PAU Applied Math ACS II (Aragón) common mistakes
PAU Applied Math ACS II (Aragón) · Understand the traps and how to avoid them.
When finding the inverse of a 3×3 matrix (or solving AX=B), the candidate computes the adjugate but forgets to divide by the determinant, or divides by det(A) without checking det(A)≠0 first, and in exam problems containing a parameter they give a single inverse formula valid even for the parameter values that make det(A)=0.
The formula A⁻¹ = adj(A)/det(A) is memorised as a procedure, and the division step is treated as cosmetic. Parameter problems are handled by mechanically applying the formula, because the candidate does not connect 'inverse exists' with 'det ≠ 0' and therefore never separates the discussion into cases.
What to do instead
Before any inversion, compute det(A) as a separate step and write the condition det(A)≠0 explicitly. For parameter matrices, solve det(A)=0, list the excluded values, and structure the answer as: 'If a ≠ ... the matrix is invertible and A⁻¹ = ...; if a = ... then rank is ... and the system is ...'. Verify one product A·A⁻¹ = I numerically as a check.
When discussing the rank of a matrix with a parameter, the candidate row-reduces and then divides by an expression in the parameter (e.g. divides a row by (a−2)) without noting that this is invalid when a=2, so the exceptional values are silently lost and the case discussion is incomplete or wrong.
Row reduction is practised on numerical matrices where pivots are never zero. The candidate carries the same habit to symbolic matrices, treating the parameter as an ordinary number, and does not track which operations assume a nonzero pivot.
What to do instead
Adopt the rule: never divide by a parameter-dependent pivot; instead factor it out and record the condition. Alternatively compute det of the largest minors and set them to zero to find the values of a where rank drops, then substitute each exceptional value back into the original matrix and compute its rank directly. End by tabulating rank versus parameter range.
In a linear programming problem the candidate sets up the constraints correctly but optimises over the wrong region: they shade the union of the half-planes instead of the intersection, or they ignore a non-negativity constraint, and then evaluate the objective only at the vertices of the drawn polygon without checking which vertices actually satisfy every inequality.
Graphing is done quickly with test points for one inequality at a time, and the feasible region is identified visually rather than by intersecting all half-planes. Non-negativity is treated as a formality rather than as two extra constraints that cut the region.
What to do instead
Write every constraint in the form ax+by ≤ c, list them numbered, and test the origin in each one, marking the allowed side with an arrow. Shade the intersection only. Then compute every pairwise intersection of boundary lines, discard points violating any constraint, and build a table of feasible vertices with the objective value. Quote the optimum with the coordinates and the value.
The candidate states that the optimum of a linear program occurs 'at a vertex' but then answers an integer-programming or unbounded-feasible-region question by rounding the continuous optimum, or claims a maximum exists when the feasible region is unbounded and the objective grows without limit along a direction of the region.
The vertex theorem is memorised for bounded polygons only, and the candidate does not distinguish bounded from unbounded feasible regions, nor continuous from integer decision variables. Rounding feels like a harmless final step.
What to do instead
Classify the feasible region first (bounded / unbounded) and check whether the objective is bounded on it by examining the extreme rays. For integer problems, state explicitly that the continuous optimum is only a bound and then test neighbouring integer points or use the appropriate enumeration. Always report the optimal value together with the point where it is attained.
When computing a limit at a point, the candidate substitutes the value, obtains 0/0, and either declares the limit does not exist or cancels factors incorrectly (e.g. cancels terms across a sum, or cancels x in (x+3)/x to get 3). For limits at infinity with radicals they subtract infinities term by term and conclude the limit is 0 or undefined.
Indeterminate forms are treated as an answer rather than as a signal to transform the expression, and algebraic cancellation is applied without the restriction that it requires a common factor of the whole numerator and denominator. Radical limits are handled by intuition about 'infinity minus infinity' instead of multiplying by the conjugate.
What to do instead
Adopt a decision tree: substitute; if 0/0, factor and cancel only genuine common factors, or use L'Hôpital after verifying differentiability, or use the conjugate for radicals. For x→∞ with radicals, divide by the highest power or multiply by the conjugate and divide. State the algebraic step explicitly in the answer so the marker sees the transformation.
In a continuity or differentiability question with a piecewise function, the candidate checks only that the two branches give the same value at the join point (continuity) and then concludes differentiability, without comparing the one-sided derivatives, or they compute the derivative of each branch and forget to evaluate the lateral derivatives at the join.
Continuity and differentiability are conflated because the implication 'differentiable ⇒ continuous' is remembered in the wrong direction, and the definition of derivative at a point as a limit of difference quotients is not used.
What to do instead
For each join point x=a, do three separate checks and write them out: (1) lateral limits of f equal f(a) → continuity; (2) lateral limits of the difference quotient [f(a+h)−f(a)]/h exist and are equal → differentiability; (3) if they differ, report a corner with the two slopes. Never infer differentiability from continuity.
When computing a definite integral or the area between curves, the candidate integrates from the left intersection to the right intersection without checking whether the curves cross in between, so a region where the upper and lower functions swap is counted with cancelling signs and the area comes out too small or zero.
Area is treated as 'integral of f minus g', a formula valid only when f ≥ g throughout the interval. The candidate finds intersection points but does not test the sign of f−g on each subinterval.
What to do instead
Always find all intersection points in the interval, sort them, and test the sign of f−g at one interior point of each subinterval. Split the integral at every crossing and integrate |f−g|, i.e. sum the pieces with the correct order. For areas involving the y-axis or a curve given as x=g(y), integrate with respect to y and state the variable of integration.
In probability problems the candidate adds probabilities of non-mutually-exclusive events, or multiplies P(A)P(B) for events that are not independent, and in conditional-probability or Bayes problems they compute P(A|B) as P(B|A) or forget to divide by the total probability of the conditioning event.
The addition and multiplication rules are recalled without their hypotheses, and conditional probability is read as a symmetric relation. Tree diagrams are drawn but the branch probabilities are multiplied along a path and then not summed over the paths leading to the conditioning event.
What to do instead
Before computing, write the hypotheses explicitly: are the events disjoint? independent? Then use P(A∪B)=P(A)+P(B)−P(A∩B) and P(A∩B)=P(A)P(B|A). For Bayes, draw the tree, compute the total probability of the observed event as the sum over all paths, and divide the relevant path by that total. Check that the answer lies in [0,1].
In a normal-distribution problem the candidate uses the value z as if it were the probability, or forgets the continuity correction when approximating a discrete binomial variable by a normal, or uses the population standard deviation where the sample standard error σ/√n is required.
Standardisation is memorised as z=(x−μ)/σ and the table lookup step is skipped under time pressure. The binomial approximation is applied as a formula without the ±0.5 adjustment, and the distinction between the distribution of X and of the sample mean is blurred.
What to do instead
Write the standardisation explicitly, then state the table value with the correct inequality direction and complement: P(X≤x)=P(Z≤z)=Φ(z), P(X≥x)=1−Φ(z). For binomial→normal use μ=np, σ=√(np(1−q)) and correct with P(X≤k)≈Φ((k+0.5−μ)/σ). For sample means use σ/√n and say so. Sanity-check the result against the 68-95-99.7 rule.
In a confidence interval or hypothesis test the candidate computes the interval or the statistic correctly but then interprets it wrongly: they say 'there is a 95% probability that the true mean lies in this interval', or they accept H0 because the sample mean is close to the hypothesised value without comparing the p-value or the statistic to the critical value, and they never state the conclusion in terms of the original problem.
The frequentist meaning of confidence and of the p-value is not understood, so the computed numbers are attached to the intuitive but incorrect statement. The decision rule is treated as optional because the arithmetic already feels like the answer.
What to do instead
Use fixed phrasing: 'We are 95% confident that the interval (a,b) contains the population mean', and 'If H0 were true, a statistic as extreme as the observed would occur with probability p; since p<α we reject H0'. Always write the decision (reject / fail to reject) and a sentence interpreting it in the context of the problem, including the direction of the alternative hypothesis.