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100+ Free European Baccalaureate Mathematics (3-period course) Practice Questions

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2026 Statistics

Key Facts: European Baccalaureate Mathematics (3-period course) Exam

135 min

Total exam duration

OSGES European Schools Regulations

0–10

Grading scale

OSGES Marking Scheme

5.0 / 10

Minimum passing score

OSGES European Baccalaureate Rules

OSGES

Administering body

Office of the Secretary-General of the European Schools

S7 (Year 13)

Target education level

European Schools Secondary Cycle

100

Practice questions in this bank

OpenExamPrep

The European Baccalaureate Mathematics (3-period course) is administered by the Office of the Secretary-General of the European Schools (OSGES). Official assessment involves written/oral components graded on a 0-10 scale (passing score 5.0). Local questions on OpenExamPrep are an English-language MCQ study adaptation designed for syllabus revision.

Sample European Baccalaureate Mathematics (3-period course) Practice Questions

Try these sample questions to test your European Baccalaureate Mathematics (3-period course) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Evaluate the limit as x approaches infinity: lim_{x -> ∞} (6x^2 - 5x + 1) / (2x^2 + 3x - 4).
A.3
B.0
C.2
D.Infinity
Explanation: For rational functions where the degree of the numerator equals the degree of the denominator, the limit as x -> ∞ is the ratio of the leading coefficients. Dividing numerator and denominator by x^2 yields lim_{x -> ∞} (6 - 5/x + 1/x^2) / (2 + 3/x - 4/x^2) = 6 / 2 = 3.
2What is the derivative f'(x) of the polynomial function f(x) = 4x^3 - 5x^2 + 7x - 9?
A.12x^2 - 10x + 7
B.12x^2 - 5x + 7
C.4x^2 - 10x + 7
D.12x^3 - 10x^2 + 7x
Explanation: Applying the power rule d/dx(x^n) = n*x^(n-1) to each term gives: d/dx(4x^3) = 12x^2, d/dx(-5x^2) = -10x, d/dx(7x) = 7, and d/dx(-9) = 0. Combining these terms results in f'(x) = 12x^2 - 10x + 7.
3Find the derivative of the exponential function f(x) = e^(3x - 2).
A.e^(3x - 2)
B.3 * e^(3x - 2)
C.(3x - 2) * e^(3x - 3)
D.3 * e^(3x)
Explanation: Using the chain rule for exponential functions d/dx(e^u) = u' * e^u, where u = 3x - 2 and u' = 3, we get f'(x) = 3 * e^(3x - 2).
4If f(x) = ln(5x^2 + 2), what is f'(x)?
A.1 / (5x^2 + 2)
B.10x / (5x^2 + 2)
C.5x / (5x^2 + 2)
D.10x^2 / (5x^2 + 2)
Explanation: The derivative of ln(u) with respect to x is (1/u) * du/dx. Here u = 5x^2 + 2, so du/dx = 10x. Therefore, f'(x) = 10x / (5x^2 + 2).
5Find the slope of the tangent line to the curve y = x^3 - 3x + 2 at x = 2.
A.9
B.5
C.12
D.4
Explanation: The slope of the tangent line is given by the derivative dy/dx = 3x^2 - 3. Evaluating at x = 2 gives dy/dx|_(x=2) = 3(2)^2 - 3 = 12 - 3 = 9.
6What is the equation of the tangent line to y = x^2 - 4x + 5 at the point where x = 3?
A.y = 2x - 4
B.y = 2x - 1
C.y = 4x - 10
D.y = x - 1
Explanation: At x = 3, y = 3^2 - 4(3) + 5 = 2. The derivative is y' = 2x - 4, so at x = 3, slope m = 2(3) - 4 = 2. Using point-slope form: y - 2 = 2(x - 3) => y = 2x - 4.
7At what value of x does the quadratic function f(x) = x^2 - 6x + 8 achieve its local minimum?
A.x = 3
B.x = 6
C.x = 4
D.x = -3
Explanation: Setting the derivative f'(x) = 2x - 6 equal to zero gives 2x - 6 = 0 => x = 3. Since f''(x) = 2 > 0, x = 3 is a local minimum.
8Calculate the derivative of f(x) = (2x + 1) * e^x.
A.(2x + 1) * e^x
B.(2x + 3) * e^x
C.2 * e^x
D.(2x + 2) * e^x
Explanation: By the product rule d/dx(u * v) = u'v + uv', with u = 2x + 1 and v = e^x: f'(x) = 2 * e^x + (2x + 1) * e^x = (2 + 2x + 1)e^x = (2x + 3)e^x.
9Find f'(x) for the rational function f(x) = (3x - 1) / (x + 2).
A.7 / (x + 2)^2
B.5 / (x + 2)^2
C.(3x + 5) / (x + 2)^2
D.1 / (x + 2)^2
Explanation: Using the quotient rule d/dx(u/v) = (u'v - uv') / v^2: u = 3x - 1, u' = 3, v = x + 2, v' = 1. f'(x) = [3(x + 2) - (3x - 1)(1)] / (x + 2)^2 = (3x + 6 - 3x + 1) / (x + 2)^2 = 7 / (x + 2)^2.
10If f(x) = (3x^2 - 4)^4, what is the derivative f'(x)?
A.12x * (3x^2 - 4)^3
B.24x * (3x^2 - 4)^3
C.6x * (3x^2 - 4)^3
D.24x^2 * (3x^2 - 4)^3
Explanation: Applying the chain rule d/dx(u^4) = 4u^3 * u': u = 3x^2 - 4, so u' = 6x. Thus f'(x) = 4 * (3x^2 - 4)^3 * 6x = 24x * (3x^2 - 4)^3.

About the European Baccalaureate Mathematics (3-period course) Practice Questions

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