Castilla-La Mancha PAU Applied Mathematics for Social Sciences II (Matemáticas Aplicadas a las Ciencias Sociales II - UCLM 2026) common mistakes
Castilla-La Mancha PAU Applied Mathematics for Social Sciences II (Matemáticas Aplicadas a las Ciencias Sociales II - UCLM 2026) · Understand the traps and how to avoid them.
When solving a 3×3 system by Gaussian elimination, the candidate multiplies a row by a number but forgets to apply the same operation to the independent-term column, so the final solution satisfies the transformed equations but not the original ones.
The candidate treats the augmented column as a label rather than as part of every row operation, usually because they write the system as a coefficient matrix plus a separate column and then operate only on the coefficient block. Under exam time pressure the augmented column is the first thing to be dropped.
What to do instead
Always write the full augmented matrix with a vertical bar and perform each elementary operation across the entire row, including the last entry. After obtaining the solution, substitute it back into the original (not transformed) equations; if any equation fails, the error is in the elimination, not in the arithmetic of the final division.
In a linear programming problem the candidate evaluates the objective function only at the vertices they can see clearly, or evaluates it at the intersection of every pair of lines including pairs whose intersection lies outside the feasible region, and reports the best of those values as the optimum.
The candidate remembers 'evaluate at the vertices' but not the precondition that the vertex must belong to the feasible region. Points outside the region often give larger (or smaller) objective values, so the wrong answer looks plausible and is not double-checked against the constraints.
What to do instead
Before evaluating, test each candidate vertex in all the inequality constraints; discard any point that violates one. Then build a small table with columns x, y, z and one row per feasible vertex, and state explicitly that by the fundamental theorem of linear programming the optimum over a bounded non-empty region occurs at a vertex. If the region is unbounded, also check whether the objective is bounded in the improving direction.
The candidate computes a one-sided limit correctly but declares the two-sided limit equal to it without checking the other side, or writes that the limit is infinite when in fact the left and right limits have opposite signs (or one is finite and the other infinite).
Piecewise-defined functions and functions with a denominator that vanishes are usually tested precisely at the break point, and candidates carry over the expression used on one side. The notation lim f(x) is read as 'plug in the value' rather than as a two-sided requirement.
What to do instead
At every point of interest, write two separate lines: the limit as x approaches from below and from above, each with the correct branch of the definition. Conclude that the two-sided limit exists only if both one-sided limits are finite and equal; otherwise state that the limit does not exist and say why (different values, or divergence to opposite infinities).
When asked about monotonicity or extrema, the candidate solves f'(x)=0, lists those roots as the intervals of increase and decrease, and never checks the sign of f' between the roots or the points where f' is undefined.
Solving an equation feels like a complete procedure, so the sign analysis is treated as optional commentary. The candidate also conflates 'critical point' with 'extremum' and assumes every root of f' is a turning point.
What to do instead
Draw a sign chart: mark on a number line every point where f'=0 or f' does not exist, pick one test value in each resulting interval, and record the sign of f' there. Then translate: positive means increasing, negative means decreasing, and a sign change at a critical point means an extremum (plus to minus is a maximum, minus to plus is a minimum). A root with no sign change is an inflection point with horizontal tangent, not an extremum.
The candidate finds a critical point of a word problem, obtains a numerical value, and presents it as the answer without verifying that it is the maximum (or minimum) requested, and without checking it lies in the admissible domain of the problem.
Optimisation word problems require translating a constraint into a relation between two variables and then reducing to one variable; once the algebra is done the candidate is relieved and stops. The second-derivative test or endpoint comparison is skipped, and negative or out-of-range values are accepted.
What to do instead
After obtaining the critical point, do three things explicitly: (1) state the domain of the single-variable function implied by the problem (lengths positive, quantities non-negative, etc.) and discard critical points outside it; (2) apply the second-derivative test or compare function values at the critical point and at the endpoints of the domain; (3) answer the question actually asked (the dimensions, the price, the quantity), not merely the value of the auxiliary variable.
When computing a definite integral the candidate finds an antiderivative and substitutes the upper limit only, or subtracts in the order lower minus upper, producing a negative area for a region above the axis.
The bar notation F(x)] from a to b is written quickly and the lower-limit evaluation is dropped or its sign lost. Candidates also believe an area can be negative and therefore do not notice the error.
What to do instead
Write the evaluation in full as F(b) − F(a) with both substitutions visible before simplifying. Sanity-check the sign: a region lying above the x-axis must give a positive result. For regions that cross the axis, split the integral at the zeros of the function and take the absolute value (or negate) the parts below the axis before adding.
For the area between two curves the candidate integrates the difference in the wrong order, or integrates from the smaller x-intercept to the larger without locating the actual intersection points, so the limits of integration do not correspond to the enclosed region.
The candidate substitutes the intersection points of one curve with the axes for the intersection of the two curves. The habit of 'integrate the upper minus the lower' is applied without first determining which function is on top on each subinterval.
What to do instead
Solve f(x)=g(x) first and use those solutions as the limits. On each interval between consecutive intersection points, test one interior point to see which function is larger, and integrate (larger − smaller). If the curves cross inside the interval, split the integral at the crossing point and add the absolute values of the pieces.
When computing the correlation coefficient or the regression line, the candidate uses the standard deviations or the covariance computed with n in one place and n−1 in another, or mixes the marginal means with the conditional means, producing a slope that does not pass through the point of means.
Textbooks and calculators differ on whether the denominator is n or n−1, and the candidate copies formulas from different sources without noticing. The property that the least-squares line passes through (x̄, ȳ) is not used as a check.
What to do instead
Fix one convention at the start of the exam (for this exam, use the population formulas with denominator n unless told otherwise) and write it down. Then always verify that the regression line satisfies ȳ = a + b·x̄, and that the sign of the slope equals the sign of the covariance and of r. If the check fails, the arithmetic is wrong.
Given P(A), P(B) and P(A∩B) or a two-way table, the candidate computes P(A|B) as P(A∩B)/P(A) instead of P(A∩B)/P(B), or claims independence because P(A∩B) is small rather than because it equals P(A)·P(B).
The conditional probability is read as 'A given B' but the denominator is chosen by habit from the numerator's first factor. Independence is confused with 'the events do not overlap' or 'the events are unrelated in the real world', which is a conceptual rather than arithmetic error.
What to do instead
Write the definition with the conditioning event underlined: P(A|B) = P(A∩B)/P(B), the denominator being the probability of what is given. To test independence, compute P(A)·P(B) and compare it with P(A∩B) numerically; equality (within rounding) means independent, inequality means dependent. For Bayes' theorem, draw the tree or the partition first and label each branch with its conditional probability.
With a normal distribution the candidate uses the value x directly in the standard normal table instead of the standardised value z=(x−μ)/σ, or forgets the continuity correction when a binomial is approximated by a normal, and reports a probability greater than 1 or a complement with the wrong tail.
The table is indexed by z but the problem is stated in the original units, and the standardisation step is done mentally and skipped. Continuity correction (using x±0.5) is taught briefly and easily forgotten; tail errors arise because the symmetry of the normal is assumed rather than checked.
What to do instead
Always write the standardisation explicitly: z=(x−μ)/σ, showing the substitution. For a binomial approximated by N(np, √(np(1−p))), adjust the endpoint by 0.5 in the direction of the inequality (P(X≤k) uses k+0.5, P(X≥k) uses k−0.5). Then sketch the bell curve, shade the required region, and check that the answer lies in [0,1] and matches the shaded area (more than half, less than half, or near 1).