PAU Applied Math ACS II (Galicia) common mistakes
PAU Applied Math ACS II (Galicia) · Understand the traps and how to avoid them.
When computing an inverse or solving a system by row reduction, the candidate divides only the first entry of a row by the pivot instead of every entry in that row, then carries the wrong matrix forward and reports a plausible but incorrect inverse or solution.
The candidate learned the sequence of row operations as a list of gestures ("divide row 1 by the pivot") without internalising that a row operation acts on the whole equation, so attention narrows to the pivot cell. Under exam time pressure the eye fixates on the number being eliminated and the rest of the row is copied unchanged.
What to do instead
Before every exam, do ten full Gauss-Jordan reductions writing each operation above the arrow (e.g. F2 -> F2 - 3F1) and re-reading the entire resulting row aloud. Adopt the habit of checking the product A·A⁻¹ = I (or substituting the solution back into every original equation) as a compulsory last step, since a single unmodified entry is immediately exposed.
The candidate computes a 3×3 determinant by Sarrus' rule but forgets that the two added diagonal products must be subtracted, or expands a determinant along a row without applying the alternating signs, producing the negative of the correct value.
Sarrus' rule and cofactor expansion are visually similar patterns of diagonals and signs, and candidates memorise the picture rather than the algebra. The sign pattern (+ - +) is easily reversed because the first term is always positive and the mind generalises 'positive first' to the whole expansion.
What to do instead
Write the sign matrix explicitly above the determinant before expanding, and verify with a property check: swapping two rows must flip the sign, and a determinant with two equal rows must be zero. Practise by computing one determinant three ways (Sarrus, cofactor along a row, cofactor along a column) and confirming identical results.
In a limit question the candidate substitutes the value into the expression, obtains 0/0, and writes that the limit does not exist, instead of recognising the indeterminate form and factorising, rationalising or applying L'Hôpital's rule.
Early limits were taught by direct substitution, and 0/0 looks like a legitimate arithmetic answer. The candidate has not built the reflex that an indeterminate form is a signal to transform the expression, not a conclusion, and often does not check whether the function is continuous at the point.
What to do instead
Train a fixed decision tree: substitute first; if the result is a real number, that is the limit (continuity); if it is k/0 with k≠0, the limit is infinite; if it is 0/0 or ∞/∞, factorise or use L'Hôpital. Drill twenty mixed limits where the first step is always to classify the form, and require the classification to be written on the paper before any manipulation.
Asked for the tangent line to a curve at a point, the candidate computes f'(a) and then writes y = f'(a), or writes y = f'(a)(x − a) omitting the constant f(a), so the line passes through the origin rather than the point of tangency.
The derivative is the headline object of the topic, and the candidate conflates 'the slope at a point' with 'the line at a point'. The point-slope formula is recalled partially because the candidate never checks that the answer must satisfy two conditions: it passes through (a, f(a)) and has slope f'(a).
What to do instead
Always write the template y − f(a) = f'(a)(x − a) and fill in the three numbers explicitly before simplifying. Self-check by substituting x = a and confirming y = f(a), and by confirming the slope equals f'(a). Do five tangent-line problems where the check is part of the required working.
In an optimisation problem the candidate finds f'(x) = 0, solves for x, and declares that point the maximum or minimum without computing f''(x) or checking the endpoints of the interval, so a local minimum or a boundary extremum is reported as the global answer.
The derivative-equals-zero procedure is over-learned as 'the' method for extrema, while the second-derivative test and endpoint evaluation are treated as optional extras. In social-science contexts the domain is often a closed interval (a budget, a production range), and the true optimum frequently sits at an endpoint where the derivative is not zero.
What to do instead
Use a written three-step protocol: (1) state the domain of the variable, (2) list critical points from f'(x)=0 and any points where f' does not exist, (3) evaluate f at all of these plus both endpoints and compare values. Require the comparison table to appear in the answer; the second derivative then only classifies interior points.
When computing a definite integral the candidate omits the limits of integration, integrates and substitutes in the wrong order, or drops the minus sign from ∫ xⁿ dx with a negative exponent, reporting a negative area for a region above the axis.
The antiderivative is computed as the main task and the evaluation step is treated as clerical, so sign errors and order errors slip through. The candidate also forgets that area between curves requires identifying which function is upper and subtracting in the correct order, and that a definite integral can legitimately be negative while an area cannot.
What to do instead
Write the definite integral with limits on every line until the final substitution, and always compute as F(b) − F(a) with b the larger x-value. For area problems, sketch the region, mark upper and lower functions, and if the computed value is negative, stop and re-examine the order. Add a sanity check: compare the result with a rough rectangle estimate from the sketch.
Given P(A), P(B) and P(A|B), the candidate computes P(A∩B) by multiplying P(A)·P(B) instead of P(B)·P(A|B), or in a Bayes problem inverts the conditional and uses P(A|B) where P(B|A) is required.
Independence is taught early and the multiplication rule P(A∩B)=P(A)P(B) is the formula most strongly associated with 'and', so it is applied even when the events are not independent. Conditional probability notation is read casually, and the direction of the conditioning is not treated as part of the meaning.
What to do instead
Before any calculation, write the conditional probability definition P(A|B)=P(A∩B)/P(B) and rearrange to P(A∩B)=P(B)·P(A|B), which makes clear that the conditioning event is the multiplier. Test independence explicitly by checking whether P(A|B)=P(A); if not, never use the product of marginals. In Bayes problems, draw the two-stage tree and label each branch with the conditional in the direction of the arrow.
For a normal distribution the candidate uses the raw value x with the standard normal table instead of standardising with z = (x − μ)/σ, or mishandles the complement, reporting P(X > a) as Φ(z) rather than 1 − Φ(z).
The table is a lookup exercise and the standardisation step is seen as an optional preliminary. Symmetry of the normal curve is understood qualitatively but not translated into the operational rules Φ(−z)=1−Φ(z) and P(X>a)=1−P(X≤a), so tail probabilities come out inverted.
What to do instead
Force the habit of writing z = (x − μ)/σ on its own line with the numbers substituted, before touching the table. Draw a small sketch of the curve, shade the required region, and label it with the probability expression. Then convert the shaded region into Φ form using the complement rule, and check that the answer lies between 0 and 1 and is less than 0.5 for a far tail.
In bivariate data the candidate computes the correlation coefficient but then interprets it as proof of causation, or uses the regression line to predict far outside the observed range of x without comment, and confuses the regression of y on x with the regression of x on y.
The arithmetic of the correlation coefficient and the regression line is mastered, but the interpretive cautions are treated as exam trivia rather than as part of the answer. The two regression lines share the same r but have different slopes, and candidates do not distinguish which variable is being predicted.
What to do instead
Prepare a standard interpretive sentence to attach to every correlation answer: strength and direction of the linear association, with an explicit statement that association does not imply causation and that predictions are reliable only within the observed range. Always state which variable is dependent before computing the slope b = cov(x,y)/var(x). Check that the regression line passes through the point of means.
In a linear programming problem the candidate graphs the constraints, shades the feasible region, but then evaluates the objective function only at the vertices that are easiest to read, or optimises over the wrong side of a constraint because the inequality direction was not tested with a point.
The graphical work is visually persuasive and the candidate trusts the picture without verifying which half-plane satisfies each inequality. Vertices formed by the intersection of two constraint lines are laborious to compute, so the candidate substitutes only the intercepts, which may lie outside the feasible region.
What to do instead
For each constraint, substitute the origin (0,0) and record whether the inequality holds, shading the correct side accordingly. Then list every vertex of the feasible polygon by solving the pairs of lines that meet there, and build a table of vertex coordinates with the objective value. State the optimal vertex and the optimal value explicitly, and confirm that the chosen vertex satisfies all constraints.