In the Milliy sertifikat maths paper tasks 36-45 are open: each has one condition and two questions - a) worth 1.5 points and b) worth 1.7. You do not choose from options, you write the number. The parts are marked separately, so a correct a) alone earns 1.5 points. Together the ten tasks are worth 32 points, nearly a third of the 100 raw points. Below: the topics, how to write an answer, worked examples and the mistakes that cost the most.
How the open part is built
In total 10 × (1.5 + 1.7) = 32 points. A closed task (1-35) is worth 1.3 or 2.2, so one open task (3.2 points) is worth more than two medium MCQs. Part b) usually goes one step beyond a): if a) asks for a root, b) asks for the product or sum of the roots.
| Task | Topic (from the specification) | Points |
|---|---|---|
| 36-37 | Equations and inequalities | 1.5 + 1.7 |
| 38 | Functions | 1.5 + 1.7 |
| 39-40 | Calculus (derivative, integral) | 1.5 + 1.7 |
| 41-44 | Geometry (plane and solid) | 1.5 + 1.7 |
| 45 | Modelling (word problem) | 1.5 + 1.7 |
How it is marked
- a) and b) are marked independently: if one is right and the other wrong, you get the points for the right one.
- Only the final number is checked, not your working. A small slip in an intermediate step makes the answer wrong.
- Wrong answers lose nothing - even when time runs out, write your answer.
- The answer can be an integer, a decimal or a fraction. Read the instructions of the official sample paper beforehand to see how answers are written on the real exam.
How to write the answer
- Write only what is asked: for «the product of the roots» give one number, not the roots.
- Uzbek papers use a decimal comma: 4,8. On Trylab both 4,8 and 4.8 are accepted.
- A fraction: 16/3. Trylab accepts 16/3, 5.3333 or 5.33 - a rounded answer counts while it is within 1% of the key.
- Do not round intermediate results: round only at the end, to the requested accuracy.
- If the condition names a unit (cm, m², %), work in it, but usually only the number goes in the answer.
Worked example: absolute value equation
Condition: |x - 1| + |x - 4| = 7. a) Find the smallest root. b) Find the product of the roots.
Solution: the points 1 and 4 split the line into three intervals. For x < 1: -2x + 5 = 7, x = -1. For 1 ≤ x ≤ 4 the expression equals 3, not 7 - no root. For x > 4: 2x - 5 = 7, x = 6.
Answer: a) -1; b) -1 · 6 = -6.
Worked example: rational inequality
Condition: (x + 2)(x - 3) / ((x - 1)(x - 5)) ≤ 0. a) Find the number of integer solutions. b) Find their sum.
Solution with the interval method: points -2, 1, 3, 5. Zeros of the denominator (1 and 5) are excluded. Signs from right to left: +, -, +, -, +. Solution: [-2; 1) ∪ [3; 5).
Integer solutions: -2, -1, 0, 3, 4. Answer: a) 5; b) -2 - 1 + 0 + 3 + 4 = 4.
Worked examples: calculus and geometry
- f(x) = x³ - 3x² - 9x + 5 on [-2; 4]. f′(x) = 3x² - 6x - 9 = 3(x - 3)(x + 1), critical points -1 and 3. f(-2) = 3, f(-1) = 10, f(3) = -22, f(4) = -15. a) Maximum 10. b) Minimum -22. Forgetting the endpoints is the classic mistake.
- A right triangle with legs 6 and 8. a) Hypotenuse: √(36 + 64) = 10. b) Height to the hypotenuse: 6 · 8 : 10 = 4.8.
- Modelling: a product cost 50,000 so‘m, the price rose 20%, then fell 20%. a) After the rise: 60,000. b) At the end: 60,000 · 0.8 = 48,000 - it does not return to the start.
The most common mistakes
- Writing the a) answer in b) or the other way round: read the question to the end.
- Not checking the domain: logarithms, roots and denominators create extraneous roots.
- After a substitution (t = (x - 1)²) keeping a negative t or forgetting to substitute back.
- Giving the point (x) when the value of the function is asked.
- Rounding an intermediate result and moving the final answer by more than 1%.
- Losing a sign: a minus before brackets, opening an absolute value.
Timing and strategy
- Keep 50-60 minutes for the open part, about 80-90 of the 150 minutes for the MCQs.
- Go through all the a) parts first: they are easier and worth 1.5 each.
- If a task holds you more than 8-10 minutes, move on and come back at the end.
- Check by substitution: putting a root back into the equation takes 30 seconds.
