Karnataka II PUC Basic Mathematics (KSEAB) common mistakes
Karnataka II PUC Basic Mathematics (KSEAB) · Understand the traps and how to avoid them.
Expands a 3×3 determinant along the second or third row (or any column) using the same +, −, + signs learned for the first row, so the cofactor of a21 is taken as +|minor| instead of −|minor|.
Nearly all practice in this paper expands along R1, so the sign pattern gets memorised as a property of the matrix rather than as a (−1)^(i+j) tag attached to a position. When a question forces expansion along R2, R3 or a column (common when a row has two zeros), every second term gets the wrong sign and the answer is off by twice the error terms.
What to do instead
Before expanding, write the 3×3 sign grid (+ − + / − + − / + − +) above the determinant and read the sign off the position of the entry, not off the order in which you write the terms. Cross-check by expanding along a different line and confirming the same value, since det A = det Aᵀ.
Uses 10C4 = 210 for 'In how many ways can 4 posts of Secretary, Treasurer, etc. be filled from 10 people?' and 10P4 = 5040 for 'In how many ways can a committee of 4 be formed from 10 people?' - i.e. applies one formula to both questions.
The keyword heuristic ('choose/select = combination') works until the wording separates choosing from assigning. Words like 'committee', 'team', 'group' signal nCr; 'posts', 'arrangement', 'words', 'seating' signal nPr, and questions that mix both ('select 3 and appoint a leader') are the ones that get merged into a single nCr.
What to do instead
Before computing, answer one written test: does swapping two of the selected items produce a different outcome? Yes → nPr, no → nCr. Write that one word on the answer script. For mixed questions, split the work into an explicit nCr (selection) step multiplied by a factorial or nPr (arrangement) step, and check the answer is larger when order is added.
Applies S∞ = a/(1 − r) to a geometric series whose ratio is greater than 1 (e.g. 3 + 6 + 12 + …), or uses the same infinite-sum formula when the question asks for the sum of the first 8 terms, and accepts a negative total.
The formula list is memorised without the condition attached, and a negative sum of positive terms does not read as obviously wrong when the working is rushed. The distinction between Tn, Sn and S∞ also blurs when a question asks for 'the sum' without naming which.
What to do instead
Compute r = (second term)/(first term) as the first line and write |r| < 1 or |r| > 1 beside it. Use a/(1 − r) only when |r| < 1; otherwise use Sn = a(rⁿ − 1)/(r − 1). Treat a negative sum of positive terms as a stop signal. For AP term-counting, use n = (last − first)/d + 1 and verify by recomputing the last term from a + (n − 1)d.
Tests perpendicularity with m1 = m2 and parallelism with m1m2 = −1, or finds the slope of a line perpendicular to ax + by + c = 0 by keeping the same slope m = −a/b.
The two conditions are learned as an unordered pair and get swapped under exam pressure. Separately, converting ax + by + c = 0 to slope form drops the minus sign, so the perpendicular's slope comes out as the same fraction with the wrong sign.
What to do instead
Never quote the test alone; derive each slope into y = mx + c form first and write both slopes explicitly side by side. Then use the coefficient tests, which have no sign ambiguity: parallel when a1b2 − a2b1 = 0, perpendicular when a1a2 + b1b2 = 0. For the perpendicular slope itself, flip and negate: if m = 2/3, the perpendicular slope is −3/2.
Reads centre and radius straight from x² + y² + 2gx + 2fy + c = 0 as centre (g, f) and radius √(g² + f² + c), and applies the formulas without first dividing by 2 when the equation is 2x² + 2y² − 8x + 6y − 1 = 0.
The formulas are stated with g and f already defined as half-coefficients, but students substitute the visible coefficients directly. The sign of c under the radical is also misremembered, since it appears as +c in the equation and −c in the radius formula.
What to do instead
Normalise first: divide the whole equation by the coefficient of x² and y² so it reads x² + y² + 2gx + 2fy + c = 0. Then halve literally: g = (coefficient of x)/2, f = (coefficient of y)/2, centre = (−g, −f), radius = √(g² + f² − c). If r² comes out negative, the equation represents no real circle - that itself is a check on the arithmetic.
Substitutes x = a directly into a rational or radical limit, obtains 0/0, and writes the answer as 0 or 'limit does not exist' instead of factorising, rationalising or cancelling.
Direct substitution succeeds on most limit items in this paper, so it becomes the automatic first and only step. 0/0 gets read as an output value rather than as a signal that algebra must happen first, and the indeterminacy is not treated differently from a genuine zero.
What to do instead
Make 0/0 a written red flag: the moment it appears, write 'factorise/rationalise' and do that before anything else. After cancelling (x − 3) from (x² − 9)/(x − 3), state that the cancellation holds for x ≠ 3, then substitute. Check the answer numerically by evaluating at x = 2.99; if the expression heads to 6, an answer of 0 is immediately exposed.
Writes the antiderivative of x³ as x⁴/4 with no '+ C', and for a definite integral with substitution u = g(x) keeps the original x-limits in the u-integral.
In definite integrals the constant cancels, so it is dismissed as cosmetic and then dropped in indefinite ones where it carries a mark. With substitution, the limits are copied down from the previous line out of habit, and the old x-values are then plugged into an antiderivative written in u.
What to do instead
Adopt two fixed lines: close every indefinite integral with '+ C' before starting the next step, and after choosing u = g(x) immediately compute u1 = g(a), u2 = g(b) in the same line - or convert the antiderivative back to x before applying the original limits. Never carry an x-limit into a u-integral; mixing the two is the single largest source of wrong definite-integral answers.
Computes variance by the assumed-mean or step-deviation method as σ² = Σfd²/n and omits the correction term, or forgets the factor h² when class widths are not 1.
When the assumed mean happens to equal the true mean, the correction term (Σfd/n)² is zero, so practice runs 'confirm' the incomplete formula. With step deviation the deviations are in class-units, and the conversion back to real units by h² is a separate step that is easy to leave out.
What to do instead
Write the formula in full on the same line as the table: σ² = h²[Σfd²/n − (Σfd/n)²], with a column for fd and a column for fd² so both sums physically exist. Sanity checks: σ² must be ≥ 0, σ must be smaller than the range, and a negative raw σ² means the correction term was subtracted or an fd sign is wrong. Report σ as the positive square root.
Shades a constraint such as 2x + 3y ≥ 12 on the origin side, or finds the corner points of the feasible region and then picks the optimum at the vertex with the largest x-value without evaluating the objective function at every vertex.
The origin test is remembered as a rule ('always shade towards the origin') rather than as a calculation, and multiplying an inequality by a negative number without flipping the sign silently reverses the half-plane. The word 'maximum' is also confused with 'largest coordinate'.
What to do instead
Substitute (0,0) into the constraint and write the truth value on the graph: 0 ≥ 12 is false, so shade away from the origin. Then build a table with one row per vertex of the feasible region, including intercepts on the axes, and one column for Z = ax + by; the optimum is the extreme value in that column, not the extreme coordinate. Flag unbounded regions and note ties as multiple optimal solutions.
In bill discounting, computes the banker's discount as FV × r × t and then reports the proceeds, using the same rate on the face value when the question asks for true discount - and substitutes time in months straight into formulas that require years.
Banker's discount and true discount are both 'interest on the bill', but one is charged on the face value (amount due) and the other on the present value, so the two look interchangeable until both are computed. Time in months and rates given as a percentage both get copied verbatim into a formula that expects years and a decimal.
What to do instead
Convert before substituting: t = months/12 and r = R/100. Then work in a fixed order: PV = FV/(1 + rt), TD = FV − PV, BD = FV·r·t, and remember BD − TD equals the interest on TD, so BD > TD always. For annuities, set i = R/100 per conversion period and n = number of periods (not years) for half-yearly or quarterly compounding, and check that an annuity-due answer exceeds the ordinary-annuity answer by one period of interest.