UEE Mathematics common mistakes
UEE Mathematics · Understand the traps and how to avoid them.
When solving an inequality, the candidate multiplies or divides both sides by a negative expression (or by a term whose sign is unknown) and keeps the inequality direction unchanged, e.g. turning -2x > 6 into x > -3, or clearing a denominator like (x-3) without splitting cases.
Equation-solving habits are overlearned: for equalities multiplying by anything is safe, so the reflex carries over. The sign-flip rule is memorised as a slogan but not linked to the order structure of the reals, so it is dropped under time pressure or when the coefficient is symbolic rather than a visible negative number.
What to do instead
Before any division, write the sign of the divisor explicitly above the operation. If the divisor contains a variable, split into cases (positive, negative, zero) or move everything to one side and factor rather than cross-multiplying. Finish by testing one value from each interval in the original inequality.
The candidate expands (a+b)^2 as a^2+b^2, or cancels a term across a sum, e.g. writing (x^2+3x)/x = x^2+3 or (x+2)/(x+2+5) = 1/5.
Pattern-matching to the shortest visible string; the distributive law is applied to a non-linear operation. Illegal cancellation persists because it often produces a plausible-looking answer and is rarely checked by substitution.
What to do instead
Force a one-line expansion of (a+b)^2 = a^2+2ab+b^2 every time before simplifying. For cancellation, only cancel factors of a product, never terms of a sum: factor the numerator and denominator completely, then cancel. Verify with a numerical substitution such as x = 2 in both the original and simplified expressions.
The candidate states the domain or range of a function from its formula alone, ignoring restrictions: giving the domain of f(x)=1/(x-2) as all reals, of sqrt(x-3) as all reals, or of ln(x) as including x=0.
Domain questions are treated as a formality after the 'real' work of differentiating or graphing. The candidate also conflates the algebraic form with the graph they have in mind, where the asymptote or endpoint is visually present but not translated into a condition.
What to do instead
Adopt a checklist applied to every function: denominators nonzero, even roots non-negative, logarithm arguments strictly positive, and inverse-trig arguments in [-1,1]. Write the domain as a set before doing anything else, and mark it on the sketch as open or closed endpoints.
When finding an inverse function, the candidate swaps x and y but forgets to restrict the domain, so f(x)=x^2 is inverted to f^{-1}(x)=sqrt(x) over all reals, or a periodic function is inverted without choosing a principal branch.
The swap-and-solve algorithm is learned mechanically; the one-to-one requirement is understood abstractly but not connected to the step where a choice must be made. The candidate also assumes the inverse has the same domain as the original.
What to do instead
Before inverting, state whether f is one-to-one on the given domain; if not, restrict to a monotone interval and say which. After finding the inverse, verify f(f^{-1}(x))=x on the inverse's domain and confirm the range of f equals the domain of f^{-1}.
In geometry, the candidate uses a length, area or volume scale factor as if it were linear: scaling a solid by factor k and claiming the volume scales by k, or using the ratio of areas directly as the ratio of lengths.
Similarity ratios are memorised as a single number without the exponent structure. Diagrams are not to scale, so the candidate cannot sanity-check the magnitude of the answer, and the phrase 'ratio of' is read as interchangeable across dimensions.
What to do instead
Attach the exponent to the scale factor at the moment it is written: lengths k, areas k^2, volumes k^3. When a problem gives a ratio of areas or volumes, take the square or cube root first and label the result as the linear scale factor before substituting.
The candidate computes an angle in a triangle with the sine rule when the triangle is obtuse and takes the acute value from the calculator, or applies the cosine rule with the wrong included angle, producing a triangle that cannot close.
Inverse sine returns only the principal value in [-90°,90°], and the candidate does not check whether the supplementary angle is admissible. The included angle is also misidentified because the formula is recalled as b^2 = a^2+c^2-2ac cos B without tracking which vertex B is.
What to do instead
After any inverse-sine result, test the supplement: check whether the angles sum to less than 180° with the known angle, and prefer the cosine rule when the given information is side-angle-side or side-side-side, since it has no ambiguity. Label vertices on a sketch before substituting so the included angle is unambiguous.
When solving sin x = k or cos x = k over an interval, the candidate gives only the principal solution, missing the second solution in the same period and all solutions in later periods (e.g. sin x = 0.5 on [0, 360°] answered as 30° only).
The calculator returns one value and the candidate treats it as the answer. The symmetry of the unit circle is known in theory but not operationalised, and the interval is read as a hint rather than a constraint to be exhausted.
What to do instead
Use the standard templates: for sin x = k, x = alpha + 360°n or 180° - alpha + 360°n; for cos x = k, x = ±alpha + 360°n; for tan x = k, x = alpha + 180°n. Then list n values that land inside the stated interval and count them against the expected number of solutions.
The candidate computes P(A and B) as P(A)·P(B) for events that are not independent, or computes P(A or B) as P(A)+P(B) without subtracting P(A and B), leading to probabilities above 1.
The multiplication and addition rules are memorised in their simplified independent and mutually exclusive forms, and the qualifying conditions are dropped. Word problems rarely say 'independent' explicitly, so the candidate defaults to the easy formula.
What to do instead
Before computing, classify the events explicitly: are they independent, mutually exclusive, or neither? Use P(A∪B)=P(A)+P(B)-P(A∩B) always, and P(A∩B)=P(A)P(B|A) in general, reducing to the product only when independence is established. Sanity-check that the final probability lies in [0,1].
When differentiating a composite function, the candidate omits the inner derivative, writing d/dx sin(3x) = cos(3x) or d/dx (2x+1)^5 = 5(2x+1)^4, and similarly forgets the chain factor in integration by substitution.
The rule is recalled as a transformation of the outer function only; the inner function is perceived as part of the argument rather than a separate factor. Practice with simple inner functions such as x builds the habit that then fails on 3x.
What to do instead
Say the chain rule aloud as 'outer derivative times inner derivative' and physically write the inner derivative before simplifying. For integrals, change the variable fully (including dx) and convert the limits, then differentiate the antiderivative as a check.
The candidate uses the derivative to conclude a function is increasing on an interval without checking the sign of f' throughout, or treats f'(a)=0 as sufficient for a local extremum, and confuses the location of an extremum with its value.
The necessary and sufficient conditions are compressed into 'set f'=0 and solve'. Sign analysis is skipped because it feels like extra work, and the question 'where' versus 'what is the maximum value' is misread.
What to do instead
Build a sign table for f' with the critical points and any discontinuities as boundaries, and state the monotonicity interval from the signs. Classify each critical point by sign change (or second derivative), and always answer the question asked: give the x-value for location, f(x) for the extremum value.