Tamil Nadu HSE First Year Mathematics common mistakes
Tamil Nadu HSE First Year Mathematics · Understand the traps and how to avoid them.
Asked for the domain of f(x) = √(x − 2)/(x − 5), writes only 'x ≥ 2' and stops; the exclusion x ≠ 5 is lost, and the range is then worked out from this wrong domain.
Candidates learn 'domain = solve the inequality under the square root' as a single template, so whichever restriction appears first in the expression becomes the whole answer; the denominator rule is treated as a separate topic rather than a simultaneous condition.
What to do instead
Before solving, list every restriction from the expression in order: radicand ≥ 0, denominator ≠ 0, log argument > 0, inverse-trig argument in [−1, 1]. Solve each one separately, intersect them on a number line, and write the final domain in interval notation (here [2, 5) ∪ (5, ∞)).
Solving sin θ = 1/2, writes θ = π/6 and stops, or appends only 2nπ, dropping the second family of solutions that the general solution must contain.
The principal value produced by the standard table feels like 'the answer', and the form θ = nπ + (−1)ⁿα is memorised as a string of symbols rather than as a description of two intersection points on the unit circle.
What to do instead
Draw the unit circle for every trig equation, mark both intersections with the horizontal line y = 1/2, read off α and π − α, then attach the period: nπ + (−1)ⁿα for sin θ = sin α, 2nπ ± α for cos θ = cos α, nπ + α for tan θ = tan α. Close the answer with 'n ∈ ℤ'.
In a mathematical induction proof, writes 'Assume P(k) is true for all k ≥ 1' and then derives P(k+1), never verifying P(1).
The base case is treated as a formality that can be skipped, and 'assume for all k' is convenient shorthand that silently converts the argument into a circular one - exactly the error the examiner is testing for.
What to do instead
Fix a rigid four-line layout before any algebra: (1) verify P(1) numerically; (2) 'Assume P(k) is true for a fixed integer k ≥ 1'; (3) derive P(k+1) using only P(k); (4) 'Hence by the principle of mathematical induction, P(n) is true for all n ≥ 1.' Never write 'for all k' in line 2.
To find the term independent of x in (2x + 1/x²)⁹, writes T_{r+1} = ⁹C_r (2x)^{9−r}(1/x²)^r but then solves 9 − r + 2r = 0, treating the second factor's exponent as +r instead of −2r.
The general term is reproduced from memory as 'n − r and r', and the denominator of the second bracket disappears when the expression is copied quickly, so the exponent bookkeeping is never actually performed on the given expression.
What to do instead
Collect exponents in the same line you write the general term: T_{r+1} = ⁹C_r 2^{9−r} x^{9−r} · x^{−2r} = ⁹C_r 2^{9−r} x^{9−3r}. Only then set 9 − 3r = 0, giving r = 3. Also record 0 ≤ r ≤ 9, r integer, and re-read whether the question wants the term, its coefficient, or the coefficient of a specified power.
Finds the distance from (1, 2) to the line 3x − 4y + 6 = 0 as (3 − 8 + 6)/5 = 1/5 with the sign carried on the numerator, and in a variant case reports a negative distance.
The formula is recalled as 'substitute into the numerator and divide by √(a² + b²)', with the modulus bars dropped as decoration, so the candidate has no signal that a distance can never be negative.
What to do instead
Write the formula with the bars physically attached: d = |ax₁ + by₁ + c| / √(a² + b²). Evaluate the numerator inside the bars first, take the modulus, then divide. As a check, note the perpendicular distance is 0 only when the point satisfies the line equation - if you get 0 for a point that is visibly off the line, an arithmetic slip has occurred.
For a 3×3 matrix A with |A| = 5, writes |2A| = 2 × 5 = 10 instead of 8 × 5 = 40; the same candidates write (AB)⁻¹ = A⁻¹B⁻¹.
Determinant rules are memorised as a flat list without the reason behind them. Scalar multiplication of a determinant is confused with multiplying the whole matrix by k, and the fact that each of the n rows is scaled (giving kⁿ) is never visualised.
What to do instead
Derive |kA| = kⁿ|A| once by hand for a 2×2 example, then keep one revision card of all order-sensitive rules together: |kA| = kⁿ|A|, |AB| = |A||B|, |Aᵀ| = |A|, |A⁻¹| = 1/|A|, (AB)⁻¹ = B⁻¹A⁻¹, adj(AB) = adj(B)·adj(A). Test each new problem against that card.
Evaluates lim_{x→0} (sin 3x)/(5x) by direct substitution, obtains 0/0, and concludes either that the limit is 0 or that it does not exist.
Substitution is the first reflex, and the form 0/0 is read as a final value rather than as an indeterminate signal. The standard limit lim_{x→0} (sin x)/x = 1 is not connected to a denominator whose coefficient differs from the sine's argument.
What to do instead
Force an intermediate written step: 'substitution gives 0/0 ⇒ indeterminate, not an answer'. Then match the sine's argument to the denominator: (sin 3x)/(5x) = (3/5)·(sin 3x)/(3x) → 3/5. Drill the whole family - tan x/x, (1 − cos x)/x², (aˣ − 1)/x, log(1 + x)/x - so the pattern-matching is automatic.
Differentiates y = ln(sin x) as 1/sin x, or y = sin(x²) as cos(x²)·2x written as cos(2x), because the inner derivative is omitted whenever the outer function is trigonometric or logarithmic.
The chain rule is applied reliably when the inner function is a visible polynomial; when attention is drawn to a 'difficult' outer function, the inner derivative is dropped as an afterthought.
What to do instead
Write the substitution line before differentiating: 'let u = sin x, y = ln u ⇒ dy/dx = (1/u)(du/dx) = (1/sin x)(cos x) = cot x.' Keep the u-line for at least ten two-step derivatives daily (e^{tan x}, √(1 + x²), sin³x, ln(cos 2x)) until the inner factor appears without prompting.
For ∫₀² x(x² + 1)³ dx, substitutes u = x² + 1, finds the antiderivative in u, then plugs the original limits x = 0 and x = 2 into the u-expression, producing a wrong value.
The indefinite-integral routine - find the antiderivative, then substitute the given limits - is carried over unchanged, and the limits are not treated as part of the substitution step.
What to do instead
The moment you write 'put u = ...', write the transformed limits directly underneath: 'x = 0 ⇒ u = 1, x = 2 ⇒ u = 5'. Either finish entirely in u with the new limits, or convert back to x before using the original limits. Never mix a u-antiderivative with x-limits; mark the limits with the variable name to make the mismatch visible.
Given P(A) = 0.5 and P(B) = 0.4, writes P(A ∪ B) = 0.9 and in the next line asserts A and B are independent, without noticing that independence would force P(A ∩ B) = 0.2 and hence P(A ∪ B) = 0.7.
The addition theorem and the independence product rule are stored as two separate formulas and applied one at a time; the candidate never tests whether the two can hold simultaneously for the same pair of events.
What to do instead
Keep the two statements side by side on the rough sheet: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) always; P(A ∩ B) = P(A)P(B) only under independence. Compute P(A ∩ B) by both routes and state which one the question licenses. Explicitly note that mutually exclusive events (P(A ∩ B) = 0) are not independent unless one event has probability zero.