NEB Class 12 Mathematics common mistakes
NEB Class 12 Mathematics · Understand the traps and how to avoid them.
When solving a matrix equation such as AX = B, the candidate writes X = BA⁻¹ (or multiplies both sides on the right by A⁻¹) instead of X = A⁻¹B.
They carry over the scalar habit that multiplication commutes, and they do not check that the inverse is conformable on the left. In matrix problems the order decides whether the product is defined and whether the answer is correct.
What to do instead
Before calculating, write the dimensions of A, X and B. Multiply both sides on the left by A⁻¹: A⁻¹AX = A⁻¹B, so X = A⁻¹B. Check by substituting back into AX.
In counting problems, the candidate uses nPr for a committee/selection question or nCr for a seating/arrangement question, and double-counts when letters or objects repeat.
They decide by keyword ('choose' vs 'arrange') rather than by whether order changes the outcome; repeated items are treated as distinct.
What to do instead
Apply a swap test: interchange two selected items-if this gives a different outcome, order matters and use permutations; if not, use combinations. For repeated letters divide by the factorial of each repetition count.
Solving sin θ = 1/2, the candidate gives only θ = π/6 (or 30°) and stops, missing θ = 5π/6 and all general solutions.
They treat inverse sine as a single-valued operation and forget that the equation asks for all angles in the given interval or general solution.
What to do instead
Draw the unit circle, mark both intersections with the horizontal line y = 1/2, then add the period. For sin θ = sin α write θ = nπ + (-1)^n α; for cos θ = cos α write θ = 2nπ ± α; for tan θ = tan α write θ = nπ + α, n ∈ Z.
The candidate simplifies sin⁻¹(sin 5π/6) to 5π/6, or tan⁻¹(tan π) to π, ignoring the principal ranges.
They assume inverse and direct functions always cancel, but sin⁻¹ has range [-π/2, π/2] and tan⁻¹ has range (-π/2, π/2).
What to do instead
First reduce the angle to the principal range using identities: sin(5π/6) = sin(π/6), so sin⁻¹(sin 5π/6) = π/6. For tan, subtract π until the angle lies in (-π/2, π/2).
Given 9x² + 16y² = 144, the candidate reads a² = 9 and b² = 16 without dividing by 144, then uses the wrong eccentricity formula or calls the major axis vertical.
They confuse coefficients with standard-form denominators and do not check which denominator is larger.
What to do instead
Divide through to get x²/16 + y²/9 = 1. Since 16 > 9, a² = 16, b² = 9, and the major axis is along x. For an ellipse use b² = a²(1 - e²); for a hyperbola x²/a² - y²/b² = 1 use b² = a²(e² - 1). Write the standard form before reading off parameters.
The candidate uses m₁m₂ = 1 for perpendicular lines, or drops the absolute value when using the point-to-line distance formula, or uses the wrong sign in the angle formula.
They memorize fragments of formulas and do not connect slope product to rotation by 90°, and distance is a magnitude so it cannot be negative.
What to do instead
Derive quickly: perpendicular slopes satisfy m₁m₂ = -1; parallel lines satisfy m₁ = m₂. For distance from (x₁, y₁) to ax + by + c = 0 use |ax₁ + by₁ + c|/√(a² + b²). For the angle between lines use tan θ = |(m₁ - m₂)/(1 + m₁m₂)| and check whether the acute angle is requested.
The candidate uses a × b when the question asks for work, projection or angle, or writes a · b as a vector; in triangle area problems they forget the factor 1/2.
Dot and cross products are remembered as two formulas rather than as different outputs: dot gives a scalar and measures parallel alignment, cross gives a vector and measures perpendicular area.
What to do instead
Before computing, classify the required output: scalar (work, projection, cos angle) → dot product; vector (torque, normal, area vector) → cross product. Area of triangle = ½|a × b|, parallelogram = |a × b|. Check the final type.
On seeing lim x→2 (x² - 4)/(x - 2), the candidate substitutes x = 2, obtains 0/0, and writes the answer as 0, or applies L'Hôpital even when the limit is not indeterminate.
They treat 0/0 as a value rather than an indeterminate form, and they do not check one-sided behaviour or differentiability conditions.
What to do instead
If substitution gives 0/0, factor, rationalize or use a standard limit before concluding. For this example cancel (x - 2) to get x + 2 → 4. Use L'Hôpital only for 0/0 or ∞/∞ and only after verifying the functions are differentiable near the point; check left and right limits separately.
After a u-substitution in a definite integral, the candidate keeps the original x-limits, or omits +C in an indefinite integral, or drops the negative sign when substituting dx.
They treat the substitution as a formal trick and do not track that the variable and limits must change together; indefinite integrals are families of antiderivatives.
What to do instead
Write u = ..., du = ... including the sign, change the limits from x-values to u-values before evaluating, and never substitute back inside a definite integral. For indefinite integrals, finish with '+C' and differentiate the answer as a check.
The candidate assumes independent events have P(A ∩ B) = 0, or computes P(A|B) as P(A ∩ B)/P(A) instead of dividing by P(B).
They conflate mutually exclusive (cannot both happen) with independent (one does not change the other's probability), and they memorize the conditional formula without identifying the given event.
What to do instead
Write the definitions: mutually exclusive → P(A ∩ B) = 0; independent → P(A ∩ B) = P(A)P(B). Conditional: P(A|B) = P(A ∩ B)/P(B), provided P(B) > 0; identify B as the event after the vertical bar. Test with a two-event Venn diagram or table.