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100+ Free Sachsen-Anhalt Abitur Mathematics Practice Questions

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2026 Statistics

Key Facts: Sachsen-Anhalt Abitur Mathematics Exam

3 Sachgebiete

Every Prüfungsteil of the Sachsen-Anhalt Mathematik Abitur covers Analysis, Analytische Geometrie and Stochastik

Hinweise zu den schriftlichen Abiturprüfungen 2026 Mathematik, Anlage 2

255 / 300 min

Written Bearbeitungszeit at grundlegendem and erhöhtem Anforderungsniveau, including Auswahlzeit

Hinweise zu den schriftlichen Abiturprüfungen 2026 Mathematik, Anlage 2

80 / 100 BE

Maximum Bewertungseinheiten on the written paper at grundlegendem and erhöhtem Anforderungsniveau

Hinweise zu den schriftlichen Abiturprüfungen 2026 Mathematik, Anlage 2

06.05.2026

Written Mathematik date in the 2026 main session (Leistungskurs and Grundkurs); Nachprüfung 05.06.2026

Ministerium für Bildung des Landes Sachsen-Anhalt, Terminplan Abiturprüfung 2025/2026

0–15 Punkte

Grading scale for each Prüfungsfach, from Note 1 (15–13 Punkte) to Note 6 (0 Punkte)

Verordnung über die gymnasiale Oberstufe, Sachsen-Anhalt

Kein GTR/CAS

Only a wissenschaftlicher Taschenrechner is permitted in Prüfungsteil 2 — graphing calculators and computer-algebra systems are not approved Hilfsmittel

Hinweise zu den schriftlichen Abiturprüfungen 2026 Mathematik, Anlage 2

Free 100-question English-language MCQ study bank for Sachsen-Anhalt Abitur Mathematik, mapped to the Fachlehrplan Sachgebiete Analysis (~40%), Analytische Geometrie (~30%) and Stochastik (~30%), with worked-calculation explanations. Official exam: German free-response Klausur, 255 min (gA) / 300 min (eA), 80/100 BE, written date 06.05.2026; wissenschaftlicher Taschenrechner allowed in Prüfungsteil 2 only, no GTR/CAS. Not an official-format simulation; no fee for regular school candidates.

Sample Sachsen-Anhalt Abitur Mathematics Practice Questions

Try these sample questions to test your Sachsen-Anhalt Abitur Mathematics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Using the limit definition of the derivative, f'(x0) = lim(h→0) [f(x0+h) − f(x0)] / h, find f'(3) for f(x) = x².
A.6
B.9
C.3
D.18
Explanation: [f(3+h)−f(3)]/h = [(9+6h+h²)−9]/h = (6h+h²)/h = 6+h. As h→0 this tends to 6, so f'(3) = 6, matching the power rule f'(x)=2x at x=3.
2Geometrically, what does the Differenzenquotient [f(x0+h) − f(x0)] / h represent for a function graph?
A.The slope of the secant line (Sekante) through the points (x0, f(x0)) and (x0+h, f(x0+h))
B.The slope of the tangent line at x0
C.The y-intercept of the function
D.The area under the graph between x0 and x0+h
Explanation: The Differenzenquotient compares the change in function value to the change in x between two distinct points, which is exactly the slope of the secant line joining them. Only in the limit h→0 does it become the tangent slope (Differentialquotient).
3Let f(x) = x³ − 2x. Using the derivative rules, compute f'(2).
A.10
B.12
C.14
D.2
Explanation: f'(x) = 3x² − 2 by the Potenz- and Faktorregel. At x=2: f'(2) = 3·4 − 2 = 12 − 2 = 10.
4Evaluate the limit lim(x→2) (x² − 4) / (x − 2).
A.4
B.0
C.The limit does not exist
D.2
Explanation: Factor the numerator: (x²−4) = (x−2)(x+2), so the expression simplifies to (x+2) for x≠2. Taking the limit as x→2 gives 2+2 = 4.
5For f(x) = −2x⁴ + 3x, what happens to f(x) as x → +∞?
A.f(x) → −∞
B.f(x) → +∞
C.f(x) → 0
D.f(x) oscillates without a limit
Explanation: For large |x|, the leading term −2x⁴ dominates the lower-degree term 3x. Since the degree is even and the leading coefficient is negative, both ends of the graph fall, so f(x) → −∞ as x → +∞.
6Let f(x) = 5x³ − 4x² + 7. Compute f'(1).
A.7
B.15
C.23
D.11
Explanation: f'(x) = 15x² − 8x by the Potenz-, Faktor- and Summenregel. At x=1: f'(1) = 15 − 8 = 7.
7Let f(x) = x² · eˣ. Using the Produktregel, compute f'(1) exactly.
A.3e
B.2e
C.e
D.
Explanation: Produktregel: f'(x) = 2x·eˣ + x²·eˣ = eˣ(2x + x²). At x=1: f'(1) = e·(2+1) = 3e.
8Let f(x) = (2x − 1)⁴. Using the Kettenregel, compute f'(1).
A.8
B.4
C.-8
D.2
Explanation: Kettenregel: f'(x) = 4(2x−1)³ · 2 = 8(2x−1)³. At x=1: (2·1−1)=1, so f'(1) = 8·1³ = 8.
9What is the slope of the tangent line to f(x) = x² − 3x + 2 at x = 4?
A.5
B.8
C.6
D.3
Explanation: f'(x) = 2x − 3. At x=4: f'(4) = 8 − 3 = 5, which is the slope of the tangent (Anstieg der Tangente) at that point.
10For f(x) = x³ − 3x², the critical points are at x=0 and x=2. Which classification is correct?
A.x = 0 is a local maximum and x = 2 is a local minimum
B.x = 0 is a local minimum and x = 2 is a local maximum
C.Both x = 0 and x = 2 are local maxima
D.Both x = 0 and x = 2 are inflection points, not extrema
Explanation: f'(x) = 3x² − 6x = 3x(x−2), giving critical points x=0 and x=2. f''(x) = 6x − 6: f''(0) = −6 < 0 (local maximum), f''(2) = 6 > 0 (local minimum), by the hinreichende Bedingung for extrema.

About the Sachsen-Anhalt Abitur Mathematics Practice Questions

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