CHSE Odisha Business Mathematics & Statistics common mistakes
CHSE Odisha Business Mathematics & Statistics · Understand the traps and how to avoid them.
When computing the inverse of a 2×2 matrix, the candidate swaps the diagonal entries a and d but forgets to put a minus sign on the off-diagonal entries b and c, writing the adjugate as [[d, b], [c, a]] instead of [[d, -b], [-c, a]].
The formula is memorised as 'swap the diagonal, keep the rest', which is only half of the adjugate rule; the sign-flip on the off-diagonals is the step most often dropped because it is easy to say but easy to skip under time pressure, and a wrong adjugate still produces a plausible-looking answer when multiplied back.
What to do instead
Always verify by multiplying A × A⁻¹ and checking that you get the identity matrix; if the off-diagonal products are not zero, the sign flip was missed. Write the adjugate as a two-step ritual: first swap a and d, then negate b and c, and only after that divide by the determinant.
The candidate computes the determinant of a 3×3 matrix by expanding along the first row but applies the alternating sign pattern as +, +, + or +, −, + across the columns instead of the correct +, −, + for row 1, or negates the whole expansion when the first entry is negative.
The cofactor sign pattern (−1)^(i+j) is understood in principle but is applied mechanically; when the leading entry is negative, candidates often absorb the minus into the cofactor twice, or they treat the second term's sign as fixed rather than dependent on position.
What to do instead
Before expanding, write the 3×3 sign grid [[+, −, +], [−, +, −], [+, −, +]] on the rough sheet and point to each entry as you expand. Expand along the row or column containing the most zeros to reduce the number of signed terms.
While solving a system of linear equations by Cramer's rule or matrix inversion, the candidate computes the determinant of the coefficient matrix as zero and then still divides by it, or concludes 'no solution' without distinguishing between 'no solution' and 'infinitely many solutions'.
The candidate treats det = 0 as a computational dead end rather than a case to classify; the distinction requires checking whether the numerators (or the consistency of the augmented matrix) also vanish, which is an extra step that is skipped when the arithmetic feels finished.
What to do instead
When det(A) = 0, stop and test consistency explicitly: compute the determinant of the coefficient matrix with the constant column substituted for each variable column; if all such determinants are also zero, the system has infinitely many solutions, otherwise none. State the classification in words before moving on.
The candidate differentiates a product or quotient by differentiating each factor separately, writing d/dx(uv) = u′v′ or d/dx(u/v) = u′/v′, or applies the chain rule only to the outer function and forgets to multiply by the derivative of the inner function.
The power rule and sum rule are over-learned and are generalised incorrectly to products and quotients; the chain rule's inner derivative is invisible when the inner function is simple, so the omission goes unnoticed until a composite like sin(3x²) appears.
What to do instead
Before differentiating, label the structure out loud: 'this is a product of two functions', 'this is a composite with inner function 3x²'. Write the rule template first (u′v + uv′, or (u′v − uv′)/v², or f′(g(x))·g′(x)) and then substitute; never differentiate term by term across a product or quotient.
When finding maxima and minima, the candidate sets f′(x) = 0, solves for the critical points, and then declares the largest value found to be the maximum without applying the second-derivative test or checking the endpoints of a closed interval.
The first-derivative step is the most practised and feels like the whole method; the second-derivative test is treated as optional confirmation, and endpoint checking is forgotten because the function is often given without an explicit interval in the candidate's memory.
What to do instead
Adopt a fixed three-line routine: (1) solve f′(x) = 0 for stationary points, (2) evaluate f″ at each and record the sign, (3) if an interval [a, b] is given, also evaluate f(a) and f(b) and compare. Write the conclusion as 'maximum value = … at x = …', naming both the value and the location.
In integration by substitution, the candidate changes the integrand from x to u but leaves the limits of integration in terms of x, or forgets to replace dx by du/g′(x), producing an answer that is dimensionally inconsistent or off by a constant factor.
The substitution is performed on the function but not on the differential or the limits, because these are written in separate places on the page and the candidate's attention is on the algebraic simplification; the missing du factor is invisible when g′(x) happens to be 1.
What to do instead
Write the full substitution chain in one line: u = g(x), du = g′(x) dx, so dx = du/g′(x). For definite integrals, either convert the limits at the same time (u₁ = g(a), u₂ = g(b)) or write 'back-substitute to x before applying original limits' explicitly. Always differentiate your antiderivative to check it returns the integrand.
The candidate computes the mean, median and mode of a grouped frequency distribution using the class marks as if they were raw observations, ignoring that the mean requires Σfx/Σf with midpoints and that the median and mode require interpolation formulas involving cumulative frequency or the modal class boundaries.
The ungrouped formulas are learned first and are simpler; grouped data look like a table of numbers, so the candidate applies the raw-data mean and reports a value that is close but wrong, and the median/mode formulas are the least-practised part of the chapter.
What to do instead
Before any calculation, write 'grouped data' at the top and list the midpoints x, frequencies f, and cumulative frequencies. Use mean = Σfx/Σf, median = L + [(N/2 − cf)/f] × h, and mode = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h, identifying L, cf, f₁, f₀, f₂, h explicitly for each problem.
The candidate confuses standard deviation with variance, reporting the variance as the standard deviation (or vice versa), and forgets to take the square root at the final step, or uses n instead of n − 1 in the denominator for a sample without noticing which is required.
The two quantities are computed in the same table and the last operation (square root) is a single step that is easily skipped when the arithmetic is long; the n versus n − 1 distinction is glossed over because many textbooks use n for population data and candidates do not check which the question implies.
What to do instead
Label the final answer explicitly as 'variance σ² = …' and 'standard deviation σ = √variance = …', and compute the square root as a separate written line. Decide at the start whether the data is a population (divide by n) or a sample (divide by n − 1) and write that choice down before starting the table.
When computing the correlation coefficient, the candidate substitutes into the formula but mixes up Σxy with (Σx)(Σy)/n, or computes r for a regression problem and then uses it as the regression coefficient b, reporting the slope as r rather than r·(σy/σx).
The correlation and regression formulas look structurally similar and both involve the same sums; the candidate recognises the sums and reaches for the formula that comes to mind first, and the distinction between r (a pure number between −1 and 1) and b (which carries units) is not reinforced by practice.
What to do instead
Before computing, write which quantity is asked for and its range: r must lie in [−1, 1] and b need not. Use b_yx = r·(σy/σx) and the regression line y − ȳ = b(x − x̄); after computing r, check it lies between −1 and 1, and after computing b, check that the regression line passes through (x̄, ȳ).
In probability, the candidate adds probabilities of events that are not mutually exclusive without subtracting the intersection, or multiplies probabilities of events that are not independent, and misreads 'at least one' as 'exactly one'.
The addition and multiplication rules are learned as bare formulas (P(A∪B) = P(A) + P(B), P(A∩B) = P(A)P(B)) without the conditions attached; the phrase 'at least one' is processed as a single event rather than its complement, which is the standard shortcut.
What to do instead
For every problem, first write whether the events are mutually exclusive and whether they are independent, then choose the rule: P(A∪B) = P(A) + P(B) − P(A∩B) in general, and P(at least one) = 1 − P(none). Draw a two-circle Venn diagram or a tree diagram before substituting numbers.