I N D U BRIDGING THE GAP TO A/S LEVEL MATHS C T I O N

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1 I N D U BRIDGING THE GAP TO A/S LEVEL MATHS C T I O N

2 INTRODUCTION TO A LEVEL MATHS The Mathematics Department is committed to ensuring that you make good progress throughout your A level or AS course. The aim of the booklet is to provide you with the groundwork on which you can build your skills before you embark on the course. It is vitally important that you spend some time working through the questions in this booklet over the summer - you will 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. You do not necessarily have to do every question, but enough to ensure you understand the topic thoroughly. The answers are given at the back of the booklet. The biggest hindrance to the transition from GCSE is usually lack of confidence with number and algebra. We will test you at the start of September to check how well you understand these topics, so it is important that you have looked at all the booklet before then. If you do not pass this test you will be demonstrating that you are not suitable for the course. We hope that you will use this introduction to give you a good start to your AS work and that it will help you enjoy and benefit from the course more. Sources for further help are indicated throughout the booklet. You may also find the following book useful AS-Level Maths Head Start Published by CGP Workbooks ISBN: Cost: 4.9 Useful websites are CONTENTS Quick Revision of Fractions and Negative Numbers Chapter Removing brackets Chapter Linear equations 7 Chapter Simultaneous equations Chapter 4 Factors Chapter Change the subject of the formula 6 Chapter 6 Solving quadratic equations 9 Chapter 7 Indices Chapter 8 Surds 4 Practice Induction Test

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5 Chapter : REMOVING BRACKETS To remove a single bracket, we multiply every term in the bracket by the number or the expression on the outside: Examples ) (x + y) = x + 6y ) -(x - ) = (-)(x) + (-)(-) = -4x + 6 To expand two brackets, we must multiply everything in the first bracket by everything in the second bracket. We can do this in a variety of ways, including * the smiley face method * FOIL (Fronts Outers Inners Lasts) * using a grid. Examples: ) (x + )(x + ) = x(x + ) + (x + ) or (x +)(x + ) = x + + x + x = x + x + or x x x x x (x +)(x + ) = x + x + x + = x + x + ) (x - )(x + ) = x(x + ) - (x +) = x + x 4x - 6 = x x 6 or (x - )(x + ) = x 6 + x 4x = x x 6 or x - x x -4x x -6 (x +)(x - ) = x + x - 4x - 6 = x - x - 6

6 EXERCISE A Multiply out the following brackets and simplify.. 7(4x + ). -(x - 7). a 4(a - ) 4. 4y + y( + y). -x (x + 4) 6. (x - ) (x - 4) 7. (x + )(x + ) 8. (t - )(t - ) 9. (x + y)(x 4y) 0. 4(x - )(x + ). (y - )(y + ). ( + x)(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 - ) = (x )(x ) = 4x x + 9 (x - )(x + ) = x = x 9 EXERCISE B Multiply out. (x - ). (x + ). (7x - ) 4. (x + )(x - ). (x + )(x - ) 6. (y - )(y + ) More help on expanding brackets is available by downloading this video tutorial: 6

7 Chapter : LINEAR EQUATIONS When solving an equation, you must remember that whatever you do to one side must also be done to the other. You are therefore allowed to 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, you should collect all the letters onto the same side of the equation. If the equation contains brackets, you should start by expanding the brackets. A linear equation is an equation that contains numbers and terms in x. A linear equation does not contain any x or x terms. More help on solving equations can be obtained by downloading the leaflet available at this website: Alternatively a video explanation is available here: Example : Solve the equation 64 x = Solution: There are various ways to solve this equation. One approach is as follows: Step : Add x to both sides (so that the x term is positive): 64 = x + Step : Subtract from both sides: Step : Divide both sides by : 9 = x = x So the solution is x =. Example : Solve the equation 6x + 7 = x. Solution: Step : Begin by adding x to both sides 8x + 7 = (to ensure that the x terms are together on the same side) Step : Subtract 7 from each side: 8x = - Step : Divide each side by 8: x = -¼ Exercise A: Solve the following equations, showing each step in your working: ) x + = 9 ) x = ) 4x = 4) 7x = -9 ) + x = 8 x 6) 7x + = 4x 7

8 Example : Solve the equation (x ) = 0 (x + ) Step : Multiply out the brackets: 6x 4 = 0 x 6 (taking care of the negative signs) Step : Simplify the right hand side: 6x 4 = 4 x Step : Add x to each side: 9x 4 = 4 Step 4: Add 4: 9x = 8 Step : Divide by 9: x = Exercise B: Solve the following equations. ) (x 4) = 4 ) 4( x) = (x 9) ) 8 (x + ) = 4 4) 4 (x + ) = 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 + = Solution: Step : Multiply through by (the denominator in the fraction): y + 0 = Step : Subtract 0: y = Example : Solve the equation (x + ) = Solution: Step : Multiply by (to remove the fraction) x + = Step : Subtract from each side x = 4 Step : 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. 8

9 Example 6: Solve the equation x + x + + = 4 Solution: Step : Find the lowest common denominator: The smallest number that both 4 and divide into is 0. Step : Multiply both sides by the lowest common denominator 0( x + ) 0( x + ) + = 40 4 Step : Simplify the left hand side: 4 0 ( x + ) 0 ( x + ) + = 40 4 (x + ) + 4(x + ) = 40 Step 4: Multiply out the brackets: x + + 4x + 8 = 40 Step : Simplify the equation: 9x + = 40 Step 6: Subtract 9x = 7 Step 7: Divide by 9: x = Example 7: Solve the equation x x x + = 4 6 Solution: The lowest number that 4 and 6 go into is. So we multiply every term by : ( x ) ( x) x + = Simplify x + ( x ) = 4 ( x) Expand brackets x + x 6 = x Simplify x 6 = 8 + 0x Subtract 0x x 6 = 8 Add 6 x = 4 Divide by x = 4.8 Exercise C: Solve these equations ) ( ) x + = ) x x = + 4 y y ) + = 4) 4 x x =

10 Exercise C (continued) ) 7x x y y + y + = 6) + = 6 7) x x + x + = 8) 0 = 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 + and the third is n +. Therefore n + (n + ) + (n + ) = 96 n + = 96 n = 9 n = So the numbers are, and. Exercise D: ) Find consecutive even numbers so that their sum is 08. ) 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. ) Two girls have 7 photographs of celebrities between them. One gives 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. 0

11 Chapter : SIMULTANEOUS EQUATIONS An example of a pair of simultaneous equations is x + y = 8 x + y = In these equations, x and y stand for two numbers. We can 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. We do this by making the coefficients of y the same in both equations. This can be achieved by multiplying equation by, so that both equations contain y: x + y = 8 0x + y = = To eliminate the y terms, we subtract equation from equation. We get: 7x = 4 i.e. x = To find y, we substitute x = into one of the original equations. For example if we put it into : 0 + y = y = Therefore the solution is x =, y =. Remember: You can check your solutions by substituting both x and y into the original equations. Example: Solve x + y = 6 x 4y = Solution: We begin by getting the same number of x or y appearing in both equation. We can get 0y in both equations if we multiply the top equation by 4 and the bottom equation by : 8x + 0y = 64 x 0y = 4 As the SIGNS in front of 0y are DIFFERENT, we can eliminate the y terms from the equations by ADDING: x = i.e. x = Substituting this into equation gives: 6 + y = 6 y = 0 So y = The solution is x =, y =. If you need more help on solving simultaneous equations, you can download a booklet from the following website: Alternatively, you can download the video tutorial:

12 Exercise: Solve the pairs of simultaneous equations in the following questions: ) x + y = 7 ) x + y = 0 x + y = 9 x + y = -7 ) x y = 4 4) 9x y = x + y = -6 4x y = 7 ) 4a + b = 6) p + q = a 4b = 4 p + q = 4

13 Chapter 4: FACTORISING Common factors We can factorise some expressions by taking out a common factor. Example : Factorise x 0 Solution: taking 6 6 is a common factor to both and 0. We can therefore factorise by outside a bracket: x 0 = 6(x ) Example : Factorise 6x xy Solution: is a common factor to both 6 and. Both terms also contain an x. So we factorise by taking x outside a bracket. 6x xy = x(x y) Example : Factorise 9x y 8x y Solution: 9 is a common factor to both 9 and 8. 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 y 8x y = 9x y(xy ) Example 4: Factorise x(x ) 4(x ) Solution: There is a common bracket as a factor. So we factorise by taking (x ) out as a factor. The expression factorises to (x )(x 4) Exercise A Factorise each of the following ) x + xy ) 4x xy ) pq p q 4) pq - 9q ) x 6x 6) 8a b a b 4 7) y(y ) + (y )

14 Factorising quadratics Simple quadratics: Factorising quadratics of the form x + bx + c The method is: Step : Form two brackets (x )(x ) Step : Find two numbers that multiply to give c and add to make b. These two numbers get written at the other end of the brackets. Example : Factorise x 9x 0. Solution: We need to find two numbers that multiply to make -0 and add to make -9. These numbers are -0 and. Therefore x 9x 0 = (x 0)(x + ). General quadratics: Factorising quadratics of the form ax + bx + c The method is: Step : 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. Step : 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. Solution: We need to find two numbers that multiply to make 6 - = -7 and add to make. These two numbers are -8 and 9. Therefore, 6x + x = 6x - 8x + 9x identical) = x(x 4) + (x 4) (the two brackets must be = (x 4)(x + ) 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 = = ( + )( ) 6x = ( x) = (x + )(x ) Also notice that: and x 8 = ( x 4) = ( x + 4)( x 4) x 48xy = x( x 6 y ) = x( x + 4 y)( x 4 y) Factorising by pairing We can factorise expressions like pairing: x + xy x y using the method of factorising by both x + xy x y = x(x + y) (x + y) (factorise front and back pairs, ensuring brackets are identical) = (x + y)(x ) If you need more help with factorising, you can download a booklet from this website: 4

15 or you can watch a video tutorial: Exercise B Factorise ) ) ) x x x 6 + 6x 6 x + x + 4) ) x x (factorise by taking out a common factor) x + x 6) y + 7y + 7) 7y 0y + 8) 0x + x 0 9) 4x 0) x x xy + y ) 4x x + 8 ) 6m 8n ) 4y 9a y 4) 8( x + ) ( x + ) 0

16 Chapter : CHANGING THE SUBJECT OF A FORMULA We can use algebra to change the subject of a formula. Rearranging a formula is similar to solving an equation we must do the same to both sides in order to keep the equation balanced. Example : Make x the subject of the formula y = 4x +. Solution: y = 4x + Subtract from both sides: y = 4x Divide both sides by 4; y = x 4 y So x = is the same equation but with x the subject. 4 Example : Make x the subject of y = x Solution: Notice that in this formula the x term is negative. y = x Add x to both sides y + x = (the x term is now positive) Subtract y from both sides x = y Divide both sides by y x = ( F ) Example : The formula C = is used to convert between Fahrenheit and 9 Celsius. We can rearrange to make F the subject. ( F ) C = 9 Multiply by 9 9C = ( F ) (this removes the fraction) Expand the brackets 9C = F 60 Add 60 to both sides 9C + 60 = F Divide both sides by 9C + 60 = F 9C + 60 Therefore the required rearrangement is F =. Exercise A Make x the subject of each of these formulae: ) y = 7x ) y = x + 4 x ) 4y = 4) 4(x ) y = 9 6

17 Rearranging equations involving squares and square roots Example 4: Make x the subject of x + y = w Solution: Subtract y from both sides: involving x) Square root both sides: x + y = w x = ± w y x = w y (this isolates the term Remember that you can have a positive or a negative square root. We cannot simplify the answer any more. Example : Make a the subject of the formula t = a 4 h Solution: Multiply by 4 Square both sides Multiply by h: Divide by : a t = 4 h a 4t = h a 6t = h 6t h = a 6t h = a Exercise B: Make t the subject of each of the following ) wt P = ) r P = wt r ) V = π t h 4) P = t g ) w( v t) Pa = 6) g r = a + bt 7

18 More difficult examples Sometimes the variable that we wish to make the subject occurs in more than one place in the formula. In these questions, we collect the terms involving this variable on one side of the equation, and we 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. We 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 If you need more help you can download an information booklet on rearranging equations from the following website: tom.pdf or you can view a video tutorial on this topic by following this link: Exercise C Make x the subject of these formulae: ) ax + = bx + c ) ( x + a) = k( x ) ) x + y = x x x 4) = + a b 8

19 Chapter 6: 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 Note that not all quadratic equations can be solved by factorising. The quadratic formula can always be used however. Method : 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 : Solve x x + = 0 Factorise (x )(x ) = 0 Either (x ) = 0 or (x ) = 0 So the solutions are x = or x = Note: The individual values x = 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 = Method : Using the formula Recall that the roots of the quadratic equation b ± x = b 4ac a ax bx c + + = 0 are given by the formula: Example : Solve the equation x = 7 x Solution: First we rearrange so that the right hand side is 0. We get We can then tell that a =, b = and c = -. Substituting these into the quadratic formula gives: x + x = 0 ± 4 ( ) ± 0 x = = (this is the surd form for the 4 solutions) If we have a calculator, we can evaluate these roots to get: x =.8 or x = -. If you need more help with the work in this chapter, there is an information booklet downloadable from this web site: Alternatively you can view a video tutorial: 9

20 EXERCISE ) Use factorisation to solve the following equations: a) x + x + = 0 b) x x 4 = 0 c) x = x ) Find the roots of the following equations: a) x + x = 0 b) x 4x = 0 c) 4 x = 0 ) Solve the following equations either by factorising or by using the formula: a) 6x - x 4 = 0 b) 8x 4x + 0 = 0 4) Use the formula to solve the following equations to significant figures. Some of the equations can t be solved. a) x +7x +9 = 0 b) 6 + x = 8x c) 4x x 7 = 0 d) x x + 8 = 0 e) x + 4x + 4 = 0 f) x = x 6 0

21 Chapter 7: 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 basic rules of indices: m n m n ) a a = a e.g. = m n m n ) a a = a 8 6 e.g. = ) ( a m ) n = a mn e.g. ( ) = 0 Further examples y y = y 4 7 4a 6a = 4a (multiply the numbers and multiply the a s) 6 8 c c = 6c (multiply the numbers and multiply the c s) ( ) 7 7 4d 4d d = = 8d d (divide the numbers and divide the d terms i.e. by subtracting the powers) Exercise A Simplify the following: ) ) ) 4) ) 6) b b = (Remember that c c = b c bc = n n = 6 ( 6 ) 8 8n n = d d = 9 b = b ) 7) ( a ) = 4 8) ( d ) =

22 More complex powers Zero index: Recall from GCSE that 0 a =. This result is true for any non-zero number a. Therefore ( ) Negative powers = =.04 = 4 A power of - corresponds to the reciprocal of a number, i.e. Therefore and = 0. = = = 4 a = (you find the reciprocal of a fraction by swapping the top bottom over) a This result can be extended to more general negative powers: This means: 4 = = 9 = = = = = a n =. n a Fractional powers: Fractional powers correspond to roots: / a = a / a = a / 4 4 a = a In general: / n a n = a Therefore: / 8 = 8 = / = = / = 0000 = 0 m / n / n A more general fractional power can be dealt with in the following way: a = ( a ) So ( ) = = = / / / = = = / / = = = = 6 m

23 Exercise B: Find the value of: ) ) / 4 / 7 ) ( ) / 9 4) ) 6) 7) 8) 9) / 7 8 / 0) ( 0.04 ) / ) ) / / Simplify each of the following: ) 4) a a x / / x 4 ) ( x y ) /

24 Chapter 8: SURDS Simplifying Surds Surds are simplifies by finding a factor of the surd that is a square number. For example: 7 = 6 = 6 = 6 We simplify expressions involving surds in a similar way to algebraic expressions by combining similar surds. For example: + = This will sometimes involve writing individual surds in their simplest form first. For example: 7 8 = 6 = 6 = 4 4 Exercise A. Simplify the following: a) 8 e) b) 7 f) c) 48 g) d) 44 h) 7. Simplify the following and express each as a single surd: a) d) 00 b) 7 + e) 7 + c) 8 f) Expand and simplify the following: a) ( )( + ) e) ( + )( ) b) ( )( ) f) ( 4 )( ) c) ( )( ) g) ( 7 )( 7 + ) d) ( )( + ) h) ( )( + ) 4

25 Rationalising the Denominator By using equivalent fractions, multiplying both numerator and denominator by the same thing, the denominator is transformed into a number that does not involve surds. Example: Rationalise the denominators of the following fractions 7 a) b) 7 + Solution: ( ) 7( ) a) = = b) = = ( + ) ( ) Look carefully at what the denominators have been multiplied by. For a single surd you can simply multiply by the same surd, for a pair of surds change the sign between them. Exercise B. Rationalise the denominator in each of the following fractions. a) e) 7 b) f) c) + g) + d) h) Calculator not allowed.. Calculate ( ) 4 ( 9) a) 6 AS Mathematics Practice Induction Test b) ( ) 4 ( ) 6 ( ). If a =, b =, c =, d = 0. evaluate; a) a + b b) a + b c) ac + d d) ( a + c + d) e) ( a + b)( a b) a + b f) 6 c + d. Give the answers to the following questions in the form of fractions in their simplest form. a) 6 + b) c)

26 d) 6 e) 7 f) 4 7 g) ( 4) h) 8 i) 4 j) k) Simplify the following expressions 4 4 a) x x y b) 7 xy + yx c) 4 7 8x y d) e) f) ( x 4 ) 4x y 6 ( y ) g) 4 y y 4x 9y h) 4 7 8x x x i) 6. Simplify the following a) 7 b) 7 c) d) e) Rationalise the denominator of the following fractions a) b) c) + + d) 7. Evaluate the following expressions b) c) a) d) Expand and simplify the following a) ( x + )( x ) b) (x + )( x + x ) c) 9 6 ( x + ) 9. Factorise each of the following a) b + ab b) x + x xy + yx c) ( ) ( + ) d) x + x + e) x + 7x + 6 f) x x 8 0. Solve the following pairs of simultaneous equations. 4x + y = x y = 8 a) b) x y = 7 x + y =. Make x the subject of the following formulae a) y = x 7 b) 7 = x y c) y = 7 x x 6

27 SOLUTIONS TO THE EXERCISES CHAPTER : Ex A ) 8x + ) -x + ) -7a + 4 4) 6y + y ) x 4 6) 7x 7) x + x + 6 8) t t 0 9) 6x + xy y 0) 4x + 4x 4 ) 4y ) + 7x x Ex B ) x x + ) 9x + 0x + ) 49x 8x + 4 4) x 4 ) 9x - 6) y 9 CHAPTER Ex A ) 7 ) ) ½ 4) ) -/ 6) -7/ Ex B ).4 ) ) 4) ½ Ex C ) 7 ) ) 4/7 4) / ) 6) 7) 9/ 8) Ex D ) 4, 6, 8 ) 9.87, 9.6 ) 4, 48 CHAPTER ) x =, y = ) x = -, y = ) x = 0, y = - 4) x =, y = ) a = 7, b = - 6) p = /, q = 4/ CHAPTER 4 Ex A ) x( + y) ) x(x y) ) pq(q p) 4) q(p q) ) x (x - ) 6) 4a b (a b ) 7) (y )(y + ) Ex B ) (x )(x + ) ) (x + 8)(x ) ) (x + )(x + ) 4) x(x ) ) (x - )(x + ) 6) (y + )(y + 7) 7) (7y )(y ) 8) (x )(x + ) 9) (x + )(x ) 0) (x )(x y) ) 4(x )(x ) ) (4m 9n)(4m + 9n) ) y(y a)(y + a) 4) (4x + )(x 4) CHAPTER Ex A y + 9y + 0 ) x = ) x = 4y ) x = (4 y + ) 4) x = 7 Ex B rp rp V P g Pag ) t = ) t = ± ) t = ± 4) t = ) t = v 6) w w π h w Ex C c a + k y + ab ) x = ) x = ) x = 4) x a b k y = b a t = ± r a b CHAPTER 6 ) a) -, - b) -, 4 c) -, ) a) 0, - b) 0, 4 c), - ) a) -/, 4/ b) 0.,. 4) a) -.0, -.70 b).07, c) -.0,.4 d) no solutions e) no solutions f) no solutions CHAPTER 7 Ex A ) b 6 ) 6c 7 ) b c 4 4) -n 8 ) 4n 6) d 7) a 6 8) -d Ex B 7

28 ) ) ) / 4) / ) 6) /7 7) 9 8) 9/4 9) ¼ 0) 0. ) 4/9 ) 64 ) 6a 4) x ) xy CHAPTER 8 Ex A a) b) c) 4 d) e) f) 7 g) h) 6 a) b) 9 c) d) 0 e) 7 f) a) b) 6 c) 8 d) e) 9 f) 4 6 g) h) Ex B 7 ( + ) 7 0 a) b) c) d) + e) f) ( 7 ) g) 4 h) 9 77 SOLUTIONS TO PRACTISE INDUCTION TEST a) b) 6 a) b) 7 c) d) e) f) / a) 4/8 b)84/ c) 6/ d) 8/ e) / f) 0 g) /6 h) 4 i) /6 j) 4 k) 67/60 4a) 8 x y b) 0 xy c) 4 4 x d) / e) y f) x g) y h) a) b) 6 c) d) e) + 6a) b) ( ) c) ( + ) d) 4 + 7a) b) 8 c) 4 d) 4 8a) x + x 6 b) x + x c) 4x + x + 9 x i) x y 9a) b ( + a) b) xy ( y + x) c) ( x )( x + ) d) ( x + )( x + ) e) ( x + )( x + ) f) ( x + 4)( x ) 9 x = x = 0a) b) 4 y = y = a) = y x = (7 y + ) x b) c) x = orx = y + 0 y + 7 8