Karnataka II PUC Mathematics (KSEAB) common mistakes
Karnataka II PUC Mathematics (KSEAB) · Understand the traps and how to avoid them.
When testing whether a relation is reflexive, symmetric, or transitive, the candidate checks only one or two sample ordered pairs and then declares the property holds, instead of testing the defining condition for every element of the set.
Textbook worked examples use tiny sets (two or three elements), so students pattern-match on the visible pairs and never internalize that reflexivity requires (a, a) for every a in A and transitivity requires checking every composable chain. The quantifier structure ('for all') is the actual content and it gets skipped.
What to do instead
Before answering, list the full set A and build the complete relation as a set of ordered pairs. Then run three mechanical scans: (1) tick off (a, a) for each a in A; (2) for every pair (a, b) present, check (b, a) is also present; (3) for every pair of pairs (a, b) and (b, c), verify (a, c) is present. Show the scan on paper; a single missing pair is enough to disprove the property, and examiners award method marks for the counterexample.
The candidate writes the principal value of an inverse trigonometric expression using the general solution or the wrong branch, e.g. giving sin^-1(sin 5π/6) = 5π/6 or tan^-1(tan 3π/4) = 3π/4.
Students memorize the identity sin^-1(sin x) = x without the range restriction. They forget that sin^-1 maps into [-π/2, π/2], tan^-1 into (-π/2, π/2), and cos^-1 into [0, π], so arguments outside those intervals must be reduced first.
What to do instead
Always reduce the inner angle into the principal range before applying the inverse. For sin^-1(sin θ), rewrite θ = π - α or θ = -π - α so the new angle lies in [-π/2, π/2]; for cos^-1(sin θ) convert to cos first. Drill the four reduction identities sin^-1(sin θ) = π - θ when θ ∈ [π/2, 3π/2], etc., and state the principal range explicitly at the start of each such problem.
In matrix problems the candidate computes AB and BA, finds they differ, and still concludes the matrices commute, or assumes AB = BA when simplifying an expression or cancelling a factor.
Scalar algebra habits carry over: students treat multiplication as commutative and freely cancel A from AB = AC. Matrix multiplication is non-commutative and cancellation requires A to be invertible, which is rarely checked.
What to do instead
Treat AB and BA as different objects until proven equal. Never cancel a matrix factor unless you have shown it is invertible (det ≠ 0). When a problem asks for commutativity, compute both products fully and present them side by side. For expressions like (A + B)^2, expand as A^2 + AB + BA + B^2, not A^2 + 2AB + B^2.
The candidate expands a 3×3 determinant by a row or column but drops the alternating signs, adding all three cofactors, or expands along a row after already using it in a row operation without adjusting the determinant.
The sign pattern (+ - +, - + -, + - +) is easy to misapply, and row/column operations change the determinant value (adding a multiple of one row to another preserves it, but swapping rows or multiplying a row by a scalar does not). Students mix these rules.
What to do instead
Write the cofactor sign factor (-1)^(i+j) explicitly next to each term before multiplying. Prefer expanding along the row or column with the most zeros; create zeros using only the operation R_i → R_i + kR_j (which preserves det). If you swap rows, multiply the result by -1; if you factor a scalar out of a row, track it outside. Verify the final determinant by a second expansion along a different line.
When testing continuity or differentiability at a point, the candidate computes only the two one-sided limits of f and stops, or computes only the derivative formula and never checks that the function is continuous there first.
Continuity needs three things (left limit, right limit, value at the point all equal), and differentiability at a point requires the left-hand and right-hand derivatives to exist and agree, plus continuity. Students conflate 'the formula is differentiable everywhere else' with 'differentiable at the join point'.
What to do instead
Write a fixed checklist for each point x = a: (1) compute f(a); (2) compute lim x→a^- f(x) and lim x→a^+ f(x); (3) compare all three. For differentiability, compute LHD = lim h→0^- [f(a+h) - f(a)]/h and RHD similarly, and state that LHD = RHD is required. For piecewise functions with |x - a| or max/min terms, always split at the join and evaluate both sides symbolically.
On maxima–minima and rate-of-change problems the candidate finds f'(x) = 0, solves for x, and declares that critical point the maximum or minimum without checking the second derivative or the endpoints of the interval.
The first-derivative test is over-learned, and students forget that a stationary point can be an inflection (f'' = 0 and no sign change), and that on a closed interval the absolute extremum can occur at an endpoint where f' ≠ 0.
What to do instead
After solving f'(x) = 0, build a sign chart for f' on either side of each critical point (or evaluate f'' there). Then evaluate f at every critical point and at both endpoints of the given interval and compare values. For word problems (maximize volume, minimize cost), first write the objective as a function of a single variable using the constraint, then state the domain of that variable before differentiating.
The candidate integrates a rational or algebraic integrand by substitution but forgets to change the limits of integration, or forgets to substitute back to the original variable, leaving the answer in terms of u.
The substitution step is mechanical and the limits conversion is a separate bookkeeping task. Under exam time pressure, students either convert limits and then also substitute back (double work, wrong result) or do neither, and the final answer is in the wrong variable or evaluated at the wrong bounds.
What to do instead
Adopt one convention and always follow it: when you write u = g(x), immediately write du = g'(x) dx and convert the limits to u-values (u = g(a), u = g(b)) before integrating. Then never substitute back. If you instead keep x-limits, you must substitute back to x before evaluating. Mark the chosen path at the top of the solution so you do not mix them.
In differential equations the candidate separates variables and integrates but omits the constant of integration, or drops the absolute value when integrating 1/y, writing ln y instead of ln|y|, and then loses the solution y = 0 or a negative branch.
Indefinite integration is done quickly and the '+ C' is treated as cosmetic. The absolute value in ∫(1/y) dy = ln|y| + C is a standard formula that gets simplified away, and the singular solution (e.g. y = 0 for dy/dx = ky) is forgotten entirely.
What to do instead
Write '+ C' in the same line as the integral, before any rearrangement, and apply the initial condition only after C appears. Keep |y| until the final step, then resolve the sign using the initial condition. Separately check whether dividing by a factor (like y) excluded a constant solution, and list it if the problem asks for the general solution. For homogeneous equations, state the substitution y = vx and dy/dx = v + x dv/dx explicitly.
In vector and 3D-geometry problems the candidate computes a·b and a×b but confuses which answers perpendicularity, parallelism, area, and volume, and computes the shortest distance between skew lines using the wrong formula.
There are many vector products and each answers a different geometric question. Students memorize formulas without the geometric meaning, so they apply the cross product for a perpendicularity test (which is a dot-product question) or use |a×b| for the volume of a parallelepiped (which needs the scalar triple product).
What to do instead
Attach a meaning to each operation: a·b = 0 means perpendicular, a×b = 0 means parallel, |a×b| is the area of the parallelogram, and |a·(b×c)| is the volume of the parallelepiped. For skew lines, use the distance formula |(a2 - a1)·(b1×b2)| / |b1×b2|, where a1, a2 are points on the lines and b1, b2 are direction vectors, and verify the lines are not parallel before applying it. Write the formula down before substituting numbers.
In linear programming the candidate graphs the constraints, shades the feasible region, and then evaluates the objective function at the vertices of the wrong region (e.g. the unbounded side or an infeasible corner), or ignores the non-negativity constraints x ≥ 0, y ≥ 0.
Shading conventions are ambiguous and students shade the region satisfying each inequality individually rather than the intersection. Non-negativity is assumed rather than enforced, so vertices with negative coordinates get tested and sometimes give a spurious optimum.
What to do instead
For each constraint, test the origin (0, 0) to decide which side to shade, then shade the common intersection. Explicitly add x ≥ 0, y ≥ 0 as constraints and restrict the region to the first quadrant. List every corner point of the feasible region by solving pairs of boundary lines, then tabulate the objective value at each corner. State whether the optimum is a maximum or minimum and whether the region is bounded; for unbounded regions, note that a maximum may not exist.