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100+ Free VWO Wiskunde B Practice Questions

Netherlands VWO Wiskunde B — Centraal Examen and Schoolexamen practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: VWO Wiskunde B Exam

3 Hours

CE Exam Duration

CvTE Examenblad 2026

5.5 / 10

Standard Passing Grade

Dutch Secondary Education Act

5 Subdomains

Core Syllabus Domains (B1, B2, C, D, E)

CvTE Examination Syllabus

CE + SE

50% School Exam + 50% Central Exam

CvTE Regulations

Exam Mode

Graphic Calculator Requirement

CvTE Equipment Policy

N-Term

National Scaling Term (0.0 to 2.0)

CvTE Grading System

STEM Gateway

Prerequisite for Dutch Engineering & Physics

Dutch University Admissions

100 MCQs

English Adaptation Question Count

OpenExamPrep

Official VWO Wiskunde B 2026 is CE (3-hour written exam) plus school SE. These 100 English MCQs provide rigorous calculus, geometry, and vector calculation practice aligned to CvTE VWO Wiskunde B subdomains.

Sample VWO Wiskunde B Practice Questions

Try these sample questions to test your VWO Wiskunde B exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Solve for x: 2 \ln(x) = \ln(3x + 4).
A.x = -1
B.x = 2
C.x = 4
D.x = 4 and x = -1
Explanation: Rewrite as \ln(x^2) = \ln(3x + 4), which implies x^2 = 3x + 4 or x^2 - 3x - 4 = 0. Factoring gives (x - 4)(x + 1) = 0. Since the logarithm domain requires x > 0, the extraneous solution x = -1 is discarded, leaving x = 4.
2Simplify the expression e^{3\ln(x) - \ln(2x)} for x > 0.
A.x^2 / 2
B.x^3 / 2
C.2x^2
D.x^2 - 2
Explanation: Using logarithm rules, 3\ln(x) - \ln(2x) = \ln(x^3) - \ln(2x) = \ln(x^3 / (2x)) = \ln(x^2 / 2). Therefore, e^{\ln(x^2 / 2)} = x^2 / 2.
3Find the equation of the horizontal asymptote of f(x) = \frac{6x^2 - 5x + 1}{2x^2 + 7}.
A.y = 0
B.y = 3
C.y = 6
D.No horizontal asymptote
Explanation: To find horizontal asymptotes, calculate \lim_{x \to \pm\infty} f(x). Divide numerator and denominator by x^2 to get \lim_{x \to \infty} \frac{6 - 5/x + 1/x^2}{2 + 7/x^2} = \frac{6}{2} = 3. Thus, y = 3.
4Determine the domain of the real-valued function f(x) = \sqrt{9 - x^2}.
A.x \ge 3
B.-3 \le x \le 3
C.x \le -3 or x \ge 3
D.All real numbers
Explanation: For the square root to be defined in real numbers, the radicand must be non-negative: 9 - x^2 \ge 0, which means x^2 \le 9, leading to -3 \le x \le 3.
5Find the inverse function f^{-1}(x) for f(x) = 2^{x+1} - 3.
A.f^{-1}(x) = ^2\log(x+3) - 1
B.f^{-1}(x) = ^2\log(x-3) + 1
C.f^{-1}(x) = \log_2(x+1) + 3
D.f^{-1}(x) = 2^{x-1} + 3
Explanation: Set y = 2^{x+1} - 3. Solve for x: y + 3 = 2^{x+1} \implies x + 1 = ^2\log(y+3) \implies x = ^2\log(y+3) - 1. Swapping x and y yields f^{-1}(x) = ^2\log(x+3) - 1.
6Evaluate the limit: \lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2}.
A.0
B.2
C.4
D.Undefined
Explanation: Direct substitution yields 0/0. Factor both numerator and denominator: \frac{(x-2)(x+2)}{(x-2)(x-1)}. Cancel the common factor (x-2) for x \ne 2 to get \frac{x+2}{x-1}. Evaluating at x = 2 gives \frac{2+2}{2-1} = 4.
7Solve the exponential equation: 4^x - 5 \cdot 2^x + 4 = 0.
A.x = 0 and x = 2
B.x = 1 and x = 4
C.x = 2 and x = 4
D.x = -1 and x = 2
Explanation: Substitute u = 2^x, so 4^x = (2^x)^2 = u^2. The equation becomes u^2 - 5u + 4 = 0, factoring as (u - 1)(u - 4) = 0. Thus u = 1 or u = 4. Since 2^x = 1 \implies x = 0, and 2^x = 4 \implies x = 2.
8Find the asymptotes of f(x) = \frac{x^2 + 3x - 1}{x + 1}.
A.Vertical: x = -1; Oblique: y = x + 2
B.Vertical: x = 1; Oblique: y = x + 3
C.Vertical: x = -1; Horizontal: y = 1
D.Vertical: x = -1; Oblique: y = x - 1
Explanation: Vertical asymptote occurs at x = -1 where denominator is 0. Perform polynomial division: (x^2 + 3x - 1) / (x + 1) = x + 2 - \frac{3}{x+1}. As x \to \pm\infty, \frac{3}{x+1} \to 0, so the oblique asymptote is y = x + 2.
9Solve the logarithmic equation: ^3\log(x) + ^3\log(x - 6) = 3.
A.x = 3
B.x = 9
C.x = 9 and x = -3
D.x = 27
Explanation: Combine logarithms: ^3\log(x(x-6)) = 3. Convert to exponential form: x(x-6) = 3^3 = 27. Thus x^2 - 6x - 27 = 0 \implies (x - 9)(x + 3) = 0. The domain requires x > 6, so x = -3 is rejected, leaving x = 9.
10Solve the absolute value equation: |2x - 7| = 3x - 1.
A.x = 1.6 only
B.x = -6 only
C.x = 1.6 and x = -6
D.No solution
Explanation: Case 1: 2x - 7 = 3x - 1 \implies x = -6. Check: |2(-6)-7| = |-19| = 19, but 3(-6)-1 = -19, so x = -6 is extraneous. Case 2: 2x - 7 = -(3x - 1) = -3x + 1 \implies 5x = 8 \implies x = 8/5 = 1.6. Check: |2(1.6)-7| = |-3.8| = 3.8, and 3(1.6)-1 = 3.8. Valid solution is x = 1.6.

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