Catalonia PAU Mathematics II (Matemàtiques II - CIC 2026) common mistakes
Catalonia PAU Mathematics II (Matemàtiques II - CIC 2026) · Understand the traps and how to avoid them.
When asked to discuss the system AX=B as a function of a parameter k, the candidate computes det(A), sets it to zero, solves for k, and then declares 'if k≠2 the system is compatible determined' without ever checking whether the coefficient matrix and the augmented matrix have the same rank at the special values of k (or forgetting to substitute k back into the augmented matrix and compute the rank there).
Textbook drills on Cramer's rule train the reflex 'det≠0 ⇒ unique solution', so the singular cases feel like the only work left. The candidate treats rank analysis of the augmented matrix as a formality and stops at the determinant, which silently loses the compatible-indeterminate case that the exam is specifically testing.
What to do instead
Adopt a fixed three-line protocol and write it out every time: (1) rank(A) for the generic case; (2) substitute each critical k into the augmented matrix and compute rank(A|B) by Gaussian elimination, not by determinant; (3) compare ranks and, if equal and less than the number of unknowns, solve the reduced system to give the parametric family. Practise on systems where the critical value gives rank(A)=rank(A|B)=2 with 3 unknowns, so the indeterminate answer becomes automatic.
In a 3×3 determinant or an inverse, the candidate expands along a row or column and gets the cofactor sign pattern wrong (writing + − + / + − + instead of the checkerboard), or, when computing the adjugate, forgets to transpose the cofactor matrix before dividing by the determinant.
The sign rule and the transpose step are procedural details that are easy to hold in working memory but are not checked by any internal consistency test, so an error propagates silently into the inverse and then into the final solution of the system. Under time pressure the candidate writes the cofactors directly into the adjugate position.
What to do instead
Verify the inverse by multiplying A·A⁻¹ and checking for the identity before using it. For determinants, cross-check with Sarrus' rule on 3×3 (cheap and independent of the cofactor expansion) and with the property that swapping two rows flips the sign. Write the cofactor matrix, then physically rewrite it transposed on a new line before dividing.
Given three points or a point and two direction vectors, the candidate computes the plane by writing the general equation ax+by+cz=d and solving for a,b,c,d, but normalises incorrectly or loses the constant term, ending with a plane that passes through only two of the three points; alternatively, when asked for the relative position of two planes, the candidate compares only the normal vectors and answers 'parallel' when the planes are actually coincident or 'intersecting' when they are parallel.
The normal-vector comparison answers a question about directions, not about the affine position, and the constant term is exactly what distinguishes parallel from coincident planes. Candidates also under-use the cross product, so the normal is obtained by an error-prone linear system rather than directly.
What to do instead
Always compute the normal as the cross product of two vectors in the plane, then fix d by substituting one of the given points, and finally verify by substituting the other two points into the resulting equation. For relative positions, compare the rank of the coefficient matrix with the rank of the augmented matrix (same criterion as linear systems): rank 2/2 intersecting in a line, 1/1 coincident, 1/2 parallel and distinct.
When computing a distance or an angle, the candidate substitutes the wrong point: for the distance from a point P to a plane π, they plug P into the plane's equation but then also project P onto the plane using the plane's own normal starting from a point of the plane rather than from P, or when computing the distance between two skew lines they use the direction vector of one line instead of the cross product of both directions in the numerator of the formula.
The distance formulas look alike (point–plane, point–line, line–line) and share the same structure of a scalar triple product over a norm, so candidates memorise one template and apply it everywhere. The line–line formula in particular requires |(P₂−P₁)·(v₁×v₂)|/|v₁×v₂|, and dropping the cross product gives a meaningless number that is rarely sanity-checked.
What to do instead
Derive each formula once from the projection idea and keep a one-line derivation next to it: distance = |projection of the connecting vector onto the normal/direction|. Then do a plausibility check: a distance cannot exceed the distance between any two specific points you can compute directly (e.g. the distance between the two given points on the lines), and an angle between 0° and 90° for line–plane must use sin, not cos.
In a limits/continuity question with a piecewise function, the candidate computes the one-sided limits at the join point but then equates the limit to the function value using the wrong branch, or concludes 'continuous' after checking only that the two one-sided limits agree, without checking that the common value equals f(a).
Students conflate 'the limit exists' with 'the function is continuous', and the piecewise definition makes it easy to evaluate f(a) from the branch that does not contain a. The condition f(a)=lim is the one that actually fails in most exam items.
What to do instead
Write the three conditions explicitly for every join point a: lim(x→a⁻)f(x), lim(x→a⁺)f(x), and f(a), each on its own line with the branch used. Continuity requires all three equal. When a parameter appears, solve the equation from the two one-sided limits first, then substitute into the third condition to check consistency.
In an optimisation problem, the candidate sets up the objective function and constraint correctly but then solves f′(x)=0 and reports that critical point as the answer, without verifying it lies in the domain imposed by the constraint, and without using the second derivative (or an interval sign table) to confirm it is a maximum rather than a minimum.
The algebraic step f′(x)=0 is the most practised part, so it becomes the endpoint of the reasoning. The geometric/contextual constraints (lengths positive, perimeter fixed, x within a physical range) and the max/min classification are treated as obvious and skipped.
What to do instead
End every optimisation with an explicit closing paragraph: state the feasible interval for the variable, restrict the critical points to it, evaluate the objective at the endpoints and at interior critical points, and justify the extremum with f″(x) or a sign change. If the problem asks for dimensions, convert the optimal x back into the requested quantities and check units and that the perimeter/area constraint is satisfied.
When computing a definite integral of a function defined piecewise or involving an absolute value, the candidate integrates over the whole interval with a single expression, ignoring the point where the definition changes (e.g. integrating |x−2| from 0 to 4 as if it were x−2 throughout). Relatedly, in an area-between-curves problem they subtract in the wrong order and report a negative area.
The integrand is read as a formula rather than as a function with a domain, and the split point is not visually obvious once the absolute value has been 'simplified'. The sign issue arises because candidates trust the antiderivative rather than the geometry.
What to do instead
Before integrating, locate the zeros/sign changes of the integrand and split the integral at those points, writing each sub-integral with its own expression. For areas, sketch or at least determine which curve is above on each subinterval and integrate (upper − lower); if the result is negative, that is a signal that the order is wrong, not an answer to report.
In a Bayes/conditional-probability problem, the candidate computes P(A∩B) as P(A)·P(B) even though the events are not stated independent, or confuses P(A|B) with P(B|A) when the question is phrased with a causal story (test positive given disease vs. disease given positive test).
The multiplication rule for independent events is the first formula learned and is over-applied; the reversal error comes from reading the conditional in the order the events appear in the narrative rather than the order in the notation. The base-rate information is psychologically easy to discount.
What to do instead
Write the conditional probability definition P(A|B)=P(A∩B)/P(B) at the top of the solution and only replace P(A∩B) by P(A)P(B) after explicitly justifying independence. For diagnostic-test problems, draw the two-branch tree with prior probabilities on the first branches and label the target branch before computing; then check that the answer is plausible given the base rate (a rare disease usually gives a low posterior even with a good test).
When approximating a binomial by a normal (or using the normal in a sampling/intervals problem), the candidate applies the distribution without the continuity correction, or standardises with the sample standard deviation s instead of σ/√n, producing an interval that is too narrow.
The standardisation z=(x−μ)/σ is heavily drilled, so candidates copy the template and forget that the binomial is discrete (needing ±0.5) and that the sampling distribution of the mean has standard deviation σ/√n, not σ. The distinction between the population and the sample standard deviation is blurred by notation.
What to do instead
Write the model explicitly before standardising: XB(n,p) ⇒ μ=np, σ=√(npq), and P(X≤k)≈P(Z≤(k+0.5−μ)/σ). For means, state X̄N(μ, σ/√n) and use the standard error in the denominator. Sanity-check the interval width: it must shrink as n grows, and for a 95% interval the half-width should be about 1.96·σ/√n.
When asked for a confidence interval for the mean, the candidate uses z=1.96 for every confidence level and every sample size, ignores whether σ is known or estimated from the sample, and reports the interval without stating the confidence level or rounding consistently with the data.
1.96 is memorised as 'the 95% number' and the link to the quantile of the normal (or t) distribution is lost, so 90% and 99% questions are answered with the same value. The known/unknown σ distinction is often glossed over in the statement, and candidates do not look for it.
What to do instead
Build the interval from the formula x̄ ± z_{α/2}·σ/√n and look up the quantile from the confidence level: 1.645 (90%), 1.96 (95%), 2.576 (99%); use the t-quantile when σ is replaced by s and n is small. Always write the interval as (lower, upper) with the confidence level attached, and check that x̄ lies inside it.