Bremen Abitur Math GK common mistakes
Bremen Abitur Math GK · Understand the traps and how to avoid them.
When differentiating a composite function, the candidate differentiates only the outer function and drops the inner derivative - writing d/dx sin(3x²) = cos(3x²) or d/dx e^(x²) = e^(x²).
Chain rule is learned as a pattern ('derivative of sin is cos') and under timed conditions the candidate pattern-matches the outer function and stops, treating the argument as a single variable. Nested functions with polynomial or exponential inner parts are where the omission is most likely, because the inner derivative is not visually salient.
What to do instead
Before writing any answer, explicitly name the inner function u and write the chain as outer′(u)·u′ on the page, even for one-line answers. Drill a mixed set of 20 composites (trig, exponential, log, power, inverse trig) and mark each inner derivative in a different colour until the step becomes automatic. On the exam, do a final pass over every derivative checking that each composite has a visible inner-derivative factor.
The candidate computes a definite integral or area but substitutes the limits in the wrong order or forgets to change limits after substitution, producing a sign-flipped or wrong-valued answer, e.g. evaluating ∫₀² with F(0) − F(2).
The upper-minus-lower convention competes in memory with the substitution rule's 'evaluate at old limits', and under pressure the candidate writes the antiderivative values in the order they appear on the page. When using u-substitution the candidate often back-substitutes inconsistently or plugs original limits into the u-antiderivative.
What to do instead
Adopt a fixed written template: bracket the antiderivative, write '= [F(x)]_a^b = F(b) − F(a)', and physically copy b above a before evaluating. For substitution, always either convert the limits to u-values in the same line or write 'back-substitute to x, then apply original limits' - never mix. Practise five substitution integrals where the limits are converted, checking that a positive integrand gives a positive result.
When finding eigenvectors, the candidate solves det(A − λI) = 0 for λ correctly but then solves (A − λI)x = 0 by row-reducing A itself rather than A − λI, or picks a single eigenvector for a repeated eigenvalue and misses the second independent one.
Row-reduction of A is a well-practised reflex, and substituting the numeric λ into every diagonal entry is error-prone mental arithmetic. For repeated eigenvalues the candidate stops once one free variable yields a vector, not checking the nullity of A − λI.
What to do instead
Write A − λI explicitly with λ substituted before any row operation, and compute its nullity to know in advance how many independent eigenvectors to expect (nullity = multiplicity for diagonalisable cases). For each eigenvalue, state the dimension of the eigenspace before choosing vectors. Verify each candidate v by computing Av and confirming it equals λv - a 30-second check that catches both errors.
The candidate computes a cross product or scalar triple product and reports a magnitude or a signed value when a vector (or the opposite orientation) was required, or takes the dot product where the cross product is needed for a normal vector.
Area, volume and normal-vector problems all invoke 'cross product' in the candidate's notes, and the operations get conflated. Sign conventions in the triple product (a·(b×c) versus (a×b)·c) and the direction of a×b are also easy to reverse under time pressure.
What to do instead
Write the geometric quantity required first - 'normal vector =', 'area = ½|a×b|', 'volume = |a·(b×c)|' - then perform the operation that matches it. Compute cross products with the determinant expansion written out in full (i, j, k rows) rather than mentally, and check orientation with the right-hand rule or by confirming a·(a×b) = 0 and b·(a×b) = 0.
The candidate computes P(A|B) as P(A)P(B), or uses P(A|B) = P(A∩B)/P(B) but substitutes P(A) for P(A∩B) when events are not independent, and applies the multiplication rule for independent events to dependent sampling without replacement.
The formula P(A∩B) = P(A)P(B) is over-learned from independence examples and gets applied as a default. Condition notation is also read backwards: candidates compute P(B|A) when P(A|B) is asked, especially in Bayes-type word problems where the given and target events are described in prose.
What to do instead
Before computing, write the two events in words and label which is given ('given B') and which is the target. Use the general rule P(A∩B) = P(A|B)P(B) unless independence is explicitly justified, and for without-replacement problems draw the tree with changing denominators. For Bayes problems, write the partition and total probability denominator explicitly before dividing.
The candidate carries out a hypothesis test and states the conclusion as 'the null hypothesis is true' or 'proven', or interprets a p-value as the probability that H₀ is true, rather than reporting a decision about evidence at the stated significance level.
Everyday language treats 'not rejected' as 'confirmed', and the p-value's definition (probability of data at least as extreme, given H₀) is easily inverted in memory to 'probability H₀ is true'. The candidate also forgets to state the test statistic, degrees of freedom and the decision rule in context.
What to do instead
Use a fixed conclusion sentence: 'At the 5% significance level there is (insufficient/sufficient) evidence to reject H₀; we cannot conclude H₀ is true, only that it is not contradicted.' Write the p-value definition once in the margin as a reminder. Always report the test statistic, its distribution and degrees of freedom, and phrase the conclusion in terms of the original variables.
When computing a definite integral with substitution, the candidate keeps the antiderivative in the substituted variable but evaluates it at the original x-limits, or changes the limits to u-values and then substitutes x back before evaluating, producing a numerically wrong value with correct-looking work.
The two valid procedures (change limits and stay in u, or keep limits and back-substitute) are each individually simple, but mixing them is easy because the back-substitution step feels like part of 'finishing the problem'. The error is silent: the algebra is all correct, only the evaluation is wrong.
What to do instead
Choose one convention and write it at the top of the solution. If changing limits, compute the new limits immediately after defining u and cross out the old ones on the integral sign. If keeping limits, write 'back-substitute, then evaluate at x = a and x = b' as a literal step. Sanity-check the sign and rough magnitude against an area estimate.
The candidate states a limit exists and equals a finite number when the one-sided limits disagree, when the function is unbounded, or when the expression is an indeterminate form such as 0/0 that was never resolved; alternatively, the candidate differentiates a function to check continuity.
Limit, continuity and differentiability are studied in adjacent chapters and their definitions blur together. Indeterminate forms tempt the candidate to cancel incorrectly or to declare 'undefined' as if it were a limit value, and the distinction between a limit existing and a function being defined at a point is subtle in notation.
What to do instead
For every limit question, first compute the left-hand and right-hand limits separately and compare them explicitly. Classify the form before doing algebra, and only then choose a technique (factor and cancel, rationalise, or compare growth rates). Write the formal criteria as a three-line checklist: limit exists, value equals f(a), and the two-sided limit matches f(a), for continuity; derivative from the limit definition for differentiability.
The candidate interprets a p-value as the probability that the null hypothesis is true, or concludes that a non-significant result proves the null hypothesis, and describes a confidence interval as having a 95% chance of containing the true parameter for this particular sample.
The conditional direction of the p-value is reversed in ordinary speech ('the probability the result is due to chance'), and textbook phrasing is compressed. The correct statement is conditional on the null hypothesis and concerns data, not hypotheses, which is the opposite of what intuition supplies.
What to do instead
Write out the definition in full sentences with the conditional made explicit: the p-value is the probability of data at least as extreme as observed, assuming the null model. Practise rewriting wrong interpretations into correct ones, and phrase confidence intervals as a property of the procedure over repeated sampling, not of the single computed interval.