JC Mathematics common mistakes
JC Mathematics · Understand the traps and how to avoid them.
When finding the range of a function defined piecewise or with a restricted domain, the candidate differentiates to locate a stationary point and reports that value as the maximum or minimum without checking the endpoints of the stated domain.
School practice drills 'differentiate, set to zero, solve' as the universal extremum routine, so the candidate treats the interior critical point as the whole answer and forgets that a restricted domain forces an endpoint comparison; the habit is reinforced because most textbook quadratics have their vertex inside the interval.
What to do instead
Adopt a fixed three-step checklist for every range question: (1) state the domain explicitly, (2) list interior critical points, (3) evaluate the function at both endpoints and at every critical point, then compare. Do five restricted-domain examples (including a monotone function on a closed interval where the stationary point lies outside) and mark the endpoint values on a sketch before writing the answer.
When transforming a graph, the candidate applies a horizontal stretch or translation in the wrong order or with the wrong direction, writing the image of y = f(x) under 'translate 2 left then stretch by factor 3 parallel to the x-axis' as y = f(x/3 + 2) instead of y = f((x + 2)/3).
Function notation hides the operation: candidates substitute x mechanically into the bracket without tracking that a horizontal stretch by factor k replaces x by x/k, and that a left translation by a must replace x by x + a. Because horizontal changes feel counterintuitive (adding moves left), the candidate reverses them under time pressure.
What to do instead
Always rewrite the transformation as an explicit substitution chain: start from y = f(x), write 'new x-coordinate X relates to old x by X = ...', substitute for x in terms of X, and only then simplify. Verify with a single test point (e.g. track where the point (0, f(0)) lands) and check the resulting equation reproduces that point.
In induction proofs, the candidate writes the base case, then states the inductive hypothesis and immediately asserts the n = k + 1 case is true 'by the hypothesis', without algebraically reducing the (k + 1) expression to the required form.
The candidate memorises the shape of an induction proof (base, assume, conclude) as a template and reproduces the labels without performing the linking algebra; the crucial step - showing the k + 1 statement follows from the k statement - is exactly the part that requires genuine manipulation, so it gets skipped or hand-waved.
What to do instead
Force the linking step to be visible: write the target expression for n = k + 1 first, then write the n = k expression, then show the explicit algebraic bridge (factor out, add the extra term, or substitute the hypothesis) line by line. Practise three summation identities where the bridge is 'add the (k+1)th term to both sides and simplify to the closed form'. Never write 'by induction' as a substitute for that line.
In three-dimensional vector work, the candidate computes a dot product or cross product correctly but then misinterprets what it measures - for example using the dot product of two direction vectors to test whether lines are perpendicular when the question asks for the angle, or reporting the magnitude of the cross product as the distance from a point to a plane.
The formulas for dot and cross product are learned as computational procedures detached from their geometric meanings, so the candidate selects an operation by pattern-matching on keywords rather than by asking what geometric quantity is needed. Under exam pressure the first familiar formula wins.
What to do instead
Build a one-page 'geometric meaning' sheet: dot product gives cos of the angle and tests perpendicularity; cross product gives a normal and its magnitude gives a parallelogram area; point-to-plane distance uses the projection of a vector onto the normal. Before computing, write one sentence naming the geometric quantity required and the formula that produces it, then compute.
When solving equations involving complex numbers, the candidate equates real and imaginary parts only after dividing by a complex number without rationalising, or takes a square root of a complex number and reports only one of the two roots.
Real-number habits (divide freely, take the positive root) are carried into the complex setting where division by a non-real number is not a valid single-step operation and where every non-zero complex number has two square roots. The candidate does not pause to check whether the operation is closed or single-valued in the complex field.
What to do instead
Impose two rules: never divide by a complex expression without first multiplying numerator and denominator by its conjugate, and whenever extracting a root of a complex number, solve z^2 = w by writing z = a + bi and equating parts, which automatically yields both roots. Check by squaring each root back to w.
When applying the chain rule to a composite function such as ln(sin(3x)) or e^(x^2), the candidate differentiates the outer function correctly but omits one or more inner derivatives, writing the derivative of ln(sin 3x) as 1/(sin 3x) or as cos 3x / sin 3x.
The chain rule is understood as 'differentiate the outside' rather than as a product of derivatives along the composition chain; with three nested layers the candidate's working memory drops the innermost factor, especially when the inner function is linear and 'looks trivial'.
What to do instead
Use explicit layer labelling: write u = 3x, v = sin u, y = ln v, then dy/dx = (dy/dv)(dv/du)(du/dx) and substitute. After any composite derivative, count the layers in the original expression and confirm the answer has that many multiplicative factors. Do ten three-layer examples until the count becomes automatic.
When evaluating a definite integral by substitution, the candidate changes the variable of integration but leaves the original limits in place, or substitutes back to x and then applies limits that were already transformed.
The substitution step is treated as an algebraic convenience applied to the integrand only, while the limits are copied forward from the previous line; the candidate does not register that the limits are values of the old variable and must be mapped through the substitution.
What to do instead
Adopt a rigid layout: write 'let u = g(x), so du = g'(x) dx; when x = a, u = g(a); when x = b, u = g(b)' before integrating, and physically cross out the old limits. Then integrate entirely in u and evaluate at the new limits, never returning to x. If you do return to x, erase the transformed limits and re-substitute the originals.
When solving a first-order separable differential equation, the candidate separates variables but omits the constant of integration, or includes it only on one side, and then fails to apply the initial condition to determine it.
Indefinite integration is practised heavily as a standalone skill where '+ C' is a reflex, but inside a differential equation the candidate focuses on the separation algebra and treats the integration as a means to an end; also, adding C to both sides looks redundant, so the candidate drops it entirely rather than combining into one constant.
What to do instead
Write the general solution with a single constant C on one side and state explicitly 'C absorbs constants from both integrations'. Then, as a separate labelled step, substitute the initial condition to solve for C before presenting the particular solution. Check the final answer by differentiating it back and confirming it satisfies the original equation.
In hypothesis testing, the candidate computes a p-value or test statistic correctly but states the conclusion as 'the probability that the null hypothesis is true', or accepts the null hypothesis rather than failing to reject it.
Everyday language treats a small p-value as evidence the null is false and a large one as evidence it is true, and the candidate imports this symmetric intuition into a framework where the p-value is a conditional probability of the data given the null, not of the null given the data.
What to do instead
Memorise the exact sentence template and use it verbatim: 'Since p < alpha, there is sufficient evidence at the alpha level to reject H0 in favour of H1' or 'there is insufficient evidence to reject H0; we do not conclude H0 is true.' Write the definition of the p-value as P(data at least this extreme | H0 true) at the top of every practice attempt until the phrasing is automatic.
In combinatorial counting, the candidate cannot decide between permutations and combinations and defaults to nPr whenever order 'feels' involved, or double-counts arrangements that contain identical items by treating them as distinguishable.
The permutation/combination distinction is taught as a formula choice rather than as a modelling decision about whether two selections differing only in order count as different outcomes; identical-object overcounting arises because the candidate lists arrangements positionally without dividing by the factorial of repeats.
What to do instead
Before computing, write one sentence describing a single outcome and ask 'if I swap two chosen items, is that a different outcome?' - yes means permutation, no means combination. For identical items, use the multinomial count n!/(n1! n2! ...) and verify on a small case by listing all arrangements by hand. Build a decision tree on one card and consult it during practice.