PSEB Class 12 Maths common mistakes
PSEB Class 12 Maths · Understand the traps and how to avoid them.
Writing the principal value of sin⁻¹(sin 5π/6) as 5π/6 (or of cos⁻¹(cos 7π/6) as 7π/6) instead of reducing the angle into the principal branch and answering π/6 (respectively 5π/6).
The candidate treats sin⁻¹ and sin as exact cancelling inverses for every input, ignoring that sin⁻¹ has range [−π/2, π/2] and cos⁻¹ has range [0, π]; the exam routinely sets arguments outside these ranges precisely to test the restriction.
What to do instead
Before answering, write the range of the inverse function beside the question. Then rewrite the argument as θ = kπ ± α or 2kπ ± α with α inside the principal range, apply the odd/even or supplementary identity, and only then evaluate. Do three practice conversions such as sin⁻¹(sin 5π/6), cos⁻¹(cos 7π/6), tan⁻¹(tan 3π/4) each time writing the range first.
Treating |kA| as k|A| instead of kⁿ|A| for an n×n matrix, and cancelling matrices in an equation such as AB = AC to conclude B = C.
Scalar algebra habits carry over: the candidate applies the same rules to determinants and matrix products that hold for numbers, forgetting that matrix multiplication is not commutative and not cancellative, and that a scalar multiplying a matrix multiplies every one of the n rows in the determinant.
What to do instead
Verify with a concrete 2×2 example: take A = [[1,2],[3,4]], compute |2A| = 4|A| = 4(−2) = −8 and compare with 2|A| = −4, so the formula is fixed in memory. For cancellation, replace AB = AC by A(B − C) = 0 and state that this forces B = C only when A is invertible (|A| ≠ 0); otherwise give a counterexample with a singular A.
Testing continuity of a piecewise function at the join x = a by computing only f(a) and one one-sided limit, or testing differentiability at the join by differentiating each piece with the formula and comparing the resulting expressions at a.
The candidate believes a formula valid on an open interval may simply be evaluated at the endpoint, but the derivative at the join must be computed from the limit definition because the two pieces generally have different derivatives there and the function may not even be continuous first.
What to do instead
Use a fixed four-line template: LHL = lim(x→a⁻) f(x), RHL = lim(x→a⁺) f(x), f(a); continuous only if all three are equal. Only after continuity is established, compute f′(a) by first principles from each side, i.e. lim(h→0)[f(a+h) − f(a)]/h and the left version, and require equality. Rehearse on f(x) = |x| and on a split definition with x² and x.
Finding local maxima/minima only from f′(x) = 0 and ignoring the endpoints when the question asks for absolute maximum or minimum on a closed interval [a, b]; also declaring a stationary point an extremum without checking the sign change of f′.
The candidate conflates the local-extremum procedure with the global one, and f′(x) = 0 is treated as sufficient; for closed intervals the extreme values can occur at endpoints where f′ need not vanish, and a stationary point can be a point of inflection.
What to do instead
For interval questions always run the closed-interval algorithm: list all c in (a, b) with f′(c) = 0 or f′ undefined, then tabulate f at each such c together with f(a) and f(b), and choose the largest and smallest values. For local questions use the first-derivative sign table (or f″ at the stationary point) and state the sign change explicitly.
Dropping the constant of integration in indefinite integrals, and in definite integrals by substitution either keeping the old x-limits after changing the variable or substituting back incorrectly so the limits are applied to the wrong variable.
Under time pressure the '+C' looks cosmetic and the limit change is treated as an optional step; the candidate also substitutes u = g(x) but then evaluates the antiderivative in u between the original x-values, which is dimensionally wrong and gives a meaningless number.
What to do instead
Write '+C' as part of the answer line before simplifying, and make it a habit in every indefinite integral. For substitution, write three explicit lines: u = g(x), du = g′(x) dx, new limits u(a) and u(b); then integrate entirely in u and substitute the new limits. Recheck by differentiating the indefinite result back to the integrand.
Setting up an area between two curves as ∫(lower − upper) or forgetting to split the region at an intersection point, and reporting a negative area without taking the modulus.
The candidate integrates in the order the functions appear in the question rather than the order on the graph, and misses that when the curves cross inside the interval the 'upper' function changes, so a single integral evaluates a signed difference rather than the total area.
What to do instead
Always sketch first, find all intersection points by solving f(x) = g(x), and write the region as ∫(top − bottom) dx on each sub-interval between consecutive intersections. If the integral comes out negative, that is a signal the subtraction order is wrong; take absolute values per sub-interval and add. Check the answer is positive and plausible against the sketch.
Separating dy/dx = f(x)g(y) as dy/g(y) = f(x) dx and dividing by g(y) without recording the constant solutions g(y) = 0; in the integrating-factor method, applying the formula before rewriting the equation in the standard form dy/dx + Py = Q.
Division by g(y) is done mechanically and the lost solutions are invisible afterwards; in the linear-equation method the candidate reads P and Q off an equation whose coefficient of dy/dx is not 1, so the integrating factor e^∫P dx is computed with the wrong P.
What to do instead
After separating, immediately write 'if g(y) = 0 then y = c is a solution' and check whether it satisfies the equation (and any initial condition). For linear equations, first divide through so the coefficient of dy/dx is exactly 1, then state P(x) and Q(x) on separate lines before computing the integrating factor e^∫P dx.
Writing the angle between two lines as cos θ = (b₁·b₂)/(|b₁||b₂|) without the modulus and then reporting an obtuse angle, or mixing direction ratios of lines with normal vectors of planes in the same formula, and confusing scalar projection a·b̂ with the vector projection.
The formula for the acute angle between lines needs |b₁·b₂|, but the candidate copies the dot-product formula verbatim; for planes the angle is between normals, and the candidate substitutes direction vectors, and for projection the candidate drops the unit vector b̂ and gives a scalar times |b|.
What to do instead
Label each vector before substituting: 'direction of line 1', 'normal to plane 1'. Use cos θ = |b₁·b₂|/(|b₁||b₂|) for lines and cos θ = |n₁·n₂|/(|n₁||n₂|) for planes, and state θ ∈ [0, π/2]. For projections write scalar projection = a·b/|b| and vector projection = (a·b/|b|²) b, keeping the two formulas side by side.
Solving an LPP by testing only the intersection points of constraint lines while ignoring the non-negativity constraints x ≥ 0, y ≥ 0, or shading the wrong side of an inequality so an infeasible point is evaluated as a corner.
The candidate graphs each line but decides the feasible side by guesswork, and the coordinate axes, which are themselves constraints, are left out of the corner list; intersection points of lines can lie outside the first quadrant and are then wrongly included.
What to do instead
For every inequality, substitute the origin (0, 0) to determine the shaded side (unless the line passes through the origin, then use (1, 0)). Draw and label x = 0 and y = 0 as constraints, list every corner of the shaded region including axis intercepts, then tabulate Z at each corner and state the maximum/minimum with its point.
Reversing the conditional in Bayes-type problems, computing P(B|A) when P(A|B) is asked, or multiplying P(A)P(B) for events that have not been shown independent, and applying total probability without first defining a complete set of mutually exclusive events.
The notation P(A|B) is read as 'A given B' but substituted in the wrong slot, and the candidate assumes independence whenever two events look unrelated; without a clearly defined partition E₁, E₂, … the denominator of Bayes' theorem is assembled from the wrong terms.
What to do instead
Before any calculation, define the partition events in words and write the target conditional as P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σ P(Eⱼ)P(A|Eⱼ), reading each factor aloud as 'probability of Eᵢ times probability of A given Eᵢ'. Test independence separately by checking P(A∩B) = P(A)P(B) rather than assuming it; a two-branch tree diagram with probabilities on the branches makes the direction of each conditional explicit.