Madrid PAU Applied Mathematics for Social Sciences II common mistakes
Madrid PAU Applied Mathematics for Social Sciences II · Understand the traps and how to avoid them.
When solving a system by Gaussian elimination or when checking whether a matrix is invertible, the candidate stops at a row of the form (0 0 … 0 | b) with b ≠ 0 and declares the system has a unique solution or 'no problem', instead of declaring it inconsistent.
Rote application of row reduction without interpreting the final echelon form; candidates focus on arithmetic and never translate the last row back into an equation, so the contradiction 0 = b goes unnoticed.
What to do instead
Before writing the answer, always re-read the final matrix row by row as equations. Train a fixed checklist: a pivot-less row with nonzero constant means no solution (inconsistent); pivot-less row with zero constant means infinitely many solutions with n − rank free parameters; full column rank means unique solution. Apply it explicitly in every practice problem, writing the sentence 'Row 3 reads 0 = 5, therefore incompatible' before the final answer.
In a system with a parameter a (e.g., a 3×3 system whose determinant is a polynomial in a), the candidate computes the determinant, sets it equal to zero, and then either ignores the special values of a or studies only the determinant without separately analysing the cases where the system is consistent with infinitely many solutions.
The determinant only distinguishes unique solution from the singular case; candidates confuse 'det = 0' with 'no solution' and forget that singular systems can be compatible indeterminate, requiring a separate rank discussion.
What to do instead
Adopt a rigid three-branch scheme for parameter problems: (1) det ≠ 0 → unique solution by Cramer or inverse; (2) for each value of a with det = 0, substitute that value back into the original system and reduce it to decide compatible indeterminate vs incompatible; (3) state the ranks of coefficient and augmented matrices in each branch. Practise writing the final answer as a case table so no branch can be skipped.
In linear programming, the candidate evaluates the objective function only at the vertices obtained from the intersection of constraint boundary lines, but does not first verify which of those intersection points actually satisfy all the inequalities, and reports an infeasible point as the optimum.
Candidates treat every pairwise intersection of boundary lines as a vertex of the feasible region, forgetting that the feasible region is the intersection of half-planes and that many intersection points lie outside it.
What to do instead
Draw the feasible region first (shading the excluded side of each constraint), then read the vertices off the drawing, and only afterwards solve the 2×2 systems for those vertices. Always substitute each candidate vertex into every inequality as a verification step before evaluating the objective. If the region is unbounded, additionally check whether the objective can grow without bound along the open direction.
When asked for the maximum of a function on a closed interval or in an optimization word problem, the candidate finds f′(x) = 0, solves for the critical points, and declares the largest of those values the maximum, without evaluating the function at the endpoints of the interval or checking the domain restriction.
Candidates memorise 'optimum at f′ = 0' from single-variable calculus and do not connect it to the closed-interval theorem; the endpoint candidates are invisible because they are not roots of the derivative.
What to do instead
Use a written protocol: list all candidates = critical points inside the domain ∪ endpoints ∪ points where f′ does not exist; evaluate f at each; compare. In word problems, first write the domain imposed by the context (lengths positive, quantities nonnegative) and check whether the optimum lies inside it. Verify with the second derivative or a sign table only to classify, not to replace the comparison of values.
In limits, the candidate substitutes the value into the expression, obtains an indeterminate form such as 0/0 or ∞/∞, and then either writes 'does not exist' or cancels terms incorrectly (e.g., cancelling across a sum), instead of resolving the indeterminacy by factoring, rationalising, or applying L'Hôpital's rule correctly.
Indeterminate forms look like a final answer to a candidate who has not internalised that 0/0 is a signal to transform the expression; algebraic cancellation errors compound the problem.
What to do instead
Treat every 0/0 or ∞/∞ as a trigger, not an answer. Drill the three standard techniques and when each applies: factor and cancel for polynomials, multiply by the conjugate for radicals, L'Hôpital only after confirming differentiability and the 0/0 or ∞/∞ hypothesis, and divide by the highest power for limits at infinity. After transforming, substitute again and state the resolved value explicitly.
When computing a definite integral that requires substitution or integration by parts, the candidate changes the variable in the integrand but keeps the original limits of integration in x, or forgets the differential dx and the constant factor from the substitution.
The mechanics of the antiderivative absorb all attention, and the change of limits is treated as optional bookkeeping rather than a necessary part of the substitution.
What to do instead
Adopt one of two disciplined routines and never mix them: either (a) change the limits when you change the variable (compute the new limits u(a), u(b) and never return to x), or (b) find the antiderivative in x and only then apply the original limits. Always write 'u = …, du = … dx' explicitly so the differential and constant are accounted for. Check the sign and magnitude of the result against the geometric area when possible.
In area problems between two curves, the candidate computes a single integral of the difference of the functions over an interval where the curves cross, obtaining a small number or zero, instead of splitting the interval at the intersection points and integrating the absolute difference.
The formula 'area = ∫(f − g)' is applied mechanically; the candidate does not check which function is on top, and a sign change inside the interval silently cancels area.
What to do instead
Always begin by finding the intersection points and sketching which curve is above. If the upper/lower order changes at an interior point c, split the integral into ∫(upper − lower) on each subinterval. As a sanity check, the area must be nonnegative and must roughly match the sketch; a result of zero for two visibly distinct curves signals a missed split.
In conditional probability, the candidate computes P(A|B) by writing P(A)·P(B) or P(A)/P(B), or confuses P(A|B) with P(B|A), especially in Bayes-type problems about tests with false positives.
The notation P(A|B) is read as a product or a ratio of the wrong events; the definition is memorised as a formula without the conditioning event in the denominator, and the base-rate information is ignored.
What to do instead
Write the definition explicitly every time: P(A|B) = P(A∩B)/P(B), and identify the conditioning event B before touching numbers. For test/Bayes problems, build a two-way table or tree with the population base rate, then read the required conditional probability off the table. Check that the answer is plausible: with a rare disease, a positive test does not imply a high probability of disease.
When approximating a binomial by a normal distribution, the candidate uses the mean np and standard deviation np(1−p) (variance instead of standard deviation) or omits the continuity correction, and computes P(X ≤ k) as P(Z ≤ (k − np)/σ) without adjusting k.
The formulas for mean and variance of the binomial are recalled but the square root is dropped, and the continuity correction is perceived as an optional refinement rather than a required step when a discrete variable is modelled continuously.
What to do instead
Write the standardisation in full each time: Z = (X − np)/√(np(1−p)), and for P(X ≤ k) use the corrected value k + 0.5, for P(X ≥ k) use k − 0.5, for P(a ≤ X ≤ b) use a − 0.5 and b + 0.5. Also state the validity condition np > 5 and n(1−p) > 5 before applying the approximation, and keep the sign convention straight with a quick sketch of the normal curve.
When constructing a confidence interval for a mean, the candidate uses the sample standard deviation in place of the population standard deviation without switching to the t distribution, or misreads the confidence level (using z = 1.96 for 90% or z = 1.645 for 95%), and reports the interval as [x̄ − E, x̄ + E] with the wrong margin of error E = z·σ instead of z·σ/√n.
The formula is recalled in fragments; the role of √n in the standard error and the correspondence between confidence level and critical value are not consolidated, and the small-sample case is not distinguished from the large-sample case.
What to do instead
Memorise the interval as x̄ ± z_{α/2}·σ/√n and, when σ is unknown and n is small, x̄ ± t_{n−1,α/2}·s/√n. Keep a reference table of critical values (90%: 1.645, 95%: 1.96, 99%: 2.576) and always write the confidence level next to the critical value you use. Interpret the result in words: 'we are 95% confident the population mean lies in this interval', and check the interval is centred on x̄ with the correct half-width.