Mathematics — Complete Formula Sheet

$n$: Composite natural number being factorised · $p_1,p_2,\dots ,p_k$: Distinct prime factors of n, in increasing order · $a_1,a_2,\dots ,a_k$: Powers (natural numbers) of the primes

Every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the factors. This uniqueness drives every other result in the chapter.

Example: 84 = 2² × 3 × 7 and 120 = 2³ × 3 × 5. Common primes are 2 and 3, so HCF = 2² × 3 = 12.

Take every prime that appears in ANY of the numbers. For 84 and 120: LCM = 2³ × 3 × 5 × 7 = 840.

$a,b$: Two positive integers

True for any two positive integers. Use it to find the LCM when the HCF is given, or to find the second number.

Rearranged form of the product rule. Given HCF(336, 54) = 6, LCM = 336 × 54 ÷ 6 = 3024.

Co-prime numbers share no prime factor, so their LCM is simply their product. Example: HCF(8, 15) = 1, LCM = 120.

The LCM is always an exact multiple of the HCF. If LCM ÷ HCF is not a whole number, no such pair of numbers exists.

For 6, 72 and 120: HCF = 6, LCM = 360, HCF × LCM = 2160 but 6 × 72 × 120 = 51 840. Find HCF and LCM of three numbers only by prime factorisation.

By uniqueness of prime factorisation, 4ⁿ = 2²ⁿ contains no 5, so 4ⁿ can never end in 0 for any natural number n.

$p$: A prime number · $a$: A positive integer

a is a positive integer. This is the key step inside every irrationality proof of √2, √3, √5.

2 dividing both a and b contradicts that a and b are co-prime. The same proof works for √3, √5, √7 (√p for any prime p).

$r$: A rational number · $s$: An irrational number

Sum or difference of a rational and an irrational number is irrational. Product and quotient of a NON-ZERO rational and an irrational number are irrational. So 5 − √3, 3√2 and 2/√7 are all irrational.

Highest exponent n is called the degree. Leading coefficient aₙ ≠ 0.

Zero of p(x) = root of p(x) = 0 = x-coordinate of the graph where it crosses the x-axis.

For ax² + bx + c = 0. Note the MINUS sign.

For ax² + bx + c = 0. Positive if the leading and constant terms share sign.

Any scalar multiple of this is also a quadratic with the same zeroes.

For ax³ + bx² + cx + d = 0. Same negative sign pattern as quadratic.

The MIDDLE symmetric function of the roots. Sign is positive.

Note the MINUS sign (odd degree).

Analogue of division algorithm for integers. r(x) may be zero (⇒ g(x) is a factor).

Quick way to check whether a linear factor works. Cornerstone of factorisation.

$x,y$: The two unknowns (variables) · $a_1,b_1,a_2,b_2$: Coefficients of x and y in the two equations · $c_1,c_2$: Constant terms

a₁, b₁, c₁, a₂, b₂, c₂ are real numbers with a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0. The graph of each equation is a straight line.

Quickest two points to plot: (−c/a, 0) on the x-axis and (0, −c/b) on the y-axis. Put y = 0, then x = 0 (a, b ≠ 0).

A solution must satisfy BOTH equations. Graphically it is a common point of the two lines.

The lines intersect at exactly one point, so the pair is consistent. c₁/c₂ does not matter here.

One equation is a multiple of the other, so both describe the same line and every point on it is a solution. The pair is dependent (and consistent).

The lines are parallel and distinct, so they never meet: the pair is inconsistent. A pair is consistent if it has at least one solution (unique or infinitely many).

Express one variable in terms of the other from one equation, substitute into the other to get an equation in one variable, solve, then back-substitute.

Multiply the equations so that one variable has equal coefficients, then subtract (or add if the signs are opposite) to remove it. In practice multiply by the smallest numbers that do the job.

In substitution or elimination, if both variables cancel, a true statement means coincident lines and a false statement means parallel lines.

$t$: Tens digit · $u$: Units digit

t = tens digit, u = units digit. Note that (10u + t) − (10t + u) = 9(u − t).

Let x be the PRESENT age. Write each condition using the same present-age variables, then solve the pair.

Any degree-2 polynomial equation reduces to this form.

Gives both roots at once. Works even for irrational or complex roots.

Decides the NATURE of roots — see next three rows.

Graph crosses x-axis at two distinct points.

Graph touches x-axis at exactly one point. Roots = −b/(2a) (repeated).

Graph does not touch the x-axis. Roots are complex conjugates.

Same as for the polynomial ax² + bx + c. Use to form a new equation with given roots.

Multiply by any non-zero scalar to get an equivalent equation.

Basis of both the formula derivation and vertex-form of a parabola.

$a$: First term of the AP · $d$: Common difference (a₂ − a₁ = a₃ − a₂) · $n$: Position (natural number)

a = first term, d = common difference. Find nth term without listing all previous terms.

Constant difference between consecutive terms defines an AP.

Use when 'd' is known but the last term is not.

l = last term = aₙ. Use when a and l are given.

a, b, c are in AP iff b is the arithmetic mean of a and c.

l = last term. Useful to find, say, '5th term from the end'.

Rearranged nth-term formula. n MUST be a positive integer for a valid AP term.

Sum of first n natural numbers and first n odd numbers. Useful shortcuts.

For polygons with the same number of sides, BOTH conditions are needed. A square and a rectangle have equal angles but are not similar; a square and a rhombus have proportional sides but are not similar.

The symbol ~ means 'is similar to'. The order of letters fixes the correspondence A ↔ P, B ↔ Q, C ↔ R. Congruent triangles are similar with ratio 1.

$D$: Point where the parallel line cuts side AB · $E$: Point where the parallel line cuts side AC

A line drawn parallel to one side of a triangle, cutting the other two sides at distinct points, divides those sides in the same ratio.

Part-to-whole forms that follow from AD/DB = AE/EC. Use them when a whole side (AB or AC) is given.

If a line divides two sides of a triangle in the same ratio, it is parallel to the third side. Used to PROVE two lines parallel.

If corresponding angles of two triangles are equal, their corresponding sides are proportional and the triangles are similar.

Two pairs of equal angles are enough, because the third angles are then equal by the angle sum property (180°). The most-used criterion in proofs.

If all three pairs of corresponding sides are in the same ratio, the corresponding angles are equal and the triangles are similar.

One pair of equal angles AND the sides including those angles in the same ratio. A non-included angle does not work.

Follows directly from proportional sides: if every side is multiplied by k, so is their sum. Valid for ΔABC ~ ΔPQR.

AD and PM are corresponding medians (or corresponding altitudes). Prove it by showing ΔABD ~ ΔPQM: SAS for medians, AA for altitudes.

O is the point where diagonals AC and BD meet. ΔAOB ~ ΔCOD by AA, using alternate angles.

$h_1,h_2$: Heights of the two vertical objects (m) · $s_1,s_2$: Lengths of their shadows (m)

Vertical objects at the same place and time: the sun's rays make the same angle with the ground, so the right triangles are similar (AA).

$H$: Height of the lamp above the ground (m) · $h$: Height of the person (m) · $d$: Distance of the person from the foot of the lamp-post (m) · $s$: Length of the person's shadow (m)

Large triangle: lamp-post and the ground up to the shadow's tip. Small triangle: the person and the shadow. They share the angle at the shadow's tip, so they are similar (AA).

$(x_1,y_1)$: Coordinates of the first point · $(x_2,y_2)$: Coordinates of the second point · $d$: Distance (units)

Direct application of Pythagoras in the plane. Distance is always positive.

Special case with (x₁, y₁) = (0, 0). Same as the magnitude of the position vector.

P divides AB in the ratio m:n internally (from A to B). Weights cross-over: m goes with x₂, n with x₁.

Section formula with 1:1 ratio. Average the coordinates.

Average the coordinates of the three vertices. G divides each median in the ratio 2:1.

Absolute value ⇒ area is always positive. Zero means the three points are collinear.

Three points collinear ⇔ triangle area is zero.

Two trisection points divide AB in ratios 1:2 and 2:1 respectively. Use section formula twice.

In a right triangle, opposite to angle θ over the hypotenuse. Value between 0 and 1 for 0° < θ < 90°.

Adjacent (next to θ, not the hypotenuse) over hypotenuse.

Sine over cosine. Undefined when cos θ = 0 (θ = 90°).

cosec, sec, cot are just reciprocals. Learn all six together.

The most-used identity. Follows directly from Pythagoras.

Divide the Pythagorean identity by cos²θ.

Divide the Pythagorean identity by sin²θ.

'Co' functions swap under 90° − θ. Symmetry of a right triangle explains this.

Memorise this row; cos is the reverse (cos 0° = 1, cos 90° = 0).

tan = sin/cos; the 90° value is undefined because cos 90° = 0.

$\theta$: Angle of elevation (degrees) · $h$: Height of the object above eye level (m) · $d$: Horizontal distance from the observer to the foot of the object (m)

The angle between the line of sight and the horizontal when the object is ABOVE the eye. The line of sight is the line from the observer's eye to the object.

Rearranged forms. Find h when the distance is known; find d when the height is known.

$l$: Slant length — hypotenuse of the right triangle (m) · $h$: Vertical height (m) · $d$: Horizontal distance (m)

l is the hypotenuse: the length of a kite string, a ladder, a rope or the line of sight itself. θ is measured from the horizontal.

The angle between the line of sight and the horizontal when the object is BELOW the eye. The depression of B seen from A equals the elevation of A seen from B, because the two horizontals are parallel.

$H$: Total height of the object (m) · $h_o$: Height of the observer's eyes above the ground (m) · $d$: Horizontal distance from the observer to the object (m)

If the angle is measured from the observer's eyes, d tan θ gives only the part ABOVE eye level. Add the eye height.

cot is the reciprocal: cot 30° = √3, cot 45° = 1, cot 60° = 1/√3.

Needed for string, ladder and slant-distance problems.

$s$: Length of the shadow (m) · $h$: Height of the vertical object (m)

θ is the sun's altitude (angle of elevation of the sun). The higher the sun, the shorter the shadow.

$x$: Distance between the two observation points (m) · $h$: Height of the object (m) · $\alpha ,\beta$: Angles of elevation from the farther and nearer points

Angles α (farther point) and β (nearer point), α < β, observed at two points in line with the foot. Also gives the distance between two ships seen from a lighthouse, or the change in a shadow's length.

A handy check: if the elevation changes from 30° to 60° after walking x towards the tower, the tower is (√3/2)x high and you are now x/2 from its foot (BD = distance of the nearer point from the foot B). Show the full working in the exam.

For two points on opposite sides of the object's foot (or two poles of equal height h seen from a point between them). x is the total distance between the two points.

$h$: Height of the building (m) · $H$: Height of the tower (m) · $d$: Horizontal distance between building and tower (m)

From the top of a building of height h: β is the angle of depression of the tower's foot, α is the angle of elevation of the tower's top. The horizontal through the building's top splits the tower into h and d tan α.

A tangent meets the circle at exactly one point, called the point of contact. A tangent is the limiting case of a secant when its two points of intersection coincide.

No tangent passes through a point inside the circle. There is exactly one tangent at each point of the circle, and exactly two tangents from a point outside it.

$O$: Centre of the circle · $A$: Point of contact · $XY$: Tangent at A

XY is the tangent at A and O is the centre. The line through A perpendicular to the tangent passes through the centre.

$PA$: Length of the tangent from P to the point of contact A (cm) · $OP$: Distance of the external point P from the centre O (cm) · $r$: Radius, r = OA (cm)

Triangle OAP is right-angled at the point of contact A, so OP (centre to external point) is the hypotenuse. All lengths in the same unit (cm).

$P$: External point · $A,B$: Points of contact of the two tangents

PA and PB are the two tangents from the external point P, touching the circle at A and B.

From the same congruence ΔOAP ≅ ΔOBP. OP bisects the angle between the tangents and also the angle between the radii OA and OB.

In quadrilateral OAPB the angles at A and B are 90° each, so the other two angles add up to 360° − 180° = 180°.

AB is the chord joining the points of contact. Since PA = PB, ∠PAB = 90° − ½∠APB, and ∠OAB = 90° − ∠PAB = ½∠APB.

M is the point where OP meets AB. Both P (PA = PB) and O (OA = OB) are equidistant from A and B, so OP is the perpendicular bisector of AB.

Both tangents are perpendicular to the diameter AB, so they are parallel. The distance between two parallel tangents equals the diameter 2r.

$R$: Radius of the larger circle (cm) · $r$: Radius of the smaller circle (cm)

AB is a chord of the larger circle that touches the smaller circle. The radius of the smaller circle to the point of contact is perpendicular to AB, so the chord is bisected at the point of contact.

All four sides of ABCD touch the circle. Opposite sides have equal sums. A parallelogram circumscribing a circle is therefore a rhombus.

Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre. Join O to the vertices and points of contact; the eight angles at O pair up equally and total 360°.

$x,y,z$: Tangent lengths from vertices A, B, C (cm) · $r$: Radius of the inscribed circle (cm) · $s$: Semi-perimeter of the triangle (cm)

x, y, z are the tangent lengths from A, B, C. The semi-perimeter is s = x + y + z. Splitting ΔABC into ΔOAB, ΔOBC and ΔOCA gives area = ½ r (AB + BC + CA) = rs. Combine with Heron's formula to find an unknown tangent length.

$C$: Circumference (cm) · $r$: Radius (cm) · $d$: Diameter, d = 2r (cm)

Recalled from earlier classes. Use it when the circumference is given and you must find the radius first.

Every sector and segment area in this chapter is a part of this. If the diameter is given, halve it first.

$l$: Length of the arc (cm) · $\theta$: Central angle of the sector (degrees) · $r$: Radius (cm)

θ is the angle (in degrees) that the arc subtends at the centre. The arc is that fraction of the whole circumference. Unit: cm.

$A$: Area of the sector (cm²) · $\theta$: Central angle (degrees) · $r$: Radius (cm)

A sector is the region between two radii and an arc. Its area is the fraction θ/360° of the circle's area. Unit: cm².

$l$: Length of the arc (cm) · $r$: Radius (cm)

Follows by dividing the sector-area formula by the arc-length formula: A/l = r/2. Useful when the arc length is given instead of the angle.

Sectors with central angles 90° and 180°.

θ is the angle of the minor sector. The major sector has angle 360° − θ.

$P$: Perimeter of the sector (cm) · $r$: Radius (cm) · $\theta$: Central angle (degrees)

Two radii plus the arc. A common slip is to give only the arc length. Unit: cm.

The segment is the region between the chord AB and the minor arc. O is the centre and θ = ∠AOB.

The CBSE syllabus limits segment problems to these three central angles. For 60°, OA = OB = r and the base angles are also 60°, so the triangle is equilateral with side r. For 90°, it is a right isosceles triangle with legs r. For 120°, draw OM ⟂ AB: ∠AOM = 60°, so OM = r cos 60° = r/2 and AM = r sin 60° = (√3/2) r, giving AB = √3 r and area = ½ × AB × OM = (√3/4) r².

Find the minor segment first, then subtract from the area of the whole circle.

The minute hand turns 360° in 60 minutes; the hour hand turns 360° in 12 hours. The length of the hand is the radius of the sector it sweeps.

$l$: Length · $b$: Breadth · $h$: Height

Volume = length × breadth × height. TSA sums the 6 rectangular faces.

Special cuboid with all sides equal. Six square faces of side a.

CSA = curved surface only. TSA adds the two circular ends.

l = slant height = √(r² + h²). CSA is a sector when unrolled.

One number to know per formula. Sphere has no edges or vertices.

Half a sphere. TSA = curved half sphere + flat circular base.

Direct Pythagoras. Use before applying CSA = πrl.

A cone with the top sliced off. R, r are the two radii; h is height; l = √(h² + (R−r)²).

When metal is melted and recast, volume stays the same; surface area changes.

The mid-point of a class. When finding the mean, each class is represented by its class mark.

$\bar{x}$: Mean of the data · $f_i$: Frequency of the i-th class · $x_i$: Class mark of the i-th class

Multiply each class mark by its frequency, add, and divide by the total frequency. Best when fᵢ and xᵢ are small numbers.

$a$: Assumed mean (one of the class marks) · $d_i$: Deviation of the class mark from a

a is the assumed mean, usually a class mark near the centre of the table. The dᵢ are smaller than the xᵢ, so the products are easier.

$h$: Class size (common factor of the dᵢ) · $u_i$: Step deviation of the i-th class

Use when all the dᵢ have a common factor h, usually the class size. All three methods give the same mean.

The mode lies in this class. Find it before applying the mode formula.

$l$: Lower limit of the modal class · $h$: Class size · $f_1$: Frequency of the modal class · $f_0$: Frequency of the class preceding the modal class · $f_2$: Frequency of the class succeeding the modal class

f₀ and f₂ are the frequencies of the classes just before and just after the modal class. If the modal class is the first class, f₀ = 0.

Running total of the frequencies up to and including the k-th class. The last cumulative frequency equals n = Σfᵢ.

n = Σfᵢ is the total frequency. Compare n/2 with the cumulative frequency column, not the frequency column.

$l$: Lower limit of the median class · $n$: Total frequency, Σfᵢ · $cf$: Cumulative frequency of the class preceding the median class · $f$: Frequency of the median class · $h$: Class size

cf is the cumulative frequency of the class PRECEDING the median class, and f is the frequency of the median class itself.

An approximate relation between the three measures of central tendency. Use it when two measures are known and the third is asked.

For classes 10–19, 20–29, … the gap is 1, so subtract 0.5 from every lower limit and add 0.5 to every upper limit: 9.5–19.5, 19.5–29.5, … The class size stays 10.

$E$: The event whose probability is required · $P(E)$: Probability of E (a number from 0 to 1, no unit)

Valid only when all outcomes are equally likely: a fair coin, a fair die, a well-shuffled pack, a ball drawn at random.

A negative probability or one greater than 1 means a counting error. Recheck the total and the favourable cases.

Getting a number less than 7 on a die is sure (probability 1). Getting 8 on a single die is impossible (probability 0).

$\overline{E}$: Complement of E, the event 'not E'

Ē (read 'not E') happens exactly when E does not. Use this when 'not E' is easier to count, or when P(E) is given and P(not E) is asked.

An elementary event has only one outcome. The probabilities of all the elementary events of an experiment add up to 1. Use it to find one missing probability.

Two coins: HH, HT, TH, TT. Three coins: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. 'One head and one tail' has 2 outcomes (HT, TH), not 1.

Outcomes for two dice are ordered pairs (a, b). There are 6 doublets: (1, 1), (2, 2), …, (6, 6).

Sum 7 has the most ways (6); sums 2 and 12 have 1 way each. Check: 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36.

Spades ♠ and clubs ♣ are black; hearts ♥ and diamonds ♦ are red. Each suit has A, 2 to 10, J, Q, K. Face cards are J, Q and K only; aces are not face cards.

⌊N/k⌋ is the whole-number part of N/k. Multiples of 7 from 1 to 100: ⌊100/7⌋ = 14. Useful for cards or discs numbered 1 to N.