Core Mathematics 3

Core Mathematics 3

1. Algebraic fraction

   • You can treat algebraic fractions in exactly the same way as numerical ones. You can cancel them down by finding factors that are common to both the numerator and the denomimator
     • Example 1
       • Find the simplest forms of the fractions
       • a) 16 / 20
         • = [4*{4}] / [5*{4}]
           • {} shows that anything inside gets cancelled out or down to a lower number. Ignore them in any functions eg. when times things together
           • what is inside [] shows what is above or below the fraction or what is being times, added or subtracted from each other
           • = 4/5
       • b) [x+3] / [2x+6]
         • = [1*{(x+3)}] / [2*{(x+3)}]
           • = 1/2
       • c) [x+2] / [3x+8]
         • There are no common factors
     • When your algebratic expression has fractions in the numerator or denomiator, it is sensible to multiply both by the same 'number' to create an equivalent fraction
       • Example 2
         • Simplify [1/2x+1] / [1/3x+2/3]
           • = [(1/2x+1)*6] / [(1/3x+2/3)*6]
             • = [3x+2] / [2x+2]
               • = [3(x+2)] / [2(x+2)]
                 • = 3/2 or = 1.5
       • Example 3
         • Simplify the expressions
         • a) [x^2 -1] / [x^2 +4x+3]
           • = [{(x+1)} (x-1)] / [{(x+1)} (x+3)]
             • = [x-1] / [x+3]
         • b) [(x- 1/x ) *x] / [(x+1) *x]
           • = [x^2 -1] / [x(x+1)]
             • = [(x-1){(x+1)}] / [x{(x+1)}]
               • = [x-1] / x
   • You multiply fractions together by finding the product of the numerator and dividing by the product of the denominator
     • Example 4
       • Calculate :
       • a) = [1/2] * [3/5]
         • = [1*3] / [2*5]
           • = 3/10
       • b) [a/b] * [c/d]
         • = [a*c] / [b*d]
           • = ac / bd
     • When ther factors common to both the numerator and the denominator cancel down first
       • Example 5
         • Simplify the following products
         • a) (3/5) * [5/9]
           • = [{3}/{5}] / [{5}/{9}]
             • = [1*1] / [1*3]
               • = 1/3
         • b) [a/b] / [c/d]
           • = [{a}/b] / [c/{a}]
             • = [1*c] / [b*1]
               • = c/b
         • c) [(x+1) / 2] * [3 / (x^2 -1)]
           • = [{(x+1)} /2] * [3 / {(x+1)}(x-1)]
             • = 3 / 2(x-1)
     • To divide by a fraction multiply by the reciprocal of that fraction
       • Example 6
         • Divide 5/6 by 1/3
           • [5/6] / [1/3]
             • = [5/{6}] * [{3}/1]
               • = [5*1] / [2*1]
                 • = 5/2
       • Example 7
         • Simplify :
         • a) [a/b] / [a/c]
           • = [{a}/b] * [c/{a}]
             • = [1*c] / [b*1]
               • = c/b
         • b) [(x+2) / (x+4)] / [(3x+6) / (x^2-16)]
           • = [(x+2) / (x+4)] * [(x^2 -16) / (3x+6)]
             • = [{(x+2)} / {(x+4)}] * [({(x+4)}(x-4)) / 3{(x+2)}
               • = (x-4) / 3
   • You need to have the same denominator when you add or subtact numeric or algebratic fractions
     • Example 8
       • Add 1/3 to 3/4
         • = 1/3 + 3/4
           • = 4/12 + 9/12
             • = 13/12
     • Example 9
       • Add 1/3 to 5/12
         • = 1/3 + 5/12
           • = 4/12 + 5/12
             • = 9/12
               • = 3/4
     • Example 10
       • Simplify the following fractions :
       • a) (a/x) + b
         • = (a/x) + (b/1)
           • = (a/x) + (bx/x)
             • [a + bx] / x
       • b) [3] / [x+1] - [4x] / [x^2 -1]
         • = 3 / [x+1] - [4x] / [(x+1)(x-1)]
           • = [3(x-1) / [(x+1)(x-1)] - [4x] / [(x+1)(x-1)]
             • = [3(x-1) - 4x] / [(x+1)(x-1)]
               • = [-x-3] / [(x+1)(x-1)]
                 • = [-1(x+3)] / [(x+1)(x-1)]
   • You can divide two algebraic fractions. As with ordinary numbers, an improper fraction has a larger numerator than denominator. An improper algebraic fraction is one whose numerator has a degree equal to or larger than the denominator, e.g. [x^2 +5x+8] / [x-2] and [x^3 -8x+5] / [x^2 +2x+1] are both improper fractions because the numerator has a larger degree than the denominator. You can change them into mixed number fractions either by long division or by using the remainder theorem
     • Example 12, 13 and 14
       • Look in the textbook for the examples as I can't put them on the mind map

2. Functions

   • A mapping transforms one set of numbers into a different set of numbers. The mapping can be described in words or through an algebratic equation. It can also be represented by a Cartesian graph
     • Example 4
       • Draw mapping diagrams and graph for the following operations
       • a) 'Add 3' on the set (-3, 1, 4, 6, x)
       • b) 'Square' on the set (-1, 1, -2, 2, x)
         • Answers in the textbook for a) and b)
   • A function is a special mapping such that every element of set A (the domain) is mapped to exactly one element of set B
     • There are three types functions
       • One-to-one function
         • Look at textbook for examples
       • Many-to-one function
       • Not a function
     • Example 2
       • Given that the function g(x) = 2x^2 +3
       • a) The value of g(2)
         • g(2) = 2(2)^2 +3 = 11
       • b) The value of a such that g(a) = 35
         • g(a) = 2a^2 +3 = 35
           • 2a^2 = 32
             • a^2 = 16
               • a = +/- 4
     • You can write functions in two different ways
       • f(x) = 2x+1
       • f : x = 2x+1
         • = should be an arrow
     • Example 3
       • Which of the following mappings represent the function? Give reasons for your answers
       • a) Every element of Set A gets mapped to two elements in Set B. Therefore this isn't a function. It is not a function. It is an operation (square root)
         • Look at textbook for the question
       • b) Every value of x gets mapped to one value of y. This is, therefore, a function. It can be written in the form. y = ax + b for some constant a and b. This type of function is called one-to-one because every element of the range comes exactly from one element in the domian
       • c) On the sketch of y = 1/x, you can see that x=0 doesn't get mapped anywhere. Therefore not all elements of Set A get mapped to elements in set B. Hence it is not a function
       • d) This is a function. It is called a many-to-one function because two elements of the domain get mapped to one element in the range
   • Many mappings can be made into functions by changing the domain
     • Consider y=x^1/2
       • If the domain is all of the real numbers (xER), then this isn't a function because values of x less than 0 don't get mapped anywhere
       • If you restrict the domain to x>= 0, all of set A gets mapped to exactly one element of set B
       • We can write this function f(x) = x^1/2, with domain (xER, x>=)
     • Example 4
       • Find the range of the following functions:
       • a) f(x) = 3x-2, domain {x=1, 2, 3, 4}
         • The range of f(x) is {1, 4, 7, 10}
           • f(x) is one-to-one
       • b) g(x) = x^2, domain {xER, -5<=x<=5
         • Range of g(x) is 0<=g(x),=25
           • g(x) is many-to-one
       • c) h(x) = 1/x domain {xER, 0<x<=3}
         • Range of h(x) is h(x)>=1/3
           • h(x) is one-to-one
       • State if the function are one-to-one or many-to-one
     • Example 5
       • The function f(x) is defined by :
         • f(x) = {5-2x, x<1}
         • f(x) = {x^2 +3 x>=1}
       • a) Sketch f(x) stating its range
         • Graph is in the textbook pg 18
         • f(1) = 5-2(1) = 3
           • The range is the value that y takes and therefore f(x) > 3
       • b) Find the values of a such that f(a) = 19
         • The positve point is where x^2 +3 = 19
           • x^2 = 16
             • x = 4
         • The negative point is where 5-2x=19
           • -2x = 14
             • x = -7
               • The two values of a are 4 and -7
   • You can combine two or more basic functions to make a new more complex function
     • fg(x) means apply g first, followed by f. fg(x) is called a composite function
     • Example 6
       • Given f(x) = x^2 and g(x) = x+1, find:
         • a) fg(1)
           • fg(1) = f(1+1)
             • = 2^2
               • = 4
         • b) fg(3)
           • fg(3) = f(3+1)
             • = 4^2
               • = 16
         • c) fg(x)
           • fg(x) = f(x+1)
             • = (x+1)^2
     • Example 7
       • The functions f and g are defined by f(x) = 3x + 2 and g(x) = x^2 + 4. Find:
         • a) The functions fg(x)
           • fg(x) = f(x^2 + 4)
             • = 3(x^2 + 4) + 2
               • = 3x^2 + 14
         • b) The functions gf(x)
           • gf(x) = g(3x + 2)
             • = (3x + 2)^2 + 4
               • = 9x^2 + 12x + 8
         • c) The functions f^2(x)
           • f^2(x) = f(3x + 2)
             • = 3(3x + 2) + 2
               • = 9x + 8
         • d) The values of b such that fg(b) = 62
           • fg(x) = 3x^2 + 14
             • If fg(b) = 62
               • 62 = 3b^2 + 14
                 • 3b^2 = 48
                   • b^2 = 16
                     • b = +/- 4
     • Example 8
       • The functions m(x), n(x) and p(x) are defined by m(x) = 1/x, n(x) = 2x + 4, p(x) = x^2 - 2. Find in terms of m, n, and p the functions
         • a) 2/x + 4
           • 2/x + 4 = 2(1/x) + 4
             • = 2m(x) + 4
               • = nm(x)
         • b) 4x^2 + 16x + 14
           • 4x^2 + 16x + 14 = (2x + 4)^2 - 2
             • [n(x)]^2 - 2
               • = pn(x) - 2
         • c) {1} / {4x + 12}
           • = {1} / {2(2x + 4) + 4}
             • = {1} / {2n(x) + 4}
               • = {1} / nn(x)
                 • = mnn(x)
                   • = mn^2(x)
   • The inverse function performs the opposite operation to the function. It takes element of the range and maps them back into elements of the domain. For this reason inverse functions only exist for one-to-one functions
     • The inverse of f(x) is written f^-1(x)
     • ff^-1 (x) = f^-1 f(x) = x
     • For many straightforward functions, the inverse can be found using flow chart
     • Example 9
       • The inverse of the functions h(x) = 2x^2 - 7
         • let y = h(x)
           • ¬ means square root and what is inside {} is square rooted
           • Look at textbook for flowchart
           • y = 2x^2 - 7
             • y+7 = 2x^2
               • ¬ {(y + 7)/2}
     • Example 10
       • Find the inverse of the function f(x) = 3/{x-1} (xER, x =/ 1)
         • =/ means not equal to
         • Let y = f(x)
           • y = 3/{x-1}
             • y(x - 1) = 3
               • yx - y = 3
                 • yx = 3 + y
                   • x = {3 + y} / y
                     • f^-1 (x) = {3 + x} / x
     • Example 11
       • The function f(x) is defined by f(x) = ¬{x - 2} (xER, x>= 2). Find the function f^-1 (x) in a similar form stating its domain
         • The range of the function is f(x) >= 0. Therefore the domain of the inverse function is x>= 0.
           • Look at the textbook for the graph
           • y = ¬{x - 2}
             • x = y^2 + 2
               • The inverse function is f^-1(x) = x + 2 {xER, x >= 0}
                 • Look at graph in the textbook
     • Example 12
       • If g(x) is defined as g(x) = 2x - 4 (xER, x >= 0)
       • a) Calculate g^-1 (x)
         • g^-1 (x) = {x + 4} / 2 {x >= -4}
           • Look at the textbook for the flowchart
       • b) Sketch the graphs of both functions on the same set of axes
         • Look at textbook for the answer on the graph
       • c) What is the connection between the graphs?
         • The graph of g^-1 (x) and g(x) are mirror images of each other in the line y = x
     • Example 13
       • The function f(x) is defined by f(x) = x^2 -3 (xER, x>= 0)
       • a) Find f^-1 (x) in similar terms
         • Let y = f(x)
           • y = x^2 -3
             • y + 3 = x^2
               • ¬{y + 3} = x
                 • f^-1(x) = ¬{x + 3}
       • b) Sketch f^-1 (x)
         • Look at textbook for answer
       • c) Find values of x such that f(x) = f^-1 (x)
         • When f(x) = f^-1(x)
           • f(x) = x
             • x^2 - x - 3 = 0
               • So x = {1+ ¬{13}} / 2

3. Algorithms

   • You need to be able to carry out a bubble sort
     • A common data processing task is sorting an unordered list into alphabetical and numerical order
     • In a bubble sort we compare adjacent items
     • This is the bubble sort algorithm
       • 1) Start at the beginning of the list. Pass through the list and compare adjacent values. For each pair of values:
         • If they are in order, leave them
         • If they are not in order, swap them
       • 2) When you get to the end of the list, repeat step 1
       • 3) When a pass is completed without any swaps,the list is in order
     • Example 7
       • Use a bubble sort to arrange this list into ascending order: 24 18 37 11 15 30
         • Look at the textbook for the comparisons
         • First pass: 18 24 11 15 30 37
           • Second pass: 18 11 15 24 30 37
             • Third pass: 11 15 18 24 30 37
               • Final pass: 11 15 18 24 30 37
               • The fourth pass produces no swaps, so the list is in order
     • Example 8
       • Use a bubble sort to arrange these letters into alphabetical order
         • After 1st pass: A G L O I R H M T
           • After 2nd pass: A G I L H M O R T
             • After 3rd pass: A G I L H M O R T
               • After 4th pass: A G I H L M O R T
                 • After 5th pass: A G H I L M O R T
                   • Final pass: A G H I L M O R T
     • Example 9
       • Use a bubble sort to arrange these numbers into descending order: 39 57 72 39 17 24 48
         • Look at the textbook for the comparisons
         • After 1st pass: 57 72 39 39 24 48 17
           • After 2nd pass: 72 57 39 39 48 24 17
             • After 3rd pass: 72 57 39 48 39 24 17
               • Note the 48 is in between the 39s don't treat them as one term
               • After 4th pass: 72 57 48 39 39 24 17
                 • Final pass: 72 57 48 39 39 24 17
   • You need to be able to carry out a quick sort
     • As its name suggests, a quick sort is quick and efficient
     • You select a pivot and then split the items into two sub-lists: those less than the pivot and those greater than the pivot
     • Pivots are selected in each sub-list to create further sub-lists
     • Here is the quick sort algorithm, used to sort a list into ascending order
       • 1) Choose the item at the mid-point of the list to be the first pivot
       • 2) Write down all the items that are less than the pivot, keeping their order, in a sub-list
       • 3) Write down the pivot
       • 4) Write down the remaining items (those greater than the pivot) in a sub-list
       • 5) Apply steps 1 to 4 to each sub-list
       • 6) When all items have been chosen as pivots, stop
     • The number of pivots has the potential to double at each pass. There is 1 pivot at the first pass, there could be 2 from the second, 4 at the third, 8 at the fourth, and so on
     • Example 10
       • The old pivots are in speech bubbles e.g. "2" New pivots are brackets (2)
       • Use a quick sort to arrange the numbers below into ascending order: 21 24 42 29 23 13 8 39 38
         • 21 24 42 29 (23) 13 8 39 38
           • 21 (13) 8 "23" 24 42 (29) 39 38
             • (8) "13" (21) "23" (24) "29" 38 (39) 42
               • "8" "13" "21" "23" "24" "29" (38) "39" (42)
                 • 8 13 21 23 24 29 38 39 42
     • Example 11
       • Use a quick sort to arrange the list below into descending order: 37 20 17 26 44 41 27 28 50 17
         • 37 20 17 26 44 (41) 27 28 50 17
           • 44 (50) "41" 37 20 17 (26) 27 28 17
             • "50" (44) "41" 37 (27) 28 "26" 20 (17) 17
               • "50" "44" "41" 37 (28) "27" "26" 20 (17) "17"
                 • "50" "44" "41" (37) "28" "27" "26" (20) "17" "17"
                   • 50 44 41 37 28 27 26 20 17 17