Guinea Bac Math common mistakes
Guinea Bac Math · Understand the traps and how to avoid them.
Computing a limit by direct substitution and, when a 0/0 indeterminate form appears, writing that the limit does not exist instead of factoring, conjugating, or applying the rate-of-growth comparison.
Candidates learn that division by zero is forbidden and stop there; they have not internalised that an indeterminate form signals the need for an algebraic rewrite, and they confuse 'undefined at the point' with 'no limit at the point'.
What to do instead
Adopt a reflex routine: substitute first; if the result is 0/0 or ∞/∞, rewrite before concluding. For rational functions, factor numerator and denominator and cancel; for radicals, multiply by the conjugate; for trigonometric limits, use sin(x)/x → 1. Then verify with a table of values on both sides of the point and state the conclusion with the correct notation (lim x→a f(x) = L).
Applying the intermediate value theorem or the theorem of continuous monotone bijection without first proving continuity on the closed interval, and using an open interval where a closed one is required.
The hypotheses feel like formalities, so candidates jump from 'f(a) < 0 < f(b)' to 'there is a root' without checking that f is continuous on [a, b]; they also forget that the sign change must be evaluated at the endpoints of a closed interval.
What to do instead
Write the three-line template every time: (1) f is continuous on [a, b] because it is a sum/product/composition of continuous functions; (2) f(a) and f(b) have opposite signs, with the computed values shown; (3) by IVT there exists c in ]a, b[ with f(c) = 0. For uniqueness, add that f is strictly monotone, justified by the sign of f'.
Treating exp and ln as ordinary powers: writing ln(a + b) = ln a + ln b, ln(a·b) = ln a · ln b, e^(x+y) = e^x + e^y, or (ln x)' = 1/x without restricting to x > 0.
The visual similarity between ln and a plain letter, and between e^x and x^n, leads candidates to reuse power rules; the domain restrictions of ln and of the derivative 1/x are treated as decoration rather than as conditions of validity.
What to do instead
Memorise the two correct lists side by side: ln(ab) = ln a + ln b and ln(a/b) = ln a − ln b, with no rule for ln(a + b); e^(x+y) = e^x·e^y and e^(x−y) = e^x/e^y. Before any ln manipulation, write the condition x > 0. For derivatives, state the interval: (ln x)' = 1/x on ]0, +∞[, (ln|u|)' = u'/u where u ≠ 0.
Solving an equation involving ln or exp by squaring or cross-multiplying and then keeping extraneous solutions, for example accepting x = −2 from ln(x) = ln(4) − ln(x) type manipulations, or forgetting to convert ln A = B into A = e^B.
Candidates apply algebraic reflexes from polynomial equations; they do not systematically track the domain of validity, and they treat the final line of calculation as the answer without re-checking it in the original equation.
What to do instead
Start every logarithmic equation by writing the domain as a system of inequalities (each argument > 0). Solve, then test each candidate in the original equation or against the domain system, and strike out those that fail with an explicit note. For ln A = B, use A = e^B directly rather than exponentiating both sides blindly.
Computing a definite integral without checking the sign of the integrand, or integrating 1/x from a negative to a positive bound, thereby producing a negative area or an integral across a singularity.
The integral is treated as a purely formal antiderivative calculation; the geometric meaning (signed area) and the continuity hypothesis on the closed interval are set aside, and 1/x looks integrable everywhere because its antiderivative ln|x| exists.
What to do instead
Before integrating, sketch or sign-study the function on the interval. If f changes sign, split at the zero and add the absolute values for an area, or keep the signs for a signed integral. State that f must be continuous on [a, b]; if a bound is a singularity (e.g. 0 for 1/x), the integral is improper and must be handled as a limit or declared divergent.
Confusing a sequence with a function of a real variable: using l'Hôpital's rule on u_n, or concluding that a sequence converges because it is bounded, or that it converges because it is monotone.
The tools of the analysis chapter are over-generalised; candidates do not distinguish 'bounded', 'monotone', 'convergent' and 'adjacent', and they forget that the monotone convergence theorem needs both monotonicity and boundedness.
What to do instead
Keep a decision tree: to prove convergence, either exhibit an explicit limit, or prove the sequence is monotone and bounded (state both), or use adjacent sequences, or use a recurrence of the form u_{n+1} = f(u_n) with a fixed point and an inequality on |f'|. L'Hôpital applies only to functions of a real variable, so replace u_n by f(x) = ... and take the limit as x → +∞, then conclude for the sequence.
Manipulating complex numbers without separating real and imaginary parts, writing inequalities such as z < z' or |z| = z, and mishandling conjugates in quotients.
Candidates carry over ordering and absolute-value reflexes from the real line; the two-dimensional nature of C is not yet operational, and the modulus is confused with the number itself.
What to do instead
Impose the rule that C is not ordered: never write z < z'. For any equation, write z = x + iy with x, y real and identify real and imaginary parts as two real equations. For quotients, multiply numerator and denominator by the conjugate of the denominator and simplify using z·z̄ = |z|². Check that |z| is a non-negative real number and that |z| = z only when z is a non-negative real.
Solving a system of linear equations by substitution or by adding equations without tracking equivalence, then presenting a partial solution set or losing solutions when a parameter appears.
The Gaussian elimination logic is done informally on the side, and the discussion of the number of solutions (unique, none, infinitely many) depending on a parameter is skipped because it requires a case analysis.
What to do instead
Use the pivot method on the augmented matrix, writing each row operation explicitly. At the end, read the rank: if a row reads (0 0 ... 0 | c) with c ≠ 0, there is no solution; if the rank of the coefficient matrix equals the number of unknowns, the solution is unique; otherwise express the solutions with free parameters. For a parameter m, isolate the values of m that make a pivot vanish and treat them as separate cases.
In plane geometry, using a direction vector as if it were a normal vector (or the reverse) to write a line or plane equation, and applying a transformation without checking whether it preserves lengths or angles.
The vector formalism is memorised as formulas without the underlying orthogonality picture; the distinction between 'parallel to the line' and 'orthogonal to the line' collapses under exam pressure.
What to do instead
Anchor each formula to its meaning: a line with direction vector u has normal vector n with u·n = 0; a line through A with normal n is the set of M with n·AM = 0; a line through A with direction u is M = A + t·u. For transformations, classify first (translation, rotation, homothety, reflection) and state what is preserved: translations and rotations preserve lengths and angles, homotheties preserve angles and ratios of lengths, reflections reverse orientation.
In space geometry, computing a scalar product or a cross product incorrectly to find a normal vector, and then writing a plane equation with the wrong constant term because the point substitution is skipped.
The cross-product formula is remembered as a jumble of coordinates, and candidates stop once they have a normal vector, forgetting that the plane equation must still be calibrated on a known point.
What to do instead
Write the cross product as a determinant with the basis vectors in the first row and expand carefully, then verify orthogonality by dotting the result with each original vector (both must give 0). With normal n = (a, b, c) and a point A(x_A, y_A, z_A), write a(x − x_A) + b(y − y_A) + c(z − z_A) = 0 and expand only at the end; check by substituting A back into the final equation.
Confusing P(A ∩ B) with P(A | B), and adding probabilities of non-disjoint events without subtracting P(A ∩ B).
The notation is visually close, and conditional probability is often introduced late, so the formula P(A | B) = P(A ∩ B)/P(B) is recalled without its interpretation as 'restricting the sample space to B'.
What to do instead
Before computing, write the sample space and the events in words. Use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) always, and note that the subtraction vanishes only when A and B are disjoint (justify it). For conditional probability, redraw the situation as a reduced universe B, or use a tree diagram with branches weighted by conditional probabilities, and check independence by testing P(A ∩ B) = P(A)·P(B) rather than assuming it.
In combinatorics, using arrangements where combinations are required (or the reverse), and forgetting to divide by the factorial of repeated elements in permutations with repetitions.
The formulas are memorised as symbols without the decision criterion 'does order matter?', and the anagram-type correction is easily omitted because the initial count already looks plausible.
What to do instead
Ask two questions before choosing a formula: does order matter, and are repetitions allowed? Order matters → arrangements (n!/(n−k)!); order does not matter → combinations (n choose k). For words or anagrams with repeated letters, count total permutations of the multiset and divide by the factorial of each repetition count. Cross-check small cases by listing them by hand.
In arithmetic, concluding that a number is prime after testing only a few small divisors, or applying divisibility rules without justifying them, and confusing gcd with lcm in problems involving periods or simultaneous events.
Primality testing by trial division is done unsystematically, and the gcd/lcm pair is remembered as a formula rather than through prime factorisation; the link 'gcd divides lcm' and the product relation are not used as a check.
What to do instead
To test n for primality, divide by all primes p with p² ≤ n and stop at the first divisor or when p² > n; state the bound explicitly. For gcd and lcm, decompose both numbers into prime factors and read off the minimum and maximum exponents. In word problems, decide by meaning: a common period or a simultaneous return corresponds to the lcm; a partition into equal largest groups corresponds to the gcd. Verify with gcd(a, b)·lcm(a, b) = a·b.