PAU Applied Mathematics for Social Sciences II (Cantabria) common mistakes
PAU Applied Mathematics for Social Sciences II (Cantabria) · Understand the traps and how to avoid them.
When asked whether a matrix has an inverse, the candidate computes only the determinant of the coefficient matrix, finds it nonzero, and stops, without checking that the matrix is square or, in a parameter problem, without stating the excluded parameter values as a set.
Textbook exercises usually pair 'invertibility' with a square matrix, so the squareness condition gets absorbed into routine practice. In parameter problems the candidate is focused on solving det = 0 and forgets that the answer is the complement of those roots, and that a singular case must be discussed separately.
What to do instead
Adopt a fixed checklist before computing anything: (1) write the order of the matrix and confirm rows = columns; (2) compute det A as a polynomial in the parameter; (3) solve det A = 0 and write the solution set explicitly; (4) state 'A is invertible for all real k except k = ...' as a full sentence. In exam answers, always report the singular values separately and say what happens to the system at those values.
In Rouché–Frobenius discussions, the candidate compares the rank of the coefficient matrix with the rank of the augmented matrix but then mislabels the system: rank(A) = rank(A|b) = number of unknowns is called 'infinite solutions', or a consistent indeterminate system is declared incompatible because a free variable appeared.
The three-way classification (incompatible / determined / indeterminate) is memorised as a list rather than as a comparison of three quantities, so candidates collapse two of the conditions. The appearance of a parameter in the solution is confused with non-uniqueness of the system itself.
What to do instead
Write the three quantities side by side every time: r = rank(A), r* = rank(A|b), n = number of unknowns. Then apply: r < r* incompatible; r = r* = n unique; r = r* < n infinite with n − r parameters. Practise by constructing one 3×3 system of each type and labelling them, so the mapping becomes mechanical rather than recalled.
When solving a matrix equation of the form AX = B or AXB = C, the candidate multiplies on the wrong side, writing X = BA⁻¹ instead of X = A⁻¹B, or treats matrix division as if it existed.
Scalar algebra habits carry over: in ordinary numbers order does not matter, so 'move A to the other side' feels symmetric. Matrix multiplication is not commutative, and the candidate has not internalised that left and right inverses act differently.
What to do instead
Always derive the step explicitly: AX = B ⇒ A⁻¹(AX) = A⁻¹B ⇒ X = A⁻¹B, showing the multiplication applied to both sides on the left. For AXB = C, isolate in two steps: A⁻¹C B⁻¹. Before substituting numbers, check the orders of all factors to confirm the product is defined, which catches side errors immediately.
In linear programming, the candidate sets up the constraints correctly but then optimises over the wrong region: it uses x ≥ 0, y ≥ 0 as the only bounds, ignores a constraint written as 'at least' versus 'at most', or evaluates the objective only at the origin and one intercept instead of at every vertex of the feasible region.
The modelling step and the solving step are treated as separate tasks. Once the candidate starts drawing lines, attention shifts to the graph and the translation of the wording into inequality direction is no longer re-checked. Vertices are found by eye from the sketch rather than by solving the intersecting pairs of lines.
What to do instead
After writing the constraints, translate each back into a sentence and test one point (often the origin) to confirm the inequality direction. Then solve every pairwise intersection of boundary lines, discard points violating any constraint, and evaluate the objective at all remaining vertices, presenting them in a table. State the optimal value and the point, and mention the corner-point theorem as justification.
When computing a limit at a point, the candidate substitutes the value, obtains 0/0, and either writes 'the limit does not exist' or applies L'Hôpital's rule to a quotient that is not yet an indeterminate form, or applies it without verifying differentiability near the point.
The 0/0 case is treated as a final answer rather than as a signal to factor, rationalise, or use an equivalent infinitesimal. L'Hôpital is over-generalised from a memorised rule into a universal tool, and candidates do not check the hypotheses before differentiating numerator and denominator.
What to do instead
Follow a decision tree: substitute first; if 0/0 or ∞/∞, try algebraic simplification (factor, common denominator, conjugate) before L'Hôpital; use L'Hôpital only for genuine indeterminate quotients and only after noting numerator and denominator are differentiable near the point. For piecewise functions at the join, compute both one-sided limits separately and compare them with the function value before concluding continuity.
When asked for the domain of a function built from a logarithm, a square root, or a denominator, the candidate writes the domain of the outer expression only, for example giving x > 0 for ln(x² − 4) instead of solving x² − 4 > 0, or forgetting that a denominator must be nonzero after simplifying a common factor.
Domain questions are answered by pattern-matching on the visible function symbol rather than by imposing each structural condition in turn. Simplifying a rational expression first, then reading the domain off the simplified form, silently removes the excluded point.
What to do instead
List the restrictions of the original, unsimplified expression as a system of inequalities: radicand ≥ 0, logarithm argument > 0, denominator ≠ 0. Solve the system and intersect the solution sets. Never simplify before recording the domain; if you simplify afterwards, keep the original excluded points in the final answer and state them explicitly.
In optimisation and graph-analysis problems, the candidate finds stationary points by setting f′(x) = 0 but never checks the second derivative or the sign change, and, more seriously, omits checking the endpoints of the interval or the boundary of the feasible region, so a constrained maximum or minimum is missed.
The procedure 'derivative equals zero' is memorised as the complete method for extrema. The distinction between local and global extrema, and the fact that a continuous function on a closed interval attains extrema at endpoints, is understood in theory but not applied in practice.
What to do instead
Use a written protocol: (1) state the interval or constraint set; (2) compute f′ and solve f′ = 0 within it; (3) classify each stationary point by the sign of f″ or a sign table; (4) evaluate f at all stationary points and at both endpoints; (5) compare values and answer the question actually asked (maximum, minimum, or both). For word problems, define the variable and the function before differentiating, and confirm the answer lies in the physical domain.
When computing a definite integral, the candidate omits the dx, forgets to change the limits when using substitution, or applies Barrow's rule across a point where the integrand is not continuous, producing a finite number for what is actually a divergent improper integral.
The mechanics of finding an antiderivative absorb all attention, and the bookkeeping steps (differentials, limit transformation) are seen as cosmetic. Discontinuities inside the interval are not scanned for because most textbook integrals are well behaved.
What to do instead
Before integrating, scan the interval for points where the integrand is undefined or unbounded and split the integral there, treating each piece as a limit. When substituting, write u = g(x), du = g′(x) dx, and convert both limits in the same line, then never return to x. Always close the calculation with the evaluation bracket and the numerical or exact value, and state units for area problems.
In area problems between curves, the candidate computes the integral of the difference without first finding the intersection points, or takes the difference in the wrong order so that a region below the axis yields a negative area that is reported as the answer.
The integral is treated as automatically giving area, and the geometric meaning of a negative integral is forgotten. Intersection points are assumed to be the obvious values 0 and 1 from the exercise style rather than solved from f(x) = g(x).
What to do instead
Always begin by solving f(x) = g(x) to obtain the integration limits, and sketch enough to see which curve is above on each subinterval. Integrate (upper − lower) over each subinterval, splitting where the curves cross. If an integral comes out negative, treat it as a signal that the order is reversed, not as a final answer, and report areas as positive quantities.
In probability problems, the candidate confuses P(A|B) with P(B|A) or with P(A ∩ B), adds probabilities of non-mutually-exclusive events without subtracting the intersection, and treats sampling without replacement as if draws were independent.
Conditional probability notation is read left to right as a description rather than as a formal operation, and the multiplication rule is recalled in the wrong direction. The independence assumption is the default in simpler exercises and is not re-examined when the wording says 'without replacement'.
What to do instead
Write the formula before substituting: P(A|B) = P(A ∩ B)/P(B), and identify which event is the conditioning one from the phrase 'given that'. For unions use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and check whether the intersection is empty. For sampling without replacement, build a tree diagram or use the hypergeometric count, and verify that the total probability sums to one.