European Baccalaureate Mathematics (5-period course) common mistakes
European Baccalaureate Mathematics (5-period course) · Understand the traps and how to avoid them.
When finding the inverse of a function such as f(x) = (2x − 1)/(x + 3), the candidate swaps x and y and then solves for y but forgets to state the restricted domain and range, or declares the inverse exists without checking that f is one-to-one on its stated domain.
Inverse-function questions are usually drilled as a mechanical swap-and-solve procedure, so the domain check feels like an optional extra. Candidates also confuse the domain of f with the domain of f⁻¹, and do not connect the horizontal-line test to the algebraic restriction that makes the inverse valid.
What to do instead
Before any algebra, test injectivity on the given domain (horizontal-line reasoning for a graph, or show f(a) = f(b) implies a = b). After solving for the inverse, explicitly write dom(f⁻¹) = range(f) and range(f⁻¹) = dom(f), and verify with one composite: f(f⁻¹(x)) = x for x in the stated domain. Practise on functions with restricted domains, e.g. f(x) = x² − 4x + 7 for x ≥ 2.
In limit questions the candidate substitutes the value directly, obtains an indeterminate form such as 0/0, and then either writes 'the limit does not exist' or cancels terms without stating that the cancellation is valid only where the cancelled factor is non-zero.
Direct substitution works for most routine limits, so the candidate has no decision procedure for the indeterminate case and treats 0/0 as a final answer rather than a signal to transform the expression. Cancelling a common factor is learned as a trick without the accompanying domain caveat.
What to do instead
Adopt a fixed routine: substitute; if you get a determinate value, stop; if you get 0/0 or ∞/∞, factorise or rationalise, simplify, then substitute again. Write the caveat explicitly ('valid for x ≠ 2, and the limit concerns x → 2'). For one-sided limits, test a value just below and just above the point. Check with a numerical table of values approaching from both sides.
When differentiating a composite or product function, the candidate applies only the outer derivative - for example writing d/dx[sin(3x²)] = cos(3x²), or d/dx[x²eˣ] = 2xeˣ - and omits the inner derivative or one term of the product rule.
The chain rule requires holding two structures in mind at once, and under time pressure the candidate pattern-matches the outer function only. The product rule is often misremembered as (u')(v') rather than u'v + uv', a confusion reinforced by the visual similarity of the two rules.
What to do instead
Say the rule aloud while writing: 'outer derivative times inner derivative' for composites, 'first times derivative of second plus second times derivative of first' for products. Before differentiating, annotate the expression: label u, v, or the inner function g(x). For quotients use the quotient rule or rewrite as a product with a negative exponent. Check one result by differentiating numerically at a point.
In definite integration the candidate computes an antiderivative correctly but substitutes the limits in the wrong order or ignores the sign, and when the integrand is negative over the interval reports the signed integral as if it were the geometric area between the curve and the x-axis.
The notation [F(x)] from a to b is read quickly and the subtraction F(b) − F(a) is inverted. The distinction between a signed integral and a geometric area is blurred because both are taught with the same diagram, and the candidate does not check whether the curve crosses the axis inside the interval.
What to do instead
Always write the bracket explicitly as F(upper) − F(lower) before substituting numbers. For area questions, first find the zeros of the integrand in the interval, split the integral at each zero, take absolute values of the pieces, and add. Sanity-check the sign: if the curve lies below the axis, the integral must be negative. Verify with a rough rectangle estimate.
Solving a trigonometric equation such as 2sin(2x − π/3) = 1 on a stated interval, the candidate finds one solution and stops, losing the second or further solutions generated by periodicity, or fails to adjust the interval bounds after the substitution u = 2x − π/3.
Algebraic equations have finitely many solutions, so the habit of 'solve and stop' carries over. The periodic structure of sine and cosine, and the effect of a horizontal stretch on the interval of interest, are not internalised as a step that must be performed every time.
What to do instead
Use a substitution u = (inner expression), solve sin u = value to get u = α + 2kπ and u = π − α + 2kπ, then convert the original interval into a u-interval and list every k that lands inside it. Finally convert back and check each solution lies in the original interval. Sketch the curve over the interval to count expected solutions before solving.
In vector geometry the candidate computes a scalar product or a cross product and then treats the resulting scalar as a vector, or writes the equation of a line as r = a + tb while substituting the direction vector b as a position and the position a as a direction.
The two objects - point and direction - appear in the same formula and are both written as column vectors, so the roles are visually indistinguishable. The scalar/vector distinction is lost because both results are written in the same bracket notation.
What to do instead
Label every vector in a problem as either a position vector (a point on the object) or a direction vector (parallel to the object) before writing any equation. Write the line as r = (position) + t(direction) with the labels visible. For scalar products, state explicitly 'this is a number' and use it for angles and perpendicularity; for cross products, state 'this is a vector' and use it for normals and areas.
When asked for the probability of a union or a conditional probability, the candidate adds P(A) + P(B) without subtracting P(A ∩ B), or computes P(A|B) as P(A ∩ B)/P(A) instead of P(A ∩ B)/P(B).
The addition rule and the conditional formula look similar and are memorised as strings rather than derived from a Venn diagram or a two-way table. The candidate does not pause to identify which event is the conditioning event, i.e. which set is being restricted to.
What to do instead
Draw a Venn diagram or a two-way table for every probability question before computing. For a union, shade the region and subtract the overlap explicitly. For conditional probability, write the formula with the conditioning event underlined, and check the answer lies between 0 and 1. Verify independence separately by testing P(A ∩ B) = P(A)P(B).
In hypothesis testing the candidate states the conclusion as 'the null hypothesis is true' or 'the probability that H₀ is true is 0.03', and confuses the significance level with the p-value.
The language of testing is asymmetric and counter-intuitive, and everyday speech treats 'not rejected' as 'proven true'. The p-value is informally described as 'the chance the result is due to luck', which is then written down as a probability about H₀.
What to do instead
Use a fixed sentence template: 'There is insufficient evidence at the α% level to reject H₀' or 'There is sufficient evidence at the α% level to reject H₀ in favour of H₁.' Never attach a probability to H₀ itself. State the p-value as 'the probability, assuming H₀ is true, of observing a result at least as extreme as the one obtained'. Compare p with α explicitly.
Working with an arithmetic or geometric sequence, the candidate uses the wrong index in the nth-term formula - writing uₙ = u₁ + (n − 1)d but substituting n for the term number when the sequence is defined from u₀ - or uses the arithmetic formula for a geometric sequence.
The formulas are visually similar and the indexing convention (starting at u₀ or u₁) varies between textbooks and between parts of the same paper. The candidate matches on shape rather than on the defining property (constant difference versus constant ratio).
What to do instead
First test the defining property: compute u₂ − u₁ and u₃ − u₂; if equal, it is arithmetic with d equal to that value; if u₂/u₁ = u₃/u₂, it is geometric with ratio r. Write the first term and the index convention at the top of your working. Then check by computing the term you want directly from the definition for small n.
In a proof by induction the candidate verifies the base case and then, in the inductive step, assumes the statement for n = k and manipulates it to 'show' it for n = k + 1 by substituting k + 1 into both sides of the assumed equality, rather than proving the implication.
The structure of the argument (assume P(k), prove P(k + 1)) is confused with a direct proof, and the candidate does not distinguish what may be assumed from what must be shown. The base case is treated as the whole proof.
What to do instead
Write the induction in three labelled parts: (1) Base case - verify P(1) numerically; (2) Inductive hypothesis - state P(k) as a full equation; (3) Inductive step - start from the left-hand side of P(k + 1), use the hypothesis to replace the P(k) part, and algebraically reach the right-hand side of P(k + 1). Finish with the conclusion sentence. Never write the target equality as a starting line.