Bulgaria DZI Mathematics common mistakes
Bulgaria DZI Mathematics · Understand the traps and how to avoid them.
When solving an inequality such as (x-1)/(x+3) ≥ 0, the candidate multiplies both sides by (x+3) and writes x-1 ≥ 0, giving x ≥ 1, without treating the sign of the denominator or recording x ≠ -3.
The habit of 'clearing denominators' is carried over from equation solving, where multiplying by an expression is harmless. In inequalities the direction flips when the multiplier is negative, and candidates rarely track the sign of a variable factor across intervals.
What to do instead
Train the sign-chart routine: move everything to one side, factor numerator and denominator completely, mark zeros of numerator (solid) and zeros of denominator (open) on a number line, then test one point per interval and read off the sign. Always state the domain restriction separately before writing the answer set.
After solving a logarithmic equation such as log₂(x-1) + log₂(x-3) = 3, the candidate reports both roots of x²-4x-5=0 without checking which satisfy x-1>0 and x-3>0, so an extraneous root is included.
The algebraic manipulation (combining logs, exponentiating) is valid only where the arguments are positive, but the transformed polynomial equation has a larger domain and silently admits values the original expression never allowed.
What to do instead
Adopt a fixed two-line protocol: write the domain conditions (all arguments > 0, base > 0, base ≠ 1) before the first step, then verify each candidate root against those conditions explicitly and strike out failures with a visible cross. Practise on equations deliberately built with one extraneous root.
The candidate computes sin(2x) or cos(x/2) by writing 2sin x or (cos x)/2, treating the function symbol as a multiplier.
Function notation is read as ordinary multiplication because the argument is not in parentheses in familiar expressions like sin 2x, and the candidate has no mental model of the double-angle identities as distinct objects.
What to do instead
Drill the identity sheet in both directions until sin 2x = 2 sin x cos x is recalled faster than the incorrect '2 sin x'. Include a numerical disproof exercise (e.g. compare sin 60° with 2 sin 30°) so the wrong rule fails visibly, then apply identities inside equation solving.
Asked for the domain of f(x) = √(x-2)/(x-5) or of a composite involving a logarithm, the candidate gives only x ≥ 2 and forgets x ≠ 5, or gives only the denominator condition.
Conditions are collected one at a time as they are noticed rather than systematically, and the square-root condition feels like 'the' domain condition once found.
What to do instead
Use a checklist written at the top of every such problem: even root → radicand ≥ 0; denominator → ≠ 0; logarithm → argument > 0 and base valid; tangent → argument ≠ π/2 + kπ. Intersect the resulting sets on a number line and write the final answer in interval notation.
In a geometric progression the candidate uses aₙ = a₁ + (n-1)d or applies Sₙ = n(a₁+aₙ)/2 to a geometric sequence, mixing the arithmetic and geometric formulas.
Both families are memorised as similar-looking formula blocks, and under time pressure the first recalled formula is applied without checking whether the sequence has a constant difference or a constant ratio.
What to do instead
Before computing, force one diagnostic step: divide consecutive terms (or subtract them) to identify the type, and write the type label above the working. Keep a two-column comparison card of arithmetic versus geometric formulas, including the special case q = 1 for the geometric sum, and solve mixed problem sets that require the identification step.
When a logarithm appears with a base other than 10 or e, such as log₃ 7, the candidate evaluates it as log 7 / log 3 but then confuses which quantity goes in the numerator, or applies log(a+b) = log a + log b.
The change-of-base formula is remembered as a vague fraction, and the product/quotient laws are over-generalised to sums because the correct laws for products look structurally similar.
What to do instead
Anchor change of base with a sanity check: log₃ 81 must come out as 4, so the larger number goes on top when the base is smaller. For the laws, memorise the slogan 'logs turn products into sums, never sums into anything' and verify each manipulation on a numerical example before using it in a proof or equation.
Solving a trigonometric equation like 2sin²x - sin x - 1 = 0 on [0, 2π), the candidate finds x = π/2 and x = 7π/6 but omits the second solution x = 11π/6 of sin x = -1/2, giving an incomplete solution set.
The candidate solves for the value of sin x and then recalls only the reference angle, forgetting that sine takes the same value at two points in each period and that the required interval may contain several periods.
What to do instead
After obtaining sin x = c, always draw the horizontal line y = c on a unit circle or graph over the stated interval and mark every intersection before writing answers. Use the general-solution form (x = α + 2kπ or x = π - α + 2kπ) and then restrict k to the required interval, listing k values explicitly.
In probability and combinatorics the candidate cannot decide between permutations and combinations, computing ⁵P₂ when the problem asks for a committee of two, or forgetting to divide by repeated factorials in arrangements of a word with duplicate letters.
The decision is made from surface wording ('choose', 'arrange') rather than from whether order matters, and the multiset case is not recognised as a distinct pattern.
What to do instead
Ask one question before every count: 'If I swap these two selected items, is the outcome different?' If no, use combinations. For arrangements with repeats, write the multiset counts and apply n!/(n₁!n₂!…) directly. Build a small decision tree and run it on ten short word problems until the branch choice is automatic.
Computing the volume or surface area of a pyramid, cone, or truncated solid, the candidate uses the slant height where the perpendicular height is required, or applies the full-cone formula to a frustum.
Textbook diagrams show the slant height prominently, and the right triangle relating height, slant height, and radius is not reconstructed at the moment of substitution.
What to do instead
Before substituting, draw the axial cross-section and label the right triangle explicitly, marking which length is perpendicular to the base. For frustums, derive the result as difference of two similar solids rather than recalling a separate formula, and check units and magnitude against a rough estimate.
In analytic geometry the candidate finds the equation of a line through a given point parallel or perpendicular to a given line, but uses the same slope for the perpendicular case, or drops the sign when taking the negative reciprocal.
The parallel case is practised more often, and the perpendicular condition m₁m₂ = -1 is recalled as 'flip the fraction' without the sign change.
What to do instead
Convert the given line to slope-intercept form first and write the slope down as a separate labelled quantity. Then apply the rule explicitly: parallel → same m; perpendicular → m₂ = -1/m₁, showing the reciprocal and the sign change as two visible steps. Verify by substituting the given point back into the final equation.