Kaoyan Mathematics II - independent mathematical foundations common mistakes
Kaoyan Mathematics II - independent mathematical foundations · Understand the traps and how to avoid them.
When finding eigenvectors, the candidate solves det(A − λI) = 0 for λ correctly but then solves (A − λI)x = 0 by row-reducing A itself rather than A − λI, or picks a single eigenvector for a repeated eigenvalue and misses the second independent one.
Row-reduction of A is a well-practised reflex, and substituting the numeric λ into every diagonal entry is error-prone mental arithmetic. For repeated eigenvalues the candidate stops once one free variable yields a vector, not checking the nullity of A − λI.
What to do instead
Write A − λI explicitly with λ substituted before any row operation, and compute its nullity to know in advance how many independent eigenvectors to expect (nullity = multiplicity for diagonalisable cases). For each eigenvalue, state the dimension of the eigenspace before choosing vectors. Verify each candidate v by computing Av and confirming it equals λv - a 30-second check that catches both errors.
The candidate computes a cross product or scalar triple product and reports a magnitude or a signed value when a vector (or the opposite orientation) was required, or takes the dot product where the cross product is needed for a normal vector.
Area, volume and normal-vector problems all invoke 'cross product' in the candidate's notes, and the operations get conflated. Sign conventions in the triple product (a·(b×c) versus (a×b)·c) and the direction of a×b are also easy to reverse under time pressure.
What to do instead
Write the geometric quantity required first - 'normal vector =', 'area = ½|a×b|', 'volume = |a·(b×c)|' - then perform the operation that matches it. Compute cross products with the determinant expansion written out in full (i, j, k rows) rather than mentally, and check orientation with the right-hand rule or by confirming a·(a×b) = 0 and b·(a×b) = 0.
When differentiating a composite function, the candidate differentiates only the outer function and drops the inner derivative - writing d/dx sin(3x²) = cos(3x²) or d/dx e^(x²) = e^(x²).
Chain rule is learned as a pattern ('derivative of sin is cos') and under timed conditions the candidate pattern-matches the outer function and stops, treating the argument as a single variable. Nested functions with polynomial or exponential inner parts are where the omission is most likely, because the inner derivative is not visually salient.
What to do instead
Before writing any answer, explicitly name the inner function u and write the chain as outer′(u)·u′ on the page, even for one-line answers. Drill a mixed set of 20 composites (trig, exponential, log, power, inverse trig) and mark each inner derivative in a different colour until the step becomes automatic. On the exam, do a final pass over every derivative checking that each composite has a visible inner-derivative factor.
The candidate computes a definite integral or area but substitutes the limits in the wrong order or forgets to change limits after substitution, producing a sign-flipped or wrong-valued answer, e.g. evaluating ∫₀² with F(0) − F(2).
The upper-minus-lower convention competes in memory with the substitution rule's 'evaluate at old limits', and under pressure the candidate writes the antiderivative values in the order they appear on the page. When using u-substitution the candidate often back-substitutes inconsistently or plugs original limits into the u-antiderivative.
What to do instead
Adopt a fixed written template: bracket the antiderivative, write '= [F(x)]_a^b = F(b) − F(a)', and physically copy b above a before evaluating. For substitution, always either convert the limits to u-values in the same line or write 'back-substitute to x, then apply original limits' - never mix. Practise five substitution integrals where the limits are converted, checking that a positive integrand gives a positive result.
The candidate solves a first-order linear ODE without dividing through by the leading coefficient, so the integrating factor uses P(x) from the un-normalised equation, or omits the solution constant after integrating the product equation.
The standard form y′ + P(x)y = Q(x) is memorised, but equations are often presented as x y′ + 2y = … and the candidate reads off the coefficient of y directly. The integrating factor e^∫P is treated as a formula to plug into, and the solution constant after integrating μQ is confused with the arbitrary scaling constant of μ, which may legitimately be fixed.
What to do instead
Always rewrite the ODE into y′ + P(x)y = Q(x) as an explicit first line, dividing every term by the coefficient of y′. Compute μ = e^∫P dx with the integral shown, and note that the arbitrary constant may be taken as zero only because it cancels - state this. Finish by differentiating the solution and substituting into the original (un-normalised) equation as a check.
When classifying the equilibrium of a 2×2 linear system, the candidate uses the trace and determinant but misassigns the cases - calling a saddle a node because det < 0 is confused with trace < 0, or declaring stability from the sign of the determinant alone.
The classification table (trace τ, determinant Δ, discriminant τ² − 4Δ) is memorised as a list rather than derived from eigenvalues, and the conditions for saddle (Δ < 0), stable node/spiral (τ < 0, Δ > 0) and centre (τ = 0, Δ > 0) blur together.
What to do instead
Compute the eigenvalues explicitly first, then classify: real parts both negative = stable, both positive = unstable, opposite signs = saddle, purely imaginary = centre. Use τ and Δ only as a cross-check, writing the rule beside the working ('Δ < 0 ⇒ saddle regardless of τ'). Practise ten systems where you state eigenvalues, then classification, then sketch the phase portrait.
The candidate sets up a modelling problem by substituting numbers immediately, producing an equation in the wrong variables or with mismatched units, and never states the assumptions or checks the solution against the physical constraints (non-negativity, boundary values, limiting behaviour).
Exam pressure pushes toward computation, and word problems invite premature substitution before the relationship between quantities is defined. Units (per hour versus per minute, percentage versus proportion) are silently mixed, and the resulting answer is accepted without a plausibility test.
What to do instead
Write three lines before any algebra: define every variable with its units, state the governing relationship or balance law (rate in − rate out, conservation, proportionality), and list the assumptions. Keep symbols until the final step, then substitute with units carried through. Finish by checking the solution against boundary conditions and limits (e.g. t → 0 and t → ∞) and confirming it lies in the physically admissible range.
When computing a limit that leads to an indeterminate form, the candidate applies L'Hôpital directly without first checking the form, or applies it to a quotient that is not 0/0 or ∞/∞, and differentiates numerator and denominator as if differentiating a quotient (using the quotient rule instead of separate derivatives).
L'Hôpital is over-practised as a reflex, and the quotient rule is so recently drilled that it contaminates the procedure; the candidate also skips the substitution step that would reveal the form.
What to do instead
Adopt a fixed routine: substitute the value first and write the resulting form (0/0, ∞/∞, 0·∞, ∞−∞, 1^∞…) explicitly. Only for 0/0 and ∞/∞ apply L'Hôpital, and write the derivatives of numerator and denominator as two separate expressions, never as a quotient. For other forms, transform first (common denominator, conjugate, or take logarithms for 1^∞). Verify the final limit by a quick numerical check with a value close to the limit point.
In an optimisation problem, the candidate finds f′(x)=0 and reports those critical points as the answer, without verifying that the point lies in the feasible interval, without classifying it as a maximum or minimum, and without answering the actual question asked (the dimensions, the maximum area, the minimum cost).
The algebraic step (solve f′=0) is the part that feels like 'the maths', while translating back to the problem statement and checking endpoints is treated as cosmetic; the candidate also forgets that a constrained problem can have its optimum at a boundary.
What to do instead
End every optimisation with a four-line checklist written in the answer: domain of the variable from the geometry; critical points inside that domain; classification via the second derivative or a sign table of f′; evaluation of f at the critical point and at the endpoints. Then write a sentence in words giving the requested quantity with units. If the question asks for dimensions, the final line must state the dimensions, not just the value of the function.
When computing a definite integral by substitution or by parts, the candidate changes the variable but keeps the original limits, substituting them into the new variable's expression, or forgets to return to the original variable before evaluating.
The mechanics of the antiderivative dominate attention, and the limits are copied down at the start of the line and never revisited; the candidate also treats the change of limits as an optional refinement.
What to do instead
Choose one of two disciplined formats and stick to it: either (a) change the limits at the same time as the variable and never write x again after the substitution, or (b) keep indefinite-integral notation, find the antiderivative in x, and only then apply the original limits. Mark the limits with a bracket [F(x)] from a to b so the evaluation step is visually forced. After evaluating, check the sign and rough magnitude (e.g. an area must be positive).
When computing the area between two curves or the volume of a solid of revolution, the candidate integrates from the x-coordinates of the intersection points without checking which curve is on top, and reports a negative area, or uses the function value instead of the actual distance to the rotation axis when the axis of revolution is not a coordinate axis.
The intersection points are found correctly and the integral is set up from them automatically, but the ordering of the curves and the geometry of the radius are not examined; the candidate assumes 'area = integral of the difference' regardless of sign and 'radius = f(x)' regardless of the axis.
What to do instead
Sketch the region (even roughly) and mark which curve is upper. Set up the integral as ∫(upper − lower) and, if the curves cross inside the interval, split the integral at the crossing point and add absolute values. For volumes, write the radius explicitly as a distance: for rotation about y=k, r(x)=|f(x)−k|; about x=h with horizontal slices, r(y)=|g(y)−h|. State the radius formula in words before integrating.