Tamil Nadu HSE Second Year Business Mathematics and Statistics common mistakes
Tamil Nadu HSE Second Year Business Mathematics and Statistics · Understand the traps and how to avoid them.
When asked for the inverse of a 2×2 matrix, the candidate writes the adjoint matrix but forgets to divide every entry by the determinant, or divides only the first row.
The formula A⁻¹ = (1/|A|)·adj(A) is memorised as two separate steps, and under time pressure the scalar multiplication of the adjoint is dropped or applied partially. Students also confuse adj(A) with A⁻¹ in notation.
What to do instead
Always write the three-line template first: |A|, then adj(A) (swap a and d, negate b and c), then A⁻¹ = (1/|A|)·adj(A). Before moving on, verify with A·A⁻¹ = I by multiplying one row by one column; and state explicitly that the inverse exists only when |A| ≠ 0, showing the determinant computation as its own line.
In a system of three linear equations solved by Cramer's rule or matrix inversion, the candidate mis-assigns which column is replaced by the constant vector, or mixes up cofactor signs when expanding a 3×3 determinant.
Cofactor expansion requires the alternating sign pattern + − + / − + − / + − +, and students expand along a row while forgetting the sign factor (−1)^(i+j). In Cramer's rule the replacement column is remembered vaguely rather than derived.
What to do instead
Fix a single expansion row (usually the row with a zero) and write the sign factor next to each term before computing. For Cramer's rule, label the columns x, y, z above the coefficient matrix and physically cross out the column being replaced. Check the answer by substituting the triple back into the original equations, and use the condition |A| = 0 to discuss consistency rather than forcing a division.
The candidate computes the slope of a line correctly but then applies the wrong perpendicular or parallel condition, or substitutes a point into y = mx + c with x and y interchanged.
The conditions m₁ = m₂ (parallel) and m₁m₂ = −1 (perpendicular) are learned together and get swapped under exam pressure; also the point-slope form y − y₁ = m(x − x₁) is mis-remembered as y − y₁ = m(x + x₁) or the coordinates are read in the wrong order.
What to do instead
Write the condition in words before using it: 'perpendicular means the product of slopes is −1, so the new slope is the negative reciprocal'. Always restate the given point as (x₁, y₁) with the numbers labelled, then substitute into y − y₁ = m(x − x₁) and simplify to the required form. As a check, substitute the given point back into the final equation and confirm it satisfies it.
The candidate differentiates a product or quotient by differentiating each factor and multiplying the results, i.e. writes (uv)' = u'v' or (u/v)' = u'/v'.
The power rule and constant-multiple rule are over-generalised to all operations; the product and quotient rules are recalled as formulas but not as procedures, so the structure 'first times derivative of second plus second times derivative of first' collapses.
What to do instead
Before differentiating, explicitly name u and v (or u and the denominator), write u' and v' separately, then assemble using the stated rule. For quotients, use the chant '(low d-high minus high d-low) over low squared' and verify with a simple numerical case. Practise mixed expressions such as x²·sin x and (x+1)/(x−2) until the structure is automatic.
In maxima–minima and rate-of-change problems, the candidate finds dy/dx = 0 but never tests the second derivative, or forgets to substitute the stationary value back to obtain the actual maximum or minimum value.
The procedure is treated as ending at 'solve dy/dx = 0'; the second-derivative test and the final evaluation are separate steps that are skipped when the algebra is long. Applied word problems also suffer because the variable relationships are not written down first.
What to do instead
Adopt a fixed four-line solution format: (1) express the quantity as a function of one variable, stating the constraint used; (2) set dy/dx = 0 and solve; (3) compute d²y/dx² and state its sign at each stationary point; (4) substitute back and state the maximum or minimum value with units. Never end the answer at the stationary x-value.
The candidate integrates xⁿ as x^(n+1)/(n+1) even when n = −1, and omits the constant of integration in indefinite integrals, or forgets to change the limits when using substitution in a definite integral.
The power rule is applied mechanically without checking the exceptional case; the '+ C' is treated as cosmetic because definite integrals dominate practice; and substitution is done without the discipline of transforming limits.
What to do instead
Before integrating any power, ask 'is the exponent −1?'; if so, the answer is log|x|. Write '+ C' on every indefinite integral as a rule of the format, not an afterthought. For definite integrals by substitution, either convert the limits to the new variable and mark them clearly, or integrate back to the original variable before substituting the original limits - never mix the two.
When solving a first-order differential equation, the candidate separates variables but mishandles the constant - placing it inside a logarithm, or losing it entirely - and fails to apply the initial condition to find its value.
The constant is added at an arbitrary point in the working, and the algebra of combining logarithms (log A + log B = log AB, and C = log k) is shaky. Initial conditions are treated as optional rather than as part of the answer.
What to do instead
Add the constant immediately after the first integration and only there. When logarithms appear, rename the constant as log k at the earliest opportunity so the equation becomes log y = log k + f(x), giving y = k·e^{f(x)}. Always finish by substituting the given initial condition, solving for k, and writing the particular solution explicitly.
The candidate adds probabilities for 'at least one' events, or uses the binomial formula without checking that trials are independent and the probability p is constant, and forgets the binomial coefficient.
The addition rule and multiplication rule are confused because 'or' and 'and' are not linked explicitly to the operations; the binomial distribution is memorised as p^x q^(n−x) with the nCr factor dropped, and the conditions for the model are not checked.
What to do instead
For 'at least one', use the complement: P(at least one) = 1 − P(none), and show this line explicitly. Before applying the binomial distribution, write down n, p and q and state in one sentence why the trials are independent with constant p. Write the full formula P(X = r) = nCr·p^r·q^(n−r) including the coefficient, then substitute.
In hypothesis testing, the candidate states the null and alternative hypotheses in the wrong direction, or compares the calculated statistic with the wrong critical value because the one-tailed versus two-tailed nature of the test was not decided first.
The alternative hypothesis is written as whatever the problem seems to want proved, without linking it to the claim; and the critical value is looked up before deciding the tail, so a one-tailed value is used for a two-tailed test or vice versa.
What to do instead
Follow a fixed order: (1) write H₀ as the 'no change / equals' statement and H₁ as the claim, noting the direction; (2) decide one-tailed or two-tailed from the wording of H₁ and record α; (3) look up the critical value for that tail and α; (4) compute the test statistic; (5) compare and state the conclusion in the context of the problem, not just 'reject H₀'.
When computing a weighted index number or a moving average, the candidate uses the current-year quantity where the base-year quantity is required (Laspeyres versus Paasche), or centres the moving average incorrectly for an even period.
The Laspeyres and Paasche formulas differ only in the quantity subscript, so they are visually confused; and for an even-period moving average the need for a two-stage (centred) average is a procedural detail that is easily skipped.
What to do instead
Write the formula with the subscripts spelled out - Laspeyres uses base-year quantities q₀, Paasche uses current-year quantities q₁ - and label each column of the working table with its subscript before substituting. For an even period such as 4, compute the first averages, then take a second average of consecutive pairs to centre them, and align each centred value with the correct middle year in the table.