Kyoto Entrance Mathematics common mistakes
Kyoto Entrance Mathematics · Understand the traps and how to avoid them.
When solving an inequality, the candidate multiplies or divides both sides by a variable expression (or by a negative number) without tracking the sign change, writing e.g. x² > 4 as x > 2 and dropping the x < −2 branch.
In junior-high and early high-school work inequalities are usually solved by moving terms with a positive coefficient, so the reversal rule is applied mechanically and only for literal negative constants. Once parameters or variable denominators appear, the candidate never splits into cases and treats the inequality like an equation.
What to do instead
Adopt a fixed procedure: (1) move everything to one side so the comparison is against 0; (2) factor completely; (3) build a sign table on the real line using the roots, testing one value per interval; (4) read off intervals including or excluding endpoints per the inequality type. When a denominator or a coefficient contains a parameter, write explicit cases (denominator > 0, = 0, < 0) before dividing. Verify the final answer by substituting one interior and one exterior test value.
The candidate computes a definite integral for an area or volume without checking where the curve crosses the axis or where two curves intersect, giving a positive area where the region is partly below the x-axis, or missing an intersection point that splits the region into two pieces.
Area is memorised as ∫f dx, and textbook drills often use curves that stay positive on the stated interval. The candidate substitutes limits first and integrates second, so the geometry is never inspected, and solving f(x)=g(x) feels like optional extra work.
What to do instead
Before integrating, always (a) solve f(x)=0 and f(x)=g(x) on the interval and list the break points in order, (b) sketch or sign-test each subinterval, (c) write the area as a sum of integrals with |f−g| or with the sign corrected per piece. For volumes of revolution, state the radius function and the axis explicitly before writing π∫r² dx. Make the sketch a graded step in your own practice, not a scratch note.
The candidate differentiates or integrates trigonometric, exponential and logarithmic composites while forgetting the inner derivative, writing (sin 2x)′ = cos 2x, (e^{3x})′ = e^{3x}, or (ln(2x+1))′ = 1/(2x+1), and makes the matching omission in integrals by not dividing by the inner coefficient.
Chain-rule drills are practised with powers of x, where the inner derivative is often 1 or a plain integer that the candidate absorbs unconsciously. Under time pressure the outer derivative is retrieved first and the inner factor is simply never written down.
What to do instead
Force yourself to write the substitution u = g(x), du/dx = g′(x) on a separate line before differentiating or integrating, and only then assemble the answer. For integrals, check by differentiating the result back; if a factor is missing, the check exposes it immediately. Drill a mixed set (sin ax, e^{ax}, ln(ax+b), √(ax+b)) in both directions until the inner factor appears automatically.
In vector problems the candidate confuses the position vector with the direction vector, or uses the dot product to test perpendicularity when parallelism or a specific angle is required, and in spatial problems drops one coordinate when computing a cross-product-style determinant or an inner product.
Plane vectors are usually drawn from the origin, so position and direction coincide in early examples; the distinction only bites once lines are given by a point plus a direction. Spatial computation is done mentally, and the middle term of a 3×3 determinant is the one most often lost.
What to do instead
For every line or plane, write down explicitly: a point on it and a direction/normal vector, labelled. Use a decision checklist: parallel ⇔ b₁ = k b₂; perpendicular ⇔ a·b = 0; angle from cos θ = a·b/(|a||b|). For 3×3 determinants, always expand along a row and write the middle term with its minus sign on paper rather than in your head, then verify one component by an independent method.
The candidate uses the discriminant or Vieta's formulas on a quadratic whose leading coefficient is a parameter, forgetting that a = 0 destroys the quadratic, and applies sum/product of roots without first confirming real roots exist and without tracking the condition on the parameter.
Vieta's formulas are learned as a shortcut and applied as identities valid always, but they describe real roots only when D ≥ 0, and the whole framework collapses at a = 0. Parameter problems are also usually attempted by formula-pushing rather than by separating cases.
What to do instead
Open every parameterised quadratic problem by writing the three conditions as a checklist: a ≠ 0 (else handle the linear case separately), D ≥ 0 for real roots, and the sign/location conditions on the roots. Then apply α+β = −b/a, αβ = c/a. Finish by testing the boundary values of the parameter (including a = 0 and D = 0) in the original equation.
The candidate mixes up necessary and sufficient conditions in logic and set problems, negating a statement incorrectly (writing the negation of 'all A are B' as 'no A is B' instead of 'some A is not B'), and treating 'if P then Q' as equivalent to its converse.
Everyday speech treats 'if' and 'only if' loosely, and the contrapositive is rarely used outside logic drills. The candidate also works with Venn diagrams drawn for the typical case, which hides the counterexample region that decides necessity versus sufficiency.
What to do instead
Rewrite every statement in the standard form P ⇒ Q before judging it, and produce the contrapositive ¬Q ⇒ ¬P as a matter of routine. For negations, apply the mechanical rules: ¬(∀x, P) = ∃x, ¬P and ¬(P ∧ Q) = ¬P ∨ ¬Q. When a claim of necessity or sufficiency is made, deliberately search for a counterexample to the converse or the original, and record it if found.
In counting and probability the candidate cannot decide between permutations and combinations, or between adding and multiplying cases, and produces an answer that double-counts items treated as distinguishable when they are identical (or vice versa).
The formulas nPr and nCr look interchangeable when the numbers are small, and the candidate matches on keywords ('arrange' vs 'choose') instead of on whether order matters in the sample space. Double counting is invisible because the total often looks plausible.
What to do instead
Before computing, state in one sentence whether order matters and whether repetition is allowed, then choose the tool. For multi-stage selections, split into disjoint cases and add, and multiply only across independent stages. Test small analogues by listing all outcomes by hand (e.g. 3 items instead of 6) and compare with your formula. For identical objects, divide by the factorial of the multiplicity and justify it.
The candidate manipulates complex numbers without separating real and imaginary parts, writing inequalities such as z₁ > z₂, or computing |z₁ + z₂| as |z₁| + |z₂|, and mishandles the geometric meaning of multiplication by i or by a complex number of modulus 1.
The real-number habits of ordering and of distributing absolute values over sums carry over unexamined, and the Argand diagram is treated as decoration rather than as the primary tool. Polar form is under-practised relative to rectangular form.
What to do instead
Impose the rule that any comparison or modulus question starts by writing z = a + bi with a, b real. Use |z|² = z·z̄ (conjugate) rather than splitting the modulus, and verify triangle-inequality claims with an explicit counterexample. For geometry, convert to polar form r(cos θ + i sin θ) and read multiplication as scaling by r and rotating by θ; sketch the locus for |z − α| = r and arg(z − α) = θ before algebra.
In sequences the candidate applies the arithmetic or geometric sum formula with the wrong number of terms (using n instead of n−1, or counting from a₀), or assumes a recurrence-defined sequence is arithmetic/geometric without checking the defining ratio or difference is constant.
The formulas are memorised with a generic n, and the off-by-one is invisible unless the first and last terms are written out. Recurrence relations such as a_{n+1} = 2a_n + 3 look geometric at a glance, and the candidate substitutes into the geometric formula directly.
What to do instead
Always write the first three terms explicitly from the given definition, then identify the pattern from those terms rather than from the formula sheet. For sums, write the first and last term with their indices and count the terms. For a recurrence, compute a_{n+1} − a_n and a_{n+1}/a_n symbolically to test arithmetic/geometric structure; if neither is constant, shift the sequence (b_n = a_n + c) to remove the constant term. In induction, verify the base case and use the n = k assumption explicitly in the n = k+1 step.
In integer problems the candidate divides a congruence by a common factor without adjusting the modulus (from ac ≡ bc mod m concluding a ≡ b mod m), or applies the Euclidean algorithm and stops at the last nonzero remainder without back-substituting to express the gcd as a linear combination.
Ordinary equation habits say a common factor can be cancelled, and the modulus condition (gcd(c, m) = 1) is easy to forget because it is rarely triggered in simple drills. The Euclidean algorithm is learned as a gcd-only procedure, so the back-substitution step feels optional.
What to do instead
Memorise the cancellation rule with its condition: ac ≡ bc (mod m) implies a ≡ b (mod m/gcd(c, m)). When solving linear congruences ax ≡ b (mod m), first compute d = gcd(a, m); if d ∤ b there is no solution, otherwise divide a, b and m by d. For the Euclidean algorithm, run it forward to the gcd, then substitute backwards line by line to obtain the Bezout form, and verify by multiplying out.