Saarland Abitur Technik-Mathematik common mistakes
Saarland Abitur Technik-Mathematik · Understand the traps and how to avoid them.
When asked for the relative position of two lines or a line and a plane in space, the candidate computes the determinant of the three direction/connecting vectors and stops, declaring 'they intersect' or 'they are skew' from that single number without checking the parallel/contained cases separately.
The candidate collapses the classification into one test, forgetting that the determinant being zero covers three different situations (parallel distinct, coincident, coplanar-intersecting), and that the determinant being nonzero only rules out coplanarity.
What to do instead
Use a decision tree and write it out each time: compare direction vectors first (proportional or not). If proportional, test whether a point of one satisfies the other (coincident vs parallel distinct). If not proportional, compute the mixed product; zero means they intersect in a point (find it), nonzero means they are skew. For line–plane: check n·v; if zero, test whether the line's point lies in the plane (contained vs parallel). State the case name explicitly in the answer.
When computing a distance from a point to a plane or between skew lines, the candidate substitutes into the formula without normalising, using n·PQ instead of |n·PQ|/|n|, or forgetting the absolute value, and reports a negative distance or one scaled by the norm of the normal vector.
The formula is recalled as 'dot product over something' and the denominator (the modulus of the normal, or the cross product for skew lines) is dropped because it feels like a formality; the absolute value is dropped because the arithmetic 'came out positive anyway' in practice exercises.
What to do instead
Memorise the distance formula as a fraction with a visible denominator: d = |Ax₀+By₀+Cz₀+D| / √(A²+B²+C²), and for skew lines d = |(P₂−P₁)·(v₁×v₂)| / |v₁×v₂|. After computing, sanity-check the magnitude against the coordinates (a distance cannot exceed the distance between the two named points). Write the absolute value bars in the very first line of working, before substituting numbers, so they cannot be omitted later.
When computing a limit that leads to an indeterminate form, the candidate applies L'Hôpital directly without first checking the form, or applies it to a quotient that is not 0/0 or ∞/∞, and differentiates numerator and denominator as if differentiating a quotient (using the quotient rule instead of separate derivatives).
L'Hôpital is over-practised as a reflex, and the quotient rule is so recently drilled that it contaminates the procedure; the candidate also skips the substitution step that would reveal the form.
What to do instead
Adopt a fixed routine: substitute the value first and write the resulting form (0/0, ∞/∞, 0·∞, ∞−∞, 1^∞…) explicitly. Only for 0/0 and ∞/∞ apply L'Hôpital, and write the derivatives of numerator and denominator as two separate expressions, never as a quotient. For other forms, transform first (common denominator, conjugate, or take logarithms for 1^∞). Verify the final limit by a quick numerical check with a value close to the limit point.
In an optimisation problem, the candidate finds f′(x)=0 and reports those critical points as the answer, without verifying that the point lies in the feasible interval, without classifying it as a maximum or minimum, and without answering the actual question asked (the dimensions, the maximum area, the minimum cost).
The algebraic step (solve f′=0) is the part that feels like 'the maths', while translating back to the problem statement and checking endpoints is treated as cosmetic; the candidate also forgets that a constrained problem can have its optimum at a boundary.
What to do instead
End every optimisation with a four-line checklist written in the answer: domain of the variable from the geometry; critical points inside that domain; classification via the second derivative or a sign table of f′; evaluation of f at the critical point and at the endpoints. Then write a sentence in words giving the requested quantity with units. If the question asks for dimensions, the final line must state the dimensions, not just the value of the function.
When computing a definite integral by substitution or by parts, the candidate changes the variable but keeps the original limits, substituting them into the new variable's expression, or forgets to return to the original variable before evaluating.
The mechanics of the antiderivative dominate attention, and the limits are copied down at the start of the line and never revisited; the candidate also treats the change of limits as an optional refinement.
What to do instead
Choose one of two disciplined formats and stick to it: either (a) change the limits at the same time as the variable and never write x again after the substitution, or (b) keep indefinite-integral notation, find the antiderivative in x, and only then apply the original limits. Mark the limits with a bracket [F(x)] from a to b so the evaluation step is visually forced. After evaluating, check the sign and rough magnitude (e.g. an area must be positive).
When computing the area between two curves or the volume of a solid of revolution, the candidate integrates from the x-coordinates of the intersection points without checking which curve is on top, and reports a negative area, or uses the function value instead of the actual distance to the rotation axis when the axis of revolution is not a coordinate axis.
The intersection points are found correctly and the integral is set up from them automatically, but the ordering of the curves and the geometry of the radius are not examined; the candidate assumes 'area = integral of the difference' regardless of sign and 'radius = f(x)' regardless of the axis.
What to do instead
Sketch the region (even roughly) and mark which curve is upper. Set up the integral as ∫(upper − lower) and, if the curves cross inside the interval, split the integral at the crossing point and add absolute values. For volumes, write the radius explicitly as a distance: for rotation about y=k, r(x)=|f(x)−k|; about x=h with horizontal slices, r(y)=|g(y)−h|. State the radius formula in words before integrating.
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.
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.