Cameroon GCE A-Level Pure Math with Mechanics common mistakes
Cameroon GCE A-Level Pure Math with Mechanics · Understand the traps and how to avoid them.
When finding an inverse function, the candidate swaps x and y but then leaves the equation in the form x = f(y) and calls that the answer, or writes f⁻¹(x) = 1/f(x).
The swap-and-solve procedure is memorised as 'swap the letters' without the final step of making y the subject. The notation f⁻¹ looks like a reciprocal, so the reciprocal error is a notation trap rather than a reasoning error.
What to do instead
Always finish by writing y = ... explicitly and then relabel it f⁻¹(x). Check with a numerical pair: if f(3) = 7 then f⁻¹(7) must equal 3, and 1/f(7) will not. Also state the domain of f⁻¹ as the range of f.
When solving an inequality such as (x−1)(x−4) < 0 or a rational inequality, the candidate multiplies through by a variable factor or writes x < 1 or x < 4 instead of 1 < x < 4.
Linear-inequality habits carry over: candidates treat the solution as a union of simple bounds and forget that the sign of a product depends on the number of negative factors, which changes at each root.
What to do instead
Draw a sign table with the critical values in order and test one value in each interval. For rational inequalities, never cross-multiply; move everything to one side over a single denominator and use the sign table, remembering to exclude values making the denominator zero.
In coordinate geometry of circles, the candidate reads off the centre of x² + y² + 2gx + 2fy + c = 0 as (g, f) or forgets to divide the coefficients by 2, and answers tangent or intersection questions by algebra alone without checking the discriminant condition.
The general form is memorised as a formula string without the factor of 2, and the link between tangency and a repeated root of the substituted quadratic is not made explicit in candidates' notes.
What to do instead
Always convert to completed-square form (x + g)² + (y + f)² = g² + f² − c so the centre (−g, −f) and radius are visible. For tangency, substitute the line into the circle and set the discriminant to zero; state clearly that this is the condition for exactly one intersection.
The candidate uses the arithmetic or geometric formula with the wrong n - for example treating the first term as term 0, or using n instead of n − 1 for the nth term - and misapplies the sum to infinity when |r| ≥ 1.
Textbook formulas are quoted from memory without a check against the definition, and the convergence condition is treated as a footnote rather than a prerequisite.
What to do instead
Before substituting, write down a, d or r, and the term number explicitly. Verify with a small case: the 3rd term of an AP is a + 2d, so the nth is a + (n−1)d. For sums to infinity, first show |r| < 1; otherwise state that the sum diverges. When a question asks for the least n such that a sum exceeds a value, solve the inequality in n and round up, not down.
When solving trigonometric equations in a given interval, the candidate gives only the principal value, or divides both sides by a factor such as cos x and loses the solutions where that factor is zero.
Inverse-trig functions return one value, and candidates stop there. Dividing by a variable factor feels like normal algebra, but it is invalid when the factor can be zero.
What to do instead
Use the general solution first (for example x = 180°n + (−1)ⁿ·30° for sin x = ½), then list every value inside the stated interval. For equations like sin 2x = cos x, factorise instead of dividing, so both sin-type and cos-type families are retained. Check each answer in the original equation.
The candidate differentiates or integrates expressions involving e^x, ln x, a^x or log_a x by treating them as powers or polynomials - for example writing the derivative of e^{2x} as e^{2x}, or the integral of 1/x as ln|x| only when x > 0 without the modulus.
Power-rule patterns dominate practice, so e^x and ln x are absorbed into them. The chain-rule factor for the inner function is the step most often dropped.
What to do instead
Write the standard results on one line and apply them with the inner derivative shown: d/dx e^{kx} = k e^{kx}, d/dx ln(kx) = 1/x, d/dx a^x = a^x ln a. For integration, use ∫(1/x)dx = ln|x| + c and check every antiderivative by differentiating it back.
In stationary-point and optimisation questions, the candidate finds dy/dx = 0 and stops, or identifies a maximum or minimum by inspecting the graph sketch rather than by a valid test.
The procedure is remembered as 'set derivative to zero and solve', and the second-derivative test is either omitted or applied without checking that the second derivative is nonzero at that point.
What to do instead
Complete the full sequence: solve dy/dx = 0, then evaluate d²y/dx² at each root and state the conclusion (negative = maximum, positive = minimum). If d²y/dx² = 0, use a sign table of dy/dx. For worded optimisation, define the variables, write the constraint, express the target as a function of one variable, state its domain, then optimise and answer the question asked.
When evaluating a definite integral by substitution, the candidate changes the variable but keeps the original limits, or omits the modulus/absolute-value and constant issues in areas and volumes.
Limits are copied from the question without noticing that they refer to the old variable, and the mechanics of 'change limits or substitute back' is not fixed in the routine.
What to do instead
Either convert the limits to the new variable at the same time as the substitution, or integrate fully and substitute back before applying the original limits. For areas, sketch the curve, find intersection points, and split the integral where the curve crosses the axis, taking the modulus of each piece.
In vector work, the candidate confuses a position vector with a direction vector, computes the angle between two lines using the wrong pair of direction vectors, or treats a dot product of zero as meaning the vectors are parallel.
Notation such as a·b = 0 is memorised without the geometric meaning, and the distinction between a point on a line and the line's direction is blurred when lines are given in different forms.
What to do instead
Write each line in the form r = a + tb and label a as a point and b as the direction before doing anything. Use cos θ = (b₁·b₂)/(|b₁||b₂|) with the two direction vectors only. Remember a·b = 0 means perpendicular, and parallel means b₁ = kb₂. Check the sign of the dot product against a sketch.
In kinematics, the candidate applies constant-acceleration formulas (suvat) to motion where acceleration varies with time or displacement, or mixes up displacement, velocity and distance travelled.
The suvat equations are heavily drilled and become the default, even when the question gives a = f(t) or v = f(x). Distance and displacement are treated as synonyms.
What to do instead
First identify what is given: if a is constant, suvat is valid; if a is a function, integrate (a = dv/dt or v = dv/dx·dx/dt). For distance travelled, integrate |v| over the interval, splitting where v = 0. Always state the direction convention and the units at the end.
In dynamics problems on inclined planes and connected particles, the candidate omits a force from the free-body diagram - commonly the component of weight along the plane, the normal reaction, or the tension on one side of a pulley - and then applies F = ma inconsistently to the two bodies.
The diagram is drawn hastily from the picture in the question rather than from a systematic force inventory, and the same acceleration is not enforced for connected particles.
What to do instead
Draw a separate free-body diagram for each body, list weight, normal reaction, tension, friction and applied force explicitly, and resolve along and perpendicular to the plane. Write F = ma for each body with the same magnitude of acceleration and consistent sign convention, then solve the simultaneous equations. Check that the normal reaction is positive.
In statics and moments, the candidate takes moments about a point that is not the point of contact or hinge, forgets that a force through the pivot has zero moment, or ignores the perpendicular-distance requirement and uses the slant distance directly.
The moment formula is recalled as 'force times distance' without the perpendicular condition, and the strategic choice of pivot to eliminate unknown reactions is not practised.
What to do instead
State the pivot explicitly and take moments about it, noting that any force passing through it drops out. Use the perpendicular distance, or resolve the force into components and take the component perpendicular to the rod. For a rod in equilibrium, also write the force-balance equations in two perpendicular directions, and check that the number of equations matches the number of unknowns.