Državna Matura Math B common mistakes
Državna Matura Math B · Understand the traps and how to avoid them.
When solving an equation with a denominator such as (x−2)/(x+3)=1, the candidate multiplies both sides by (x+3) and reports x−2=x+3 giving −2=3, then writes 'no solution', without first noting that x=−3 must be excluded from the domain.
The candidate treats rational expressions as if they were polynomials and has memorised 'multiply both sides by the denominator' as a purely mechanical step. The habit comes from linear equations where no value of x is forbidden, and the domain condition is never written down because it feels like extra work that does not earn marks.
What to do instead
Before touching the equation, write one line: 'uvjet: nazivnik ≠ 0' and list the excluded values. Then solve, and at the end cross-check each solution against that list. Practise on equations where the excluded value is actually the solution you would otherwise get, so the check visibly matters.
The candidate differentiates or analyses a function without restricting attention to its stated domain, for example reporting that f(x)=log(x−1) has a vertical asymptote at x=0, or giving the range of f(x)=x² on [1,3] as [0,∞).
The candidate works from the algebraic form alone and treats the domain as decorative information in the task text. Graphs drawn by hand or by a graphing tool are read off without checking which part of the picture is actually inside the domain.
What to do instead
Make 'domain first' the opening move of every function task: write D_f explicitly, then mark it on the sketch with a vertical boundary line. For range questions on a restricted interval, evaluate the function at the endpoints and at any interior stationary point, then take the min and max of those values only.
When computing the volume or surface area of a composite solid, the candidate adds the volumes of the parts but, for surface area, also adds the full surface areas of both parts, counting the shared internal face twice.
Volume is additive and the candidate correctly generalises 'add the parts', then applies the same rule to area. The internal contact surface is invisible in a written description, so nothing prompts the candidate to subtract it.
What to do instead
Always sketch the composite solid and shade the faces that are actually on the outside. Compute the exposed area face by face, or compute the sum of the separate surface areas and subtract twice the contact area. Sanity-check with a simple case, e.g. two unit cubes glued together: total area is 10, not 12.
Given an arithmetic sequence with a_3=7 and a_7=19, the candidate computes the common difference as d=(19−7)/7=12/7 instead of dividing by the difference in indices, 7−3=4.
The candidate has memorised d=a_n−a_1 over n−1 but does not generalise it, so when the two known terms are not the first and last, the candidate divides by the wrong number or by the larger index. The formula is recalled as a string rather than as a relationship between positions.
What to do instead
Write the general relation a_n=a_m+(n−m)d and derive d from it every time, substituting the indices explicitly before dividing. Then verify by stepping forward from a_3 to a_7 term by term and confirming you land on 19.
In a compound interest problem, the candidate uses the simple interest formula A=P(1+rt) when interest is compounded, or applies the annual rate once per year when the problem states monthly or quarterly compounding, without adjusting the rate and the number of periods.
Simple interest is taught first and is the formula that comes to mind under time pressure. The phrase 'compounded monthly' is read as background detail rather than as an instruction to change r and n.
What to do instead
Before substituting, write down three quantities: the periodic rate r/m, the number of periods n·m, and the compounding frequency m taken from the wording. Use A=P(1+r/m)^(mn) and check plausibility: compound interest must exceed simple interest over the same term.
The candidate computes a probability as a ratio of counts without checking that the outcomes counted are equally likely, for example giving 1/3 for 'the sum of two dice is even' by listing three possible sums instead of counting the 36 ordered pairs.
The candidate confuses 'number of distinct descriptions' with 'number of equally likely outcomes'. Listing outcomes by their value rather than by the underlying elementary events is a natural shortcut that silently breaks the classical definition of probability.
What to do instead
State the sample space explicitly and confirm every element has the same probability before dividing. For dice and coins, enumerate ordered pairs. If outcomes are not equally likely, switch to a tree diagram or to the complement rule rather than forcing the ratio.
In a proof by induction, the candidate verifies the base case and then, in the induction step, assumes the statement for n and immediately writes the statement for n+1 as if it followed, without ever using the induction hypothesis to bridge the two.
The candidate has seen model proofs where the algebra is compressed and mistakes the final line for the argument. The logical structure - that the hypothesis must be invoked - is not distinguished from the algebraic manipulation.
What to do instead
Write the induction step in three labelled lines: 'pretpostavka: S(n)', 'treba dokazati: S(n+1)', 'dokaz:'. In the third line, start from the left side of S(n+1), split off the term belonging to S(n), and substitute the hypothesis there. If the hypothesis is never substituted, the proof is incomplete.
When asked whether a function is injective or surjective, the candidate answers from the shape of the graph on a sketch - 'it passes the horizontal line test' - without stating the domain and codomain, so the same rule is applied to functions where it does not decide the question.
The horizontal line test is learned as a visual shortcut for injectivity on ℝ and is then applied outside that setting. Surjectivity has no such visual test, so it is answered by intuition rather than by solving f(x)=y for x.
What to do instead
For injectivity, solve f(x₁)=f(x₂) algebraically and derive x₁=x₂, noting where the domain matters, e.g. x² is injective on [0,∞) but not on ℝ. For surjectivity, fix an arbitrary y in the codomain and solve f(x)=y, checking whether the solution lies in the domain.
When a task gives a set in set-builder notation or as an interval and asks for a complement or intersection, the candidate works with the numbers visible in the notation and forgets the universal set or the endpoints, e.g. writing the complement of [2,5) in ℝ as (−∞,2)∪(5,∞) or as (−∞,2]∪[5,∞).
The candidate reads the interval as a pair of numbers rather than as a set of points, and the bracket types are treated as typography. The universal set is often stated once at the start of the task and then ignored.
What to do instead
Rewrite every set in the same notation before operating on it, and write the universal set at the top of the working. Handle endpoints separately: an open endpoint becomes closed in the complement and vice versa. Verify with a test point just inside and just outside each endpoint.
In a right-triangle problem the candidate substitutes a side adjacent to the angle into the sine ratio, or uses the hypotenuse as the adjacent leg in the cosine ratio, because the triangle is drawn in an unusual orientation or the angle is not at the base.
The labels 'opposite' and 'adjacent' are learned relative to the standard drawing with the angle at the lower left. When the triangle is rotated, or when the right angle is at the top, the candidate matches labels by position on the page rather than by relation to the marked angle.
What to do instead
Before choosing a ratio, mark the right angle and the given angle, then label the three sides in words relative to that angle: the side touching the angle but not the right angle is adjacent, the side across from it is opposite, and the side across from the right angle is the hypotenuse. Write SOH-CAH-TOA against those labels, not against the picture.