MAIDSTONE GRAMMAR SCHOOL FOR GIRLS DEPARTMENT OF MATHEMATICS

MAIDSTONE GRAMMAR SCHOOL FOR GIRLS DEPARTMENT OF MATHEMATICS Introduction to A level Maths INDUCTION BOOKLET

INTRODUCTION TO A LEVEL MATHS AT MGGS Thank you for choosing to study Mathematics in the sixth form at Maidstone Grammar School for Girls. You will sit two modules in Pure Mathematics (C1 and C) as well as Statistics (S1) at the end of year 1. If you have chosen to study Further Mathematics as well as Maths then, in year 1, you will study modules in D1, D and Further Pure 1. The Mathematics Department is committed to ensuring that you make good progress throughout your A level or AS course. In order that you make the best possible start to the course, we have prepared this booklet. It is vital that you spend time working through the questions in this booklet over the summer. You need to have a good knowledge of these topics before you commence your course in September. You should have met all the topics before at GCSE. Work through what you need to from each chapter, making sure that you understand the examples. Then tackle the exercise to ensure you understand the topic thoroughly. The answers are at the back of the booklet. You will need to be very organised so keep your work in a folder & mark any queries to ask at the beginning of term. In the first or second week of term you will take a test to check how well you understand these topics, so it is important that you have completed the booklet by then. The pass mark is 75%. If you do not pass this test, you may be asked to leave the course or to complete a programme of additional work in order to bring your basic algebra skills to the required standard. If you have to complete more work you will be re-tested in October. A mock test is provided at the back of this booklet. Use this introduction to give you a good start to your AS work that will help you to enjoy, and benefit from, the course. The more effort you put in, right from the start, the better you will do. Mr C Ansette (Teacher i/c Key Stage 5 Mathematics) Mrs S Squibb (Head of Mathematics) Sources for further help are indicated throughout the booklet. All topics can be found on the MyMaths website. If you need to access this, our username is maidstone and our password is tables. You may also find the following book useful AS-Level Maths Head Start Published by CGP Workbooks ISBN: 978 1 84146 993 5 Cost: 4.95

CONTENTS Reading List... 3 Section 1: FRACTIONS... 4 Section : EXPANDING... 6 Section 3: LINEAR EQUATIONS... 8 Section 4: LINEAR INEQUALITIES... 13 Section 5: SIMULTANEOUS EQUATIONS... 14 Section 7: SOLVING QUADRATIC EQUATIONS... 18 Section 8: CHANGING THE SUBJECT OF A FORMULA... 0 Section 9: INDICES... 3 Section 10: SURDS... 6 Section 11: FUNCTIONS... 30 Practice Booklet Test... 31 Solutions to the Exercises... 3 Solutions to the Practice Booklet Test... 35 READING LIST As a student who is choosing to study Mathematics at A Level, it is logical to assume that you have an interest in the subject. With that said, the following books may be of interest to you. Alex s Adventures in Numberland by Alex Bellos Cabinet of Mathematical Curiosities by Ian Stewart The Num8er My5teries by Marcus du Sautoy How Many Socks Make a Pair?: Surprisingly Interesting Maths by Rob Eastway The Curious Incident of the Dog in the Night-time by Mark Haddon The Penguin Dictionary of Curious & Interesting Numbers by David Wells The Calculus Wars by Jason Socrates Bardi The Code Book by Simon Singh 50 Mathematical Ideas You Really Need to Know by Tony Crilly

Section 1: FRACTIONS To add or subtract fractions, find the lowest common denominator of the two fractions and then rewrite the fractions accordingly. Ensure that you simplify as far as possible. Examples When multiplying fractions, it is far more efficient to cancel first; this avoids trying to simplify fractions with unnecessarily large numerators and/or denominators. To multiply with fractions, simply multiply the numerators and denominators together. Example To divide by a fraction, we simply multiply by the reciprocal of the second fraction (i.e. we flip the second fraction over ). Example

For addition and subtraction with mixed numbers, add (or subtract) the integer (whole number) parts first and then work with the fractions. Examples To multiply and divide with mixed, convert the mixed numbers to improper fractions and then calculate as normal. It should also be noted that in the study of A Level Mathematics, answers are preferred as improper fractions rather than mixed numbers. EXERCISE 1. 5. 9.. 6. 10. 3. 7. 11. 4. 8. 1. More help is available from MyMaths: Adding/Subtracting Fractions, Multiplying Fractions, Dividing Fractions, Mixed Numbers

Section : EXPANDING To remove a single bracket multiply every term in the bracket by the number or expression outside: Examples 1) 3 (x + y) = 3x + 6y ) -(x - 3) = (-)(x) + (-)(-3) = -4x + 6 To expand two brackets multiply everything in the first bracket by everything in the second bracket. You may have used * the smiley face method * FOIL (First Outside Inside Last) * using a grid. Examples: 1) (x + 1)(x + ) = x(x + ) + 1(x + ) or (x +1)(x + ) = x + + x + x = x + 3x + or x 1 x x x x (x +1)(x + ) = x + x + x + = x + 3x + ) (x - )(x + 3) = x(x + 3) - (x +3) = x + 3x 4x - 6 = x x 6 or (x - )(x + 3) = x 6 + 3x 4x = x x 6 or x - x x -4x 3 3x -6 (x +3)(x - ) = x + 3x - 4x - 6 = x - x - 6

EXERCISE A 1. 7(4x + 5). -3(5x - 7) 3. 5a 4(3a - 1) 4. 4y + y( + 3y) 5. -3x (x + 4) Multiply out the following brackets and simplify. 7. (x + )(x + 3) 8. (t - 5)(t - ) 9. (x + 3y)(3x 4y) 10. 4(x - )(x + 3) 11. (y - 1)(y + 1) 6. 5(x - 1) (3x - 4) 1. (3 + 5x)(4 x) Two Special Cases Perfect Square: Difference of two squares: (x + (a) = (x + (a)(x + (a) = x + ax + a (x - (a)(x + (a) = x a (x - 3) = (x 3)(x 3) = 4x 1x + 9 (x - 3)(x + 3) = x 3 = x 9 EXERCISE B Expand the following 1. (x - 1). (3x + 5) 3. (7x - ) 4. (x + )(x - ) 5. (3x + 1)(3x - 1) 6. (5y - 3)(5y + 3 More help is available from MyMaths: Brackets

Section 3: LINEAR EQUATIONS When solving an equation whatever you do to one side must also be done to the other. You may add the same amount to both side subtract the same amount from each side multiply the whole of each side by the same amount divide the whole of each side by the same amount. If the equation has unknowns on both sides, collect all the letters onto the same side of the equation. If the equation contains brackets, you often start by expanding the brackets. A linear equation contains only numbers and terms in x. (Not x or 3 x or 1/x et(c) Example 1: Solve the equation 64 3x = 5 Solution: There are various ways to solve this equation. One approach is as follows: Step 1: Add 3x to both sides (so that the x term is positiv(e): 64 = 3x + 5 Step : Subtract 5 from both sides: Step 3: Divide both sides by 3: 39 = 3x 13 = x So the solution is x = 13. Example : Solve the equation 6x + 7 = 5 x. Solution: Step 1: Begin by adding x to both sides 8x + 7 = 5 (to ensure that the x terms are together on the same sid(e) Step : Subtract 7 from each side: 8x = - Step 3: Divide each side by 8: x = -¼ Exercise A: Solve the following equations, showing each step in your working: 1) x + 5 = 19 ) 5x = 13 3) 11 4x = 5 4) 5 7x = -9 5) 11 + 3x = 8 x 6) 7x + = 4x 5

Example 3: Solve the equation (3x ) = 0 3(x + ) Step 1: Multiply out the brackets: 6x 4 = 0 3x 6 (taking care of the negative signs) Step : Simplify the right hand side: 6x 4 = 14 3x Step 3: Add 3x to each side: 9x 4 = 14 Step 4: Add 4: 9x = 18 Step 5: Divide by 9: x = Exercise B: Solve the following equations. 1) 5(x 4) = 4 ) 4( x) = 3(x 9) 3) 8 (x + 3) = 4 4) 14 3(x + 3) =

EQUATIONS CONTAINING FRACTIONS When an equation contains a fraction, the first step is usually to multiply through by the denominator of the fraction. This ensures that there are no fractions in the equation. y Example 4: Solve the equation 5 11 Solution: Step 1: Multiply through by (the denominator in the fraction): y 10 Step : Subtract 10: y = 1 Example 5: Solve the equation 1 (x 1) 5 3 Solution: Step 1: Multiply by 3 (to remove the fraction) x 1 15 Step : Subtract 1 from each side x = 14 Step 3: Divide by x = 7 When an equation contains two fractions, you need to multiply by the lowest common denominator. This will then remove both fractions. Example 6: Solve the equation x 1 x 4 5 Solution: Step 1: Find the lowest common denominator: The smallest number that both 4 and 5 divide into is 0. Step : Multiply both sides by the lowest common denominator Step 3: Simplify the left hand side: 0( x 1) 0( x ) 40 4 5 5 4 0 ( x 1) 0 ( x ) 40 4 5 5(x + 1) + 4(x + ) = 40 Step 4: Multiply out the brackets: 5x + 5 + 4x + 8 = 40 Step 5: Simplify the equation: 9x + 13 = 40 Step 6: Subtract 13 9x = 7 Step 7: Divide by 9: x = 3

Example 7: Solve the equation x 3 5x x 4 6 Solution: The lowest number that 4 and 6 go into is 1. So we multiply every term by 1: 1( x ) 1(3 5 x) 1x 4 4 6 Simplify 1x 3( x ) 4 (3 5 x) Expand brackets 1x 3x 6 4 6 10x Simplify 15x 6 18 10x Subtract 10x 5x 6 18 Add 6 5x = 4 Divide by 5 x = 4.8 Exercise C: Solve these equations 1) 1 ( 3) 5 x ) x x 1 4 3 3 3) y y x 3 x 3 5 4) 4 3 7 14 5) 7x 1 13 x y 1 y 1 y 5 6) 3 6 7) x 1 5x 3 x 8) 3 5 10 1 x x

FORMING EQUATIONS Example 8: Find three consecutive numbers so that their sum is 96. Solution: Let the first number be n, then the second is n + 1 and the third is n +. Therefore n + (n + 1) + (n + ) = 96 3n + 3 = 96 3n = 93 n = 31 So the numbers are 31, 3 and 33. Exercise D: 1) Find 3 consecutive even numbers so that their sum is 108. ) The perimeter of a rectangle is 79 cm. One side is three times the length of the other. Form an equation and hence find the length of each side. 3) Two girls have 7 photographs of celebrities between them. One gives 11 to the other and finds that she now has half the number her friend has. Form an equation, letting n be the number of photographs one girl had at the beginning. Hence find how many each has now. More help is available from MyMaths: Solving Equations

Section 4: LINEAR INEQUALITIES Linear inequalities can be solved using the same techniques as linear equations (for the most part). We may add and subtract the same numbers on both sides and we can also multiply and divide by positive numbers; multiplying/dividing both sides by a negative needs further explanation. Example x - 3 < 11 Here we can simply add 3 to both sides: x < 14 Next, as with linear equations we divide by : x < 7 However, if we were to have 3 x > 6, we would need to adopt a different technique. If we wish to divide or multiply by a negative number, we must reverse the direction of the inequality. Example As before, we would subtract 3 from both sides: Divide by - and subsequently reverse the inequality: We can see this working on a more basic level; it is true to state that 3 < 4 but it is incorrect if we multiply both sides by a negative and keep the sign as it was: -6 < -8 is not true. You may find it easier to rearrange the inequality: Example If we add x to both sides, we remove the hassle: We then subtract 6: Divide by two as normal: Remember that you can change this round to say Both of these techniques are acceptable and is more a matter of preference. Exercise: Solve each inequality 1) ) 3) 4) 5) 6) 7) 8) More help is available from MyMaths: Inequalities, Negative Inequalities

Section 5: SIMULTANEOUS EQUATIONS Example 3x + y = 8 5x + y = 11 x and y stand for two numbers. Solve these equations in order to find the values of x and y by eliminating one of the letters from the equations. In these equations it is simplest to eliminate y. Make the coefficients of y the same in both equations. To do this multiply equation by, so that both equations contain y: 3x + y = 8 10x + y = = To eliminate the y terms, subtract equation from equation. We get: 7x = 14 i.e. x = To find y substitute x = into one of the original equations. For example put it into : 10 + y = 11 y = 1 Therefore the solution is x =, y = 1. Remember: Check your solutions by substituting both x and y into the original equations. Example: Solve x + 5y = 16 3x 4y = 1 Solution: Begin by getting the same number of x or y appearing in both equation. Multiply the top equation by 4 and the bottom equation by 5 to get 0y in both equations: 8x + 0y = 64 15x 0y = 5 As the SIGNS in front of 0y are DIFFERENT, eliminate the y terms from the equations by ADDING: 3x = 69 + i.e. x = 3 Substituting this into equation gives: 6 + 5y = 16 5y = 10 So y = The solution is x = 3, y =. Exercise: Solve the pairs of simultaneous equations in the following questions: 1) x + y = 7 ) x + 3y = 0 3x + y = 9 3x + y = -7 3) 3x y = 4 4) 9x y = 5 x + 3y = -6 4x 5y = 7 5) 4a + 3b = 6) 3p + 3q = 15 5a 4b = 43 p + 5q = 14 More help is available from MyMaths: Simultaneous Equations 1, Simultaneous Equations - Medium, Simultaneous Equations - Hard, Simultaneous Equations - Negatives, Solving Simultaneous Equations Graphically

Section 6: FACTORISING Taking out a common factor Example 1: Factorise 1x 30 Solution: Example : 6 is a common factor to both 1 and 30. Factorise by taking 6 outside a bracket: 1x 30 = 6(x 5) Factorise 6x xy Solution: is a common factor to both 6 and. Both terms also contain an x. Factorise by taking x outside a bracket. 6x xy = x(3x y) Example 3: Factorise 9x 3 y 18x y Solution: 9 is a common factor to both 9 and 18. The highest power of x that is present in both expressions is x. There is also a y present in both parts. So we factorise by taking 9x y outside a bracket: 9x 3 y 18x y = 9x y(xy ) Example 4: Factorise 3x(x 1) 4(x 1) Solution: There is a common bracket as a factor. So we factorise by taking (x 1) out as a factor. The expression factorises to (x 1)(3x 4) Exercise A Factorise each of the following 1) 3x + xy ) 4x xy 3) pq p q 4) 3pq - 9q 5) x 3 6x 6) 8a 5 b 1a 3 b 4 7) 5y(y 1) + 3(y 1) More help is available from MyMaths: Factorising Expressions

Factorising quadratics Simple quadratics: Factorising quadratics of the form x bx c The method is: Step 1: Form two brackets (x )(x ) Step : Find two numbers that multiply to give c and add to make b. Write these two numbers at the end of the brackets. Example 1: Factorise x 9x 10. Solution: Find two numbers that multiply to make -10 and add to make -9. These numbers are -10 and 1. Therefore x 9x 10 = (x 10)(x + 1). General quadratics: Factorising quadratics of the form ax bx c One method is that of combining factors. Look at factorising on MyMaths or ask a teacher for help with our preferred method but is difficult to explain on paper. Another method is: Step 1: Find two numbers that multiply together to make ac and add to make b. Step : Split up the bx term using the numbers found in step 1. Step 3: Factorise the front and back pair of expressions as fully as possible. Step 4: There should be a common bracket. Take this out as a common factor. Example : Factorise 6x + x 1. Solution: We need to find two numbers that multiply to make 6-1 = -7 and add to make 1. These two numbers are -8 and 9. Therefore, 6x + x 1 = 6x - 8x + 9x 1 = x(3x 4) + 3(3x 4) (the two brackets must be identical) = (3x 4)(x + 3) Difference of two squares: Factorising quadratics of the form x a Remember that x a = (x + (a)(x (a). Therefore: x x x x 9 3 ( 3)( 3) 16x 5 ( x) 5 (x 5)(x 5) Also notice that: and x 8 ( x 4) ( x 4)( x 4) 3 3x 48xy 3 x( x 16 y ) 3 x( x 4 y)( x 4 y) Factorising by pairing or grouping Factorise expressions like x xy x y using the method of factorising by pairing: x xy x y = x(x + y) 1(x + y) (factorise front and back pairs, both brackets identical) = (x + y)(x 1)

Exercise B Factorise 1) x x 6 8) 10x 5x 30 ) x 6x 16 9) 4x 5 3) x 5x 10) x 3x xy 3y 4) x 3 x 11) 4x 1x 8 5) 3x 5x 1) 16m 81n 6) y 17y 1 13) 3 4y 9a y 7) 7y 10y 3 14) 8( x 1) ( x 1) 10 More help is available from MyMaths: Factorising Quadratics 1, Factorising Quadratics

Section 7: SOLVING QUADRATIC EQUATIONS A quadratic equation has the form ax bx c 0. There are two methods that are commonly used for solving quadratic equations: * factorising * the quadratic formula Not all quadratic equations can be solved by factorising. Method 1: Factorising Make sure that the equation is rearranged so that the right hand side is 0. It usually makes it easier if the coefficient of x is positive. Example 1 : Solve x 3x + = 0 Factorise (x 1)(x ) = 0 Either (x 1) = 0 or (x ) = 0 So the solutions are x = 1 or x = Note: The individual values x = 1 and x = are called the roots of the equation. Example : Solve x x = 0 Factorise: x(x ) = 0 Either x = 0 or (x ) = 0 So x = 0 or x = More help is available from MyMaths: Quadratic Equations

Method : Using the formula The roots of the quadratic equation ax bx c 0 are given by the formula: b b 4ac x a Example 3: Solve the equation x 5 7 3 x Solution: First we rearrange so that the right hand side is 0. We get We can then tell that a =, b = 3 and c = -1. Substituting these into the quadratic formula gives: x 3x 1 0 3 3 4 ( 1) 3 105 x (this is the surd form for the solutions) 4 If we have a calculator, we can evaluate these roots to get: x = 1.81 or x = -3.31 Exercise 1) Use factorisation to solve the following equations: (a) x + 3x + = 0 (b) x 3x 4 = 0 (c) x = 15 x ) Find the roots of the following equations: (a) x + 3x = 0 (b) x 4x = 0 (c) 4 x = 0 3) Solve the following equations either by factorising or by using the formula: (a) 6x - 5x 4 = 0 (b) 8x 4x + 10 = 0 4) Use the formula to solve the following equations to 3 significant figures where possible (a) x +7x +9 = 0 (b) 6 + 3x = 8x (c) 4x x 7 = 0 (d) x 3x + 18 = 0 (e) 3x + 4x + 4 = 0 f) 3x = 13x 16 More help is available from MyMaths: The Quadratic Formula

Section 8: CHANGING THE SUBJECT OF A FORMULA Rearranging a formula is similar to solving an equation always do the same to both sides. Example 1: Make x the subject of the formula y = 4x + 3. Solution: y = 4x + 3 Subtract 3 from both sides: y 3 = 4x Divide both sides by 4; y 3 x 4 y 3 So x is the same equation but with x the subject. 4 Example : Make x the subject of y = 5x Solution: Notice that in this formula the x term is negative. y = 5x Add 5x to both sides y + 5x = (the x term is now positiv(e) Subtract y from both sides 5x = y Divide both sides by 5 y x 5 Example 3: 5( F 3) The formula C is used to convert between Fahrenheit and Celsius. 9 Rearrange to make F the subject. 5( F 3) C 9 Multiply by 9 9C 5( F 3) (this removes the fraction) Expand the brackets 9C 5F 160 Add 160 to both sides 9C 160 5F Divide both sides by 5 9C 160 F 5 9C 160 Therefore the required rearrangement is F. 5 Exercise A Make x the subject of each of these formulae: 1) y = 7x 1 x 3) 4y 3 x 5 ) y 4(3x 5) 4 4) y 9

Example 4: Make x the subject of Solution: Subtract y from both sides: x y w x y w Square root both sides: x w y Remember the positive & negative square root. x w y (this isolates the term involving x) Example 5: Make a the subject of the formula t 1 5a 4 h Solution: Multiply by 4 Square both sides Multiply by h: Divide by 5: 1 5a t 4 h 5a 4t h 5a 16t h 16t h 5a 16th a 5 Exercise B: Make t the subject of each of the following 1) P wt 3r 3) V 1 3 t h 5) Pa w( v t) g ) wt P 3r 4) P t g 6) r a bt More help is available from MyMaths: Rearranging Equations

Harder examples Sometimes the subject occurs in more than one place in the formula. In these questions collect the terms involving this variable on one side of the equation, and put the other terms on the opposite side. Example 6: Make t the subject of the formula a xt b yt Solution: a xt b yt Start by collecting all the t terms on the right hand side: Add xt to both sides: a b yt xt Now put the terms without a t on the left hand side: Subtract b from both sides: a b yt xt Factorise the RHS: a b t( y x) Divide by (y + x): So the required equation is a b t y x t a b y x Example 7: Make W the subject of the formula Wa T W b Solution: This formula is complicated by the fractional term. Begin by removing the fraction: Multiply by b: bt bw Wa Add bw to both sides: bt Wa bw (this collects the W s together) Factorise the RHS: bt W ( a b) Divide both sides by a + b: W bt a b Exercise C Make x the subject of these formulae: 1) ax 3 bx c ) 3( x a) k( x ) 3) x 3 y 5x 4) x x 1 a b More help is available from MyMaths: Higher Rearranging

Section 9: INDICES Basic rules of indices y 4 means y y y y. 4 is called the index (plural: indices), power or exponent of y. There are 3 basic rules of indices: 1) ) 3) ( a ) m n m n a a a e.g. m n m n a a a e.g. m n mn Further examples 3 3 3 3 3 3 4 5 9 8 6 a e.g. 3 5 3 10 y 5y 5y 4 3 7 3 5 4a 6a 4a (multiply the numbers and multiply the a s) c 3c 6 6c 8 (multiply the numbers and multiply the c s) 7 7 4d 5 4d 3d 8d (divide the numbers and divide the d terms by subtracting the powers) 3d Exercise A Simplify the following: Remember that b b 1 1) ) 3) 4) 5) b 5b 5 3c c 5 b c bc 3 6 n ( 6 n ) 8n n 8 3 6) d d 11 9 3 7) a 4 8) 3 d

Zero index: Remember 0 a 1 For any non-zero number, a. 0 Therefore 5 1 1 5.304 1 4 Negative powers 0 3 A power of -1 corresponds to the reciprocal of a number, i.e. Therefore 5 5 1 1 0.5 4 0.5 1 1 1 4 5 5 4 This result can be extended to more general negative powers: This means: 3 1 1 3 9 Fractional powers: Fractional powers correspond to roots: In general: Therefore: 1/ 3 3 8 8 a 1/n n a 0 a 1 1 (Find the reciprocal of a fraction by turning it upside down) 4 1 1 4 16 a n a 1. n a 1 1 1 4 16 4 4 1 1/ 1/ 3 3 1/ 4 4 a a a a a a 1/ 5 5 5 1/ 4 4 10000 10000 10 m / n 1/ n A more general fractional power can be dealt with in the following way: a a So 3 3/ 3 4 4 8 / 3 1/ 3 8 8 4 7 7 3 9 3/ 3/ 3 3 5 36 36 6 16 36 5 5 5 15 m

Exercise B: Find the value of: 1) ) 1/ 4 1/ 3 7 4) 5) 5 0 18 8) 3 11) 8 7 / 3 3) 1 1/ 9 6) 7) 7 1 /3 7 9) 8 / 3 10) 1/ 0.04 1) 1 16 3/ Simplify each of the following: 13) 14) a 3a 3 x x 1/ 5 / 4 15) x y 1/ More help is available from MyMaths: Indices 1, Indices, Indices 3

Section 10: SURDS A surd is a root of a number that cannot be expressed as an integer. Surds are part of the set of irrational numbers. Example: and are surds but is not as it equals. Simplifying Surds Start to simplify surds by using two rules: and By using the multiplication rule, simplify surds by finding a factor of the number you are taking a root of which is a square number. Always try to find the largest square number factor, otherwise you will have to simplify further. Example: Exercise A Simplify 1) ) 3) 4) 5) 6)

Multiplying and Dividing with Surds The rules of algebra are true for any numeric value; these include surds. To multiply and divide using surds deal with any integers together and then deal with any surds. Examples: ( ) ( )( ) ( )( ) ( ) In this example, you could expand as usual but this is an example of the difference of two squares. Exercise B Simplify 1) ) 3) 4) 5) 6) 7) ( ) 8) ( )

Addition and Subtraction with Surds You can only add or subtract with surds if the surd is the same; sometimes if they are not the same, you may be able to simplify them so that the same surd is present. Example: Here add the and as the same surd is present but you cannot add the. By simplifying to, you can add the two surds together. These methods also work for subtraction of surds. Exercise C Simplify 1) ) 3) 4) 5) 6) 7) 8) 9) 10)

Rationalising the Denominator It is far easier to calculate with a fraction if the surd if the denominator is a rational number (i.e. not a sur(d). The process of this is known as rationalising the denominator. To do this, multiply by the surd in the denominator, doing so makes use of the fact that ( ) Example: Multiply the denominator by to rationalise it and so multiply the numerator by also: Example : Example 3: ( ) Exercise D Simplify 1) ) 3) 4) 5) 6) 7) 8) 9) More help is available from MyMaths: Surds 1, Surds (Surds covers some material which is not here; this will be covered in Core 1)

Section 11: FUNCTIONS A function is like a machine: it has an input and an output. The output is related somehow to the input. f(x) tells us the rule for finding the output for any given input. f(x) is one such function and is read as f of x. Other letters may be used for functions too. Example: If a) b) a) b), find If c) d) c) d), find Exercise A Given that and, find 1) ) 3) 4) 5) 6) Given the rule and an output generated, we can find the input. Example: If and, find. If and, find the possible values of. ` Exercise B 1. Given that and find. Given that and find the possible values of

Practice Booklet Test You may NOT use a calculator If ax b + bx + c = 0 then x = b 4ac a 1. Calculate (a) (b) (c) (d). Expand and simplify (a) (x + 3)(x 1) (b) (a + 3) (c) 4x(3x ) x(x + 5) 3. Factorise (a) x 7x (b) y 64 (c) x + 5x 3 (d) 6t 13t + 5 4. Simplify 3 4x y (a) 3 8x y (b) 3x + 3 4x 1 6 5. Solve the following equations h 1 3h (a) + = 4 (b) x 8x = 0 (c) p + 4p = 1 4 5 6. Write each of the following as single powers of x and / y 1 (a) (b) (x y) 3 5 x (c) 4 x x 7. Work out the values of the following, giving your answers as fractions (a) 4 - (b) 10 0 8 (c) 7 8. Solve the simultaneous equations 3x 5y = -11 5x y = 7 9. Rearrange the following equations to make x the subject (a) v = u 1 + ax (b) V = πx x h (c) y = 3 x 1 10. Solve 5x x 1 = 0 giving your solutions in surd form 11. Find the values of x which satisfy the following inequalities (a) (b) 1. Given and, find (a) (b) (c) 1 3 13. Simplify (a) (b) (c) (d) (e)

Solutions to the Exercises SECTION 1 1) 4) 7) 10) ) 5) 8) 11) 3) 6) 9) 1) SECTION Ex A 1) 8x + 35 ) -15x + 1 3) -7a + 4 4) 6y + 3y 5) -4x 4 6) 7x 1 7) x + 5x + 6 8) t 3t 10 9) 6x + xy 1y 10) 4x + 4x 4 11) 4y 1 1) 1 + 17x 5x Ex B 1) x x + 1 ) 9x + 30x + 5 3) 49x 8x + 4 4) x 4 5) 9x -1 6) 5y 9 SECTION 3 Ex A 1) 7 ) 3 3) 1½ 4) 5) 6) Ex B 1).4 ) 5 3) 1 4) ½ Ex C 1) 7 ) 15 3) 4/7 4) 35/3 5) 3 6) 7) 9/5 8) 5 Ex D 1) 34, 36, 38 ) 9.875, 9.65 3) 4, 48 SECTION 4 1) ) 3) 4) 5) 6) 7) 8) SECTION 5 1) x = 1, y = 3 ) x = -3, y = 1 3) x = 0, y = - 4) x = 3, y = 1 5) a = 7, b = - 6) p = 11/3, q = 4/3

SECTION 6 Ex A 1) x(3 + y) ) x(x y) 3) pq(q p) Ex B 1) (x 3)(x + ) ) (x + 8)(x ) 3) (x + 1)(x + ) 4) x(x 3) 5) (3x -1 )(x + ) 4) 3q(p 3q) 5) x (x - 3) 6) 4a 3 b (a 3b ) 6) (y + 3)(y + 7) 7) (7y 3)(y 1) 8) 5(x 3)(x + ) 9) (x + 5)(x 5) 10) (x 3)(x y) 7) (y 1)(5y + 3) 11) 4(x )(x 1) 1) (4m 9n)(4m + 9n) 13) y(y 3(a)(y + 3(a) 14) (4x + 5)(x 4) SECTION 7 1) (a) -1, - (b) -1, 4 (c) -5, 3 ) (a) 0, -3 (b) 0, 4 (c), - 3) (a) -1/, 4/3 (b) 0.5,.5 4) (a) -5.30, -1.70 (b) 1.07, -0.699 (c) -1.0, 1.45 (d) no solutions (e) no solutions f) no solutions SECTION 8 Ex A y 1 1) x 7 Ex B 3rP 1) t w 3rP ) t w ) x 4y 5 3) x 3(4 y ) 3V 3) t h Pg 4) t 9y 0 4) x 1 Pag 5) t v w r a 6) t b Ex C c 3 1) x ) a b 3a k x k 3 3) y 3 x 5y 4) ab x b a SECTION 9 Ex A 1) 5b 6 ) 6c 7 3) b 3 c 4 4) -1n 8 5) 4n 5 6) d 7) a 6 8) -d 1 Ex B 1) ) 3 3) 1/3 4) 1/5 5) 1 6) 1/7 7) 9 8) 9/4 9) ¼ 10) 0. 11) 4/9 1) 64 13) 6a 3 14) x 15) xy

SECTION 10 Ex A 1) ) Ex B 1) ) 3) 3) 4) 4) 5) 6) 5) 6) 7) 8) Ex C 1) ) 3) 4) 5) 6) 7) 8) 9) 10) Ex D 1) ) 3) 4) 5) 6) 7) 8) 9) SECTION 11 Ex A 1) 8 ) -3 3) 56 4) 0 5) 65 6) Ex B 1) - ), 3

Solutions to the Practice Booklet Test 1 (a) (b) (c) (d) ) (a) 4x + 4x 3 (b) a + 6a + 9 (c) 10x -13x 3) (a) x(x 7) (b) (y + 8)(y 8) (c) (x - 1)(x + 3) (d) (3t - 5)(t 1) x 4) (a) y (b) 10 x 3 6 5) (a) h = 5 (b) x = 0 or x = 8 (c) p = -6 or p = 6) (a) x -4 (b) x 6 y 3 (c) x 7 7) (a) (b) 1 (c) 8) x = 3, y = 4 9) (a) v u x a (b) x 3V h (c) y x y 1 10) 1 1 x 10 11) (a) x < 1.6 (b) x < -.5 1) (a) 19 (b) 3 (c) 11 13 (a) (b) (c) (d) (e)