Odisha BSE HSC Mathematics common mistakes
Odisha BSE HSC Mathematics · Understand the traps and how to avoid them.
Simplifies radical expressions by writing √a + √b = √(a+b), for example treating √2 + √8 as √10 instead of 3√2.
The learner remembers the product rule for surds and wrongly extends a similar-looking rule to addition, often under time pressure.
What to do instead
Write the product and quotient rules separately, then simplify by extracting square factors: √8 = 2√2, so √2 + √8 = 3√2. Check with decimal approximations when unsure.
Solves x² = 5x by cancelling x and writing only x = 5, omitting the root x = 0.
Cancelling a variable feels like ordinary fraction reduction, but division by x is invalid when x may be zero.
What to do instead
Transpose to x² − 5x = 0, factor x(x − 5) = 0, and apply the zero-product rule. If dividing by an expression, state the non-zero condition and check the excluded case separately.
Finds the 20th term of the AP 7, 10, 13, ... as 7 + 20×3 = 67 instead of 7 + 19×3 = 64.
The learner counts 20 jumps from the first term instead of 19 intervals, or substitutes n without writing n − 1 explicitly.
What to do instead
Always write a = 7, d = 3, n = 20 and substitute into aₙ = a + (n − 1)d. Verify by checking that n = 1 gives the first term.
In similar triangles, pairs sides by position in the diagram rather than by matching equal angles, and writes area ratio equal to the side ratio instead of its square.
Diagrams are often rotated or not labelled with vertex correspondence, so visual position is used instead of the similarity statement.
What to do instead
First state the correspondence, for example ΔABC ~ ΔDEF, then write AB/DE = BC/EF = CA/FD. Use area(ΔABC)/area(ΔDEF) = (AB/DE)² and mark equal angles on the figure.
Applies the tangent–radius theorem at a point that is not the point of contact, or assumes any radius meeting a chord is perpendicular to it.
The learner recalls the words 'radius perpendicular' but not the exact condition: perpendicular to tangent at the tangent point, or from centre to midpoint of a chord.
What to do instead
Identify and label the tangent point first, draw the radius to that exact point, and use the 90° angle there. For chords, use the perpendicular from the centre only when it bisects the chord.
In a height-and-distance problem, uses tan θ = adjacent/opposite or treats the horizontal distance as the hypotenuse.
The learner substitutes into SOH-CAH-TOA without drawing the right triangle or fixing the observer's angle.
What to do instead
Draw the right triangle, mark the angle at the observer, label the vertical side as opposite and the horizontal side as adjacent, then use tan θ = opposite/adjacent. Check that the hypotenuse is the direct line of sight.
Cancels terms across addition in a trigonometric fraction, for example writing (sin θ + cos θ)/cos θ = sin θ.
The learner treats the fraction as if addition allowed term-by-term cancellation, confusing it with factors.
What to do instead
Split the fraction correctly: (sin θ + cos θ)/cos θ = sin θ/cos θ + cos θ/cos θ = tan θ + 1. Cancel only common factors of the whole numerator and denominator, never terms joined by + or −.
Uses the slant height l as the perpendicular height h when calculating the volume of a cone, writing V = (1/3)πr²l.
Both lengths appear in the diagram, and the learner memorises one formula without distinguishing volume from surface area.
What to do instead
Label h as the perpendicular height from apex to base centre and l as the slant edge. Use l² = r² + h², then V = (1/3)πr²h; use l only in curved surface area πrl.
Calculates sector area as (θ/360)πd², substituting the diameter for the radius, or omits the θ/360 factor.
The learner mixes the circle-area formula πd²/4 with the sector formula and rushes the angle fraction.
What to do instead
Write A = (θ/360)πr² first, convert diameter to radius by r = d/2, and estimate the answer as a fraction of the whole circle. Check that a 90° sector is one quarter of πr².
For two draws without replacement, keeps the original total in the denominator for the second draw, or uses P(not E) = P(E) instead of 1 − P(E).
The learner treats dependent events as independent and confuses the complement rule with equality.
What to do instead
After the first draw, reduce both the favourable count and the total by one before finding the second probability. For complements, use P(not E) = 1 − P(E) and check that the two probabilities sum to 1.