PAU Applied Math ACS II (La Rioja) common mistakes
PAU Applied Math ACS II (La Rioja) · Understand the traps and how to avoid them.
When solving a system by Gaussian elimination, the candidate performs row operations but then reads the final augmented matrix as if it still represented the original equations, or stops at row-echelon form and reports the coefficients as the solution without back-substitution.
The mechanics of row reduction are drilled, but the meaning of the resulting matrix is not: candidates treat elimination as an end in itself and lose the link between the reduced system and the values of x, y, z. Under exam time pressure they also skip the verification step that would expose the error.
What to do instead
Adopt a fixed ritual: after reaching row-echelon form, write the equations back out explicitly (z = ..., then y = ..., then x = ...) and substitute the triple into the original first equation. Practise three complete systems per session, always finishing with back-substitution and a check, so the final step becomes automatic.
The candidate computes the determinant or inverse correctly but then answers a different question than the one asked, e.g. gives det(A) when the problem asks for the value of a parameter making A singular, or inverts a matrix when only the solution of AX = B was required.
These problems bundle several sub-questions and the candidate works mechanically from the top of the page, answering the first computation that comes to mind rather than re-reading the actual demand. The parameter-based 'discuss according to the value of k' format is unfamiliar and gets reduced to a single calculation.
What to do instead
Before writing anything, underline the requested quantity and note the type of answer expected (a number, a value of k, a classification of the system). For parameter problems, always split into cases: det = 0 versus det ≠ 0, and state the nature of the system (compatible determined, compatible indeterminate, incompatible) for each case, since that classification is what earns the marks.
In a linear programming problem the candidate writes the constraints but forgets the non-negativity conditions, or tests the objective function only at the vertices of the first quadrant that happen to appear in the drawing, missing a vertex created by the intersection of two slanted constraints.
Non-negativity is treated as an implicit convention rather than a written constraint, and the feasible region is sketched loosely by shading, so intersection points of boundary lines are never computed algebraically. Candidates also confuse the feasible region with the polygon bounded by all lines rather than by the active constraints.
What to do instead
Always list non-negativity explicitly as x ≥ 0, y ≥ 0 in the constraint set. Compute every pairwise intersection of boundary lines algebraically, then test each candidate point in all constraints to decide whether it is feasible. Evaluate the objective only at feasible vertices and state the optimal value with the point, and if the region is unbounded, justify why the optimum is still attained.
When a linear programming problem is presented as a word problem, the candidate misassigns which quantity is x and which is y, or writes an inequality in the wrong direction (for example 'at least twice as many' as x ≥ 2y when the correct relation is the reverse), producing a feasible region that answers a different problem.
The translation step from prose to algebra is done quickly and without defining variables in writing, so the direction of inequalities and the roles of the two unknowns drift. Words like 'at least', 'at most', 'no more than' and 'twice' are read loosely.
What to do instead
Begin every word problem with an explicit 'Let x = ..., y = ...' statement, then translate each sentence one at a time into a labelled inequality, checking direction against a test value. After building the model, sanity-check one obvious point against the original wording before solving.
The candidate computes a limit by direct substitution, obtains an indeterminate form such as 0/0, and either declares the limit does not exist or stops, instead of factoring, rationalising, or applying L'Hôpital's rule to resolve it.
Direct substitution works for most routine limits, so candidates treat an indeterminate result as a final answer rather than a signal to change technique. Indeterminate forms are recognised as an algebraic nuisance rather than as a cue.
What to do instead
Train a decision tree: substitute first; if the result is determinate, done; if 0/0, try factoring or rationalising; if the expression involves radicals, multiply by the conjugate; if it is ∞/∞ or 0/0 with differentiable functions, apply L'Hôpital and re-check the form after each application. Do a set of mixed limits where the technique is not announced.
For a piecewise-defined function the candidate checks continuity at the join point by evaluating the two branches but forgets to compare the one-sided limits with the function value, or concludes differentiability from continuity alone.
The three-part definition of continuity (limit exists, function defined, the two agree) is compressed into 'the pieces meet', and the theorem 'differentiable implies continuous' is misremembered as a converse. The candidate also forgets to compute the derivative of each branch separately before matching them.
What to do instead
Write the continuity check as three explicit lines: left limit, right limit, f(a); state equality. For differentiability, compute the one-sided derivatives from the definition or from each branch's derivative and require them to be equal. Remember the implication runs one way only: continuity does not give differentiability, and give a counterexample such as |x| at 0 when revising.
When optimising a real-world quantity, the candidate differentiates, sets the derivative to zero, solves, and reports the critical point as the answer without verifying that it is a maximum or minimum, and without checking the endpoints of the interval or the geometric domain restrictions.
The procedure 'f'(x) = 0 gives the optimum' is memorised as a formula, so the second-derivative test or sign analysis and the endpoint comparison are skipped. Domain restrictions arising from the geometry (a length must be positive, a radius cannot exceed a bound) are ignored.
What to do instead
End every optimisation with an explicit justification: use the second derivative or a sign table to classify the critical point, and compare with endpoint values when the domain is closed. State the domain of the variable from the geometry before differentiating, and finish by answering the question actually asked (the dimensions, not just the value of x).
The candidate sets up a definite integral for an area but does not determine where the curves intersect, so the limits of integration are guessed from the sketch, and when the curves cross inside the interval the integral is computed as a single piece, giving a signed result that understates the area.
Area problems are treated as 'integrate the difference of the functions' without the preliminary analysis of intersection points and relative position. The candidate forgets that a definite integral returns a signed quantity and that area requires absolute value or splitting at crossings.
What to do instead
Always solve f(x) = g(x) first and list the intersection points. Determine which function is above on each subinterval (test a point), then integrate the upper minus the lower on each piece and add. Check the sign and plausibility of the result: an area must be positive, and a rough rectangle estimate should be in the same order of magnitude.
In a normal-distribution problem the candidate uses the raw value instead of standardising, or standardises with the wrong sign or the wrong standard deviation (using the variance), and misreads the table by failing to use the symmetry P(Z < -z) = 1 - P(Z < z).
The standardisation formula z = (x - μ)/σ is applied from memory without checking which quantity is given, and the table is consulted mechanically. The complement and symmetry relations are not internalised, so negative z-values are looked up directly in a table that only lists positive ones.
What to do instead
Write the standardisation explicitly every time, labelling μ, σ and x before substituting, and sketch the bell curve with the required region shaded. Convert every probability to a P(Z < a) form using symmetry and complement before touching the table, and check that the answer lies between 0 and 1 and matches the shaded area's rough size.
When constructing a confidence interval for a mean, the candidate uses z instead of t for small samples with unknown population standard deviation, or uses σ/√n but substitutes the sample standard deviation without the n - 1 correction, and reports the interval without stating the confidence level or interpreting it.
The z and t cases are memorised as interchangeable formulas, and the conditions for each (known σ versus unknown σ, large versus small n) are not linked to the choice. The interpretation of confidence is also confused with a probability statement about the parameter.
What to do instead
Before computing, classify the situation: known σ or n large → z; unknown σ and small n → t with n - 1 degrees of freedom. Write the formula with the correct standard error, then state the interval and interpret it correctly ('we are 95% confident the population mean lies in this interval'), avoiding the claim that the parameter has 95% probability of being inside it.