Pancyprian Core Math common mistakes
Pancyprian Core Math · Understand the traps and how to avoid them.
When simplifying a fraction containing surds, the candidate multiplies numerator and denominator by the conjugate but then expands (a+b)(a−b) as a²+b², writing e.g. 1/(√5−√3) = (√5+√3)/8 instead of /2.
The difference-of-squares identity is memorised as a slogan ('square both, add them') rather than as a derivation; under time pressure the sign of the middle term is reconstructed from memory and the minus in (a−b)(a+b) is lost.
What to do instead
Before rationalising, always write the identity out explicitly in the margin: (a−b)(a+b)=a²−b². Then substitute numbers and check the denominator numerically on a calculator (here 5−3=2). Do ten rationalisations per practice session and verify each with a decimal approximation to two significant figures.
Solving an inequality such as 3−2x ≥ 7, the candidate divides by −2 and keeps the inequality sign, giving x ≥ −2 rather than x ≤ −2; the same slip appears when multiplying an inequality by a negative factor containing a parameter.
The rule 'flip when multiplying or dividing by a negative' is learned as a separate ritual applied only when a minus sign is visibly in front of the division, so it is missed when the negative arises inside the coefficient −2.
What to do instead
Adopt a single procedure: isolate the variable term with a positive coefficient wherever possible (move −2x to the right to get 3 ≥ 7+2x, then x ≤ −2). Alternatively, test the endpoint in the original inequality: substituting x = 0 into 3−2x ≥ 7 gives 3 ≥ 7, false, so the solution set cannot contain 0 - which immediately exposes x ≥ −2 as wrong.
When asked for the domain of f(x) = √(x−3)/(x−5) or of a logarithm, the candidate states only one restriction (e.g. x ≠ 5) and ignores the radicand condition, or writes x > 3 when x = 3 is admissible.
Domain questions are treated as 'spot the denominator' pattern-matching; students check the operation they most recently studied and stop, and they confuse the strict inequality required inside a logarithm with the non-strict one allowed under an even root.
What to do instead
Use a fixed checklist applied to every function: (1) denominators ≠ 0, (2) even-root radicands ≥ 0, (3) logarithm arguments > 0, (4) tangent arguments ≠ π/2 + kπ. Write the restrictions as a system, intersect them on a number line, and give the answer in interval notation. Practise with mixed functions so the checklist, not the last topic studied, drives the answer.
Applying a transformation to a graph in the wrong direction or order: y = f(x−2)+1 is drawn shifted two units left and one down, or y = 2f(3x) is drawn as a stretch by 3 rather than a compression by 1/3.
The candidate reasons about the written expression ('minus two means left') instead of about the coordinates: replacing x by x−2 makes the feature that occurred at x = 2 occur at x = 0. Horizontal transformations feel counter-intuitive because the change is applied to the input.
What to do instead
For every transformation, track one concrete point: take a point (a, b) on y = f(x) and solve for where it moves. For y = f(x−2)+1 the point (a,b) moves to (a+2, b+1). For y = 2f(3x), the point (a,b) moves to (a/3, 2b). Do this for three points before sketching, and state the order: horizontal inside the bracket first, then vertical outside.
In circle geometry, the candidate uses the angle in a semicircle, the alternate segment theorem, or the tangent–chord theorem with the wrong pair of points, for example claiming the angle subtended by a diameter at the circumference is 90° when the vertex is not on the circle, or quoting 'angle between tangent and chord equals the angle in the alternate segment' but naming the angle on the same side of the chord.
These theorems are memorised as pictures rather than as precise statements with conditions, so the conditions (vertex on the circumference, chord and tangent sharing the point of contact, angle in the segment on the opposite side) are dropped when the diagram is unfamiliar.
What to do instead
Rewrite each circle theorem as a full sentence with its hypotheses and test it against the given diagram before using it: name the vertex, state that it lies on the circle, and identify which arc subtends the angle. When the configuration is ambiguous, add a labelled sketch and mark the equal angles in the same colour. In proofs, cite the theorem by name and state the two lines or arcs it applies to.
Solving a trigonometric equation such as sin 2x = 1/2 over 0 ≤ x ≤ 360°, the candidate writes 2x = 30° and hence x = 15°, omitting the second solution 2x = 150° and all solutions obtained by adding 360° to 2x before halving.
The inverse-sine key gives one value, and the candidate forgets that the equation is in 2x, so the period must be added to 2x (not to x) before dividing; the range must also be rescaled, which is easy to skip.
What to do instead
Use a four-step routine: (1) rescale the interval for the inner variable (0 ≤ 2x ≤ 720°); (2) find the principal value and the second value in the first cycle using the symmetry of the sine or cosine graph (180° − θ for sine, −θ for cosine); (3) add the period to the inner variable until the rescaled interval is exhausted; (4) divide by the multiplier and check each answer lies in the original interval.
In a hypothesis test, the candidate states conclusions as 'there is a 95% probability that the null hypothesis is true', or accepts the null hypothesis, or interprets a p-value of 0.03 as '3% chance the result is due to chance'.
The p-value and the significance level are conflated with the probability of the hypotheses; textbooks' shorthand ('reject H₀ at 5%') encourages treating H₀ as an event with a probability, which it is not under the frequentist framework examined here.
What to do instead
Memorise one correct sentence pattern and use it verbatim: 'There is sufficient evidence at the 5% significance level to reject H₀, i.e. to conclude that [contextual claim].' Never say 'accept H₀' - say 'insufficient evidence to reject'. Interpret the p-value as the probability of a result at least as extreme as the observed one, assuming H₀ is true, and always restate the conclusion in the context of the problem.
Differentiating or integrating without regard to the chain rule: d/dx sin(3x²) is given as cos(3x²), and ∫cos(2x)dx is given as sin(2x)+c rather than (1/2)sin(2x)+c.
The standard derivatives of sin x and cos x are over-learned and applied to the outer function alone; the inner derivative is either forgotten or multiplied in the wrong place when the candidate works quickly from memory.
What to do instead
Say the chain rule aloud at every composite derivative: 'outer derivative, keep the inside, times the derivative of the inside.' For integrals, differentiate the answer back as a check - d/dx sin(2x) = 2cos(2x), which shows the missing factor. Do five composite derivatives and five composite integrals per session, always verifying by differentiation.
When asked to 'show that' a result or to 'prove' a statement, the candidate starts from the result and manipulates both sides simultaneously, or substitutes numbers and checks a few cases and calls it a proof; for 'show that the equation has exactly one root', they find one root and stop.
The candidate has practised computation, not exposition, and treats 'show that' as 'find the answer'; working backwards from the target is easier to write down and feels like progress, and existence is confused with uniqueness.
What to do instead
Follow the direction of the implication: start from the given, transform one side only, and finish with the stated result. For uniqueness, prove existence (intermediate value theorem or an explicit root) and then prove uniqueness (monotonicity via the derivative, or a sign argument). For 'show that', the answer is already given, so the marks are entirely in the reasoning - write every step as a connected sentence with a reason.
On a multi-step word problem, the candidate substitutes numbers into a formula before defining the variables, mixes units (e.g. cm and m, or minutes and hours) inside one equation, and gives a final answer with no units or with an implausible magnitude accepted without comment.
Under exam pressure the candidate jumps to the familiar formula; unit conversion is treated as cosmetic, and there is no habit of checking whether the answer is physically sensible, so a factor-of-60 or factor-of-100 error survives to the final line.
What to do instead
Impose a three-line discipline before any calculation: (1) write 'Let x = ...' with units; (2) convert every quantity to a single consistent system and write the converted values down; (3) after obtaining the answer, check magnitude against an estimate and attach units. In modelling questions, state the domain of the model and comment on whether the answer is realistic.