NZ Scholarship Calculus common mistakes
NZ Scholarship Calculus · Understand the traps and how to avoid them.
In a 'show that' or 'prove that' question, the candidate starts from the statement to be proved and manipulates both sides until they meet, stringing the lines together with '=' signs that have not been established (e.g. writing 'x² − 4x + 7 = (x − 2)² + 3' and then working downward from it).
School algebra rewards solving equations, where operating on both sides is legitimate, so the habit transfers. Candidates also have no internal model of what a proof must look like, and treat the printed result as a given to be simplified rather than a target to be reached.
What to do instead
Fix the direction of the argument in writing: begin 'LHS = …' and transform only one expression, or write 'we need to show …' and derive it from something already known to be true. Use '⟺' only for genuinely reversible steps, '⟹' where the implication runs one way, and close with a sentence stating exactly what has been proved.
In an induction proof, the candidate expands the n = k + 1 case from scratch and 'verifies' it matches the claimed formula, never invoking the assumption that the result holds for n = k; the base case is sometimes omitted entirely, and the inductive hypothesis is never explicitly stated.
Induction is learned as a two-step ritual and candidates collapse the steps into 'test some values and generalise', which is not a proof. The substitution step also requires algebraic rewriting (splitting a sum, factoring) that they find harder than re-deriving.
What to do instead
Use a fixed scaffold every time: define P(n), verify P(1) (or the stated starting value), write 'assume P(k) holds for some k ≥ 1', then compute the (k+1) expression as [k expression] + [new term] and substitute the hypothesis. For inequality inductions, keep the direction of the bound explicit at every step and show the added term does not break it.
The candidate cancels a common factor such as x, sin x or cos x from both sides of an equation, or multiplies/divides an inequality by an expression whose sign depends on x, without treating the case where the factor is zero or negative.
Cancelling is the fastest route to a simpler equation and it always feels like progress; the lost case produces no error message, and sign-reversal in inequalities is a one-line rule that is easily forgotten under time pressure.
What to do instead
Never divide: move everything to one side and factorise, then use the zero-product property to read off all solutions, including the ones the cancelled factor would have given. For inequalities, bring to one side, factorise, and build a sign table around the critical values (e.g. x² > x becomes x(x − 1) > 0, giving x < 0 or x > 1 - dividing by x would have silently assumed x > 0).
The candidate differentiates y² with respect to x and writes 2y, omitting the dy/dx factor in implicit differentiation; with parametric curves the derivative is computed correctly but then evaluated at the x-value of the point instead of at the corresponding parameter t.
Candidates read d/dx(y²) as if y were the independent variable, because the chain rule d/dx = (dy/dx)(d/dy) is memorised as a formula rather than as a statement about which variable is being varied. For parametric curves, x is mentally 'the input', so substituting x feels natural.
What to do instead
Write d/dx(y²) = 2y·dy/dx explicitly every single time until it is automatic, and after differentiating collect all dy/dx terms on one side before solving for it. For parametric curves, compute dy/dx = (dy/dt) ÷ (dx/dt), substitute the t-value belonging to the point (not the x-value), and check dx/dt ≠ 0 there.
The candidate applies L'Hôpital's rule without first checking that the limit is of the form 0/0 or ∞/∞, or evaluates lim(x→0) sin x / x by differentiating sin x - a circular argument, since that derivative is established from that very limit. Repeated applications are also attempted where the ratio never resolves.
L'Hôpital is the only limit technique many candidates have practised fluently, so it becomes the default. The circularity is invisible unless the derivation of the derivative has been taught alongside the limit.
What to do instead
Substitute first and classify the form in writing; only genuine 0/0 or ∞/∞ justifies the rule, and a ratio like x/√(x²+1) as x→∞ loops under repeated differentiation and must be handled by dividing through by x. For sin x / x, (cos x − 1)/x and (eˣ − 1)/x, use the standard limits, a squeeze argument or an established series expansion, and say which result you are relying on.
After substituting u = g(x) in a definite integral, the candidate keeps the original x-limits and evaluates the u-antiderivative at them - or converts the limits to u and then back-substitutes to x before evaluating, so the two changes cancel into an error. The dx is sometimes converted to du only partially.
The three objects (integrand, differential, limits) must all change together, and attention is spent on the integrand, which is the visible part, while the limits are treated as decoration carried down from the question.
What to do instead
Write the full scaffold each time: choose u = g(x), compute du = g′(x)dx, express the whole integrand and the differential in u, and state the new limits u = g(a), u = g(b). Then either stay entirely in u and evaluate, or convert back to x and use the original limits - never both. Sanity-check the sign and magnitude of the answer against a rough sketch.
The candidate evaluates an improper integral by writing ∞ straight into the antiderivative, records ∫(1/x)dx as ln x without the modulus, and announces a finite value for an integral that actually diverges (e.g. ∫ from −1 to 1 of 1/x).
The limit process is seen as a formality rather than the definition, so the boundary case is skipped; the modulus on ln|x| is dropped because it rarely matters on positive intervals and so is never practised on negative ones.
What to do instead
Set up the limit explicitly before doing anything else: ∫[a,∞) f = lim(t→∞) ∫[a,t] f, and for a singularity at c split into lim(s→c⁻) and lim(t→c⁺). Evaluate the definite integral with finite bounds first, then take the limit, and finish by stating that the limit exists (convergent, with the value) or does not (divergent). Quote ln|x| + C as the standard antiderivative of 1/x, and check whether a singularity lies inside the interval.
The candidate divides a trigonometric equation through by cos x or sin x, destroying the solutions where that factor vanishes, and reports only the principal-value answer - x = π/6 rather than x = π/6 + 2nπ or x = 5π/6 + 2nπ - omitting the second family and the period.
Dividing reduces a quadratic in cos x to a linear equation, which is quicker; the general-solution formulas are memorised as three unrelated facts and are never applied systematically, and answers are often given over an unspecified domain where completeness is easy to overlook.
What to do instead
Factorise into a product and use the zero-product property instead of dividing, so that cos x = 0 cases survive. Then apply the general-solution formulas with the period built in: tan θ = tan α gives θ = nπ + α; cos θ = cos α gives θ = 2nπ ± α; sin θ = sin α gives θ = nπ + (−1)ⁿα. If an interval is specified, enumerate every solution inside it, state how many there are, and spot-check each by substitution.
The candidate writes arcsin(sin 5π/6) = 5π/6 and composes other inverse trigonometric functions as if they inverted over all real numbers, ignoring the principal ranges and the quadrant from which an angle came.
sin and arcsin are treated as exact undoings of each other, as inverse operations usually are in algebra, and the restriction needed to make sin one-to-one is stated once in class and then not used. The same reflex produces wrong-quadrant answers when a triangle is drawn from an inverse-trig value.
What to do instead
Learn the principal ranges as part of the definition - arcsin: [−π/2, π/2], arccos: [0, π], arctan: (−π/2, π/2) - and reduce the angle into the range before applying the function: arcsin(sin 5π/6) = arcsin(1/2) = π/6. When an inverse-trig value defines a triangle, place the triangle in the quadrant its range dictates rather than the first quadrant by default, and quote the domain restriction before differentiating functions such as arcsin x.
When finding the n-th roots of a complex number, the candidate divides the argument by n and reports a single principal root, and computes the argument as arctan(y/x) even when z lies in the second or third quadrant.
De Moivre's theorem is applied to one representative of the argument, and arg z is treated as a single number rather than a value modulo 2π; arctan's own range of (−π/2, π/2) is forgotten because formula sheets give θ = arctan(y/x) without the quadrant correction.
What to do instead
Convert to polar form first, choosing the argument from the quadrant of z and adding π where needed. Then write z^(1/n) = r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, …, n − 1, expect exactly n distinct roots, and sketch them as vertices of a regular n-gon on the circle of radius r^(1/n). Verify one root by raising it back to the power n.