Maharashtra HSC Mathematics & Statistics - Arts & Science (MSBSHSE) common mistakes
Maharashtra HSC Mathematics & Statistics - Arts & Science (MSBSHSE) · Understand the traps and how to avoid them.
When differentiating a composite function such as sin(3x²) or log(x²+1), the candidate differentiates only the outer function and writes 3cos(3x²) or 1/(x²+1), dropping the inner derivative entirely.
The chain rule is taught as a formula, but under time pressure candidates pattern-match to the familiar single-function derivative and stop after one layer. Maharashtra board papers routinely set nested composites (e.g. log(sin x), e^{tan x}, √(1+x²)), so the omission is systematic rather than accidental.
What to do instead
Adopt a written two-line habit: first copy the outer derivative with the inner function left intact as a placeholder bracket, then multiply by the derivative of that bracket. Practise ten composites daily and mark each bracket with an arrow before simplifying.
For x = a(θ − sin θ), y = a(1 − cos θ), the candidate computes dy/dx as (dy/dθ)/(dx/dθ) but substitutes θ wrongly or forgets to divide, giving dy/dx = a sin θ instead of sin θ/(1 − cos θ).
Parametric differentiation looks like ordinary differentiation, so candidates differentiate y and x separately and then either stop or subtract instead of dividing. The a cancels, which candidates do not expect, so they distrust the correct answer.
What to do instead
Always write the fraction dy/dx = (dy/dθ)/(dx/dθ) with both derivatives shown before cancelling. Check the result at a test value of θ, e.g. θ = π/2 should give dy/dx = 1.
In a related-rates or maxima–minima word problem, the candidate differentiates correctly but substitutes the given numerical value before differentiating, or forgets to convert the stated rate to the correct sign (e.g. treating a decreasing radius as positive).
Candidates treat word problems as algebra problems and plug numbers early, which destroys the variable needed for the derivative. Signs of rates are glossed over because the wording ('decreasing', 'leaking', 'shrinking') is read as decoration rather than as mathematical information.
What to do instead
Follow a fixed template: declare variables, write the relation, differentiate with respect to t while all quantities are still symbolic, then substitute. Underline every rate word and assign its sign explicitly before substituting.
When finding the maximum volume or minimum cost, the candidate finds dy/dx = 0, solves for x, and stops, never evaluating the second derivative or checking endpoints of the stated domain.
The first-derivative condition is the most rehearsed step, so it becomes the whole answer in the candidate's mind. Endpoint checks are skipped because candidates assume the stationary point is automatically the required optimum.
What to do instead
After solving dy/dx = 0, write one line computing d²y/dx² at each critical point and state 'negative → maximum' or 'positive → minimum' explicitly. Then list the domain endpoints and compare values before writing the final answer with units.
For ∫₀^{π/2} sin²x dx, the candidate integrates sin²x as (sin³x)/3 or −cos²x/2, treating a power of a trigonometric function like a power of x.
The power rule for xⁿ is over-learnt and visually similar; candidates do not distinguish 'power of the variable' from 'power of a function'. The identity sin²x = (1 − cos 2x)/2 is known but not retrieved because the integral is recognised as a routine one.
What to do instead
Before integrating any trigonometric power, ask: is the exponent 1? If not, write the relevant half-angle or double-angle identity first. Memorise the three identities sin²x, cos²x, tan²x = sec²x − 1 as a reflex list.
When computing the area between a curve and the x-axis, the candidate integrates from a to b without checking whether the curve crosses the axis inside that interval, so signed areas cancel and the answer comes out too small or zero.
The definite integral is taught primarily as a signed quantity, and area problems are practised mostly with curves that stay above the axis. Candidates equate 'area' with 'integral' without the absolute-value reasoning.
What to do instead
Always solve f(x) = 0 first and sketch the sign of f on each sub-interval. Split the integral at every zero and take the absolute value of each piece: Area = Σ|∫ f dx|. State the split points in the answer.
While solving dy/dx + Py = Q, the candidate computes the integrating factor as e^{∫P dx} but forgets to multiply the right-hand side Q by the same factor, or drops the constant of integration in the final general solution.
The integrating factor is remembered as a formula attached to the left side only, and the multiplication of every term by that factor is a mechanical step easily skipped. The '+C' is dropped because candidates treat the solution as a value rather than a family of curves.
What to do instead
After finding the integrating factor, rewrite the entire equation with every term multiplied by it before integrating. End every general solution with '+ C' and, for an initial-value problem, substitute the condition immediately to fix C.
In a three-dimensional geometry problem, the candidate computes a·b and a×b but confuses which one gives the angle and which gives the area, or uses a×b for a perpendicular-distance problem where the scalar projection is required.
Both products are taught close together and both are 'dot/cross product' formulas; the geometric meaning (projection versus area/normal) is memorised verbally but not attached to a decision rule.
What to do instead
Attach each product to its output type: a·b is a scalar (angle, projection, work), a×b is a vector (normal, area of parallelogram, moment). Write the target quantity's type first - 'I need a scalar, so dot' - before choosing.
When formulating or solving a linear programming problem graphically, the candidate shades the wrong side of a constraint line or forgets x ≥ 0, y ≥ 0, then evaluates the objective function at a point that is not a vertex of the feasible region.
Shading direction is decided by eye rather than by testing a point, and the non-negativity constraints are treated as automatic rather than as lines that bound the region. The corner-point theorem is stated but not applied as a checklist.
What to do instead
For each constraint, substitute the origin (0,0); if it satisfies the inequality, shade towards the origin, otherwise away. Draw and label the axes as constraints. List every vertex with coordinates, then tabulate Z at each vertex and state the optimum with its coordinates.
In logic, the candidate writes the negation of 'all students passed' as 'no students passed' instead of 'some students did not pass', or treats p → q and q → p as equivalent when testing validity.
Everyday English makes 'not all' sound like 'none', and the contrapositive/converse distinction is a formal rule that is easily blurred under exam pressure. The truth-table method is known but not used as a verification tool.
What to do instead
For quantified statements, apply the rule mechanically: 'all A are B' negates to 'some A are not B', and 'some A are B' negates to 'no A are B'. For implications, write the truth table for p → q and its converse side by side and mark the row where they differ (p false, q true).
When computing the binomial probability P(X = r) for n = 10, p = 0.3, the candidate writes C(10, r) p^r q^{n−r} but uses q = p or computes n − r as r, or reports the mean of the distribution as npq instead of np.
The formula is recalled as a shape rather than as a meaning: p is success, q = 1 − p is failure, and the exponents must sum to n. Mean and variance formulas (np and npq) are visually similar and are confused under recall pressure.
What to do instead
Before substituting, write the four quantities explicitly: n, p, q = 1 − p, r. Check that the exponents add to n. Write the mean/variance pair as 'mean = np (one factor), variance = npq (two factors)' and verify variance < mean for every binomial answer.
When computing variance from a frequency distribution, the candidate uses the assumed-mean shortcut but forgets to multiply by the class width h, or uses N instead of N − 1 when the question specifies a sample.
The step-deviation table produces a variance in the coded variable u, and the conversion back to x (multiply by h²) is a separate step that is dropped. The population/sample distinction is not flagged in most textbook drills, so it is missed when the question states 'sample of 25'.
What to do instead
After finishing the step-deviation table, write the conversion line explicitly: Var(x) = h² × Var(u), and SD(x) = |h| × SD(u). Underline the words 'sample' or 'population' in the question and use N − 1 or N accordingly.
In bivariate data, the candidate computes Karl Pearson's coefficient r but reports a value outside [−1, 1] (e.g. 1.4) without noticing, or interprets r = 0.9 as 'x causes y'.
Arithmetic slips in the covariance and SD terms are not caught because the candidate does not know the range as a check. Correlation and causation are conflated because the textbook phrase 'highly correlated' is read as 'strongly related in a causal sense'.
What to do instead
Always check −1 ≤ r ≤ 1 as a final line; a value outside means an arithmetic error, so recompute the SDs. When interpreting, write 'r measures the strength and direction of linear association only; it does not imply causation' as a standard closing sentence.
When computing skewness, the candidate uses the mean–mode formula without checking which measure the question supplies, or reports a negative skewness for a distribution whose tail extends to the right.
Bowley's and Karl Pearson's coefficients look similar, and the sign convention (positive skew = right tail = mean > mode) is memorised as a phrase rather than as a picture. Under time pressure the candidate applies whichever formula comes to mind.
What to do instead
Sketch the curve first and mark the tail direction; then choose the formula whose inputs are given. Write the sign rule once at the top of the answer: mean > median > mode → positive skew. Verify the sign of the computed coefficient against the sketch before finalising.