DOMPT-D mock exam
OMPT-D mock exam — with sample questions & answers.
Six OMPT-D sample questions in the exam's open-ended style, every solution fully worked and open to read — no signup, no PDF hunt. Then a free full-length interactive mock at the real OMPT-D format: 21 questions, 180 minutes, scored by topic. Built for the maths-heavy programmes that typically ask for the OMPT-D — advanced STEM tracks at universities like Radboud, Maastricht and UvA.
The answers below are free for everyone. The full mock needs a free account — takes a minute, no card.
Sample OMPT-D questions with answers.
Six original questions in the exam's open-ended style, spread like the official topic weighting — two in the heaviest group (functions, equations and inequalities, 30%), one each in differentiation, geometry, trigonometry and integration. Solutions fully worked, right below each question.
1
Exponentials & logarithms · equations
Solve for x: log₂(x) + log₂(x − 6) = 4.
Worked solution
- Domain first: both logarithms must have positive arguments, so x > 0 and x − 6 > 0, i.e. x > 6.
- Combine the logarithms: log₂(x(x − 6)) = 4, so x(x − 6) = 2⁴ = 16.
- Expand and solve the quadratic: x² − 6x − 16 = 0, which factors as (x − 8)(x + 2) = 0, giving x = 8 or x = −2.
- Apply the domain: x = −2 is rejected (it is not greater than 6). Check x = 8: log₂(8) + log₂(2) = 3 + 1 = 4. ✓
Answer: x = 8.
Note the shape of an OMPT-D question: the log laws are the surface, but the marks hide in the domain check and in the quadratic — a prerequisite chapter, examined rather than assumed.
2
Functions · inverse functions
Let f(x) = (2x + 1)/(x − 3) for x ≠ 3. Find a formula for the inverse function f⁻¹(x).
Worked solution
- Write y = (2x + 1)/(x − 3) and solve for x. Multiply both sides by (x − 3): y(x − 3) = 2x + 1.
- Expand and collect the x-terms on one side: xy − 3y = 2x + 1, so xy − 2x = 3y + 1.
- Factor out x: x(y − 2) = 3y + 1, hence x = (3y + 1)/(y − 2).
- Swap the variable name: f⁻¹(x) = (3x + 1)/(x − 2), defined for x ≠ 2. Quick check: f(4) = 9/1 = 9 and f⁻¹(9) = 28/7 = 4. ✓
Answer: f⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2.
3
Differentiation · chain rule & tangent lines
Let f(x) = √(3x² + 4). Find the equation of the tangent line to the graph of f at x = 2.
Worked solution
- The point of tangency: f(2) = √(3 · 4 + 4) = √16 = 4, so the line passes through (2, 4).
- Differentiate with the chain rule: f(x) = (3x² + 4)^(1/2), so f′(x) = ½(3x² + 4)^(−1/2) · 6x = 3x/√(3x² + 4).
- The slope at x = 2: f′(2) = 6/√16 = 6/4 = 3/2.
- Point–slope form: y − 4 = (3/2)(x − 2), which simplifies to y = (3/2)x + 1.
Answer: y = (3/2)x + 1.
On the real test you would type the exact fraction 3/2, not 1.5 — the CAS checks form, and working in exact values is the habit to build.
4
Trigonometry · equations
Solve 2sin²(x) − sin(x) − 1 = 0 for 0 ≤ x < 2π. Give exact answers.
Worked solution
- Substitute s = sin(x): the equation becomes 2s² − s − 1 = 0.
- Factor: (2s + 1)(s − 1) = 0, so s = −1/2 or s = 1.
- Case sin(x) = 1: on [0, 2π) the only solution is x = π/2.
- Case sin(x) = −1/2: sine is −1/2 in the third and fourth quadrants, with reference angle π/6, giving x = π + π/6 = 7π/6 and x = 2π − π/6 = 11π/6.
Answer: x = π/2, x = 7π/6, x = 11π/6.
5
Integration · substitution
Evaluate the integral of x(x² + 1)³ with respect to x, between x = 0 and x = 2.
Worked solution
- Substitute u = x² + 1, so du = 2x dx, i.e. x dx = ½ du.
- Change the limits: x = 0 gives u = 1; x = 2 gives u = 5.
- The integral becomes ½ ∫ u³ du from u = 1 to u = 5, which is ½ · [u⁴/4] from 1 to 5.
- Evaluate: (5⁴ − 1⁴)/8 = (625 − 1)/8 = 624/8 = 78.
Answer: 78.
6
Geometry · circles & lines
A circle has equation x² + y² − 6x + 4y − 12 = 0. Find its centre and radius, and the length of the chord that the line 3x + 4y − 21 = 0 cuts off on the circle.
Worked solution
- Complete the square: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4, so (x − 3)² + (y + 2)² = 25. Centre (3, −2), radius 5.
- Distance from the centre to the line: |3(3) + 4(−2) − 21| / √(3² + 4²) = |9 − 8 − 21| / 5 = 20/5 = 4.
- Since 4 < 5, the line cuts the circle. Half the chord, by Pythagoras in the radius–distance–half-chord triangle: √(5² − 4²) = √9 = 3.
- The full chord is twice that: 2 · 3 = 6.
Answer: centre (3, −2), radius 5, chord length 6.
What these six are — and aren't. They are original questions written to the chapters of the official OMPT-D syllabus, at a realistic multi-step level; the real exam's items are confidential and these are not them. If they felt comfortable, the open question is whether you stay that accurate for 21 of them in a row, on a clock, typing every answer — which is exactly what the mock measures.
The sample PDF vs an interactive mock.
Search for an OMPT-D mock exam and most results lead to the same document: the official Sample of the OMPT-A Mock Exam — 8 questions over 3 pages, with space to write your answers by hand — re-uploaded across third-party document sites, sometimes behind a signup. It's a fair first look at the question style, but it samples the OMPT-A syllabus, not the OMPT-D calculus track.
More importantly, the real OMPT-D isn't a paper test. You type your answers into a platform that checks them symbolically, with only an on-screen scientific calculator and a countdown running. A static PDF trains none of that: no timer, no maths-entry practice, no calculator constraint, no per-topic score at the end. The official omptest.org sells its own practice modules and mock test; ours is simply free, full-length at the OMPT-D's real format, and corrected instantly.
Whatever materials you use, do at least one full-length, timed run before booking the €270 attempt. Pacing and typed entry are skills, and a PDF can't train them.