LC Higher Maths common mistakes
LC Higher Maths · Understand the traps and how to avoid them.
When solving a quadratic or higher-degree equation by cancelling a factor from both sides, the candidate divides by a variable expression (e.g. dividing x² = 3x by x to get x = 3) and loses the root x = 0.
Cancelling feels like simplification, and the candidate is used to arithmetic where dividing both sides by the same nonzero number is always safe. They do not check whether the divisor can be zero, so the loss of a solution goes unnoticed.
What to do instead
Never divide by a variable expression. Factorise to x² − 3x = 0, then x(x − 3) = 0, and read both roots. Before any division by an expression, state the condition that the expression is nonzero and check separately whether it can be zero.
On a function question, the candidate substitutes into f(x) correctly but then applies transformations in the wrong order, e.g. treating f(2x + 4) as a stretch by 2 followed by a shift of 4 units left rather than a shift that is halved.
The candidate memorises 'inside the bracket means horizontal and opposite' but does not track that the horizontal shift is measured after factoring out the coefficient of x. They work left to right as written.
What to do instead
Always rewrite f(ax + b) as f(a(x + b/a)) first, then read the shift as b/a and the stretch as 1/a. Test with a single point: pick x so the bracket equals a known input, and check the image lands where the sketch says it should.
When using the sine rule to find a missing angle, the candidate takes the inverse sine of a value and gives only the acute answer, ignoring the obtuse possibility that the triangle may allow.
The inverse sine function returns only the principal value in [−90°, 90°], and candidates treat this as the unique answer. They do not connect the ambiguous case to the geometry of the given triangle.
What to do instead
After computing sin A, write both candidates A and 180° − A, then test each against the angle sum of the triangle. If the given information is side-side-angle with the side opposite the given angle shorter than the other given side, expect two triangles and check both.
When finding the equation of a line perpendicular to a given line, the candidate uses the same slope or the negative of the slope instead of the negative reciprocal, e.g. using −2 for a line of slope 2 rather than −1/2.
The candidate conflates 'perpendicular' with 'opposite direction' and remembers only part of the rule. Under time pressure they reach for the sign change and skip the reciprocal.
What to do instead
Write the rule as m₁m₂ = −1 and check the product after choosing the slope. Convert both lines to y = mx + c form before comparing slopes so the coefficient is explicit, and substitute the given point back into the final equation as a check.
In a conditional probability or independent-events question, the candidate computes P(A)P(B) when the events are not independent, or uses P(A ∩ B) = P(A)P(B) without checking independence, and reports a probability that ignores the condition given.
The multiplication rule for independent events is the most practised formula, so it becomes a default. The candidate does not distinguish P(A|B) from P(A) and skips the step of identifying whether the sample space has changed.
What to do instead
Before calculating, state in words whether the condition changes the sample space. If it does, use P(A|B) = P(A ∩ B)/P(B) or count directly from the restricted set. Only use P(A)P(B) after explicitly justifying independence.
When asked for a confidence interval or a hypothesis conclusion, the candidate computes a sample mean and standard error but reports the interval using the sample standard deviation without dividing by √n, or draws a conclusion about the population from the sample without linking it to the interval.
The formula for the standard error of the mean is easily confused with the sample standard deviation, and the interpretation step is treated as optional. Candidates often stop once a number is produced.
What to do instead
Write the interval as x̄ ± z·(σ/√n) with the √n visible in the working, and finish with a sentence interpreting the interval in context, naming the population and the confidence level. For a hypothesis test, state the conclusion in terms of the original claim, not just 'reject H₀'.
On a measurement or volume question, the candidate substitutes lengths in different units (e.g. cm and m) into the same formula, or scales an area or volume by a linear scale factor without squaring or cubing it.
Unit conversion is treated as clerical and skipped, and the candidate applies a single scale factor to all quantities because the enlargement factor feels like one number.
What to do instead
Convert every length to a single unit before substituting, and write the unit beside each quantity. For similar figures, record the linear, area and volume ratios as k, k², k³ explicitly before using them.
When differentiating or integrating, the candidate drops the chain rule on a composite function (e.g. differentiating sin 3x as cos 3x) or omits the constant of integration, and when finding a max/min uses f″ without checking the sign correctly.
The chain rule is applied automatically only when the inner function is obvious, and the constant of integration is treated as cosmetic. Candidates also confuse f″ > 0 with a maximum because they do not connect it to concavity.
What to do instead
For every derivative, name the inner function and multiply by its derivative as a written step. Always write '+ C' for indefinite integrals. For stationary points, build a sign table for f′ or evaluate f″ and state 'f″ < 0 so concave down, hence maximum' in words.
In a proof by induction, the candidate verifies the base case and then assumes the statement for n = k and immediately writes the statement for n = k + 1 as if it were given, without showing the algebraic link between them.
The structure of induction is memorised as three headings, but the crucial step of manipulating the n = k result into the n = k + 1 form is the hardest part, so candidates skip it and present the conclusion as an assertion.
What to do instead
Write the n = k + 1 target explicitly first, then start from the n = k expression and add or multiply the extra term, showing each algebraic step until the target appears. End with 'therefore true for n = k + 1, and by induction true for all n ≥ 1'.
On a multi-step problem, the candidate answers only the first part, or carries an unrounded exact value into later parts inconsistently, giving a final answer that contradicts an earlier part of the same question.
The candidate treats each part as independent and does not re-read the full question before finishing. Rounding decisions are made locally without recording the exact form.
What to do instead
Before writing, list what each part asks and which earlier result it depends on. Keep exact surd or fraction forms in working and round only at the stated point, noting the exact value beside it. Finish by checking the final answer against the context (sign, size, units).