Kerala SSLC Mathematics cheat sheet

For a sequence with first term f and common difference d, the nth term is xₙ = f + (n − 1)d; consequently xₙ − xₘ = (n − m)d for any two terms.

Sum of the first n terms: Sₙ = (n/2)[2f + (n − 1)d] = n(f + xₙ)/2, i.e. number of terms × average of the first and last terms.

A sequence is arithmetic only if the differences of consecutive terms are equal; test by subtracting successive terms.

Standard sums: 1 + 2 + … + n = n(n + 1)/2; 1 + 3 + 5 + … + (2n − 1) = n²; 2 + 4 + … + 2n = n(n + 1).

To test whether a number k is a term, set f + (n − 1)d = k and check that the resulting n is a positive integer.

If a, b, c are three consecutive terms of an arithmetic sequence then 2b = a + c; the middle term is the average of the other two.

Sums of blocks of k consecutive terms of an arithmetic sequence again form an arithmetic sequence with common difference k²d.

Rearranging d = (xₙ − f)/(n − 1) lets you find the common difference, the number of terms, or the first term when the others are known.

The perpendicular from the centre to a chord bisects the chord; conversely the line from the centre to the midpoint of a chord is perpendicular to it.

Equal chords of a circle are equidistant from the centre, and of two unequal chords the longer lies nearer the centre; the diameter is the longest chord.

A chord at distance d from the centre of a circle of radius r has length 2√(r² − d²).

The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circle.

The angle in a semicircle is 90°; conversely, if a chord subtends a right angle at a point on the circle, that chord is a diameter.

Angles in the same segment are equal, and equal arcs subtend equal angles both at the centre and at the circumference.

In a cyclic quadrilateral the opposite angles are supplementary (sum 180°), and each exterior angle equals the interior opposite angle.

If a segment joining two fixed points subtends equal angles at two other points on the same side of it, all four points lie on one circle.

Probability of an event = number of favourable outcomes ÷ total number of equally likely outcomes; the value always lies between 0 and 1.

P(not E) = 1 − P(E); for 'at least one' questions it is usually easier to compute 1 − P(none).

Tossing n coins gives 2ⁿ equally likely outcomes; two coins give HH, HT, TH, TT, so exactly one head has probability 2/4 = 1/2.

Two dice give 36 ordered outcomes; the number of ways to obtain a total s is 6 − |s − 7| for 2 ≤ s ≤ 12, largest (6 ways) for s = 7.

Drawing without replacement changes the total: after one item is removed the denominator and the favourable count both drop by one.

For multi-stage experiments multiply the numbers of outcomes of the stages (m × n) to get the size of the sample space.

Geometric probability: for a point chosen at random in a region, P = area of the favourable part ÷ area of the whole region.

An event with probability 0 is impossible and one with probability 1 is certain; a probability cannot be negative or exceed 1.

Standard form ax² + bx + c = 0 with a ≠ 0; a second degree equation has at most two real solutions.

Quadratic formula: x = [−b ± √(b² − 4ac)] / (2a), obtained by completing the square on ax² + bx + c.

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

For the roots α and β: α + β = −b/a and αβ = c/a.

If the roots are given as p and q, the equation is x² − (p + q)x + pq = 0, or any non-zero multiple of it.

Factorising by splitting the middle term: find two numbers whose product is ac and whose sum is b, then group.

Completing the square: ax² + bx + c = a(x + b/2a)² + (c − b²/4a); isolate the square and take square roots.

In word problems, form the equation from the given condition and reject any root that contradicts the situation (negative length, non-integral count, etc.).

In a right triangle: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent = sin θ ÷ cos θ.

Values: sin 0 = 0, sin 30 = 1/2, sin 45 = 1/√2, sin 60 = √3/2, sin 90 = 1; cos runs the reverse order.

tan 0 = 0, tan 30 = 1/√3, tan 45 = 1, tan 60 = √3, and tan 90 is undefined.

Identity sin²θ + cos²θ = 1 for every θ; dividing by cos²θ gives 1 + tan²θ = sec²θ.

Complementary angles: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ.

Choose the ratio that connects the known and unknown sides: tan when both legs are involved, sin or cos when the hypotenuse is involved.

Standard triangles: 45°–45°–90° has sides in the ratio 1 : 1 : √2; 30°–60°–90° has sides in the ratio 1 : √3 : 2.

Angle of elevation is measured upward from the horizontal, angle of depression downward from the horizontal; for the same pair of points the two are equal (alternate angles).

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

Midpoint of the segment joining (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2).

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

Slope of the line through two points = (y₂ − y₁)/(x₂ − x₁); parallel lines have equal slopes, perpendicular lines have slopes whose product is −1.

Area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|; area zero means the points are collinear.

A point on the x-axis has y = 0; a point on the y-axis has x = 0; the origin is (0, 0).

Line equations: y = mx + c has slope m and y-intercept c; y = k is horizontal and x = k is vertical.

In a parallelogram the diagonals bisect each other, so proving that two segments share the same midpoint proves the quadrilateral is a parallelogram.

A tangent meets a circle at exactly one point and is perpendicular to the radius drawn to that point of contact.

The two tangent segments drawn to a circle from an external point are equal in length.

The line from the external point to the centre bisects the angle between the two tangents and also the angle between the two radii.

Length of the tangent from a point at distance d from the centre of a circle of radius r is √(d² − r²).

Tangent–chord (alternate segment) theorem: the angle between a tangent and a chord through the point of contact equals the angle the chord makes in the alternate segment.

If two circles touch, the point of contact lies on the line of centres; the distance between centres is r₁ + r₂ for external contact and |r₁ − r₂| for internal contact.

A quadrilateral circumscribed about a circle satisfies AB + CD = BC + DA, i.e. the sums of opposite sides are equal.

Construction: to draw tangents from an external point P to a circle with centre O, draw the circle with OP as diameter; its intersections with the given circle are the points of contact.

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

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

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

A cone is one third of the cylinder with the same base and height, and a sphere's volume is two thirds of the cylinder that just encloses it (4/3 πr³ against 2πr³).

When a solid is melted and recast, volume is conserved: equate the volume of the original solid with that of the new shape.

For similar solids with scale factor k, surface areas are in the ratio k² and volumes in the ratio k³.

For combined solids add the volumes, but for surface area add only exposed faces and subtract the circular faces hidden at the join.

Unit conversions: 1 litre = 1000 cm³ and 1 m³ = 10⁶ cm³; convert to one unit before calculating.

Remainder theorem: the remainder when p(x) is divided by (x − a) is p(a).

Factor theorem: (x − a) is a factor of p(x) exactly when p(a) = 0.

For p(x) = ax² + bx + c with zeros α and β: α + β = −b/a and αβ = c/a.

A quadratic polynomial with given zeros α, β is k[x² − (α + β)x + αβ] for any non-zero k; if the sum is S and the product is P, use x² − Sx + P.

Zeros of a polynomial are the values of x for which p(x) = 0, that is the x-intercepts of its graph.

The degree of a polynomial is the highest power of x with a non-zero coefficient; a quadratic has at most two zeros.

Division algorithm: dividend = divisor × quotient + remainder, with the remainder of lower degree than the divisor.

If a quadratic has zeros of opposite signs, the product c/a is negative; equal zeros occur when the discriminant b² − 4ac is zero.

Two triangles are similar if two angles match (AA), all three pairs of sides are proportional (SSS), or two sides are proportional with the included angle equal (SAS).

Corresponding sides of similar triangles are proportional: AB/PQ = BC/QR = CA/RP, with vertices written in matching order.

The ratio of the areas of two similar triangles equals the square of the ratio of any pair of corresponding sides.

The ratio of the perimeters of similar triangles equals the ratio of corresponding sides.

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

Converse of the basic proportionality theorem: if a line divides two sides of a triangle proportionally, it is parallel to the third side.

The angle bisector theorem: the bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides.

In an equilateral triangle, a line parallel to one side cuts off a smaller equilateral triangle similar to the original, with sides in the same ratio as the cut segments.

Mean of raw data = Σx/n; for a frequency table, mean = Σfx/Σf where x is the class mark (midpoint of the class).

Median of n ordered values: the middle value when n is odd, the mean of the two middle values when n is even.

Grouped median: median = l + [(n/2 − cf)/f] × h, with l the lower limit of the median class, cf the cumulative frequency before it, f its frequency, h the class width.

The median class is the first class whose cumulative frequency reaches or exceeds n/2 in the cumulative frequency table.

Mode is the most frequent value or the class with the highest frequency; the empirical relation is mode ≈ 3 median − 2 mean.

Range = maximum value − minimum value; it depends on only two observations and ignores the spread of the rest.

Class mark = (lower limit + upper limit)/2; assumed mean method: mean = A + (Σfd/Σf) where d = x − A.

The sum of the deviations of all observations from the mean is zero, i.e. Σ(x − x̄) = 0.