PAU Mathematics II (La Rioja) common mistakes
PAU Mathematics II (La Rioja) · Understand the traps and how to avoid them.
When asked to discuss the system AX=B for a parameter a, the candidate computes det(A), sets it to zero, finds the critical values, and then stops, declaring the system 'incompatible' at those values without ever computing the ranks of A and of the augmented matrix for each critical value.
The scheme 'det=0 means no solution' is memorised from drills where the coefficient matrix was square and the critical case always happened to be incompatible. It works often enough in practice that the habit is never corrected, and computing ranks at the critical values requires extra work that feels redundant once a determinant has been found.
What to do instead
Adopt a fixed three-branch script: (1) det(A)≠0 gives a unique solution via Cramer or inverse; (2) for each value with det(A)=0, write out the numerical matrices A and [A|B] with that value substituted and compute ranks by Gaussian elimination; (3) conclude rank(A)=rank([A|B])=n unique, =rank([A|B])<n infinitely many with n−rank parameters, <rank([A|B]) incompatible. Practise at least five systems where the critical case is compatible and gives infinitely many solutions, so the branch is not a surprise.
In a matrix problem, the candidate finds a matrix X satisfying an equation such as AX+B=C by writing X=A⁻¹C−B, i.e. multiplying on the wrong side or subtracting before multiplying by the inverse.
Scalar algebra is so automatic that the non-commutativity of matrix multiplication is acknowledged in theory but not applied in the manipulation. Candidates also forget that B must be moved first, so the inverse multiplies the whole right-hand side, and they treat (C−B) as if order did not matter.
What to do instead
Before touching symbols, write the target equation and isolate X with explicit side labels: AX=C−B, then left-multiply both sides by A⁻¹ to get X=A⁻¹(C−B). Check invertibility (det≠0) explicitly, and verify the answer by substituting back into the original equation. As a rule, never write a product of matrices in an order different from the one derived.
In a limit involving indeterminate forms, the candidate applies L'Hôpital's rule to a quotient that is not of type 0/0 or ∞/∞, or differentiates numerator and denominator as if the limit were a derivative of a quotient, or stops after one application while the form is still indeterminate.
L'Hôpital is learned as a universal tool and the hypothesis check is skipped. Because the mechanical differentiation usually produces something, the candidate does not notice the rule was inapplicable, and repeated application until the indeterminacy clears is not treated as part of the method.
What to do instead
Write the form explicitly (e.g. '0/0') next to the limit before each application, and re-evaluate the form after every derivative. If the form is 0·∞, ∞−∞ or 1^∞, first rewrite (product to quotient, common denominator, or exponential/logarithm for powers) to reach 0/0 or ∞/∞. Prefer equivalent infinitesimals and known limits when they are faster, and verify with a numerical table of values near the point.
When asked for differentiability or for the values of a parameter making a piecewise function differentiable at the join point, the candidate imposes continuity only, or imposes equality of the one-sided derivatives without first checking continuity, and concludes differentiability from one condition.
The theorem 'differentiable implies continuous' is known in the forward direction but not used in reverse as a necessary test. Candidates also compute the derivative formula at the join point by differentiating the branch expression, which is invalid precisely at that point.
What to do instead
Use the two-step protocol at the join point a: first require lim f(x) as x→a⁻ equals lim f(x) as x→a⁺ equals f(a) (continuity); then compute the one-sided derivatives by definition, f'(a⁻)=lim (f(a+h)−f(a))/h as h→0⁻ and similarly from the right, and require equality. Solve the continuity equations first to reduce parameters, then the derivative equations. State explicitly that continuity is necessary but not sufficient.
In an area-between-curves problem, the candidate integrates f−g over the interval between the two given x-values without locating the interior intersection points, so regions where the curves cross are counted with the wrong sign and partially cancel.
Textbook exercises often have one curve above the other throughout, so the formula ∫(f−g) is applied as a reflex. The possibility of an interior crossing is not checked because finding it requires solving f(x)=g(x), an extra step that does not look like part of 'the area formula'.
What to do instead
Always begin by solving f(x)=g(x) on the interval and splitting the integral at every interior root, taking |f−g| on each subinterval (or identifying which function is on top by testing a midpoint). For areas involving the y-axis or curves given as x=g(y), decide the variable of integration before setting up. Sketch the region and mark the crossing points on the sketch.
In a volume-of-revolution problem, the candidate uses the disk formula π∫f(x)²dx when the axis of revolution is not the x-axis and the region is not adjacent to it, omitting the offset (R(x)−r(x)) or using the wrong radius.
The formula is memorised for rotation about the x-axis with the region bounded by the curve and the axis. When the axis is y=−1 or x=2, the distance from a typical point to the axis changes, and the candidate substitutes f(x) directly instead of the actual radius.
What to do instead
Draw the region and the axis, then mark a typical slice and write the distance from the slice to the axis explicitly: radius = |f(x) − axis|. For a region between two curves use π∫(R²−r²)dx with R and r measured from the axis. For rotation about a vertical axis, integrate in y or use shells; state which method and why before computing.
Given three points, the candidate computes two vectors, takes their cross product to get a normal, and then writes the plane equation using a point not established to lie on the plane, or mixes the normal components into the point coordinates, producing a plane that does not contain the given points.
The procedure has several independent objects (two direction vectors, one normal, one point) and candidates copy them from line to line without a consistency check. Under exam pressure the normal and the point get transposed, and there is no habit of verifying the final equation against the original data.
What to do instead
Fix a template: take P as base point, u=Q−P, v=R−P, n=u×v, then write n₁(x−p₁)+n₂(y−p₂)+n₃(z−p₃)=0 and expand. Immediately substitute all three original points into the final equation; each must give 0=0. If any fails, the error is in the normal or the base point and is found in seconds.
When computing the angle between two lines, a line and a plane, or two planes, the candidate plugs the direction vectors (or normals) into the dot-product formula and reports the resulting angle without converting to the geometric angle asked for - for instance giving the angle between a line and a plane as the angle between the line's direction and the plane's normal.
All three situations reduce to 'angle between two vectors', so the formula is applied mechanically. The complement relationship for line–plane (sin instead of cos, or 90° minus the result) is the step most often dropped, and the fact that the dot product can return an obtuse angle while the geometric angle is defined as acute is forgotten.
What to do instead
Before computing, write which pair of vectors you are using and which trigonometric function the geometry demands: line–line cos θ=|u·v|/(|u||v|); line–plane sin θ=|u·n|/(|u||n|); plane–plane cos θ=|n₁·n₂|/(|n₁||n₂|). Take absolute values so the angle lies in [0°,90°]. Sanity-check the answer against a rough sketch.
In integration by parts the candidate assigns u and dv arbitrarily, or applies the formula with the wrong sign, writing ∫u dv = uv + ∫v du, or forgetting to evaluate the second integral.
The formula is memorised as a shape rather than derived, and the choice of u is made without a criterion. Sign errors arise because the minus sign is buried inside the formula and is easily lost when substituting limits.
What to do instead
Memorise the selection rule LIATE (logarithmic, inverse trig, algebraic, trig, exponential) for u, and verify by differentiating the proposed antiderivative. Derive the formula once from the product rule so the minus sign has a reason. On definite integrals, write the uv term with limits in square brackets separately from the remaining integral before substituting.
When finding the area between a curve and the x-axis, the candidate integrates f(x) from a to b without checking where f changes sign, so regions below the axis subtract from regions above and a wrong (sometimes negative) area is produced.
Area is treated as a single definite integral by reflex, and sign change is not linked to the geometry. Sketching is skipped to save time, removing the only cue that the curve crosses the axis.
What to do instead
Make a sketch mandatory for every area question. Solve f(x) = 0 first and split the integral at each root, taking the absolute value of each piece: A = ∫|f(x)| dx. Check plausibility - area must be positive and roughly comparable to a bounding rectangle. Practise a curve such as y = x^3 - x over [-1, 1] where naive integration gives zero.