Přijímačky VŠE Matematika common mistakes
Přijímačky VŠE Matematika · Understand the traps and how to avoid them.
When asked for the relative position of two lines or a line and a plane in space, the candidate computes the determinant of the three direction/connecting vectors and stops, declaring 'they intersect' or 'they are skew' from that single number without checking the parallel/contained cases separately.
The candidate collapses the classification into one test, forgetting that the determinant being zero covers three different situations (parallel distinct, coincident, coplanar-intersecting), and that the determinant being nonzero only rules out coplanarity.
What to do instead
Use a decision tree and write it out each time: compare direction vectors first (proportional or not). If proportional, test whether a point of one satisfies the other (coincident vs parallel distinct). If not proportional, compute the mixed product; zero means they intersect in a point (find it), nonzero means they are skew. For line–plane: check n·v; if zero, test whether the line's point lies in the plane (contained vs parallel). State the case name explicitly in the answer.
When computing a distance from a point to a plane or between skew lines, the candidate substitutes into the formula without normalising, using n·PQ instead of |n·PQ|/|n|, or forgetting the absolute value, and reports a negative distance or one scaled by the norm of the normal vector.
The formula is recalled as 'dot product over something' and the denominator (the modulus of the normal, or the cross product for skew lines) is dropped because it feels like a formality; the absolute value is dropped because the arithmetic 'came out positive anyway' in practice exercises.
What to do instead
Memorise the distance formula as a fraction with a visible denominator: d = |Ax₀+By₀+Cz₀+D| / √(A²+B²+C²), and for skew lines d = |(P₂−P₁)·(v₁×v₂)| / |v₁×v₂|. After computing, sanity-check the magnitude against the coordinates (a distance cannot exceed the distance between the two named points). Write the absolute value bars in the very first line of working, before substituting numbers, so they cannot be omitted later.
When solving an equation such as √(2x+5) = x − 1, the candidate squares both sides, keeps the resulting root, and never substitutes back into the original equation to discard the extraneous solution introduced by squaring.
Squaring is treated as a reversible operation, and the candidate has usually been trained on equations where every algebraic step is reversible. Extraneous roots are invisible in the squared equation, so substitution is the only place they surface, and students who are focused on 'getting x' skip the check entirely.
What to do instead
Adopt a fixed routine for any non-reversible step (squaring, multiplying by an expression containing the variable, raising to an even power): after solving, substitute every candidate into the original equation and show the substitution explicitly. Practise with equations engineered to have one genuine and one extraneous root, and record in writing why the extraneous root fails (e.g. the right side is negative for that x).
The candidate applies a function transformation in the wrong order or direction: writing the graph of f(2x − 6) as a shift right 6 then a compression by 2, or graphing f(x + 3) as a shift left by 3 on the wrong side of the axis, or treating −f(x) as a horizontal reflection.
Transformations are memorised as a list of moves without understanding that horizontal changes act on the input before the function is evaluated, so the natural left-to-right reading order of the expression does not match the order of geometric operations. The horizontal direction is also counterintuitive because adding inside the parentheses moves the graph left.
What to do instead
Rewrite every horizontal transformation in the standard form f(k(x − h)) before graphing, factoring the coefficient out of the parentheses. Verify with three test points (a zero, a point where the inside equals 1, and a negative input) rather than trusting memory, and always state the mapping of a specific point such as (0, f(0)) to its image.
When finding the inverse of a function, the candidate swaps x and y, solves for y, and reports the result as the inverse without checking whether the original function is one-to-one on its stated domain, and without restricting the domain when it is not.
The swap-and-solve algorithm is learned as a purely symbolic procedure, and the existence condition (passing the horizontal line test on the given domain) is treated as a separate textbook topic rather than a precondition for the algebra. Quadratic and trigonometric examples then produce 'inverses' that are not functions.
What to do instead
Before any algebra, state the domain and range and confirm one-to-one behaviour on that domain; if the function fails, restrict the domain explicitly and say which branch is chosen (for example x ≥ 0 for x²). After solving, verify with a composition check, f(f⁻¹(x)) = x on the inverse's domain, and confirm that the graph of the inverse is the reflection across y = x.
The candidate solves a trigonometric equation such as sin(2θ) = 1/2 by writing only θ = 15° or a single principal value, ignoring the second solution in the period, the general solution with the period added, and any restriction on the interval given in the problem.
Inverse-trig buttons on a calculator return exactly one value, so the candidate equates 'the answer' with 'the calculator output'. Reference-angle symmetry and periodicity require deliberate reasoning that the calculator conceals, and the doubled angle in sin(2θ) makes the period of the solution set half of what intuition suggests.
What to do instead
Solve in two stages: first find all solutions for the inner argument over a full period using reference angles and quadrant analysis, then divide the entire solution set (including the period term) by the coefficient of θ and filter to the required interval. Always sketch the unit circle or one period of the graph to count how many solutions the interval should contain before listing them.
The candidate applies the Law of Sines to a triangle given two sides and an included angle (SAS) or three sides (SSS), or uses the Law of Cosines when a simple right-triangle ratio applies, and in SSA cases reports only the acute solution and misses the obtuse one.
Both laws look interchangeable in formula sheets, and the decision rule (which information is given) is not tied to a concrete procedure. The ambiguous SSA case is under-practised because most textbook problems are constructed to have a single clean answer.
What to do instead
Use a decision checklist keyed to the given data: ASA/AAS or a known side–opposite-angle pair means Law of Sines; SAS or SSS means Law of Cosines. For SSA, always compute the candidate angle and then its supplement, test both against the angle sum, and state how many triangles exist. Check the final triangle by verifying the angle sum and re-deriving one side with the other law.
The candidate computes the number of arrangements of a multiset or a committee with repeated roles by treating identical items as distinguishable (using n! instead of n!/(n₁!n₂!…)), or adds counts where the cases are not mutually exclusive, producing double counting.
The multiplication principle is learned first and applied by default, and the division by factorials for identical objects feels like an arbitrary patch. Overlap between cases is hard to see without a systematic case structure, and small examples are rarely enumerated by hand to check.
What to do instead
Before computing, label each object as distinct or identical and decide whether order matters and whether repetition is allowed, writing the four-way classification explicitly. For case-based counting, define cases by a single discriminating property so they partition the set, then verify with a small analogue (three or four items) enumerated by hand or by listing.
When computing a determinant or the inverse of a 3×3 matrix, the candidate expands the cofactors using the wrong sign pattern, e.g. writing +,+,+ or applying the alternating signs starting from the wrong corner, producing a determinant that differs only in sign from the correct value.
The checkerboard sign pattern (+ − + / − + − / + − +) is memorised loosely as 'alternating' without anchoring it, and under exam pressure the candidate reproduces the signs by eye rather than from the (−1)^(i+j) rule, so a single misplaced sign flips the whole result.
What to do instead
Before every expansion, write the sign grid explicitly next to the matrix and derive each sign from (−1)^(i+j) for the row or column chosen. Then verify with an independent route: expand along a different row and confirm the same value, or check the determinant of a matrix with a zero row/column pattern. Practise five 3×3 determinants per session expanding along row 1, row 2, column 3, checking each against expansion along another line.
In a system with a parameter, the candidate solves only the generic case (dividing by an expression in the parameter) and reports one solution, omitting the special parameter values where the determinant vanishes and the system becomes incompatible or has infinitely many solutions.
Rouché–Capelli is applied mechanically as 'compute det(A) and if nonzero, Cramer', and the discussion of the degenerate cases is treated as an optional extra rather than the point of the question, so the candidate never returns to the values excluded during division.
What to do instead
Always structure the answer in three explicit stages: (1) det(A) as a polynomial in the parameter, (2) solve det(A)=0 and list those values, (3) for each such value substitute into the augmented matrix and compare rank(A) with rank(A|b). Write the final answer as a case table (parameter value → type of system → solution). Before dividing by anything containing the parameter, note the excluded values in the margin so they cannot be forgotten.