Haryana BSEH Secondary Mathematics - Standard cheat sheet

Euclid's division lemma: for positive integers a and b, there exist unique integers q and r with a = bq + r and 0 ≤ r < b.

Euclid's algorithm for HCF: repeatedly apply a = bq + r, replacing (a, b) by (b, r), until r = 0; the last non-zero divisor is HCF(a, b).

Fundamental Theorem of Arithmetic: every integer greater than 1 is a product of primes, unique up to order of factors.

HCF(a, b) × LCM(a, b) = a × b holds for exactly two positive integers; for three numbers this product rule fails.

LCM(a, b) = a × b ÷ HCF(a, b); for fractions, HCF = HCF(numerators)/LCM(denominators) and LCM = LCM(numerators)/HCF(denominators).

√2, √3, √5 and in general √p for prime p are irrational; proof is by contradiction using p | a² ⇒ p | a.

A number is rational only if its decimal expansion terminates or repeats; non-terminating non-repeating decimals are irrational.

If a prime p divides a² then p divides a (Euclid's lemma); this step is essential in irrationality proofs.

Remainder theorem: the remainder when p(x) is divided by (x − a) is p(a); p(x) is divisible by (x − a) exactly when p(a) = 0.

Division algorithm: p(x) = g(x)·q(x) + r(x) with deg r < deg g, or r(x) = 0.

For ax² + bx + c with zeros α, β: α + β = −b/a and αβ = c/a.

For ax³ + bx² + cx + d with zeros α, β, γ: α + β + γ = −b/a, αβ + βγ + γα = c/a, αβγ = −d/a.

A quadratic polynomial with given zeros α, β is k[x² − (α + β)x + αβ] for any non-zero constant k.

A polynomial of degree n has at most n real zeros; the graph crosses the x-axis at zeros of odd multiplicity and touches it at zeros of even multiplicity.

If α and β are zeros of a quadratic, (α − β)² = (α + β)² − 4αβ, useful for finding the difference of zeros.

A cubic with zeros α, β, γ is k[x³ − (α+β+γ)x² + (αβ+βγ+γα)x − αβγ].

For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0: unique solution if a₁/a₂ ≠ b₁/b₂ (intersecting lines).

Infinitely many solutions if a₁/a₂ = b₁/b₂ = c₁/c₂ (coincident lines); no solution if a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (parallel lines).

Substitution method: solve one equation for one variable, substitute into the other, then back-substitute.

Elimination method: multiply equations to equalise one coefficient, then add or subtract to eliminate that variable.

Cross-multiplication: x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁), provided a₁b₂ − a₂b₁ ≠ 0.

A pair of linear equations is consistent if it has at least one solution; it is dependent only when the lines coincide.

Graphical method: plot both lines from two points each; the intersection point's coordinates give the solution.

Word problems: convert each condition into an equation, then check the solution against both original conditions.

Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a for ax² + bx + c = 0, a ≠ 0.

Discriminant D = b² − 4ac: D > 0 gives two distinct real roots, D = 0 gives equal real roots, D < 0 gives no real roots.

Sum of roots = −b/a and product of roots = c/a; a quadratic with roots α, β is x² − (α + β)x + αβ = 0.

Completing the square: ax² + bx + c = a(x + b/2a)² + (c − b²/4a); the turning point is at x = −b/2a.

If D = 0 the root is −b/2a, and the equation is a perfect square; if D > 0 and a, b, c are rational, roots are rational only when D is a perfect square.

For equal roots, set D = 0 and solve for the unknown parameter; for real roots, use D ≥ 0.

Word problems: let the unknown be x, translate conditions into a quadratic, reject roots that are negative lengths, ages, or speeds.

Check every candidate root in the original equation; squaring or clearing denominators can introduce extraneous solutions.

nth term: aₙ = a + (n − 1)d, where a is the first term and d the common difference.

Sum of n terms: Sₙ = (n/2)[2a + (n − 1)d] = (n/2)(a + l) when the last term l is known.

Common difference d = a₂ − a₁ = aₙ − aₙ₋₁ must be the same for every consecutive pair.

Number of terms: n = (l − a)/d + 1 when the last term l is known.

Sum of first n natural numbers is n(n + 1)/2; sum of first n odd numbers is n².

If a, b, c are in AP then 2b = a + c; the arithmetic mean of a and c is (a + c)/2.

aₙ = Sₙ − Sₙ₋₁ gives the nth term from sums of an AP.

Applications: evenly spaced instalments, salary increments, and rows of logs are modelled by APs; identify a and d from the wording before applying formulas.

Basic Proportionality Theorem (Thales): a line parallel to one side of a triangle divides the other two sides in the same ratio.

Converse of BPT: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side.

Similarity criteria: AA (or AAA), SSS (ratios of all three corresponding sides equal), and SAS (two sides in ratio with the included angle equal).

If two triangles are similar, ratio of areas = square of the ratio of corresponding sides.

If two triangles are similar, ratio of perimeters = ratio of corresponding sides, medians, and altitudes.

In a right triangle, the altitude to the hypotenuse creates two triangles similar to each other and to the original.

Pythagoras theorem: in a right triangle, hypotenuse² = sum of squares of the other two sides; its converse identifies right triangles.

Corresponding angles of similar triangles are equal; corresponding sides are proportional, not equal unless the triangles are congruent.

Distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]; distance from origin is √(x² + y²).

Section formula for point dividing AB internally in ratio m:n: ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).

Midpoint formula is the section formula with m = n = 1: ((x₁ + x₂)/2, (y₁ + y₂)/2).

Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Three points are collinear exactly when the triangle area formula gives 0.

Distance of a point from the x-axis is |y| and from the y-axis is |x|.

Points equidistant from two fixed points lie on the perpendicular bisector of the segment joining them.

External division in ratio m:n gives ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n)).

Ratios: sin θ = perpendicular/hypotenuse, cos θ = base/hypotenuse, tan θ = perpendicular/base, with cosec, sec, cot as reciprocals.

Standard values: sin 30° = ½, sin 45° = 1/√2, sin 60° = √3/2; cos values run in reverse order; tan 45° = 1.

Identity: sin²θ + cos²θ = 1; dividing by cos²θ gives 1 + tan²θ = sec²θ; dividing by sin²θ gives 1 + cot²θ = cosec²θ.

tan θ = sin θ/cos θ and cot θ = cos θ/sin θ; complementary angles satisfy sin(90° − θ) = cos θ and tan(90° − θ) = cot θ.

Angle of elevation is measured upward from the horizontal to the line of sight; angle of depression is measured downward.

Heights and distances: draw a right triangle, mark the known side and angle, then choose the ratio linking them (tan for opposite/base).

For two observers or two positions, set up two right triangles sharing a common vertical height and solve simultaneously.

If the angle of elevation doubles while the height is fixed, the horizontal distance is not halved; recompute with tan of the new angle.

A tangent to a circle is perpendicular to the radius at the point of contact.

Tangents drawn from an external point to a circle are equal in length.

The lengths of two tangents from an external point subtend equal angles at the centre and are equally inclined to the line joining the point to the centre.

Perpendicular from the centre to a chord bisects the chord; equal chords are equidistant from the centre.

Angle in a semicircle is a right angle; the angle between two tangents from an external point is 180° minus the central angle between the radii.

Number of tangents from a point inside the circle is 0, on the circle is 1, and outside is 2.

Construction of a tangent at a point on a circle: draw the radius to that point, then erect a perpendicular at the point.

Construction of tangents from an external point: draw the circle on the segment joining the centre and the point as diameter; its intersections with the given circle are the points of contact.

Circle: area = πr², circumference = 2πr; sector area = (θ/360°)πr², arc length = (θ/360°)2πr.

Area of minor segment = sector area − triangle area, where the triangle has sides r, r and included angle θ.

Ring (annulus) area = π(R² − r²); distance covered by a wheel in one revolution equals its circumference.

Cylinder: curved surface 2πrh, total surface 2πr(r + h), volume πr²h.

Cone: slant height l = √(r² + h²), curved surface πrl, total surface πr(l + r), volume ⅓πr²h.

Sphere: surface 4πr², volume 4/3πr³; hemisphere: curved surface 2πr², total surface 3πr², volume 2/3πr³.

Combined solids: add surface areas only for exposed faces; volumes of joined solids add directly.

When a solid is melted and recast, volume is conserved; when a cone or sphere is immersed, the rise in water level times the container's base area equals the immersed volume.

Direct mean = Σfᵢxᵢ / Σfᵢ; assumed-mean method uses dᵢ = xᵢ − a and mean = a + Σfᵢdᵢ/Σfᵢ.

Step-deviation method: uᵢ = (xᵢ − a)/h and mean = a + h(Σfᵢuᵢ/Σfᵢ).

Median of grouped data = l + [(n/2 − cf)/f] × h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency, h the class size.

Modal class is the class with highest frequency; mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h.

Empirical relation for moderately skewed data: 3 × median = mode + 2 × mean.

Cumulative frequency curve (ogive): plotting upper limits against cumulative frequencies and joining by a smooth curve; the median is the x-value at cumulative frequency n/2.

Class mark xᵢ = (lower limit + upper limit)/2; class size h is the difference between consecutive lower limits.

The less-than and more-than ogives intersect at the median; the x-coordinate of intersection gives the median value.

Probability of event E = number of favourable outcomes / total number of equally likely outcomes.

0 ≤ P(E) ≤ 1; P(impossible event) = 0 and P(sure event) = 1.

P(E) + P(not E) = 1, so P(not E) = 1 − P(E).

A deck has 52 cards: 26 red, 26 black, 4 suits of 13, 4 aces, 12 face cards (J, Q, K).

Two dice: total outcomes 36; sums range 2 to 12, with 7 the most likely sum (6 outcomes).

A coin tossed n times has 2ⁿ outcomes; three coins give 8 equally likely outcomes.

Events are equally likely only when each outcome has the same chance; biased dice or coins break this assumption.

For 'at least one' questions, use 1 − P(none) to simplify counting.