Canary Islands PAU Mathematics II / Matemáticas II COPAU ULPGC ULL common mistakes
Canary Islands PAU Mathematics II / Matemáticas II COPAU ULPGC ULL · Understand the traps and how to avoid them.
When computing a determinant or checking the rank of a matrix that contains a parameter, the candidate substitutes one convenient value of the parameter, decides the rank from that single case, and never separates the values that make the determinant zero from those that do not.
Drills on numerical matrices train a reflex of 'compute and answer', so the candidate treats the parameter as an ordinary number. The idea that the answer is a case split (det ≠ 0 versus det = 0, and inside det = 0 further sub-cases where lower-order minors may still be non-zero) is not practised enough, and the candidate is afraid of leaving a symbolic answer.
What to do instead
Adopt a fixed written protocol: (1) compute det(A) as a polynomial in the parameter and factor it; (2) list the roots; (3) for each root substitute back into A and compute the rank by minors of decreasing order, showing which minor is non-zero; (4) only then discuss the system. Practise at least ten parameter problems where the answer must be written as 'if a ≠ …, rank = 3; if a = … rank = 2 because minor M ≠ 0'. Never write a final answer without the case split.
After applying Rouché–Frobenius and finding rank(A) = rank(A|b) = r < n, the candidate writes 'infinitely many solutions' and stops, without giving the solution set, or writes a single particular solution as if it were the general solution.
The candidate confuses the classification of the system with the solving of it, and is unsure how to express a free parameter. In exam pressure, the effort spent on ranks feels like the whole task, so the answer is truncated.
What to do instead
Train the full template: set the free variable(s) equal to a parameter t (or λ, μ), solve the remaining r×r subsystem by substitution or Cramer, and present the result as a vector (x, y, z) = (…, …, …) + t(…, …, …). For homogeneous systems always state that the trivial solution exists and add the non-trivial family. Check the final family by substituting one value of t back into the original equations.
In a relative-position problem, the candidate computes the determinant of the matrix formed by the three direction vectors (or by two directions and the segment joining a point of each line), gets zero, and immediately concludes 'the lines are parallel' or 'the lines intersect', without distinguishing the coincident, parallel, intersecting and skew cases.
The candidate memorises 'det = 0 means dependent' but not the decision tree that follows. The distinction between parallel and skew requires an extra check (whether a point of one line satisfies the other), and that step is easily forgotten because it is not a computation but a logical test.
What to do instead
Memorise the tree explicitly: for lines, first compare direction vectors u and v. If u ∥ v, test whether a point of r satisfies s: if yes, coincident; if no, parallel. If u is not parallel to v, compute det(u, v, PQ) where P ∈ r, Q ∈ s: zero means they intersect (then find the point), non-zero means they are skew. For line–plane, compare u with the normal n: if u·n = 0 test a point (parallel or contained), otherwise they cut. Write the branch you are in before computing.
When asked for the distance from a point to a line or to a plane, the candidate uses the point's coordinates directly in a formula built for a plane through the origin, or computes the distance to the line's direction vector instead of to the line, and reports a number without checking it against an obvious geometric bound.
The distance formulas look similar (|AP × u|/|u| for a line, |ax₀+by₀+cz₀+d|/√(a²+b²+c²) for a plane) and are memorised as strings rather than derived. The candidate also forgets that the plane formula requires the normal vector, not any vector in the plane.
What to do instead
Derive both once by hand so the structure is understood: point–plane is the projection of AP onto the unit normal; point–line is the area of the parallelogram divided by the base. Then always identify in writing: 'normal n = (a,b,c)', 'direction u = (…)', 'AP = …'. Sanity-check: the distance to a plane containing the origin's parallel through the point must be ≤ |AP|; a distance larger than the distance between the given points signals an error.
When computing a limit that leads to an indeterminate form, the candidate substitutes the value, writes '0/0' or '∞/∞', and then either declares the limit does not exist or applies L'Hôpital without first checking that the hypotheses (differentiability near the point, genuine 0/0 or ∞/∞) hold, and without simplifying.
L'Hôpital is over-practised as a universal tool, so algebraic routes (factoring, conjugates, notable limits, equivalent infinitesimals) are under-used. The candidate also forgets that a one-sided limit may differ, and that a limit at infinity of a rational function is decided by degrees, not by derivatives.
What to do instead
Before any derivative, classify the form and try in order: direct substitution, algebraic simplification or conjugate, division by the highest power, notable limits (sin x/x → 1, (1+1/n)^n → e). Use L'Hôpital only for genuine 0/0 or ∞/∞ and only after simplifying. For piecewise or absolute-value limits, compute both one-sided limits separately and state them. Finish by writing the value, or 'does not exist because the lateral limits are … and …'.
In a continuity-and-differentiability problem with a piecewise function containing parameters, the candidate imposes continuity at the join point, finds the parameter, and then assumes differentiability follows, without imposing equality of the one-sided derivatives.
The theorem 'differentiable implies continuous' is remembered in the wrong direction. The candidate thinks continuity is the stronger condition, so once it holds the problem seems finished.
What to do instead
Write the two separate conditions as two separate equations: continuity gives f(a⁻) = f(a⁺); differentiability gives f′(a⁻) = f′(a⁺), where each lateral derivative is computed from its own branch (using the definition if the branch is not differentiable by rules at that point). Solve the system of the two equations for the two parameters, and state explicitly that continuity is necessary but not sufficient.
In an optimisation problem the candidate differentiates, sets f′(x) = 0, solves, and reports the critical point as the answer, without verifying it lies in the domain imposed by the geometry, without checking it is a maximum rather than a minimum, and without answering the question actually asked (the dimensions, not just x).
Textbook exercises often have a single critical point, so the habit of stopping there is reinforced. The candidate also forgets to translate the constraint into the domain (lengths positive, perimeter fixed) and treats the objective function as unconstrained.
What to do instead
Follow a five-step written scheme: (1) name the variables and write the constraint; (2) express the objective as a function of one variable and state its valid interval; (3) differentiate and solve; (4) justify the extremum with the second derivative or a sign table on the interval, and check the endpoints if the interval is closed; (5) answer in the units and quantities requested. Underline the final sentence so the requested quantity is not left as 'x = 3'.
When computing a definite integral of an absolute value, a piecewise function, or a function that changes sign on the interval, the candidate integrates in one piece from a to b and, in area problems, reports the value of the integral as the area even when part of the region lies below the axis.
The candidate treats 'integral' and 'area' as synonyms. The geometric meaning of the signed integral is understood in theory but not applied when the integrand is not obviously positive.
What to do instead
Before integrating, find the zeros of the integrand in [a,b] and split the interval there, or split at the definition change of the piecewise function. For areas, compute each sub-integral and take absolute values before adding; for the area between two curves, integrate |f − g| and split where f = g. State the split points explicitly in the solution so the marker sees the reasoning.
In conditional-probability problems the candidate confuses P(A|B) with P(B|A), or multiplies P(A)·P(B) for events that are not independent, and in Bayes-type problems forgets to divide by the total probability of the observed event.
The notation is read left-to-right as 'A and then B', and independence is assumed because the multiplication rule is the only one the candidate has automated. The denominator of Bayes' formula (the partition of the sample space) is the step most often omitted.
What to do instead
Always draw a tree or a two-way table before computing, and label branches with conditional probabilities. Write the formula with the condition in words: 'probability of A given that B has occurred = P(A∩B)/P(B)'. For Bayes, compute the total probability of the observed event as the sum over all branches first, then divide. Test independence only by checking P(A∩B) = P(A)P(B) numerically, never by assumption.
In normal-distribution problems the candidate uses the value of z as the probability, or forgets the continuity correction when a binomial is approximated by a normal, or uses the wrong tail (P(Z ≤ z) when P(Z ≥ z) is required).
The standard normal table gives an area, but the candidate reads the z-score row as the answer. The correction for continuity (±0.5) is a small technical detail easily dropped, and the symmetry P(Z ≥ z) = 1 − P(Z ≤ z) is applied in the wrong direction.
What to do instead
Standardise with z = (x − μ)/σ, then draw the bell curve and shade the region asked for before touching the table. Convert every 'at least', 'at most', 'between' into an inequality on Z and, for the binomial approximation, adjust the endpoint by 0.5 in the direction that includes the values. Use symmetry explicitly: P(Z ≥ z) = 1 − P(Z ≤ z) and P(Z ≤ −z) = 1 − P(Z ≤ z). Check that the answer lies in [0,1] and is consistent with the shaded picture.