Tamil Nadu SSLC Mathematics common mistakes
Tamil Nadu SSLC Mathematics · Understand the traps and how to avoid them.
For A = {1, 2, 3} and B = {4, 5}, writes A × B as only three pairs such as (1,4), (2,4), (3,4), then counts the number of relations as 2^(3+2) = 32.
They read A × B as a matching from A to B and remember the relation-count formula as a sum of the set sizes; the full pair grid is never written out.
What to do instead
Draw a 3-by-2 grid: list A down the side and B across the top, then write all six ordered pairs. The number of relations is 2^(mn) = 2^6 = 64; the number of functions is n^m = 2^3 = 8. A relation is a function only if every element of A appears exactly once as a first coordinate.
In Euclid's division lemma writes 47 = 5 × 8 + 7 and continues with divisor 7, or writes −17 ≡ −2 (mod 5) and uses −2 as the remainder.
Long division in earlier classes permits a remainder larger than the divisor when the quotient is chosen too small, and congruence notation is treated as ordinary subtraction.
What to do instead
At every step enforce 0 ≤ r < b: 47 = 5 × 9 + 2, so the next divisor is 2. For modular arithmetic reduce to the least non-negative residue: −17 = 5(−4) + 3, so −17 ≡ 3 (mod 5). Verify each step by bq + r.
Finds the 10th term of an AP as a + 10d, or uses S∞ = a/(1−r) for a GP with r = 2.
The formulas are memorised as patterns without the n−1 shift or the convergence condition; the first term is counted as if the indexing starts at zero or at the wrong term.
What to do instead
Test any formula with n = 1: a + (n−1)d must give a. Write a, d (or r) and n before substituting. For S∞ use a/(1−r) only when |r| < 1; otherwise use S_n = a(r^n−1)/(r−1).
For 2x² − 7x + 3 = 0 writes α + β = −7/2 and αβ = 3/2, or decides the roots are real and equal because b² − 4ac = 25 without noticing the sign of a.
The formulas −b/a and c/a are applied without carrying the signs of b and c from the equation, and the discriminant is evaluated as b² + 4ac or with b not squared.
What to do instead
Write a = 2, b = −7, c = 3 first, then α + β = −b/a = 7/2 and αβ = c/a = 3/2. For nature of roots compute Δ = b² − 4ac = 49 − 24 = 25 > 0, so two distinct real roots; check by factorising (2x − 1)(x − 3).
For A = [[2, 3], [1, 4]] writes adj A = [[4, 1], [3, 2]], dropping the negative signs on the off-diagonal entries, and then substitutes the constants in the wrong column in Cramer's rule.
The 2×2 adjoint is memorised as 'swap and swap' but the sign pattern is ignored; Cramer's rule columns are replaced without tracking which variable's column was substituted.
What to do instead
Use adj A = [[d, −b], [−c, a]]; here adj A = [[4, −3], [−1, 2]]. Check A · adj A = (det A)I = 5I. In Cramer's rule write Δ, then Δx by replacing the x-column (first column) with the constants; x = Δx/Δ, y = Δy/Δ.
Given ΔABC ~ ΔPQR, writes AB/QR = BC/PR = AC/PQ, or in a BPT figure with DE ∥ BC writes AD/DB = AE/AC.
The similarity statement is read as a list of side names rather than a vertex correspondence, and the proportional segments are copied without matching the two transversals.
What to do instead
Under ΔABC ~ ΔPQR the correspondence is A↔P, B↔Q, C↔R, so AB/PQ = BC/QR = AC/PR. For DE ∥ BC in ΔABC, BPT gives AD/DB = AE/EC and also AD/AB = AE/AC; mark the two segments on each side before writing any ratio.
For A(1, 2) and B(4, 5), the point dividing AB internally in the ratio 2 : 1 is calculated as ((2·1 + 1·4)/3, (2·2 + 1·5)/3) = (2, 3).
The ratio m : n is substituted in the order the numbers appear, pairing m with the first point's coordinate instead of the second.
What to do instead
Use P = ((m x₂ + n x₁)/(m + n), (m y₂ + n y₁)/(m + n)): here x = (2·4 + 1·1)/3 = 3 and y = (2·5 + 1·2)/3 = 4, so P = (3, 4). Check that P lies between A and B and is closer to the point that has the smaller share of the ratio.
With angle of elevation 30° and vertical height 10 m, finds the horizontal distance as 10 sin 30° = 5 m instead of 10/tan 30° = 10√3 m.
SOHCAHTOA is recited but the sides are not labelled relative to the given angle; the vertical height is treated as the hypotenuse because it looks like the longest side in the sketch.
What to do instead
Draw the right triangle, mark the angle at the observer, label the vertical height as opposite and the horizontal ground as adjacent; the line of sight is the hypotenuse. Use tan 30° = opposite/adjacent = 10/x, so x = 10/tan 30°.
For a solid formed by a cylinder with a hemisphere on top, computes total surface area as 2πrh + 2πr² + πr², counting the circle where the two solids meet twice; for a frustum uses slant height h in place of l.
Each formula is applied to its own solid and added without visualising which faces are hidden at the join; the frustum's l = √(h² + (R−r)²) is confused with the vertical height.
What to do instead
Sketch the combined solid and shade only exposed surfaces: curved cylinder 2πrh, the hemisphere's curved surface 2πr², and the cylinder's lower base πr²; the upper base is hidden. For a frustum use l = √(h² + (R−r)²), CSA = π(R + r)l and volume = (1/3)πh(R² + Rr + r²).
For the data 2, 4, 6, 8, 10 finds variance 8 and reports the standard deviation as 8; for two independent events writes P(A and B) = P(A) + P(B).
Variance and standard deviation are treated as the same quantity, and the addition rule for mutually exclusive events is over-generalised to the word 'and'.
What to do instead
Standard deviation is √variance = √8 = 2√2; for grouped data take the square root after computing Σfd²/N. For 'and' with independence multiply: P(A∩B) = P(A)P(B); add only for 'or' when the events are mutually exclusive.