CAPE Pure Maths Unit 1 common mistakes
CAPE Pure Maths Unit 1 · Understand the traps and how to avoid them.
When solving a modulus equation or inequality such as |2x − 5| < 7, the candidate squares both sides or simply deletes the bars, writing 2x − 5 < 7 and stopping at x < 6 without producing the lower bound x > −1.
The bars are treated as decoration rather than as a piecewise-defined function. Candidates have practised linear inequalities far more than modulus ones, so the reflex is to isolate x in one line, and the habit of writing a single interval is never disrupted by a counterexample such as x = −5, which satisfies the original inequality but is excluded by their answer.
What to do instead
Always split into cases before solving: |A| < k means −k < A < k, and |A| > k means A < −k or A > k. Write the two cases explicitly, solve each linear inequality separately, then recombine and test one interior and one exterior value back in the original modulus statement. For equations, solve A = k and A = −k and check both roots satisfy the original equation, since squaring can introduce extraneous solutions.
When asked for the inverse of a function, the candidate writes f⁻¹(x) = 1/f(x), or finds the inverse of f(x) = x² over all reals and gives ±√x as a function, ignoring that the original is not one-to-one.
The superscript −1 is over-generalised from the exponent rules of arithmetic, and the domain restriction that makes a function invertible is treated as a technicality rather than as the condition for the inverse to exist. Quadratic functions are the most familiar example, so the square-root inverse is memorised without its restriction.
What to do instead
State the rule before computing: an inverse exists only on a domain where the function is strictly one-to-one. For each problem, identify the restricted domain (for example x ≥ 0 for f(x) = x²), swap x and y, solve for y, and then state the domain and range of the inverse explicitly. Never write 1/f(x); verify by composing f(f⁻¹(x)) = x on the stated domain.
When sketching y = (x − 2)²(x + 1) or a rational function, the candidate plots points and joins them with a smooth curve, missing that the graph touches the axis at a repeated root, changes sign at a simple root, and has a vertical asymptote where the denominator vanishes.
Curve sketching is often reduced to a table of values, which cannot reveal multiplicity, asymptotic behaviour, or sign changes between plotted points. Candidates also conflate the multiplicity of a root with the degree of the polynomial and so misjudge the local shape near the axis.
What to do instead
Build a sign-and-feature table before plotting: factor fully, list each root with its multiplicity (odd multiplicity crosses, even multiplicity touches), find the y-intercept, and locate vertical asymptotes and horizontal asymptotes from the degrees of numerator and denominator. Check the sign of the function in each interval between critical points, and mark asymptotic behaviour with dashed lines before drawing the curve.
When finding the centre and radius of a circle from x² + y² + 4x − 6y − 12 = 0, the candidate reads the centre as (4, −6) or forgets to halve the coefficients, and computes the radius as √12 rather than completing the square.
The general form is memorised as a formula without deriving it from completing the square, so the factor of one-half and the sign reversal are applied inconsistently. The constant term is also treated as r² in the wrong form, mixing up the expanded and standard equations.
What to do instead
Always complete the square from scratch: group x terms and y terms, add and subtract (coefficient/2)² for each, and rewrite as (x + 2)² + (y − 3)² = 25, reading centre (−2, 3) and radius 5. Confirm by expanding back to the original equation. Do this every time until the halving and sign reversal become automatic.
When solving a trigonometric equation such as sin 2x = 0.5 for 0° ≤ x ≤ 360°, the candidate gives only 2x = 30° and x = 15°, losing the other solutions because they did not extend the range for the doubled angle.
The inverse-sine key returns a single principal value, and candidates do not adjust the interval when the argument is a multiple of x. The symmetry of the sine and cosine curves is known in sketching but not applied when solving equations, and the general solution with period 360°/n is rarely written down.
What to do instead
Transform the interval first: if 0° ≤ x ≤ 360° and the argument is 2x, then solve over 0° ≤ 2x ≤ 720°. Find all values of 2x in that extended range using the symmetry rules (for sine, 180° − θ plus multiples of 360°; for cosine, 360° − θ plus multiples of 360°), then divide every solution by 2. List solutions in order and check the count against the expected number for the interval.
When differentiating a product, quotient, or composite function, the candidate writes d/dx[x² sin x] = 2x cos x or d/dx[sin(x²)] = cos(x²), dropping a factor or a whole term from the chain or product rule.
The rules are recalled as patterns rather than as procedures, and under exam pressure the candidate matches the outer function and stops. The chain rule in particular is applied only when the inner function is obviously non-trivial, so sin(x²) is treated like sin x.
What to do instead
Write the rule out and label the parts before differentiating: for a product, name u and v and write u′v + uv′; for a composite, name the inner function and write the derivative of the outer evaluated at the inner times the derivative of the inner. Practise a mixed set where the inner function is x², 3x + 1, and sin x, and always finish by checking that the number of factors in the answer matches the structure of the original.
When finding the equation of the tangent or normal to a curve at a point, the candidate substitutes the point into y = mx + c using the gradient of the curve at that point but forgets to evaluate the gradient at the correct x, or uses the tangent gradient for the normal without taking the negative reciprocal.
The derivative is computed correctly but the coordinate at which it must be evaluated is confused with the general expression, and the perpendicular condition m₁m₂ = −1 is remembered as a fact but not applied as an operation. Candidates also skip evaluating y at the point, so c is wrong even when m is right.
What to do instead
Follow a fixed four-step routine: evaluate y at the given x to get the point; differentiate and evaluate at the same x to get m; for the normal, replace m by −1/m; then substitute the point into y − y₁ = m(x − x₁). Check the final equation by substituting the point back in and, for the normal, by confirming the product of gradients is −1.
When integrating a function such as (3x + 1)⁵ or sin(4x), the candidate writes (3x + 1)⁶/6 or −cos(4x) without dividing by the inner derivative, or omits the constant of integration entirely.
The reverse chain rule is less practised than the forward chain rule, and the candidate applies the power rule as if the bracket were a single variable. The constant of integration is dropped because indefinite integrals are treated as a calculation to finish rather than as a family of functions.
What to do instead
For a linear inner function ax + b, integrate the outer function and divide by a: ∫(3x + 1)⁵ dx = (3x + 1)⁶/18 + c, and ∫sin(4x) dx = −cos(4x)/4 + c. Differentiate the answer mentally to confirm it returns the integrand. Always append + c and, where a definite integral follows, keep it until the limits are applied.
When finding the area between a curve and the x-axis over an interval that crosses a root, the candidate computes the definite integral in one piece and reports a small or zero value instead of splitting the interval and adding the magnitudes of the separate areas.
The definite integral is identified with area without qualification, so signed area and geometric area are conflated. The candidate also fails to sketch the curve, so the crossing point is not noticed and the negative contribution silently cancels the positive one.
What to do instead
Sketch the curve and find all roots in the interval before integrating. Split the integral at each root, evaluate each piece separately, take the absolute value of each result, and add them. State explicitly that the integral gives signed area and that geometric area requires splitting at sign changes.
When answering a related-rates or optimisation question, the candidate differentiates the given relation with respect to time without chaining through the varying quantity, or finds the stationary point but never verifies it is a maximum or minimum and never answers the question actually asked.
Related rates require the chain rule to be applied to a physical model, which is a modelling step rather than a routine calculation, and candidates jump to differentiation before writing the relationship between the variables. In optimisation, the second-derivative test or a sign table is skipped because the algebra of solving f′(x) = 0 feels like the end of the problem.
What to do instead
For related rates, write the geometric or physical relation first, differentiate the whole relation with respect to t so each variable picks up its rate, then substitute the known values and solve for the unknown rate. For optimisation, define the quantity to be optimised, express it in one variable, state the constraint interval, solve f′ = 0, then classify with f″ or a sign table and state the final answer with units in a sentence.