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100+ Free European Baccalaureate Mathematics (5-period course) Practice Questions

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Key Facts: European Baccalaureate Mathematics (5-period course) Exam

OSGES

Office of the Secretary-General of the European Schools

European Schools Governance

210 Mins

Total written exam duration (Part A + Part B)

OSGES EB Exam Regulations

Min 5.0

Minimum passing score out of 10.0

European Baccalaureate Marking Scale

5 Blocks

Analysis, Integrals/DEs, 3D Geometry, Probability, Complex Numbers

EB S6-S7 Math 5P Syllabus

100

Practice questions available in this OpenExamPrep bank

OpenExamPrep

The European Baccalaureate Mathematics (5-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 (5-period course) Practice Questions

Try these sample questions to test your European Baccalaureate Mathematics (5-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 trigonometric limit: lim (x -> 0) [ sin(3x) / x ].
A.3
B.0
C.1
D.1/3
Explanation: Using the standard limit lim (u -> 0) [ sin(u) / u ] = 1, we rewrite sin(3x)/x as 3 * [ sin(3x) / (3x) ]. As x -> 0, 3x -> 0, so the limit equals 3 * 1 = 3.
2Find the derivative of the function f(x) = x^3 * e^(2x).
A.(3x^2 + 2x^3) * e^(2x)
B.6x^2 * e^(2x)
C.3x^2 * e^(2x)
D.(3x^2 + x^3) * e^(2x)
Explanation: By the product rule d/dx[u v] = u'v + uv', with u = x^3 and v = e^(2x), we get f'(x) = 3x^2 e^(2x) + x^3 (2e^(2x)) = (3x^2 + 2x^3) e^(2x).
3Find the equation of the tangent line to the curve y = x^2 - 4x + 5 at the point where x = 3.
A.y = 2x - 4
B.y = 2x - 1
C.y = 3x - 7
D.y = x - 1
Explanation: At x = 3, y = 3^2 - 4(3) + 5 = 2. Slope m = y'(3) = 2(3) - 4 = 2. Tangent line: y - 2 = 2(x - 3) => y = 2x - 4.
4Find the critical points of the function f(x) = 2x^3 - 9x^2 + 12x + 1.
A.x = 1 and x = 2
B.x = -1 and x = -2
C.x = 0 and x = 3
D.x = 3/2 and x = 2
Explanation: Setting f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2) = 0 gives critical points x = 1 and x = 2.
5Determine the vertical asymptote of f(x) = (2x + 1) / (x - 4).
A.x = 4
B.x = 2
C.y = 4
D.y = 2
Explanation: Vertical asymptotes occur where denominator x - 4 = 0 and numerator 2(4)+1 = 9 != 0, giving line x = 4.
6Find the derivative of f(x) = ln(x^2 + 1).
A.2x / (x^2 + 1)
B.1 / (x^2 + 1)
C.x / (x^2 + 1)
D.2 / (x^2 + 1)
Explanation: By chain rule, d/dx[ln(u)] = u'/u with u = x^2 + 1 (u' = 2x), giving f'(x) = 2x / (x^2 + 1).
7Find the horizontal asymptote of f(x) = (3x^2 - 5) / (2x^2 + 1).
A.y = 3/2
B.y = -5
C.y = 0
D.x = 3/2
Explanation: Equal degrees in numerator and denominator mean horizontal asymptote is y = 3/2.
8Find the point of inflection of the curve f(x) = x^3 - 6x^2 + 9x.
A.(2, 2)
B.(1, 4)
C.(3, 0)
D.(2, 0)
Explanation: f''(x) = 6x - 12 = 0 => x = 2. f(2) = 8 - 24 + 18 = 2. Thus (2, 2) is the inflection point.
9Evaluate lim (x -> 0) [ (e^(2x) - 1 - 2x) / x^2 ] using L'Hôpital's rule.
A.2
B.1
C.4
D.0
Explanation: L'Hôpital once: lim (2e^(2x) - 2)/(2x). L'Hôpital twice: lim (4e^(2x))/2 = 4/2 = 2.
10Find the derivative of f(x) = arctan(3x).
A.3 / (1 + 9x^2)
B.1 / (1 + 9x^2)
C.3 / (1 + 3x^2)
D.1 / sqrt(1 - 9x^2)
Explanation: d/dx[arctan(u)] = u'/(1 + u^2) with u = 3x (u' = 3, u^2 = 9x^2) gives 3 / (1 + 9x^2).

About the European Baccalaureate Mathematics (5-period course) Practice Questions

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