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Sets, Vectors & Functions. IGCSE Chapter 8. Note 1 : Sets. A ∩ B is green. A. B. ∩ - intersection U – Union - ‘is a subset of’. A. B. ∩. B. A. ∩. A B. Note 1 : Sets. X. a. C - ‘is a member of’ ‘belongs to’ b є X ξ – ‘universal set’ ξ = { a,b,c,d,e }
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Sets, Vectors & Functions IGCSE Chapter 8
Note 1: Sets A ∩ B is green A B ∩ - intersection U – Union - ‘is a subset of’ A B ∩ B A ∩ A B
Note 1: Sets X a C - ‘is a member of’ ‘belongs to’ bєX ξ – ‘universal set’ ξ = {a,b,c,d,e} - ‘complement of’ b c ξ d a c b e ξ B A ‘ A A U A’ = ξ
Note 1: Sets A a n(A) - ’the number of elements in set A’ A = {x : x is an integer, 2 ≤ x ≤ 9} Reads: A is the set of elements x such that x is an integer and 2 ≤ x ≤ 9 b n(A) = 3 c The set A is {2,3,4,5,6,7,8,9} ∩ Ø or { } - ‘empty set’ Ø A for any set A
Note 1: Sets ξ T R e.g. In the venn diagram 8 5 11 ξ = {students in year 10} 9 13 15 R = {Yr 10 students who play Rugby} 12 17 V = {Yr 10 students who play Volleyball} V T = {Yr 10 students who play Tennis} a.) How many students play Rugby? b.) How many students do not play Volleyball? c.) How many students play Rugby and Tennis? d.) How many students in total? e.) How many play only 1 of these 3 sports? 40 41 14 90 31 IGCSE Ex 1 Pg 242-243
e.g. If X = {1,2,3,..10} ξ Y = {1,3,5,…19} Y X 13 1 4 Z = {x : x is an integer, 5≤x≤11} 15 3 17 Find: 2 5 7 19 10 {1,3,5,7,9} 9 a.) X ∩ Y b.) Y ∩ Z c.) X ∩ Z d.) n(X U Y) e.) n(Z) f.) X’ U Y’ 11 6 8 {5,7,9,11} {5,6,7,8,9,10} Z 15 7 Ø or { } State whether true or false: False a.) 7 εX ∩ Y b.) {5,7,9,11} Z c.) Z X U Y ∩ True IGCSE Ex 2 Pg 244 ∩ True
e.g. Draw and shade this diagram to show the following sets: a.) X∩ Y b.) (X U Y)’ c.) X’ ∩ Y ξ X Y IGCSE Ex 3 Pg 245-246
e.g. Logical Problems If A = { sheep } B = { sheep dogs } C = { ‘intelligent animals’ } D = { good pet } Express the following in set language: a.) No sheep are ‘intelligent’ animals b.) All sheep dogs make good pets c.) Some sheep make good pets A ∩ C = Ø ∩ B D A ∩ D = Ø Interpret the following statements a.) B C b.) B U C = D c.) AεD ∩ All sheep dogs are intelligent animals All sheep dogs and all intelligent animals make good pets Sheep do not make good pets
e.g. Logical Problems Of 27 students in the class, 18 play chess, 15 play piano and 7 do both. How many do neither? C P 27 students = 11 + 7 + 8 + X 7 11 8 27 – 11 – 7 – 8 = X X = 1 1 There is only 1 student who does not play either piano or chess
e.g. Logical Problems The math results from the Hockey Team show that all 16 players passed at least 2 subjects, 8 passed at least 3 subjects and 5 students passed 4 subjects or more ξ How many passed exactly 2 subjects? What fraction passed exactly 3 subjects? 3 2 3 8 8 5 4 3/16 IGCSE Ex 4 Pg 247-249
Note 2: Vectors A vector is a quantity that has both magnitude and direction Vectors can be added using scale drawings. Head to Tail method: a a b a + b Therefore: a + b = b + a b * Notice that this is the same result as b + a
Note 2: Vectors A scalar is a quantity that has magnitude but no direction. e.g. ordinary numbers, & quantities like temperature, mass & volume. We can multiply a vector by a scalar. e.g.x multiplied by 2 gives 2x x x x x x x -3x 2x e.g.x multiplied by -3 gives -3x * The negative sign reverses the direction of the vector
Note 2: Vectors The result of a – b is the same as a + -b e.g.Find 3a – b a a a a a a a a b b b b 3a – b e.g.Find -4a + 2b -4a + 2b
Note 2: Vectors d Starting from Eeach time, find vectors for the following: c e.g. Find: 2c 3c + d -c + d -d c – 4d -2c – 4d EG EA EC EF EL EK C A B E F D G J H I M L K
Note 2: Vectors d The result of a – b is the same as a + -b c e.g. Find: DE DG HJ GB BE GC CG = 2c C A B = c – 2d F E D = c– 2d = 2c + 3d = -c – d G H = 4c + 3d = -4c – 3d J I
Note 2: Vectors Write each vector in terms of a, b and c e.g. • AB DE DC HD FE BE EA DG A C 2a a = -2a a B D = -a-c b c c = -4a F = 2a+c b = b +a a a G H E = -2b ( c ) = -2b + 2a = 2b – 2a
Note 2: Vectors Write each vector in terms of a, bonly e.g. DE HD A C 2a a a B D = -3a+ 2b b c c = 4a– 2b F b a a G H E IGCSE Ex 5 Pg 251-252
Starter OA = a and OB = b e.g. Using the figure, express each of the following vectors in terms of a and/or b AP = OA OB = BQ NP = QN = a a.) AP b.) AB c.) OQ d.) PO e.) PQ f.) PN g.) ON h.) AN i.) BP j.) QA = -a + b = 2b = -2a = -2a + 2b = ½PQ = -a + b Q = 2a + PN = a + b b = a + PN = b B N b = -b + 2a A P O = -2b + a a a
Note 2: Vectors OA = a and OB = b e.g. Using the figure, express each of the following vectors in terms of a and/or b AP = 3OA OB = 1/2BQ NP = QN = 3a a.) AP b.) AB c.) OQ d.) PO e.) PQ f.) PN g.) ON h.) AN i.) BP j.) QA = -a + b = 3b = -4a = -4a + 3b = ½PQ = -2a + 3/2b Q = 4a + PN = 2a + 3/2b b = 3a + PN = a + 3/2b N b B = -b + 4a b A P O = -3b + a a a a a
Note 2: Vectors ABCDEF is a regular hexagon with AB representing the vector s and AF representing the vector t. Find the vector representing AC t s BC = ? = s + t C B s = AB + BC AC A D t = s + s + t E IGCSE Ex 6 Pg 253-255 F = 2s + t
Note 3: Column Vectors The vector AB can be written as a column vector ( ) 2 Horizontal Component (Movement in x direction) AB = 3 Vertical Component (Movement in y direction) F C B ( ) ( ) ( ) -2 5 0 -3 3 0 E A H G D
Note 3: Column Vectors We can easily add column vectors. AB + CD = AB = CD = B = ( ) ( ) ( ) ( ) ( ) + 3 7 3 7 10 C -3 -3 2 2 -1 A AB + CD D * A vector is described by its length and direction, NOT its position
Note 3: Column Vectors Similarly, subtracting column vectors. AB − CD = AB = CD = C = ( ) ( ) ( ) ( ) ( ) − 6 5 6 5 1 -2 -2 1 1 3 AB – CD B D A * A vector is described by its length and direction, NOT its position
Note 3: Column Vectors Multiplying by a scalar 2AB = Each component is multiplied by 2 AB = = ( ) 2( ) ( ) 6 12 6 1 2 1 B 2AB A * A vector is described by its length and direction, NOT its position
Note 3: Column Vectors Vectors are parallel if they have the same direction i.e. Both components must be in the same ratio e.g. Is parallel to IGCSE Ex 7 Pg 257-258 e.g. Is parallel to ( ) ( ) k( ) ( ) ( ) ( ) -10 12 5 a 6 a 4 2 1 b -2 b In general, a vector is parallel if it is a scalar multiple Is parallel to
Note 3: Column Vectors y If A has the coordinate (1,2) and B has the coordinate (6,4), find the column vector for AB B AB = A ( ) ( ) 5 -5 2 -2 Find the column vector for BA x BA =
Note 3: Column Vectors y Given the following diagram, what would be the coordinate of point A such that ABCD is a parallelogram. C D A = (1,0) ( ) ( ) 1 4 1 3 B Notice the column vector for AB = DC x A AD = BC
Note 3: Column Vectors y y = - x y = x Find the image of the vector after reflection in the following lines a.) y = 0 b.) x = 0 c.) y = x d.) y = -x x ( ) ( ) ( ) ( ) ( ) -5 3 5 -5 5 5 -3 -3 3 3 IGCSE Ex 8 Pg 258-259
Note 4: Modulus of a Vector The modulus of a vector a is written a . This represents the magnitude (or length) of the vector As shown in the diagram: B a = ( ) 5 3 3 We can find the length using Pythagoras’ Theorem A 5 |a|= √(52+32) |a|= √34 units
Note 4: Modulus of a Vector e.g. If AB = & BC = , find AC AB + BC = C B = ( ) ( ) ( ) + ( ) ( ) 1 -4 5 -4 5 3 3 1 1 4 We can find the length using Pythagoras’ Theorem A IGCSE Ex 9 Pg 260-261 |AC|= √(12+42) |AC|= √17units
Note 5: Vector Geometry Position vectors give the position of a point relative to the origin, O. In the diagram we see that OD = 2OA, OE = 4OB, OA = a and OB = b b.) Express BA in terms of a and b a.) Express OD and OE in term of a and b respectively d.) Given that BC = 3BA, express OC in terms of a and b c.) Express ED in terms of a and b C OD = 2a OE = 4b BA = -b + a ED = -4b + 2a OC = OB + BC D A a b O E B OC = b + -3b +3a OC = -2b +3a
Note 5: Vector Geometry Position vectors give the position of a point relative to the origin, O. In the diagram we see that OD = 2OA, OE = 4OB, OA = a and OB = b IGCSE Ex 10 Pg 262-264 e.) Express EC in terms of a and b C EC = -6b + 3a EC = -4b – 2b + 3a EC = EO + OC EC = -4b + OC ED = -4b + 2a From part (d) D A a e.) Show that E, D and C lie on a straight line b O E B From part (c) OC = -2b +3a EC and ED are parallel vector that pass through the same point, E
Note 6: Functions Function Notation f(x) = x2+ 3 or f : xx2 +3 We can read this as “the function f, such that, x is mapped on x2 +3 If f(x) = 7x – 5 and g(x) = 2x2 +5, find: a.) f(2) b.) f(-4) c.) g(3) d.) g(0) f(2) = 7(2) - 5 f(-4) = 7(-4) - 5 g(3) = 2(3)2+ 5 g(0) = 0 + 5 f(-4) = -33 g(2) = 23 g(2) = 5 f(2) = 9
Note 6: Functions If f : x and g : x √[5(x+2)], find: a.) x if f(x) = 40 b.) x if g(x) = 10 5x2 2 √[5(x+2)] = 10 = 40 [5(x+2)] = 102 5x2 = 2 x 40 5(x+2) = 100 5x2 = 80 (x+2) = 20 x2 = 16 x = 18 x = ± 4
Note 6: Functions (flow diagrams) We can illustrate functions using flow diagrams. e.g.The function (7x + 8)2 consists of 3 simpler functions 7x x 7x+8 (7x+8)2 multiply by 7 add 8 square
Note 6: Functions e.g. The functions h and k are defined as follows: h: x x2 + 1, k:x ax + b, whereaandbare constants. If h(0) = k(0) and k(2) = 15, find values of aandb h(0) = 02 + 1 = 1 k(2) = 15 a(2) + 1 = 15 IGCSE Ex 11 Pg 265-267 k(0) = 1 a(0) + b = 1 2a + 1= 15 2a = 14 a = 7 b = 1