 was for a full method that ends in a logarithm. Getting log₂(x + 1) from a sign slip kept the M mark; writing log₂x − 1 from the wrong order of operations did not.
• Sketch f and f⁻¹ (4 marks): f starts at (−1, 1.5), passes through (0, 2) and rises steeply; f⁻¹ is its reflection in y = x, starting at (1.5, −1) and passing through (2, 0). Marks were for the shape of f, drawing it only over the right domain, marking (0, 2), and the reflection (FT).
• Why gf exists (B1): the range of f (f(x) ≥ 1.5) lies inside the domain of g (x ≥ −1). This depended on a correct range in part (a).
• gf(x) (B1): (1 + 2ˣ)(1 + 2ˣ + k). Missing brackets lost this mark, even if they were put back in the next part.
• k when gf(3) = 126 (M1, A1): f(3) = 9, so 9(9 + k) = 126 and k = 5.

Question 9: kinematics with exponentials (7 marks)

s = e⁴ᵗ − 10e²ᵗ + 9.

• t when v is stationary (5 marks): v = 4e⁴ᵗ − 20e²ᵗ (B1) and dv/dt = 16e⁴ᵗ − 40e²ᵗ (B1). Setting dv/dt = 0 and solving for an exponential (M1), then e²ᵗ = 40/16 = 2.5 (M1 dep), gives t = ½ ln 2.5 ≈ 0.458 (A1). Setting v = 0 instead of dv/dt = 0 earned no method marks, because that finds when the particle is at rest, not when its velocity is stationary.
• Acceleration at t = ln 2 (M1, A1): a = 16(2⁴) − 40(2²) = 256 − 160 = 96 m s⁻².

Question 10: a trig equation in radians (6 marks)

Solve tan(3x + 1) = 5sin(3x + 1) for −π/3 < x < π/3.

• Rearrange (B1): write tan as sin ÷ cos, then factorise rather than divide by sin(3x + 1).
• The two cases (B1, B1): sin(3x + 1) = 0 or cos(3x + 1) = 1/5.
• Solve (M1, A2): 3x + 1 runs from 1 − π to 1 + π. That gives 3x + 1 = 0, π, 1.3694 or −1.3694, so x = −1/3, 0.714, 0.123 and −0.790 (0.714 is (π − 1)/3). A1 was for any two correct values, and A2 needed all four with no extras in the range.
• Accuracy: −0.789 was not accepted, because it is truncated rather than rounded to 3 significant figures. Working in degrees earned no method marks until the values were converted to radians.

Question 11: two circles and a common chord (9 marks)

C₁ has centre (3, 5) and passes through P(1, 4). C₂ has centre (12, 8). Their common chord is y = 19 − 3x. Accurate drawing was not accepted.

• Equation of C₁ (B2): r² = 2² + 1² = 5, so (x − 3)² + (y − 5)² = 5.
• Where the chord meets C₁ (M1, A1, A1): substitute y = 19 − 3x to get (x − 3)² + (14 − 3x)² = 5, which simplifies to 10x² − 90x + 200 = 0.
• Solve (M1 dep, A1, A1): x² − 9x + 20 = 0 gives x = 4 and x = 5, so the circles meet at (4, 7) and (5, 4).
• Radius of C₂ (A1): the distance from (12, 8) to (4, 7) is √(8² + 1²) = √65 (about 8.06). This last mark depended on all the earlier marks, and a final answer of ±√65 scored A0.
• The scheme also gave full routes by subtracting the two circle equations, by Pythagoras with the line joining the centres (which meets the chord at its midpoint (4.5, 5.5)), and by solving the two circle equations together.

Question 12: connected rates of change (4 marks)

A sphere has volume V and radius r, and dV/dt = 9 dr/dt at some instant.

• M1: the chain rule, dV/dt = (dV/dr) × (dr/dt), which means dV/dr = 9.
• M1 dep: dV/dr = 4πr², so 4πr² = 9.
• A2: r = 3 ÷ (2√π) for A1, then V = (4/3)πr³ = 9 ÷ (2√π) ≈ 2.54 cm³ for A2. Only the positive value of r counts. Rounding r early (to 0.84 or 0.85) loses the last mark through premature rounding.

Where the marks are on 0606/22

By topic, the 80 marks split like this:

Topic area	Questions	Marks
Calculus (derivatives, integral, kinematics, rates of change)	3, 9, 12	17
Functions (range, inverse, composite)	8	12
Coordinate geometry of circles	11	9
Binomial expansion	6	8
Vectors in context	7	8
Quadratics (completing the square, discriminant)	1, 2	7
Modulus functions and graphs	4	7
Trigonometric equations	10	6
Permutations and combinations	5	6
Total		80

• Question 8 is worth 12 marks, and 4 of them are for one sketch of f and its inverse.
• Question 11 is worth 9 marks, and every mark but the last comes before the radius, so a full method with a slip still scores well.
• Questions 6 and 7 are worth 8 marks each. In Question 6, the second value of a was worth 2 marks on its own.
• Accuracy rules matter on this paper. Questions 7(d) and 10 end in decimal answers that must be correct to 3 significant figures, and Question 10 is set in radians.

How to self-mark 0606/22 with the mark scheme

1. Mark the method first. Find the M marks for each part and check whether your method matches, even if your final value is wrong. In Question 9(a), four of the five marks come before t = 0.458.
2. Award A marks only with their M mark, and check “dep” marks, such as the second M in Question 12, which needs the chain rule first.
3. Use follow through carefully. In Question 7(d) the distance follows through from your positions of A and B, and in Question 6 the b and c marks follow through from your value of a. But Question 3(a) had no follow through from a wrong expression, and Question 8(c)(i) needed a correct range.
4. Check accuracy against the cover rule. Round, don’t truncate: −0.789 in Question 10 is wrong. Avoid early rounding, as in Question 12.
5. Check final-answer conditions. ±√65 in Question 11, a ”+ c” on a derivative in Question 3 and a missing second value of a in Question 6 all cost marks even with correct working. The scheme also lists cao (correct answer only) for marks where no other form is accepted.
6. Log each lost mark as missing knowledge, a method error, an accuracy slip or a careless error, and revise in that order.

Sit 0606/22 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	4	6 minutes
Question 2	3	4 minutes
Question 3	6	9 minutes
Question 4	7	10 minutes
Question 5	6	9 minutes
Question 6	8	12 minutes
Question 7	8	12 minutes
Question 8	12	17 minutes
Question 9	7	10 minutes
Question 10	6	9 minutes
Question 11	9	13 minutes
Question 12	4	6 minutes
Check rounding and calculator mode	–	3 minutes

Switch your calculator to radians before Question 10, then check every final answer is to 3 significant figures.

Other May/June 2026 0606 papers

• 0606 Paper 12 (Non-calculator)

Practise and download

• Download 0606/22 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 22 May/June 2026? +

Twelve questions: completing the square, a discriminant proof, differentiation and integration, the modulus of a quadratic, permutations and combinations, a binomial expansion, boats with velocity vectors, functions with an exponential and its log inverse, kinematics with exponentials, a trig equation in radians, two circles with a common chord, and connected rates of change.

How many marks is 0606/22 and how long is it? +

It is out of 80 marks and lasts 2 hours, and a scientific calculator is allowed. Question 8, on functions, is the biggest at 12 marks.

What was the radius of circle C2 in 0606/22 Question 11? +

Circle C1 has equation (x − 3)² + (y − 5)² = 5. Substituting y = 19 − 3x gives x = 4 or 5, so the circles meet at (4, 7) and (5, 4). The distance from (12, 8) to either point is √65, about 8.06. Writing ±√65 as the final answer lost the last mark.

What were the solutions to the trig equation in 0606/22 Question 10? +

Rearranging tan(3x + 1) = 5sin(3x + 1) gives sin(3x + 1) = 0 or cos(3x + 1) = 1/5. In the range −π/3 < x < π/3 that gives x = −1/3, 0.714, 0.123 and −0.790. The answer −0.789 was not accepted because it is truncated, not rounded.

How accurate do answers need to be on 0606 Paper 2? +

The front cover asks for non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise. For π you can use your calculator value or 3.142.

Where can I download 0606/22 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.

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