Castilla y León PAU Mathematics II common mistakes
Castilla y León PAU Mathematics II · Understand the traps and how to avoid them.
When computing an inverse or solving AX=B, the candidate divides by a matrix (writes B/A) or multiplies on the wrong side, producing X = A⁻¹B where X = BA⁻¹ was required, or writing X = B·A⁻¹ for AX = B.
Scalar algebra habits transfer unconsciously: division feels like the natural undo of multiplication, and the non-commutativity of matrix products is stated in class but rarely practised on non-symmetric examples, so the side of the multiplication is never checked.
What to do instead
Before any matrix manipulation, write the equation symbolically and 'peel off' A by multiplying both sides on the same side: AX = B gives A⁻¹AX = A⁻¹B, so X = A⁻¹B. Deliberately practise with a non-symmetric 2×2 or 3×3 A and verify by substituting back into the original equation, not into the rearranged one.
In a discussion of a system with a parameter, the candidate computes det(A)=0, finds the values of the parameter, and stops, declaring those cases 'incompatible' without substituting back to distinguish compatible determined, compatible indeterminate and incompatible.
The determinant only detects loss of uniqueness, but students memorise 'det = 0 means no solution' as a shortcut, and the substitution step is the most laborious part, so it is skipped under time pressure.
What to do instead
Adopt a fixed three-branch template: (1) det(A) ≠ 0 → unique solution by Cramer or Gauss; (2) det(A) = 0 → substitute each parameter value into the augmented matrix and row-reduce, then apply Rouché–Fröbenius (rank A vs rank augmented). Always write the conclusion as a sentence naming the parameter values and the type of system for each.
When a line and a plane, or two lines, are studied for relative position, the candidate solves for the intersection parameter and, on finding no solution, concludes 'they are parallel' without distinguishing parallel from skew (for lines) or checking whether the line lies inside the plane.
In two dimensions 'no intersection' means parallel, and that intuition is carried into 3D, where skew lines and contained lines are new cases that have no 2D analogue.
What to do instead
Classify systematically: for two lines, compare direction vectors first (proportional or not); if proportional, test one point of each line for coincidence; if not proportional, solve the 3-equation system and check consistency - inconsistent means skew. For line–plane, substitute the parametric line into the plane equation: one root = cuts, no root = parallel, identity = contained.
Computing a distance (point to plane, point to line, between parallel planes) with a formula that omits the absolute value or normalises with the wrong vector, yielding a negative distance or a value that is not the perpendicular distance.
Formulas are memorised as strings of symbols without the geometric derivation, so the normalisation step (dividing by the norm of the normal or direction vector) is dropped, and the absolute value is forgotten because arithmetic 'works' without it.
What to do instead
Re-derive each distance once from the projection idea: distance point–plane = |AP·n|/|n| with n the plane normal; distance point–line = |AP × v|/|v| with v the line direction. Keep a one-line note of which vector must be normalised, and sanity-check that the answer is non-negative and plausible against a rough sketch or coordinate estimate.
Applying L'Hôpital's rule to a limit that is not an indeterminate form (for example 0/0 assumed when the numerator tends to 3), or applying it repeatedly without re-checking the form after each step, and mixing it with algebraic simplification incorrectly.
L'Hôpital is treated as a universal limit machine; the prerequisite check of the 0/0 or ∞/∞ form is a single line that feels optional, and after differentiating once students forget to re-evaluate the form.
What to do instead
Write the substitution value explicitly above the limit before doing anything: state the form (0/0, ∞/∞, 0·∞, ∞−∞, 1^∞) and only then choose a method. After each L'Hôpital step, re-substitute and re-state the form. For 0·∞ and ∞−∞, rewrite as a quotient or combine fractions first; for 1^∞ use logarithms or the (1+1/n)^n limit.
In a continuity or differentiability question with a piecewise function and a parameter, the candidate equates the two branches at the join point but never checks that the lateral derivatives coincide, or checks derivatives but not continuity, and concludes differentiability from continuity alone.
The two conditions are taught together but examined separately, and the derivative of each branch is computed mechanically; students assume 'the pieces match' is enough, forgetting that differentiability requires both continuity and equality of lateral derivatives.
What to do instead
Use a fixed checklist at the join point x = a: compute f(a⁻) and f(a⁺) and set equal (continuity); then compute f′(a⁻) and f′(a⁺) from the branch derivatives and set equal (differentiability). Present both equations explicitly, solve the parameter system, and state that continuity is necessary but not sufficient for differentiability.
When computing a definite integral by substitution, the candidate changes the variable but keeps the original limits of integration, or forgets to return to the original variable after an indefinite substitution and mixes x and t in the same expression.
The mechanical substitution steps are rehearsed, but the bookkeeping of limits is treated as an afterthought, and under exam pressure the final antiderivative is evaluated at the old limits because they are already written down.
What to do instead
Choose one of two disciplined routines and always use it: either change the limits when changing the variable (write 'x = a → t = …' explicitly) and never return to x, or keep the limits and back-substitute before evaluating. Underline the variable in the limits each time and mark the final answer with the original variable.
In area-between-curves problems, the candidate integrates f − g over an interval without finding the intersection points, so a region where the curves cross is counted with the wrong sign and the total area is understated; or the area is reported as a negative number.
Area is treated as 'integral of the function', and the geometric meaning (integral of the upper minus the lower curve, split at crossings) is lost; the sign convention of the definite integral is not connected to the area interpretation.
What to do instead
Always begin by solving f(x) = g(x) to get the integration limits and any interior crossing points, then split the integral at every crossing and integrate |f − g| on each subinterval, deciding which curve is on top by testing a midpoint. Report a positive value and sketch the region to confirm the number of pieces.
In conditional probability, the candidate computes P(A∩B) by multiplying P(A)·P(B) without checking independence, or confuses P(A|B) with P(B|A) (the prosecutor's fallacy), or uses the total-probability denominator incorrectly in Bayes' theorem.
The multiplication rule for independent events is over-learned and applied to dependent events; the notation P(A|B) is read as 'A given B' but the conditional direction is not tracked through the formula, and the partition in the denominator is often omitted.
What to do instead
Before multiplying, state whether the events are independent and justify it (or use P(A∩B) = P(A)·P(B|A) when dependent). For conditional and Bayes problems, draw a tree with branches labelled by the conditioning event and write the formula with the full denominator P(B) = ΣP(Aᵢ)P(B|Aᵢ) before substituting numbers.
In a normal-distribution problem, the candidate uses the raw value x with the standard normal table instead of standardising z = (x − μ)/σ, or mishandles the complement and symmetry (using P(Z < z) where P(Z > z) or a two-sided probability is required), producing a probability greater than 1 or a wrong tail.
The standardisation step is mechanical and easily skipped when the numbers 'look like' table entries; the symmetry identities are memorised as isolated facts rather than derived from the bell curve, so the wrong tail is selected.
What to do instead
Write the standardisation explicitly every time: P(X ≤ x) = P(Z ≤ (x − μ)/σ), showing the arithmetic. Sketch the bell curve and shade the required region before consulting the table, then convert the shaded area into 1 − Φ(z) or 2·(1 − Φ(z)) as the sketch dictates. Verify the result lies in [0,1] and matches the shaded area's size.