Saxony Abitur Mathematics common mistakes
Saxony Abitur Mathematics · Understand the traps and how to avoid them.
When asked for the range of a function or the image of an interval, the candidate differentiates, sets f'(x)=0, and reports the resulting function values as the range, ignoring the behaviour at the interval endpoints and any vertical asymptote inside the domain.
The candidate has memorised the procedure 'extremes = derivative zero' from optimisation tasks and transfers it to range questions, where monotonicity and boundary behaviour are what matter. Local extrema are not automatically global bounds on an open or half-open interval.
What to do instead
Before computing anything, sketch the graph qualitatively: domain, limits at the domain boundaries and at any pole, and monotonicity intervals. Then argue the range explicitly: state where the function is continuous and monotone, evaluate the limits, and quote the intermediate value theorem to conclude that every value between the limits is attained. In the exam, write the sentence 'f is continuous and strictly monotone on ..., with limits ..., hence the image is ...' - that sentence is where the marks are.
In curve-discussion tasks with a parameter, the candidate computes f_a'(x) and f_a''(x) correctly but then treats the parameter as a constant when solving inequalities, e.g. concludes 'x = a is always a maximum' without checking whether the critical point lies inside the domain or whether the second derivative's sign depends on a.
Symbolic manipulation routines are trained on fixed functions; the parameter invites the candidate to carry it along mechanically. The case distinction that a parameter forces (a>0, a=0, a<0) is an extra cognitive step that is easily skipped under time pressure.
What to do instead
After every parameter-dependent result, write a one-line case table: which values of a change the number of zeros, the position of the extremum relative to the domain, or the sign of the leading coefficient. Practise by taking a finished solution and asking 'for which a does this formula break down?' - division by a, logarithm arguments, and roots are the usual triggers. In the exam, reserve two or three lines for the case distinction; it is almost always worth marks.
When applying integration by substitution or integration by parts, the candidate changes the integrand but forgets to change the limits of a definite integral (or forgets the differential dx, writing ∫f(u) instead of ∫f(u)·u'(x)dx).
The substitution is performed as a pattern-matching exercise on the integrand only; the limits are perceived as decoration rather than as part of the transformed problem. The differential is dropped because the notation du = u'(x)dx is treated as a formality.
What to do instead
Adopt a fixed written ritual: write 'u = ..., du = ... dx, limits: x=a → u=..., x=b → u=...' on its own line before touching the integral. Then substitute in one step and never mix x-limits with u-integrals. For by-parts, write the u, v, du, dv table explicitly and only then assemble. Check by differentiating your antiderivative - a 20-second verification that catches most sign and factor errors.
When a task asks for the area enclosed between two graphs, the candidate computes ∫(g(x) − f(x))dx over the interval between the intersection points without checking which function is above, and without splitting the integral at an interior intersection, producing a negative or wrongly signed area.
The formula 'area = integral of upper minus lower' is remembered, but the identification of upper and lower is treated as obvious. With more than two intersection points, the ordering can swap inside the interval, and candidates rarely test a sample point.
What to do instead
Always determine the intersection points first, sort them, and test the sign of g(x) − f(x) at one interior sample point per subinterval. Then write the area as a sum of integrals with the correct sign per piece, or integrate |g(x) − f(x)| conceptually. State the sign test in your solution - it is a legitimate, mark-earning justification and protects against the swap error.
In analytic geometry, the candidate sets up a line-plane or plane-plane intersection and stops after finding one point, or declares the line of intersection to be the normal vector of the plane, confusing the direction of the line with the normal of the plane.
The candidate conflates two different objects: a plane is given by a normal vector, a line by a direction vector. The intersection line's direction must be perpendicular to both normals (cross product), which is a different operation from reading off the normal.
What to do instead
Keep an explicit object dictionary on your scratch sheet: plane → normal vector n; line → direction vector u and a point; line in plane → u · n = 0. For an intersection line, compute u = n1 × n2, then find one point by solving the two plane equations with one coordinate set to zero, and write the line in parametric form. Verify the point lies in both planes and that u is orthogonal to both normals before moving on.
When computing distances (point to plane, point to line, skew lines) the candidate plugs numbers into a memorised formula but forgets to normalise the normal vector, or uses the normal vector of the plane as the direction in a line-distance problem, yielding answers off by a scale factor.
The Hesse normal form requires |n| = 1; candidates remember the numerator (n · (x − p)) but not the denominator. Under exam pressure the formula is recalled as a single blob, and the normalisation step is the part that gets lost.
What to do instead
Derive the distance formula once from the projection idea and write it in your own notation: distance = |n · (x − p)| / |n|. Then, for every distance problem, first write down n and compute |n| as a separate line. For skew lines, use the scalar triple product with the cross product of the two directions and divide by its length; sanity-check that the result is smaller than the distance between any two chosen points on the lines.
In a binomial setting, the candidate computes P(X ≥ k) as 1 − P(X = k) or as 1 − P(X ≤ k) inconsistently, and misreads 'at least', 'more than', 'at most' when translating the German phrasing into an inequality.
The complement rule is remembered in the form 'one minus the thing', but which 'thing' depends on the strictness of the inequality. The language-to-inequality mapping ('mindestens' = ≥, 'mehr als' = >, 'höchstens' = ≤) is treated as trivial and is not written down.
What to do instead
Before computing, translate the event into a formal inequality in one written line: P(X ≥ 3) = 1 − P(X ≤ 2). Then check with a small numerical sanity test: for n=10, p=0.5, does your expression give a value between 0 and 1 and does it move in the expected direction when k increases? For cumulative values use the table or the calculator's cumulative function deliberately, and state which function you used.
When a task introduces a new random variable Y = aX + b or a sum of independent variables, the candidate applies the expectation rule correctly but assumes Var(aX + b) = a·Var(X) + b, or adds standard deviations instead of variances for sums.
Linearity of expectation is genuine and memorable, so candidates generalise it to variance. The variance rules (Var(aX+b) = a²Var(X); Var(X+Y) = Var(X) + Var(Y) only for independent variables) are less intuitive and are often compressed into a single wrong formula.
What to do instead
Memorise the two rules as a pair with the reason: expectation is linear because averaging is linear; variance is quadratic in the scaling because it is an average of squared deviations, and the shift b cancels in the deviations. For sums, always state the independence assumption explicitly before adding variances, and convert to standard deviation only at the very end, with a square root.
In a hypothesis test, the candidate states the null hypothesis as the claim to be proven, or reports the p-value as 'the probability that H0 is true', and then draws a conclusion phrased as 'H0 is proven'.
The logic of a significance test is counterintuitive: it is a proof by contradiction in probability, not a confirmation. Everyday language ('we tested whether...') reinforces the wrong framing, and the p-value's definition is easily misremembered.
What to do instead
Write the test in a fixed four-line template: (1) H0: p = p0 versus H1: p > p0 (or <, ≠), (2) test statistic and its distribution under H0, (3) rejection region or p-value with the significance level, (4) conclusion in the form 'the data give significant evidence against H0 at level α' or 'H0 is not rejected'. Never write 'proven'. For the p-value, say 'the probability, assuming H0, of a result at least as extreme as observed'.
In modelling tasks, the candidate produces a function or a computation that answers a slightly different question than the one asked: the model's variables are not defined, units are dropped, and the final sentence does not refer back to the real-world context (e.g. reports x = 4.7 without saying what x is or whether a non-integer value is meaningful).
The candidate treats modelling tasks as disguised calculation tasks and rushes to the algebra. The context sentence at the end is perceived as decoration and is skipped, although it is where the modelling competence is assessed.
What to do instead
Use a three-part structure for every modelling task: (a) define every variable with its unit and the valid range, (b) do the mathematics, (c) write an answer sentence that names the quantity, its unit, and any rounding or feasibility comment (e.g. 'about 5 days; the model is only valid while the tank is not empty'). Practise by reading your final sentence alone and asking whether a reader who has not seen the calculation would understand the answer.