SPM Add Maths common mistakes
SPM Add Maths · Understand the traps and how to avoid them.
When finding the inverse function, candidates swap x and y but then fail to make y the subject, leaving the answer as x = 2y + 3 instead of f⁻¹(x) = (x − 3)/2; or they write f⁻¹(x) = 1/f(x), treating the inverse as a reciprocal.
The swap-and-solve procedure is memorised as 'swap x and y' and candidates stop there, believing the swap itself produces the inverse. The notation f⁻¹ looks like an exponent, so many read it as 1/f.
What to do instead
Always finish with 'y = ...' and state f⁻¹ in terms of x. Verify with a composition check: compute f(f⁻¹(x)) and f⁻¹(f(x)); both must return x. For a linear f(x) = ax + b, drill the pattern f⁻¹(x) = (x − b)/a until it is automatic, and test numerically with one value, e.g. f(2) = 7 so f⁻¹(7) must equal 2.
Candidates state the range of a composite function fg as the range of f, or they claim fg exists without checking that the range of g lies inside the domain of f, and they ignore restricted domains when writing the final range.
Composite functions are treated as algebraic substitutions only; the set conditions are seen as optional decoration. Restricted domains in the question are skimmed because the algebra 'works' regardless.
What to do instead
Before composing, write the domain and range of each function explicitly. Check range(g) ⊆ domain(f) before asserting fg exists. After forming fg(x), substitute the endpoints of the restricted domain of g (not of fg) to get the actual range, and sketch the graph to confirm. Practise questions where the restriction makes fg undefined for part of the domain.
After completing the square, candidates read off the vertex with the wrong sign, giving vertex (3, −5) for f(x) = (x + 3)² − 5, or they confuse the axis of symmetry x = −3 with the coordinate of the vertex.
The completed-square form (x − h)² + k is memorised loosely; the sign inside the bracket is copied rather than reversed. The axis of symmetry and the vertex x-coordinate are mentally merged.
What to do instead
Rewrite the bracket as (x − (−3))² − 5 before reading off h = −3. Always state the vertex as a coordinate pair and the axis as an equation. Check by substituting: at x = −3 the squared term vanishes and f(−3) = −5 must be the minimum (or maximum) value. Do three or four sign-varied examples each session until the reversal is reflex.
When a question says the quadratic equation has two distinct real roots, candidates write b² − 4ac > 0 but then solve the resulting inequality as if it were an equation, listing only the boundary values instead of the solution set, or they flip the inequality sign incorrectly.
The discriminant condition is known, but the follow-on step is a quadratic inequality, which is a separate skill. Candidates rush from 'condition' to 'answer' without drawing the number line or parabola.
What to do instead
After forming the inequality in the unknown parameter, solve it by sketching the parabola of the discriminant expression and reading off where it is above or below the axis. State the answer as a set, e.g. k < −2 or k > 5. Distinguish the three cases explicitly: > 0 two distinct roots, = 0 equal roots, < 0 no real roots, and write the condition before substituting.
In simultaneous equations with one linear and one non-linear equation, candidates substitute but forget to find the paired value of the other variable for every root, giving only x-values, or they divide through by a variable and lose a solution (e.g. dividing by x when x² = 3x).
The substitution step consumes all attention, so back-substitution is treated as optional. Cancelling a common factor feels like simplification rather than the loss of a root.
What to do instead
Adopt a fixed routine: solve for one variable, list every root, then back-substitute each root into the linear equation to get pairs, and present answers as coordinate pairs. Never divide by a variable; factorise instead, so x² = 3x becomes x(x − 3) = 0. Check each pair in both original equations.
Candidates apply log laws to expressions that are not logarithms of products or quotients, writing log(A + B) = log A + log B, or they solve log x + log(x − 3) = 1 and forget to reject the root that makes an argument non-positive.
The product and quotient laws are over-generalised because the valid forms are drilled more than the invalid ones. Domain restrictions on logarithms are treated as a technicality rather than a condition of the equation.
What to do instead
Write the two invalid identities on a card: log(A + B) ≠ log A + log B and log(A − B) ≠ log A − log B. Before solving any log equation, state the conditions: every argument must be positive. After finding roots, test each against those conditions and explicitly reject the invalid one. Convert to a single logarithm first, then exponentiate.
In coordinate geometry, candidates find the gradient of the given line and reuse it for the perpendicular line, or they substitute the wrong point into y − y₁ = m(x − x₁), using the midpoint where the endpoint was required.
The perpendicular condition m₁m₂ = −1 is known but applied late, after the gradient has already been written down. Under time pressure the point is grabbed from the most recent line of working rather than from the question.
What to do instead
Write 'm₂ = −1/m₁' as an explicit line of working before writing any equation. Underline the point the new line must pass through in the question stem. For perpendicular bisector questions, compute the midpoint first and label it clearly, then use it. Verify by substituting the point into the final equation.
In circular measure, candidates use degrees in s = rθ and A = ½r²θ, or they find the arc length and call it the perimeter of the sector, omitting the two radii.
The formulae are remembered without their radian condition, and degree measure is the default from earlier geometry. 'Perimeter of sector' is read as 'arc length' because the arc is the unfamiliar part.
What to do instead
Convert to radians as the first line of working and write 'θ in radians' beside the formula. For perimeter questions, write perimeter = arc + 2r explicitly before substituting. Sanity-check: a sector perimeter must exceed 2r, and an arc length must be less than the circumference 2πr.
In differentiation, candidates differentiate a product or quotient term by term, writing d/dx(uv) = u'v' or d/dx(u/v) = u'/v', and they forget the chain rule factor when differentiating composite functions such as (2x + 1)⁵.
The sum rule is learned first and generalised by habit. The chain rule is applied only when the word 'chain' appears in a lesson, not when a bracket is raised to a power.
What to do instead
Label u and v before differentiating any product or quotient, and write the full quotient rule with the denominator squared. For any bracket raised to a power, write the inner function and its derivative separately, then multiply. Check with a simple case: differentiate x·x by the product rule and confirm you get 2x, not 1.
In integration, candidates omit the constant of integration, or they evaluate a definite integral without changing the limits when a substitution is used, substituting back to x but keeping the u-limits.
'+ c' is treated as cosmetic because it never affects a definite integral answer. The change of limits is a step that feels optional when the antiderivative has already been converted back to x.
What to do instead
Write '+ c' immediately after every indefinite integral, before any other simplification. For substitution, either change the limits at the same time as the substitution or convert the antiderivative back to x before evaluating - never mix. Mark the limits with a bracket notation and evaluate top minus bottom in one written line.