Matematika | Középiskola » High School Math

Sets SETS A set is given if we can unambiguously decide about every element whether it is included or not. (eg: group of schools in Hungary is a set, but group of tall girls is not a set) If ’a’ is element of set B we note this relation like this : a € B If ’a’ is not an element of B : a not € B A set which has no elements is called empty set. Notation: {Ø} When a set is not empty then it has at least two elements. Two sets are equal if they have the same elements. A is a subset of B if A hasn’t got any elements which are not elements of B or every element of A is element of B as well: Notation: The empty set is a subset of every set and every set is a subset of itself. A is a real subset of B if A is a subset of B and A is not equal to B. If A is element of B and B is element of A then A equals to B. If A is element of B and B is element of C then A is element of C. A set can be finite or infinite. Finite if it has countable elements (eg.: natural numbers between

1-20) Infinite if it has uncountable elements. (eg: Natural numbers) Compliment of set A: set of things which are not elements of A. Notation: A’ or capital A with a horizontal line above it Operations with sets: 2006. Mathematics oral exam Page 1. Sets AUØ=A AUB=BUA (A U B) U C = A U (B U C) Ø=Ø A B=B A (A B) A C=A (B C) The difference of set A and set B is the set of those elements which are elements of set A but not elements of set B. Notation: A B DEFINITION: Given a line (h) and a point (F) on the plane, where the point is not the element of the line. Parabola is the set of points on the plane which are equidistant from point F and line h (Point F will be the focus point and line h will be the headline of the parabola) (Picture 1) Construction of the points of the parabola: draw a perpendicular from an optional point of the headline and get point Q. Then construct the perpendicular bisector of FQ and get point P (one element of the parabola) at the

intersection of the perpendicular from Q and the perp. bisector of FQ. FP equals to FQ (by def) Repeat this construction to get all points of the parabola. (Picture 2) Picture 1 Picture 2 THEOREM 1 The perpendicular bisectors of a triangle intersect each other in one point (which point is the centre of the circumscribed circle of the triangle) Proof: let K be the intersection point of perp. bisectors of AC and AB Suppose that AC is not parallel to AB, then point K exists. K is on the perp. bisector of AB therefore AK = KB by def K is on the perp. bisector of AC therefore AK = KC by def 2006. Mathematics oral exam Page 2. Sets If AK = KB and AK = KC then KB = KC which means that point K is on the perp.bis of BC as well. Q.ed THEOREM 2 The number of subsets in a set which has n elements is 2n. Proof 1: We can decide about every element if it is in the subset or not - two variations (chosen or not) - decision n times (because the set has n elements) - 2*222.2*22 (n times) = 2n

Proof 2: if the set has n elements we can select subset with 0 elements: in 1 way subset with 1 elements: in n ways subset with 2 elements: in (2 under n) ways subset with 3 elements: in (3 under n) ways subset with k elements: in (k under n) ways (0 under n) + (1 under n) + (2 under n) ++ (n under n)= number of subsets Adding up the two proofs we get that : (0 under n) + (1 under n) + (2 under n) ++ (n under n)= 2n APPLICATIONS: koordinate geometry (basic set, range) Constructions (eg: triangle, circle) 2006. Mathematics oral exam Page 3. Sets of numbers, cardinality of sets Take the set of natural numbers as a starting point. The set of natural numbers consists of non-negative whole numbers. Its sign is N (If we only want to refer to the positive whole numbers we sign it with N+ . Consider two basic operations in the set of natural numbers: addition and multiplication If a and b are elements of N, then a+b and ab are also elements of N so these operations do not lead out from

the set. If an operation in the set does not lead out of the set( i.e the result of the operation will also be element of the set), then the set is closed on this operation. If the result of an operation in the set is not the element of the set then the set is open on that operation. So the set of natural numbers is closed on addition and multiplication • Most important properties of these operations: a+b=b+a , ab=ba - commutative property (parts are exchangable) (a+b)+c=a+(b+c) , ab(c)=(ab)c - associative property (a+b)c=ac+bc - distributive property (addition to multiplication only) We also accept the division with remainder in the set of natural numbers. If a>b and c is element of N, then there exsits a natural number m<b for which a=cb+m is true. The remainder of the division is m (it can also be 0). If m=0, then a is divisible by b Next we introduce the prime numbers • Def.: We call p positive integer a prime number if and only if it has two divisors, 1 and itself. If a

natural number greater than 0 is not a prime than it is called a complex number. (there are infinitely many prime and complex numbers) Theorem: every complex number can be written up as the product of prime numbers • With the help of this theorem we can define the greatest common divisor, smallest common multiple of two or more positive integers, or the number of divisors of a positive integer etc. • If two positive integers have no common divisors out of 0, then they are called relative primes • Most important rules of divisibility o With 2 – only even numbers o With 3 – if the sum of the digits is divisible by 3 o With 4 – if the number formed by the last two digits is divisible by 4 o With 5 – if the numbers ends in 5 or 0 o With 6 – if it is divisible by 3 and 2 o With 8 – if the number formed by the last 3 digits is divisible by 8 o With 9 – if the sum of the digits is divisible by 9 • Other important rules of divisibility: (let a,b,c and n be elements

of the set of positive natural numbers o If n is a divisor of a and b as well then n is a divisor af a+b and a-b o If n is a divisor of a, and a is a divisor of b then n is a divisor of b o If n is a divisor of ab , and (n; a)=1, then n is a divisor of b o If a is a divisor of n, and b is also a divisor of n, and (a; b)=1, then ab is a divisor of n Investigating the operation of subtraction, we get that if a and b are elements of N, then a-b is not necessarily an element of N. Therefore the set of natural numbers is an open set on subtraction, and this way leads us to the set of negative numbers. The negative and positive integers and 0 are together called whole numbers, the sign of the set is Z. Set Z is closed on addition, multiplication and subtraction as well, and the properties of the set of natural numbers are true in this set as well (commutative, distributive, associative) On division, Z is open, because if a and b are elements of the set of whole numbers, then a/b will not

necessarily be an integer these numbers will be fractions Def.: if a and b are integers and b is not 0, the we call a/b rational number So rational numbers are numbers that can be written up as the ratio of two integers. The sign of the set of rational numbers is Q, and this set is closed on all four basic operations, since ab, a/b, a+b and a-b are all rational numbers if a and b are. (0 * X = a equation has no solution if a is not 0, because 0 X is 0 for all X. On the other hand if a =0 then all the rational numbers would be solutions of the equation, therefore we do not allow divising by 0 in the set of rational numbers. ) There are operations which lead us out of the set of rational numbers: for example taking the sqare root of a rational number may give us a number which can not be written up as the ratio of two integers. Def.: Those numbers which can not be written up as the ratio of two inetegers are called irrational numbers ( the set of irrational numbers is marked by Q*.)

Theorem: square root of 2 is irrational Proof: indirect proof • Let’s suppose that the sqare root of 2 can be written up as the ratio of two inetegers p/q where p and q are relative primes • This way we can get: 2q2=p2 that means that p2 is even therefore p is even, so p=2k • Then: 2q2=(2k)2=4k2 therefore q2 = 2k2 • This last eqation means that q2 therefore q is even. So if the square root of 2 could be written up as the ratio of two relative primes, then both number would be even. But they can not be since they are relative primes, therefore our statement was false Rational numbers – véges vagy végtelen szakaszos tizedes törtek Irrational numbers – végtelen, nem szakaszos tizedes törtek Q and Q* are disjunct sets because ther intersection is the empty set. Def.: the union of Q and Q* is the set of real numbers R • There is a one to one function between the elements of R and the numbers on the numberline, therefore we can demonstrate them on the numberline We

define the number of elements in a set with the cardinality of the set • If set A has finitely many elements (n elements) then its cardinality is n Def.: if a one to one function is between the elements of two sets, then the cardinality of the two sets are equal • E.g: positive integers N+ and positive even integers o For every element n we can find 2n in the other set and the other way around, so there is a one to one function between the elements of the two sets Def.: If the cardinality of any set H is equal to that of N+ then H has a countable cardinality • E.g Q has the same cardinality as N+ • E.g R has an uncountable cardinality Applications: basic arithmetical problems, calculation with fractions, introduction of units, construction of sqare root n. 3. Raising to a power, the power function Raising to the n-th power if n€Z and n>0 -in this case raising to a power is multiplication with identical terms -let a€R and n>2: an is a progression (sequence) with

all its terms = a an = a*aaa ↑ n times - n shows how many times the base (a) should be taken as a term in the multiplication -a1 = a by definition Identities: -multiplication of ‘powers’ with identical bases an*ak = an+k proof: due to definition of raising to a power and associative property of multiplication ↓ an*ak = (aaa)(aa) = an+k ↑ n times ↑ k times ↑ n+k times -division of powers with identical bases a€R ; n,k€Z+ if n>k: an = an-k ak ↑ by def. of raising to a power and rules of simplification if n = k an = 1 ak if n < k an = 1 ak ak-n -raising a power to a power n,k€Z+ (an)k = ank -raising a product to a power (ab)n = (ab)(ab)(ab) = (a*aa)(bbb) = an bn ↑ by definition Commutative: Associative: ↑ commutative and assoc. characteristics of multiplication ↑ by definition A*B = BA (A*B)C = A(BC) -raising a fraction to a power (b is not 0) (a/b)n = an / bn Identities with 0, negative and irrational powers: ‘Carried out’

in a way that all the previous features/identities of the process remain the same. (permanenciaelv) -raising to the power of 0 a≠0 a0 = 1 -raising to the power of –(n), where nEZ+ a-n = 1 an -power with rational exponents ap/q = q√(ap) If the base is 0, the result with positive exponents is 0. With negative exponents, there’s no sol. Power – functions xx3 if n is odd and nEZ : xxn domain: R range: R not bounded strictly monotonously increasing no extrema odd x;y intersection at the origin if n is even and nEZ xxn domain: R range: [0;∞[ ‘alulról korlátos’ at 0 ] -∞; 0] str. mon decreasing, [0; ∞[ str mon increasing minimum at (0;0) even x; y intersection at the origin xx2 Applications: Mathematics: Probability calculations Surface area, Volume calculations Physics: Calculations with radioactive decay Exponential and logarithm functions Exponential functions (power: exponens in Latin) Def: f(x)=ax (a>0 ;a≠1) is called an exponential function. If

a = 1 then f(x) is constant (1x =1) Properties of the exponential function: • Domain: set of real numbers (Df=R) • Range: set of positive real numbers (Rf=R+) • One to one function thus it has an inverse: xlogaX • Has no zero • The graph of the function intersects y axis at (0;1) • It has no extrema • It is not bounded • When a>1 it is strictly monotonously increasing (for V x1< x2 ax1 <ax2 ) • When 0<a<1 it is strictly monotonously decreasing (for V x1< x2 ax1 >ax2 ) smd exponential function 0<a<1 smi exponential function a>1 Applications: • Compound interest (kamatos kamat) • Pressure (nyomás számítás) • Hangintenzitás • Oldódási idő 2006. Mathematics oral exam Page 1. Exponential and logarithm functions Logarithm Def: The logab is a unique real exponent, indicating the power to which the base must be raised to produce the number b while b>0 ; a>0 ; a≠1 (alogab =b) Properties of function: • Domain: set

of positive real numbers (Df=R+) • Range: set of real numbers (Rf=R) • It is a one to one function thus it has an inverse function (x ax) • Zero: x=1 • The graph of the function intersects the x axis at point (1;0) • When a>1 it is smi • When 0<a<1 it is smd • It has no extrema a>1 smi log function 0<a<1 smd log function loga1=0 logaa=1 Identities: 1. logabc=logab+logac 2. loga(b/c)= logab-logac 3. logabn=nlogab 4. logab= logcb/ logca Theorem: logabc=logab+logac Proof: 1. alogabc=bc ↑ by definition 2. alogab+logic =alogab x alogac=bc ↑identities for power logabc a ↓↓ logab+logic =a ↕ because f(x)= ax is smi logabc=logab+logac Q. e d 2006. Mathematics oral exam Page 2. Linear and quadratic functions, equations Linear function: A function f is called a linear function, if f: HR, H≠ Ø, H C R and there exist real numbers a and b such that f(x) = ax+b a∈R, b∈R and a≠0 When a>0, the function is strictly monotone increasing,

when a<0 it is strictly monotone decreasing When H=R, the graph of a linear function is a line, whose slope is a. and the y intercept is (0;b) When b =0 the linear function is a direct variation Quadratic function: A function f is called a quadratic function, if f: HR, H≠ Ø, H C R and there exist real numbers a,b and c such that f(x)=ax2+bx+c, where a, b, c ∈R, a ≠ 0 When H=R, the graph of the quadratic function is a parabola, whose axis is parallel to the y axis. Extremum: The function f has a maximum at value x0 in the domain, if for every element x of the domain f(x) ≤ f(x0) The function f has a minimum at value x0 in the domain, if for every element x of the domain f(x) ≥ f(x0) The discriminant of the quadratic equation: D= b2 – 4ac The value of D discriminant, or points out three different cases: • When D<0 the equation has no root • When D=0 the equation has one (double) root: x=-b/2a and x1=x2. The factor form is a(x- x1)2 =0 • When D>0 the equation

has 2 distinct real roots. Thus the discriminant determines the number and type of roots. The factor form of the quadratic formula: a(x- x1) (x- x2)=0 Proof: Theorem: the solutions of the quadratic function are x1, 2 = − b ± b 2 − 4ac where ax2+bx+c, and a, b, c ∈R, a ≠ 0 2a Proof: The standard form of a quadratic equation is given by ax2+bx+c, and a, b, c ∈R, a ≠ 0 Solving this equation by completing the square gives a formula for finding all real number solutions of any quadratic equation. ax2+bx+c Factor the equation: b c x+ )=0 a a Rewrite the expression in parentheses in order to get a perfect square: b2 c b a (( x + ) 2 − 2 + ) = 0 a 2a 4a Add the two last fractions: b b 2 − 4ac a(( x + ) 2 − )=0 a 4a 2 When b2 – 4ac≥0 you can continue as follows: a(x 2 + b 2 b 2 − 4ac 2 ) −( ) )=0 2a 2a Factor the left side by using the identity a2-b2=(a+b)(a-b) a (( x + b b 2 − 4ac b b 2 − 4ac a( x + + )( x + − )=0 2a 2a 2a 2a The product equals zero

if either factor is zero. We know that a≠0 thus b b 2 − 4ac x+ + = 0 or 2a 2a b b 2 − 4ac − =0 2a 2a The two roots of the equation: x+ − b − b 2 − 4ac − b + b 2 − 4ac or x 2 = x1 = 2a 2a Summarizing the solutions: − b ± b 2 − 4ac 2a This is the equation of Quadratic formula. It gives us the roots of the quadratic eqution, is a ≠ 0 and b2 – 4ac≥0. Q. e d x1, 2 = Applications: In Math: Trigonometric, logarithm, exponential, and square root equations that can be originated with quadratic equation, solving extremum problems, simplifying fractions In Physics: uniformly accelerating motion, projection, free fall, circular motion, s-t functions 11. Function analysis using basic methods and differential calculus BASIC FUNCTION ANALYSIS LINEAR FUNCTIONS: Linear functions are most commonly in the form f ( x) = ax + b , where a, b ∈ R . • The sign of a shows whether we have an increasing or decreasing function. • The absolute value of a shows how much the

function is distorted (stretched or compressed) horizontally. • The value of b shows how much we have to translate our function vertically, in the opposite direction as the sign of b. b • The x intercept will be at point x0 = − . a • There is no extremum. QUADRATIC FUNCTIONS: Quadratic functions are most commonly in the form f ( x) = ax 2 + bx + c , where a ≠ 0 and a, b, c ∈ R . Let us convert this to the complete square form: 2  c b  b2 c  2 b f ( x) = ax + bx + c = a x + x +  = a  x +  − 2 + = 2a  a a a  4a   2 2  b  b 2 − 4ac  b  b 2 − 4ac  − a x − = + = a  x +     2a  2a  4a 4a 2    The sign of a shows whether it is an upright or a downward parabola. The absolute value of a shows how much the parabola is contracted or stretched (horizontally). b The vertex of the parabola will be at the value x0 = − . 2a  b b 2 − 4ac  . The

coordinates of the vertex will be (x; y ) =  − ;− 4a   2a 2 • • • • • • − b ± b 2 − 4ac The x intercept(s) – if there are any – will be at point(s): x1, 2 = . 2a b ): The extremums will be the following (in both cases at value x0 = − 2a o minimum if a > 0 o maximum if a < 0 1 VIEWPOINTS OF BASIC FUNCTION ANALYSIS: • • • Domain : Range: Monotonity: • • Zero values: Extremums: • Bounds: • Convexity: • Parity: • Periodicity: The set of all input values of a function. The set of all possible output values of a function. A function is monotonic increasing [decreasing] when for any x1 < x 2 it is true that f ( x1 ) ≤ f ( x 2 ) [ f ( x1 ) ≥ f ( x 2 ) ]. A function is strictly monotonic increasing [decreasing] when for any x1 < x 2 it is true that f ( x1 ) < f ( x 2 ) [ f ( x1 ) > f ( x 2 ) ]. The values of the range, where f ( x) = 0 . The function has a minimum [maximum] at x1 [ x 2 ] if

there is no smaller [greater] value of the range than f ( x1 ) [ f ( x 2 ) ]. A local extremum is an extremum within an interval, though this might coincide with the (global) extremum. A function is upper (lower) [or both] bounded if there exist a K number that f ( x) ≤ K ( f ( x) ≥ K ) [ f ( x) ≤ K ]. A function is convex if a segment connecting any two points of the function is above the function, else it is concave. A function is even if f (− x) = f ( x) , that is, it is symmetrical to the y axis. A function is odd if f (− x) = − f ( x) , that is, it is symmetrical to the origin. A function is periodical if there exists a p number that f ( x − p) = f ( x) = f ( x + p) . FUNCTION TRANSFORMATIONS: Transformations of the function value: f ( x) + c The function is shifted upwards (if c > 0 ) or downwards (if c < 0 ) along the y axis by |c| units. The function is stretched (if c > 1 ) or compressed (if 0 < c < 1 ) along the y axis c ⋅ f (x) c-times. The

function is reflected over the x axis. − f (x) Transformations of the variable: f ( x + c) The function is shifted left (if c > 0 ) or right (if c < 0 ) along the x axis by |c| units. f (c ⋅ x) The function is compressed (if c > 1 ) or stretched (if 0 < c < 1 ) along the x axis 1 -times. c The function is reflected over the y axis. f (− x) DIFFERENTIAL CALCULUS Differentiation expresses the rate at which a quantity (y) changes with respect to the change in another quantity (x), on which it has a functional relationship [that is, y is a function of x, and differentiation expresses the rate at which y changes in respect to the change in x]. 2 DEFINITIONS: def: Differentia quotient: (also: Difference quotient) Let f be a real function with domain Df. Let x0 be an inner point of the domain. The differentia quotient of function f at point x0 is f ( x) − f ( x0 ) , where x ∈ D f {x0 } x − x0 [that is: the differentia (difference) quotient is the slope of the

line connecting points ( x0 ; f ( x0 )) and ( x; f ( x)) ]. def: Differential quotient: If the finite real limit of the differentia quotient of function f in point x0 exists [that is, function f can be differentiated in point x0] the differential quotient of function f at point x0 is f ( x) − f ( x0 ) lim x x0 x − x0 [that is: the differential quotient is the slope of the tangent drawn to function f at its point ( x0 ; f ( x0 )) ]. THEOREMS: 1) Continuity: Differentiability implies continuity, but not vice versa [that is, continuity is a necessary, but not sufficient condition of differentiability]. 2) Rules of differentiation: If functions f(x) and g(x) are differentiable in point x0, then the following differential quotients exist in the same point, and the following relations are true: • • • • • (c ⋅ f )′ = c ⋅ f ′ , where c is a constant ( f ± g )′ = f ′ ± g ′ ( f ⋅ g )′ = f ′ ⋅ g + f ⋅ g ′ ′ f ( f ′ ⋅ g − f ⋅ g ′) ,

where g ( x ) ≠ 0   = 0 g2 g ( f  g )′ = ( f ′  g ) ⋅ g ′ [chain rule] 3) The derivative of some basic functions: • f ( x) = c n • f ( x) = x f ( x) = a x f ( x) = e x • f ( x) = log a x • f ( x) = ln x • • f ′( x) = 0 f ′( x) = nx n −1 • • f ( x) = sin x f ( x) = cos x f ′( x) = a x ln a f ′( x) = e x 1 1 f ′( x) = ln a x 1 f ′( x) = x • f ( x) = tan x • f ( x) = cot x 3 f ′( x) = cos x f ′( x) = − sin x 1 f ′( x) = cos 2 x 1 f ′( x) = − 2 sin x 4) Monotonity: If function f (x) can be differentiated on the interval [a; b] , and the derivative function of f (x) is positive [negative] on this interval, then f (x) is strictly monotonic increasing [decreasing] on the interval. 5) Convexity: If function f (x) can be differentiated twice on the interval [a; b] , and the second derivative function of f (x) is positive [negative] on this interval, is convex then f (x) [concave] on the

interval. 6) Extremum: If function f (x) can be differentiated on the interval [a; b] , and at point x0 it has a local extremum then it is true that f ′( x0 ) = 0 (it is a necessary, but not sufficient condition). 7) Extremum: If function f (x) can be differentiated on the interval [a; b] , and at point x0 f ′( x0 ) = 0 and the derivative changes signs then there is a local extremum of the function at point x0. 8) Theorem: The function f ( x) = x n (where n is a positive whole number) can be differentiated at all real values of x, and that will be: ( x n )′ = nx n −1 Proof (by means of total induction): • for n=1, it is true ( x ′ = 1 ) • Lets suppose that the same is true for n, and let us show that it is true for n+1 as well. The condition of induction: ( x n )′ = nx n −1 • We know that x n +1 = x ⋅ x n , and by using the rule of product differentiation, we get: ( x ⋅ x n )′ = x ′ ⋅ x n + x ⋅ ( x n )′ = 1 ⋅ x n + x ⋅ n ⋅ x n −1 = (n + 1) ⋅

x n • We have proved the formula from n to n+1, thus it will be true for all positive whole power (in fact, it will be true for all real powers). 4 FUNCTION ANALYSIS USING DIFFERENTIAL CALCULUS 1. Let us differentiate function f (x) at least twice (thrice, if possible) thus we get f ′(x) , f ′′(x) and f ′′′(x) . 2. Let us find the zero values of f ′(x) and f ′′(x) 3. The function will be increasing on an interval if f ′( x) > 0 and decreasing if f ′( x) < 0 on the same interval. 4. The function will be convex (increasingly increasing or decreasing) on an interval if f ′′( x) > 0 and concave (decreasingly increasing or decreasing) if f ′′( x) < 0 on the same interval. 5. The (local) extremums (minimums and maximums): the zero values of f ′(x) , such that they satisfy the sufficient conditions of an extremum [below]. 6. The inflection points: those zero values of f ′(x) that did not satisfy the sufficient conditions of an extremum,

that is, are not extremums; and the zero values of f ′′(x) , such that they satisfy the sufficient conditions of an inflection point [below]. The necessary condition of an extremum: f ′( x) = 0 . The sufficient condition of an extremum: Function f ′(x) changes signs before and after its zero value, OR f ′′( x) ≠ 0 (minimum if f ′′( x) > 0 and maximum if f ′′( x) < 0 ). The necessary condition of an inflection point: f ′′( x) = 0 . The sufficient condition of an inflection point: Function f ′′(x) changes signs before and after its zero value, OR f ′′′( x) ≠ 0 . Note 1: A function may have an extremum at an inner point if it is not differentiable at that given point. Note 2: A function may have an extremum at the endpoints of its domain if is an interval. APPLICATIONS OF DIFFERENTIAL CALCULUS 1. Physics (elaborated): • The velocity(time) function [ v(t ) ] is the first derivative of the displacement(time) function [ s (t ) ], where we

differentiated by t. • The acceleration(time) function [ a(t ) ] is the first derivative of the velocity(time) function [ v(t ) ], thus the second derivative of the displacement(time) function [ s (t ) ], where we differentiated by t. • v(t ) = s ′(t ) , and also a(t ) = v ′(t ) = s ′′(t ) . 2. Economics: The Marginal Utility function is the first derivative of the Total Utility function. The flexibility of functions (how much does a function change in respective to the change of one [or more] of its components). 3. Error calculus (in sciences) 4. (Math problems regarding differentiation, for example function plotting, finding extremums, inflection points and the tangent of a function.) 5 13. Right angle triangles Right-angle triangles Definitions: In a right-angled triangle the ratio of the sides is equal to the ratio of the sin of the opposite angles. a:b:c=sinα:sinβ:sinγ While γ=90° , sinγ=1 , so a:b:c=sinα:sinβ:1 Construct right triangles with one angle

α. These triangles are similar, because they have two congruent angles. Therefore, the ratios of the corresponding sides are equal, and these ratios only depend on the value of α. These ratios are called trigonometric ratios, and they have the following definitions: sinα= a/b or length of opposite leg/length of hypotenuse cosα= b/c or length of adjacent leg/length of hypotenuse tanα= a/b or length of adjacent leg/length of opposite leg cotα= b/c or length of opposite leg/length of adjacent leg Theorems The mostly used theorem is the Pithagorean Theorem The sum of the squares drawn on the legs of a right angle triangle is equal to the square drawn on the hypotenuse. In algebra: a2+b2=c2 where a and b are the legs of the right angle triangle, and c is the hypotenuse. So the Pithagorean Theorem says that in a right angle triangle the sum of the squares of the legs is equal to the square of the hypotenuse. Statement: For proving the Pithagorean Theorem it is enough to

prove the Cosine Theorem In any triangle we can get the square of one side, if we subtract two times the other two sides plus the cosine of their enclosed angle from the sum of the two sides square. c2=a2+b2-2abcosγ while we examine right angle triangle, γ=90°, so cosγ=0, so c2=a2+b2 2006. Mathematics oral exam Page 1. 13. Right angle triangles Proof: Let the triangle look like on the picture. We get 3 vectors, thus the following statement is true: c=a–b The square of the equation is also true: c2 = (a – b)2 c2 = a2 – 2ab + b2 Because of the definition of the scalar product : a2 = a a cos0° = a 2 1 = a 2 = a2 similarly: b2 = b2 and c2 = c2 a b = a b cosγ = a b cosγ Substituting into the equation above: c2 = a2 + b2 – 2ab cosγ We used that a is not equal to 0, b is not equal to 0 and c is not equal to 0, which are true, because they are the sides of the triangle. q.ed Converse of the Pithagorean theorem If the sum of the

squares of the length of two sides of a triangle is equal to the square of the length of the third side, then the triangle is a right triangle. Proof -Construct a triangle ABC with sides a, b, and c. -The equation a2+b2=c2 is given. -Construct a right triangle A’B’C’ with legs a and b. -Its hypotenyuse is c’. -Applying the Pythagorean theorem we obtain a2+ b2= (c’)2. From the two equations it follows that c2=(c’)2. Since c>0 and c’>0, we can conclude that c=c’. Thus the two triangles are congruent, and triangle ABC is a right triangle, as well 2006. Mathematics oral exam Page 2. 13. Right angle triangles q.ed Thales theorem The two endpoints of a diameter of a circle and az arbitrary point on the circle (not on the diameter) can be connected to form a right-triangle. The diameter is the triangle’s hypotenuse Proof: -Construct a circle with centre O. -Find the diameter and call it AB. -Choose an arbitrary point on the circle (not on the diameter). Call

this point C -Draw OC radius. Triangle AOC and triangle BOC are isosceles, thus the angles on their bases are equal α and β. α + β + (+β)=180o 2α + 2β = 180o 2(α + β)=180 α + β =90o So the triangle is right angled. q.ed Converse of the Thales theorem The circumcentre of a right triangle is the midpoint of the hypotenuse. Proof: -Construct a right triangle ABC. -We have to prove that midpoint F is equidistance from all vertices: - Reflect the triangle around midpoint F to get a parallelogram AC’BC. AC’BC is a rectangle because it has a right angle. Applying theorem that the diagonals of the rectangle bisect each other and have equal length  AF=FB=FC So we get that midpoint F is equidistant from A and B and C. q.ed 2006. Mathematics oral exam Page 3. 13. Right angle triangles Theorem of altitude: The altitude of the hypotenuse is the geometrical mean of the two parts of the hypotenuse divided by the altitude. m=√pq Proof: ABC ~ CTB ~ ATC , because their

angles are equal, so m/p = q/m m2 = pq so m = √pq q.ed 2006. Mathematics oral exam Page 4. 13. Right angle triangles Theorem of the legs: Any leg of the right angle triangle is the geometrical mean of the hypotenuse and the orthogonal projection of the leg to the hypotenuse. a = √cp b = √cq Proof: ABC ~ CTB ~ ATC , because their angles are equal So a/c = p/a a² = cp , so a = √cp and b/q = c/ b b² = cq , so b = √cq q.ed The theorem of the legs is a possible proof for the Pithagorean Theorem: a² + b² = cp + cq a²+b²=c(p+q) a²+b²=c² Application of right triangles: Area, Volume calculations In Physics: - Mechanics, dividing forces into components - Slope, lever - Pythagorean numbers Constructing tangent (geometry). Trigonometry Calculating distances in geometry. 2006. Mathematics oral exam Page 5. Perpendicular bisector Theorem: The perpendicular bisectors of a triange intersect each other in one point. Proof: The locus of points equidistant from A and B

is the perpendicular bisector of segment AB. The locus of points equidistant from C and B is the perpendicular bisector of segment BC e and f are not in the same line they are not parallel therefore intersect each other Their intersection point is at the same distance from A and B + from C and B as well it is the same distance from A and C M is on the perpendicular bisector of segment AC since M is at equal distance from all verteces, it is the centre of the circumscribed circle if the triangle has • an obtuse angle (tompaszögű háromszög), then the centre of the circumscribed circle is outside the triangle • acute angles only, then the centre is inside the triangle • a right angle, then the centre is on the hypotenuse (Thales theorem) Angle bisector The locus of points equidistant from two given arms of an angle is an angle bisector Theorem: The angle bisectors of a triangle intersect each other in one point Proof: Line fα is an angle bisector of angle BAC:

all points in f are equidistant from sides AB and AC Line fβ is the angle bisector of angle ABC: all points in e are equidistant from sides AB and BC The two bisectors intersect each other in the triangle This point N is at the same distance from sides AB and BC and from AC and BC as well it is equidistant from sides AC and BC as well N is on the angle bisector of ACB since N is at equal distance from all sides it is the centre of the inscribed circle Theorem: If the area of a triangle is T, and the circumference is C then the radius of the inscribed circle is 2T/C Proof: • If we draw perpendiculars from the centre of the inscribed circle to the sides, the length of these segment will be the radius of the inscribed circle • Therefore the area of the three triangles ABO, ACO, BCO is (c⋅r)/2, (a⋅r)/2 and (b⋅r)/2 • The sum of these areas is eyual to the area of the original triangle (ABC) • r (a+b+c)=2T therefore r=2T/ (a+b+c) = 2T/ C Theorem: An angle bisector

divides the opposite side in the ratio of the neighbouring sides b1/b2 = a/c We enlarge side c with side a at vertex B connecting D and C we get an isoceles tringle The external angle of B in DBC triangle is β, therefore the angles on the base are β/2 therefore they are congruent the angle bisector of B is parallel with DC Applying the párhuzamos szelők tétele we get the original statement An internal angle bisector and the external angle bisectors of the two other verteces meet in one point, which is the centre of the ‘hozzáírható’ circle. The ‘hozzáírható circle is the circle that touches the extension of two sides and the third side of the triangle. Altitude An altitude of a triangle is a segment drawn from a vertex of a triangle perpendicular to the line containing the opposite side of the triangle Theorem: The altitudes intersect each other in one point. Proof: Draw lines parallel to each of the sides through the opposite vertices to form a new

triangle A’B’C’ The quadrilaterals (ABCB’) etc. are paralellograms thus A is the midpoint of B’C’, B is the midpoint of A’C’, and C is the midpoint of A’B’ The sides of the triangle are the midlines of the new triangle From the construction of the triangles we know that the altitudes are the perpendicular bisectors of the sides of the new triangle Applying the theorem that the perpendicular bisectors are concurrent we can conclude that the altitudes of the original triangle are concurrent as well Midlines A midline is the segment connecting the two midpoints of the sides of the triangle Theorem: A midline of a triangle is parallel to the third side and its length is the half of the length of the third side. Proof: F is the midpoint of AC and G is the midpoint of BC Triangle CFG is similar to CAB (having an angle and the ratio of the neighboring sides (1:2) in common) Thereofre the length of FG is the half of AB, and because of the similarity the

corresponding angles are equal, AB is parallel to FG Medians A median of a triangle is a segment from a vertex of a triangle to the midpoint of the oppposite side. Theorem: The medians of a triangle meet in a single point which is two thirds of the distance from each vertex to the midpoint of the opposite side. This point is called the centroid of a triangle, which is the centre of gravity of a triangle. Proof: E is the midpoint of AC and F is the midpoint of BC, point G is the midpoint of AB Medians BE and AF meet at point S EF is half as long as AB and is parallel to AB Angles FAB and AFE, and ABE and BEF are pairs of alternate angles, therefore they are congruent ABS and FES triangles are similar the ratio of similitud is 2:1 (because of the ratio of AB and EF The proof is the same for the other medians as well, therefore point S is 2/3 of the distance from each vertex to the midpoint of the opposite side. Euler-line The orthocentre (Q), the centroid (G) and the

centre of the circumscribed circle (O) lay on a single line, the Euler-line. The distance between the circumcentre and the centroid is half of the distance of the centroid and the orthocentre. Proof: Applications: In mathematics: • finding the area of the triangle (using the radius of the inscribed circle) In physics: • finding the point of support of a homogen triangle in everyday life • we want to build a railway station at equal distance from three villages 16. Cyclic, circumscribable and symmetric quadrilaterals Cyclic, circumscribable and symmetric quadrilaterals Definitions: Polygon: a closed plane figure bounded by straight sides Quadrilateral: four sided polygon, if the four vertexes are coplanar, it is called plane quadrilateral, and otherwise it is non plane quadrilateral. Inscribed circle: a circle drawn into a polygon such that the sides are all tangents to it. Circumscribed circle: a circle drawn around a polygon, such that it contains every vertex of the

figure. Chord: a straight line joining two points on the circumference. The longest chord in a called a diameter. The diameter passed through the centre Tangent: if a straight line and a circle have only one point of contact, then that line is called a tangent. A tangent is always perpendicular to the radius drawn to the point of contact Classification of plane quadrilaterals: A quadrilateral is plane quadrilateral if every vertex is coplanar. Concave polygon: A polygon such that there is a straight line that cuts it in four or more points Convex polygon: A polygon such that no side extended cuts any other side or vertex; it can be cut by a line in at most two points On the base of the parallel sides: - Have two parallel sides: trapeziums Those trapeziums that have perpendicular line of symmetry to the base are isosceles trapeziums. - Two-two sides are parallel: parallelograms On the base of side lengths: - Two-two opposite sides are equal: parallelograms o Both sides are equal:

rhombus - Two-two neighboring sides are equal: kites Special quadrilaterals: - Cyclic quadrilaterals - Circumscribable quadrilaterals - Symmetric quadrilaterals 2006. Mathematics oral exam Page 1. 16. Cyclic, circumscribable and symmetric quadrilaterals 1. Cyclic quadrilaterals Definition: A quadrilateral is cyclic if it can be inscribed in a circle, that is, if its four vertices belong to a single, circumscribed, circle. Theorem: A quadrilateral is cycle if and only if the sum of any of the two opposite angles is 180°. Proof: I. In every cyclic quadrilateral the sum of any of the two opposite angles is 180°. AOC angle =2β (Since it is the central angle of the circumferential angle of ABC) COA angle=2δ (Since it is the central angle of the circumferential angle of ADC) O-point: 2β+2δ=360o ⇓ β+δ=180o Q.ed II. If the sum any of the two opposite angles is 180° in a quadrilateral, than it is cyclic. C” D 180°- 180°-α  theorem above C 180°- C’ A

α C δ 180°-α  theorem above B First case: Let’s assume that C’ is inside the circumscribed circle of ABD triangle. Now DC’B angle is 180°- α since DAB angle is α. The line DC’ intersects the circle in point C We know from the proven theorem above, that ABCD is now cyclic, so C’CB angle is 180°- α. In the triangle BCC’, DC’B angle is an exterior angle, and we know it is the sum of the two interior angle not neighboring it. From this aspect, C’BC angle should be zero, that can only be, if C’≡C. Second case: Similarly to the first case, but now assume that C” is out of the circle. C”B intersects the circle in point C. DCB angle is 180°- α, which is an exterior angle of triangle DCC”, so again C”DC angle should be zero. That can only be if C”≡C Q.ed 2006. Mathematics oral exam Page 2. 16. Cyclic, circumscribable and symmetric quadrilaterals 2. Circumscribable quadrilaterals Definition: A quadrilateral is circumscribable if it has an

inscribed circle (that is, a circle tangent to all four sides). Theorem: A quadrilateral is circumscribable if and only if the sums of the length of its pair opposite sides are equal. Proof: I. In every circumscribable quadrilateral the sums of the length of its pair opposite sides are equal. Using the theorem, that the lengths of the tangents drawn from the same point are equal, we can set up a pair of equations: AB AD + DC = x + y + k + z + BC = x + k + y + z We know that addition is commutative, so AB + DC = AD + BC Q.ed II. If in a convex quadrilateral, the sums of the length of its pair opposite sides are equal, than it is circumscribable. From the given condition, a + c = b + d. Let a be the longest side, from that b and d are convergent. If there were two equal sides ( a and c ), from the condition and from that a is the longest, they cannot be in front of each other. Side a and the extended line of b and d define a circle k. Let’s make an assumption that c does not

touch k Now we have two cases: First is when c intersects k Second: c and k has no point of intersection c c’ c’ c d’ b’ b d k b’ b d d’ k a a Move line c parallel to herself until touches k. The quadrilateral that we get now is circumscribable, so a + c’ = b’ + d’. In the first case, c’ < c and b’ > b, d’ > d. From that, a + c > a + c’ = b’ + d’ > b + d which contradicts to the given a + c = b + d condition. With the same method, we get to a contradiction from case two as well. Now c’ > c, b’ < b and d’ < d. Then a + c < a + c’ = b’ + d’ < b + d, which also contradicts to the given a + c = b + d condition. Q.ed 2006. Mathematics oral exam Page 3. 16. Cyclic, circumscribable and symmetric quadrilaterals 3. Symmetric quadrilaterals Linear symmetry - Isosceles trapezium • The two sloping sides are the same length • There is one line of symmetry • Its non-parallel sides are equal • Two pairs of

adjacent angles are equal. - Kite • One pair of opposite angles equal and two pairs of adjacent sides equal. • One line of symmetry. It has no rotational symmetry Rotational symmetry - Parallelogram • Opposite sides are equal and parallel, • Opposite angles are equal. • A parallelogram has no lines of symmetry but it does have rotational symmetry. Linear and Rotational symmetry - Rhombus • • • • The opposite angles are equal The opposite sides are parallel All four sides are the same length. Two lines of symmetry (the diagonals), and a rotational symmetry around the intersection point of diagonals - Square • all four sides are equal • all four angles are right angles. • four lines of symmetry and rotational symmetry - Rectangle • • • the opposite sides are equal all four angles are right angles two lines of symmetry and rotational symmetry 4. Applications - Regular polygons Area and volume calculation Maximum area problems Architecture, regular shapes,

Center of mass in physics 2006. Mathematics oral exam Page 4. 21. Circle and parabola in the coordinate system Circle and parabola in the coordinate system The Coordinate-system: In the Coordinate-system of Descartes the 2 axes are perpendicular to each other. These are called x (abscissa) and y (ordinate) axes. Using this system we can define any point in the plane, only by giving its coordinates (form: P(x;y)). In the same way, we can define the equation of lines or plane figures, like circles or paraboles. The circle Definition o f circle: A c ircle i s t he s et of e lements on a pl ane, w hich ar e at e qual di stance from a point. Equation of a circle when the centre and radius are given Theorem: The equation of a circle in the coordinate-system with centre C(u;v) and radius r: (x-u)2 + (y-v)2 = r2 (Note: if u = 0 and v = 0 (the origin is the centre), then equation is: x2 + y2 = r2) the Proof: Let C(u;v) be centre of the circle, and P(x,y) be the element of the circle.

By the definition of circle the length of CP is constant, and equal to r. The distance of points C and P can be written as follows: Both sides are positive, so raising to the square is an equivalent transformation. (x-u)2 + (y-v)2 = r2 For this reason the solutions of this equation are only those (x;y) pairs of numbers, which are the coordinates of a circle with centre C(u;v) and radius r. Quadratic equations with 2 variables and the circle The equation of the circle can be factorized as follows: (x-u)2+ (y-v)2 = r2 x2-2ux+u2+y2-2vy+v2-r2=0 2 x + y2 – 2ux –2vy + (u2 + v2 – r2) = 0 In general: Ax2+Ay2+Bx+Cy+D=0 where A is the coefficient of x and y2, B=-2u, C=-2v and D=(u2 + v2 – r2) (D is constant) 2 For this reason a quadratic equation belongs to a circle if and only if the coefficients of x2 and y2 are equal and the equation has no xy term. 2006. Mathematics oral exam Page 1. 21. Circle and parabola in the coordinate system Relative position of a circle and a line

The number of common points of a circle and a line depends on the number of solutions of the set of equations, consisting of the equation of the circle and that of the line: In case of:  2 solutions: they intersect each other;  1 solution: the line is the tangent of the circle;  No solution: they do not have a common point. Relative position of 2 circles Similarly to the relative position of a circle and a line:  2 solutions: they intersect each other;  1 solution: they have 1 common point (1 circle can be inside the other);  No solution: no common point (1 can be inside the other). Tangent to a circle from an outer point 1 solution: Using Thales-circle: Given the centre of the circle C(u;v), its radius and an outer point P(x;y), the centre of the Thales circle will be the midpoint of CP, and its radius is (the length of OP)/2. The intersection point of the original circle and the Thales circle are the intersection points of the tangents from P. From these points

and the given point P the equation of the 2 tangents can be calculated. The parabola Definition of parabola: A parabola is the set of points on t he plane, which are equidistant from a gi ven l ine – the d irectrix – and a gi ven point – the f ocus. T he di stance of t he f ocus point and t he directrix is called the parameter, the midpoint of the shortest section between the focus and the directrix is the vertex. The equation of the simplest parabola Theorem: If the vertex of a parabola is the origin, its symmetry axis is the y axis and its focus is F(0;p/2), then the equation of this parabola is: x2 = 2py Proof: In this case the equation of the directrix is: y = -p/2 By definition, P(x;y) is the element of the parabola if and only if its distance from the focus and from the directrix is the same: As both sides are positive, raising to the square is an equivalent transformation: x2 +y2 – py + (p/2)2 = y2 + py + (p/2)2 x2 = 2py Q.ed 2006. Mathematics oral exam Page 2.

21. Circle and parabola in the coordinate system Equation of parabolas with parallel axes to the coordinate axes: • • • • • • • If the parabola x2 = 2py is shifted with vector v(u;v), then its equation will be: (x-u)2 = 2p(y-v) 2 If the parabola x = 2py is rotated about the x axis, then its equation will be: x2 = -2py 2 If the parabola x = -2py is shifted with vector v(u;v), then its equation will be: (x-u)2 = -2p(y-v) If the vertex of a parabola is the origin and its focus is F(p/2;0), then its equation is: y2 = 2px If the parabola y2 = 2px is shifted with vector v(u;v), then its equation will be: (y-v)2 = 2p(x-u) If the parabola y2 = 2px is rotated about the x axis, then its equation will be y2 = -2px If the parabola y2 = -2px is shifted with vector v(u;v), then its equation will be (y-v)2 = -2p(x-u) . Quadratic functions and the parabola If a parabola’s axis is perpendicular to the ordinate axis, then its equation can be written as: y = ax2 + bx + c where a=p/2,

b=-u/p, c=(u2/2p + v), therefore a, b and c are Real numbers, and a≠0. For this reason, the function of quadratic equations is a parabola, which’s axis is parallel to the y axis. (If the parabola’s axis is parallel to the x axis, then it is not a funtion!) Application: Maths: accurate constructions, calculations, areas Other: Vectorgraphics, engineering, architecture, physics: v vs. T graph, projectile motion 2006. Mathematics oral exam Page 3. 23. Area calculation using basic methods and integral calculus BASIC AREA CALCULATION AREA OF POLYGONS: We regard a positive number associated to every polygon, which represents its (surface) area. The following properties must stand: • We need a unit. Let us agree that the area of a square with sides being one (length) unit is our one unit of area. • Identical plane figures must have equal areas. • If we divide a polygon into two polygons, then the sum of the areas of the new polygons must be equal to the area of the original

polygon. (Thus, we can divide it into any finite number of polygons.) Area of a rectangle: (a difficult but rather nice proof, it is not obligatory) • The area of a square with sides n ( n ∈ Ζ + ) is n 2 , because we can divide it into n 2 square units, if we construct lines parallel to the sides, one unit in distance. • The area of a rectangle with sides n and m ( n, m ∈ Ζ + ) is nm, because we can divide it into nm square unit, following the previous sequence of thought. 1 1 • The area of a square with sides ( n ∈ Ζ + ) is 2 , because we can divide the square n n 2 unit into n identical squares, as in the previous two cases. • The area of a rectangle with sides a and b ( a, b ∈ R + ). Let us construct segments 1 parallel to the sides, in distance. Let m and k be numbers ( m, k ∈ N + ) that n 1 1 1 1 m ≤ a < (m + 1) and k ≤ b < (k + 1) n n n n • From the property of area derives that the area of ABCD is bigger than the area of AB’C’D’, but

smaller than AB”C”D”. If we denote the area of ABCD with t, we get 1 1 1 1 m k ≤ t < (m + 1) (k + 1) n n n n 1 1 1 1 • and from m and k, it is clear that m k ≤ ab < (m + 1) (k + 1) n n n n • From these two inequalities, we get m k 1 1 1 a b 1 a + b +1 . t − ab < (m + 1)(k + 1) 2 − mk 2 = 2 + 2 + 2 < + + = n n n n n n n n n t − ab has to be smaller than any positive number, because a + b + 1 is a constant, and we can choose n to be arbitrarily great, thus t = ab. Thus, the area of a rectangle is T = a ⋅ b , where a and b are the sides. • Area of a parallelogram: Let us construct a perpendicular segment to one of the sides, such that it shall contain one of the vertices of the other side. Using this method, we have cut the parallelogram into pieces which fit together to form a rectangle. Thus, the area of a parallelogram is T = a ⋅ ma , where a is a side, and ma is the altitude to a. 1 It is trivial, that we can calculate the area of any

regular quadrilateral with the help of the preceding two methods. Area of a triangle: If we complete a triangle to a parallelogram, we have doubled its area, therefore the area is half the area of the parallelogram. a ⋅ ma Thus, the area of a triangle is: T = , where a is a side, and ma is the altitude for side a. 2 Theorem: (with an easy proof) The area of a triangle can be calculated if two sides and the enclosed angle are known, such that: a ⋅ c ⋅ sin β . T= 2 Proof: • Let us suppose that we know sides a and c and their enclosed angle β, in addition to a ⋅ ma that, the formula T = is known. 2 • • ma , thus we can conclude that ma = c ⋅ sin β , c irrespectively of β being an acute or an obtuse [because sin β = sin (π − β ) ] or even a right [because then sin β = 1 and the two sides serve as altitudes] angle. a ⋅ c ⋅ sin β Substituting this into the known formula we get: T = . 2 In the right angle triangle ∆ ABT sin β = Area of polygons: Every

convex polygon can be divided into triangles, if we connect one vertex with every other vertex. Every concave polygon can be divided into a convex polygon with the lines of the sides lengthened. Then, we can follow the previous sequence of thought AREA OF THE CIRCLE: The area of the circle is T = r 2π . 2 INTEGRAL CALCULUS Integral calculus expresses the area under a curve, that is, the area under a specific function. DEFINITIONS: Let f (x) be a limited function defined on [a; b] , the units: a = x0 < x1 < . < x n −1 < x n = b n S = ∑ mi ( xi − xi −1 ) , def: Lower sum: i =1 where mi is the lower limit of the function on interval [xi-1, xi]. n S = ∑ M i ( xi − xi −1 ) , def: Upper sum: i =1 where M i is the upper limit of the function on interval [xi-1, xi]. def: Definite integral: We regard function f defined on [a; b] integrable. As we refine the units of partitioning, the upper and lower sums tend towards the same number. If there exists only

one number, which is not smaller than any of the lower sums, and not greater than any of the upper sums of function f¸ then this number is called the definite integral of function f on interval [a; b] . def: Integral function: If function f can be integrated on the interval [a; b] , then function F is the integral function of f: f : [a; b] R def: Primitive function: A primitive function of function f (x) (provided that it exists) is a function, whose derivative is f (x) . def: Indefinite integral: The set of primitive functions of f (x) . THEOREMS: 1) Rules of indefinite integral: If functions f(x) and g(x) are integrable on interval [a; b] , then the following differential quotients exist in the same point, and the following relations are true: • a ∫ f ( x)dx = 0 a • b ∫ a a f ( x)dx = − ∫ f ( x)dx b 3 • • b b a a ∫ cf ( x)dx = c ∫ f ( x)dx b b a • b ∫ ( f ( x) ± g ( x))dx =∫ f ( x)dx ± ∫ g ( x)dx a a c b c a a b ∫

f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx 2) If function f (x) is continuous at every point of interval [a; b] , then F (x) , the integral function of f (x) is differentiable at every point of [a; b] , and F ′( x) = f ( x) 3) The primitive functions of a given function may only differ in a constant. 4) Newton-Leibniz theorem: if f (x) is a continuous function defined on interval [a; b] , and F (x) is a primitive function of it, then: b ∫ f ( x)dx = F (b) − F (a) a 5) The primitive functions of some basic functions: • f ( x) = c F ( x) = cx • • f ( x) = x n where n ≠ −1 1 f ( x) = x • f ( x) = a x • f ( x) = e x F ( x) = • f ( x) = log a x n +1 x , n +1 • • • 1 (x ln x − x ) ln a f ( x) = ln x F ( x) = x ln x − x f ( x) = sin x F ( x) = − cos x f ( x) = cos x F ( x) = sin x • f ( x) = tan x F ( x) = − ln cos x • f ( x) = cot x F ( x) = ln sin x F ( x) = F ( x) = ln x 1 x a ln a F ( x) = e x F ( x) = 6) The volume of a body of

revolution: b V = π ∫ ( f ( x) ) dx 2 a APPLICATIONS OF INTEGRAL CALCULUS 1. 2. 3. 4. 5. Finding the area under a function. Area, surface area and volume calculation. Calculations with bodies of revolution. Finding the centroid of figures. Finding the inertial momentum of bodies (of revolution) rotating around an axis (for example a solid plate). 6. Using it in sciences like mathematics, geography (land survey) and physics 4 Combinatory, Probability Combinatory Permutation: in how many ways we can arrange n elements P=n!=1*234n 1!=1 0!=1 Proof by total induction: n=1: 1!=1 possibility: a n=2: 2!=2 possibilities: ab, ba n=3: 3!=6 possibilities: abc, acb, cab, bac, bca, cba Now, we have to prove that if the statement stands for n=k, then it is true for n=k+1 as well Let’s t ake n el ements. W e can ar range t hem i n n ! w ays If we ad d o ne m ore el ement, t his element can fit into each order in n+1 ways. Therefore n!*(n+1)=(n+1)! Repeated p ermutation: i n h ow m any

ways c an we ar range n elements i f t here a re i dentical elements as well Pi=n!/(k1!*k2!k3!kn!) Variation: choosing k elements from n elements (order is important) Vnk=n*(n-1)(n-2)(n-(k-1))=n!/(n-k)! Repeated v ariation: c hoosing k e lements f rom n e lements ( order i s i mportant a nd w e can choose one element more than once) Vink=nk Combination: choosing k elements from n elements (order is not important) We can choose k elements from n elements n!/(n-k)! times, but since order is not important, we have to divide by k! (we can arrange the chosen element this many ways). Cnk=n!/((n-k)!*k!) Repeated combination: choosing k e lements from n e lements (order is not important and we can choose one element more than once) Cink=Cn+k-1n-1 Theorems in combinatory: Statement: (Cnk)= (Cnn-k) Proof: n!/((n-k)!*k!)=n!/(k!(n-k)!) Statement: (Cnk)+ (Cnk+1)= (Cn+1k+1) Proof by examining the two cases in which the extra element is either chosen or not Binomial coefficients: 1 11 121 1331

14641 n=0 n=1 n=2 n=3 n=4 (sum of the coefficients) 1 2 4 8 16 Binomial theorem: (a+b)n=(0n)anb0+(1n)an-1b1++(nn)a0bn Statement: (0n)+(1n)+(2n)++(nn)=2n Proof: (1+1)n= we use the binomial theorem Statement: (0n)-(1n)+(2n)-+/- (nn)=0 Proof: (1-1)n= we use the binomial theorem Probability p(A)= good events / all events p(A)+p(Ā)=1 1st axiom: 0≤p(a)≤1 2nd axiom: p(certain event)=1; p(impossible event)=0 3rd axiom: p(AUB)=p(A)+p(B)-p(A∩B)  p(AUB)=p(A)+p(B) if p(A∩B)=0. The events are independent if and only if p(A∩B) = p(A)*p(B). Conditional probability of an event: p(A│B) = p(A∩B)/p(B) Total probability of an event: p(A) = ∑p(A│Bk)*p(Bk) = ∑p(A∩Bk) Binomial distribution of an event: p(A)=(kn)*p(A)kp(Ā)n-k Expected value: E(x)=∑xi*pi Application of combinatory and probability - weather forecast - infrastructure: installing traffic lights, bus stops, etc. - economy: insurance companies: risk calculation banks: risk calculation and interest determination

stock markets: expected price of shares - games: pools, lottery, card games - biology: dominant and regressive properties - demography: life-expectancy - mathematics: binomial theorem (for bracket expansion) Definition of root, nth root, Identities Square root of a non-negative real number ’a’ (a≥0) is a unique non-negative number (√a) whose square is ‘a’ a≥0 and √a≥0 √a2 = a (a2 is always a non-negative number) The absolute value of a real number ‘a’ is a = ‘a’, when a≥0 ‘-a’, when a<0 Nth root For any positive even ‘n’ and non-negative ‘a’ n√a is a non-negative real number whose nth power is ‘a’. For any even ‘n’ and negative ‘a’, the nth root is undefined, since the power of any real number with even exponent is always non-negative. For any odd ‘a’ (n>1), n√a is a real number whose nth power is ‘a’. 3 4 √27=3 √256=4 5 √-32=-2 Nth root of a power a. n √a*b=n√a n√b n>1 and ‘n’ is

an integer - if ‘n’ is even  ‘a’ and ‘b’ are non-negative real numbers. - if ‘n’ is odd  ‘a’ and ‘b’ are arbitrary real numbers. Proof n √a*b = n√a n√b /raising to the nth power by definition the left side is a * b nth power of the right side: (n√a * n√b)n = (n√a)n (n√b)n = a b - if ‘n’ is odd and the nth power of the two sides of the equation are equal, then the two sides are equal if ‘n’ is even and both sides of the equation are non-negative and the nth power of the two sides of the equation are equal, then the two sides of the equation are equal. The nth root of a quotient n√a/b = n√a/n√b, where n>1 and ‘n’ is an integer - if ‘n’ is even  ‘a‘ is a non-negative real number, while ‘b’ is a positive real number. If ‘n’ is odd  ‘a’ and ‘b’ are arbitrary real numbers, b ≠0. Proof n√a/b = n√a/n√b / raising to the nth power by definition the nth power of the left side is

a/b nth power of the right side: (n√a/n√b)n = (n√a)n / (n√b)n = a/b The nth power of a quotient is equal to the quotient of the nth powers of the numerator and the denominator. - if ‘n’ is odd and the nth power of thw two sides of the equation are equal then the two sides of the equation are equal. If ‘n’ is even and both sides of the equation are non-negative and the nth powers of the two sides of the equation are equal  the two sides of the equation are equal. Root of a power (k√a)n = k√an, where k>1 and ‘k’ is an integer and n>1 and ‘n’ is an integer. - if ‘k’ is odd, then ‘a’ is an arbitrary real number if ‘k’ is an even number, then ‘a’ is a non-negative real number. Proof k - √an = k√(a * a . * a) = k√a k√a k√a = (k√a)n └―――――――┘ n times - we used the identity of the root of a product to find the root of a power first find the root, then raise to power. Properties of x√x function -

D is R+ including 0 R is R+ including 0 Zero at x=0 X intercept: x=0 Y intercept y=0 smi function no max min at x=0 with value 0. Note: if ‘n’ is odd, the Domain and the Range of the function are R. Applications: - Hero’s formula - Quadratic formula - Compound interest