GCE A/L Combined Mathematics common mistakes
GCE A/L Combined Mathematics · Understand the traps and how to avoid them.
When solving trigonometric equations like sin 2θ = 1/2, the candidate finds only θ = π/12 and 5π/12, omitting the solutions obtained by adding π to 2θ (giving θ = 7π/12 and 11π/12) and any others within the required interval.
The candidate treats the double-angle as a single variable and forgets that sin x = 1/2 has infinitely many solutions; they solve only the principal-value branch and do not use the general solution or the given domain to find all angles.
What to do instead
Always write the general solution first: for sin 2θ = 1/2, 2θ = π/6 + 2nπ or 2θ = 5π/6 + 2nπ, then divide by 2 and list all θ in the specified interval by testing integer values of n. Practise with a unit-circle sketch and check that the number of solutions matches the expected periodicity.
When differentiating a product such as x² sin x, the candidate writes 2x cos x, multiplying the derivatives instead of applying the product rule; similarly, for a quotient they differentiate numerator and denominator separately.
The candidate has memorised the power rule and assumes differentiation is always a simple operation on each term; they do not recognise the structure as a product (or quotient) requiring the product (or quotient) rule.
What to do instead
Before differentiating, explicitly label the function as a product u(x)v(x) or quotient u/v and write down the rule. For x² sin x, set u = x², v = sin x, then (uv)' = u'v + uv' = 2x sin x + x² cos x. Drill a mixed set of 10 expressions and state the rule used before each derivative.
When integrating a rational function like 1/(x²−1) or 1/(x²+1), the candidate writes ln|x²−1| or ln(x²+1) as the answer, incorrectly applying the ln rule for 1/x.
The candidate overgeneralises the standard result ∫(1/x)dx = ln|x| to any denominator, ignoring that the derivative of the denominator is not a constant multiple of the numerator; they do not check by differentiating.
What to do instead
Before integrating, check whether the numerator is a constant multiple of the derivative of the denominator. If not, use partial fractions for 1/(x²−1) = ½[1/(x−1) − 1/(x+1)] or the arctan formula for 1/(x²+1). Always differentiate the answer mentally to verify.
In a vector problem, the candidate computes the dot product of two vectors and then treats the scalar result as a vector, or adds a scalar to a vector (e.g., writing a·b + c as a vector sum).
The candidate does not distinguish between scalar and vector quantities and has not internalised that the dot product returns a scalar while the cross product returns a vector; notation is used loosely.
What to do instead
Annotate every quantity as scalar or vector before calculating. Remember: a·b is a scalar (projection length times length), a×b is a vector perpendicular to both. Never add a scalar to a vector; if an expression mixes them, it is undefined. Practise classifying expressions like (a·b)c, a·(b×c), |a|b as scalar or vector.
When finding the equation of a tangent to a curve y = f(x) at x = a, the candidate uses f(a) as the gradient and writes y − f(a) = f(a)(x − a), instead of using f′(a) as the gradient.
The candidate confuses the function value (the y-coordinate of the point) with the derivative (the slope of the tangent); they have memorised the point-slope form but not which quantity goes where.
What to do instead
Write the tangent recipe explicitly: point = (a, f(a)), gradient m = f′(a), equation y − f(a) = f′(a)(x − a). Before substituting, compute f(a) and f′(a) separately and label them 'point' and 'gradient'. Check with a simple curve such as y = x² at x = 1: point (1,1), gradient 2, tangent y − 1 = 2(x − 1).
In a statics problem involving a ladder leaning against a wall, the candidate takes the normal reaction at the wall to be equal to the weight of the ladder, or omits friction at the wall when the wall is described as rough.
The candidate applies 'forces balance' globally without resolving horizontally and vertically, and assumes every surface is smooth unless explicitly told otherwise; they do not draw a complete free-body diagram.
What to do instead
Draw a free-body diagram showing every contact force: weight at the centre of mass, normal reaction perpendicular to each surface, friction along each surface opposing impending motion. Then resolve horizontally and vertically and take moments about a point to eliminate unknowns. Label each force with its direction and point of application before writing equations.
When calculating the probability of A or B, the candidate uses P(A) + P(B) without subtracting P(A ∩ B), even when the events are not mutually exclusive (e.g., drawing a card that is a heart or a king).
The candidate memorises the addition rule only for mutually exclusive events and does not check the condition; they fail to identify the overlap event and its probability.
What to do instead
Always write P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and explicitly test whether A ∩ B is non-empty. For a standard deck, P(heart or king) = 13/52 + 4/52 − 1/52 = 16/52. Draw a Venn diagram for each problem and shade the union before computing.
When finding the centre and radius of a circle from x² + y² + 2gx + 2fy + c = 0, the candidate writes centre = (g, f) instead of (−g, −f), or radius = √(g² + f² + c) instead of √(g² + f² − c).
The candidate memorises the formula without deriving it by completing the square, so the signs and the constant term are misremembered; they do not check by substituting the centre back into the equation.
What to do instead
Derive it once: (x + g)² + (y + f)² = g² + f² − c, so centre (−g, −f) and radius √(g² + f² − c). For every problem, complete the square yourself rather than quoting the formula, and verify that the radius squared is positive. Test with x² + y² − 4x + 6y − 12 = 0: centre (2, −3), radius 5.
When solving an inequality such as (x − 1)/(x + 2) < 0, the candidate multiplies both sides by (x + 2) without considering that x + 2 may be negative, and so reverses the inequality incorrectly or loses the sign case.
The candidate treats inequalities like equations and multiplies by an expression of unknown sign; they do not use a sign chart or consider critical values where the expression is zero or undefined.
What to do instead
Never multiply an inequality by an expression whose sign is unknown. Instead, bring all terms to one side, find the critical values (zeros of numerator and denominator), and test the sign in each interval on a number line. For (x − 1)/(x + 2) < 0, critical values x = 1 and x = −2 (excluded), giving −2 < x < 1. Always state excluded values where the denominator is zero.
In a dynamics problem with a particle on an inclined plane connected by a string over a pulley to a hanging mass, the candidate uses the same acceleration for both particles without checking the direction of motion, or omits the component of weight along the plane (mg sin θ).
The candidate applies F = ma to the system as a whole without resolving forces along the direction of motion for each particle, and assumes tension is equal to the hanging weight rather than solving the coupled equations.
What to do instead
Draw separate free-body diagrams for each particle, choose a positive direction for each (consistent with the string constraint), and write an equation of motion for each: for the hanging mass Mg − T = Ma, for the mass on the plane T − mg sin θ = ma (if moving up the plane). Solve the simultaneous equations for T and a. Check the special case θ = 90° reduces to the Atwood machine.