Maths A Level Summer Assignment & Transition Work

Maths A Level Summer Assignment & Transition Work The summer assignment element should take no longer than hours to complete. Your summer assignment for each course must be submitted in the relevant first lesson in September. Failure to do so will automatically place you on your first warning during the induction period. As a guideline, the transition work element will take hours a week over the summer holiday. This booklet has been produced to help you prepare for A-Level Maths. It is essential that you begin the course having maintained and developed your skills in the higher level algebra content from the GCSE syllabus. You also need to understand the volume of work that will be required for A-level Maths. Completion of this booklet will provide evidence that you have the commitment and work ethic required. Additionally, this work will prepare you for the LaSWAP algebra test on enrolment day in August. This will be required if your GCSE maths grade does not meet our automatic entry criteria. Please check the LaSWAP website for the maths entry requirements when you get your GCSE grades. INSTRUCTIONS: there are parts to this booklet. Part A: Summer Assignment (submitted in st lesson of the year) This must be completed in full on A4 paper and handed in to your teacher during the first lesson of the year. You must show all workings and set your work out clearly and logically, clearly labelling each question and sub-question. Do not attempt to complete this on a printout of the assignment there is not enough room for your full workings and solutions so it must be handed in on A4 paper. The assignment will be marked by your teacher and graded A E. It should take no more than hours to complete, once you have done any necessary revision in advance. The transition work in this booklet will prepare you for your summer assignment. You will also be given a -hour baseline test in the first lesson of the year. Part B: Transition Work This booklet has been produced by our eam board Pearson Edecel to ensure that all A-Level Maths students across the country begin A level with the algebra skills necessary for the course. You are required to complete all of this work over the summer. As well as being compulsory, it will help with any necessary preparation for the summer assignment, the LaSWAP algebra test at enrolment and the baseline test in the first lesson. Complete the answers to the eercises on A4 squared paper, with each topic and question number clearly labelled. You must show all working (final answers alone are not accepted). If you are stuck, use the Key Points and Eamples to remind yourself of how to answer the questions. Answers to all of the questions are at the back of this booklet. You should use these answers to correct all of your work in a different coloured pen. If you have a different answer you are epected to make corrections, again showing all of your methods. Correcting your work is a vital skill in A-level mathematics as it contributes to improving your understanding in key areas To summarise, in September you will be assessed in ways, after meeting our entry criteria:. Part A: Summer Assignment will be formally marked and graded by your teacher.. Part B: Transition Work should be self-marked and handed in to your teacher.. -hour Baseline Test in your first lesson, which will be formally marked and graded by your teacher, forming part of your Induction assessment. Parts A & B will help you prepare for this. DBO/GAT 08

Part A: Summer Assignment ( hours work) ) Show that ( ) ( + 5) simplifies to a + b + c + d where a, b, c and d are integers to be found. () ) Functions f and g are such that f() = 4 5 and g() = (a) Find the value of g(-5). (b) Find gf() in the simplest form. () () Find the inverse function g ().. ) Sketch the graph of f() = + 9 + 5, writing the coordinates of the turning point and the coordinates of any intercepts with the coordinate aes. 4) By completing the square, write the function f() = 8 in the form f() = a( + b) + c () () where a, b and c are rational numbers to be found. () 5) The graph shows the speed, in metres per second, of a car as it approaches a set of traffic lights. 0 Speed (m/s) 0 0 0 5 0 5 0 Time (seconds) 5 Use the graph to estimate the distance travelled by the car in the first 0 seconds. () DBO/GAT 08

6). A computer uses the iteration n 5 n to find one solution for a quadratic equation. 9 (a) What quadratic equation is being solved? () (b) Using this iterative formula, with 0 = 0, find a solution to this quadratic equation to d.p. () 7) Solve 6 () 8). If a city s population grows by 0% every year, how many years will it take for this city to double in population? 9) The first 5 terms of a quadratic sequence are -4-5 - 5 6 Find the epression, in terms of n, for the nth term of this sequence. 0) The following shows the results of a Higher Education Open Day survey of 00 people. 79 wanted to attend Maths. 55 wanted to attend Physics wanted to attend Chemistry 4 wanted to attend both Maths and Physics 6 wanted to attend Physics and Chemistry 8 wanted to attend Maths and Chemistry 6 do not want to attend any of these subjects (they want to choose different subjects altogether) Find the probability that a randomly selected person from the survey wants to attend all three of these subjects. () () (4) ) Using the quadratic formula, find all the solutions to 5 = 0. () DBO/GAT 08

). The graph shows the speed of a motorbike over a period of.5 minutes. 0 Velocity (m/s) 5 0 5 0 0 40 60 80 00 (a) Between which times was the acceleration the greatest? (b) Find the value of this acceleration. Time (seconds) ) The graph shows the speed, in micrometres per second, of a snail crossing a garden path during a hot summer day. () () Speed ( μm/s ) Work out the average deceleration of the snail in the first 600 seconds. () DBO/GAT 08 4

4. Show that can be written as () 5. ABD is a right angled triangle. All measurements are given in centimetres. C is the point on BD such that CD = AD = BD = Work out the eact area, in cm, of the shaded region. 6. (a) Find the value of... cm () (b) Find the value of... ()... () (c) Solve = DBO/GAT 08 =... () 5

7. (a) Simplify a 4 a (b) Simplify (b ) 7... () (c) Write down the value of 0... () (d) Write down the value of 4... ()... () DBO/GAT 08 6

Part B: Transition Work (Answers at end but you must show working) Epanding brackets & simplifying epressions A LEVEL LINKS Scheme of work: a. Algebraic epressions basic algebraic manipulation, indices and surds Key points When you epand one set of brackets you must multiply everything inside the bracket by what is outside. When you epand two linear epressions, each with two terms of the form a + b, where a 0 and b 0, you create four terms. Two of these can usually be simplified by collecting like terms. Eamples Eample Epand 4( ) 4( ) = 8 Multiply everything inside the bracket by the 4 outside the bracket Eample Epand and simplify ( + 5) 4( + ) ( + 5) 4( + ) = + 5 8 = 5 Epand each set of brackets separately by multiplying ( + 5) by and ( + ) by 4 Simplify by collecting like terms: 8 = 5 and 5 = Eample Epand and simplify ( + )( + ) ( + )( + ) = ( + ) + ( + ) = + + + 6 = + 5 + 6 Epand the brackets by multiplying ( + ) by and ( + ) by Simplify by collecting like terms: + = 5 Eample 4 Epand and simplify ( 5)( + ) ( 5)( + ) = ( + ) 5( + ) = + 0 5 = 7 5 Epand the brackets by multiplying ( + ) by and ( + ) by 5 Simplify by collecting like terms: 0 = 7 DBO/GAT 08 7

Practice Epand. a ( ) b (5pq + 4q ) c (y y ) Epand and simplify. a 7( + 5) + 6( 8) b 8(5p ) (4p + 9) c 9(s + ) 5(6s 0) d (4 ) ( + 5) Epand. a (4 + 8) b 4k(5k ) c h(6h + h 5) d s(4s 7s + ) Watch out! When multiplying (or dividing) positive and negative numbers, if the signs are the same the answer is + ; if the signs are different the answer is. 4 Epand and simplify. a (y 8) 4(y 5) b ( + 5) + ( 7) c 4p(p ) p(5p ) d b(4b ) b(6b 9) 5 Epand (y 8) 6 Epand and simplify. a (m + 7) b 5p(p + 6p) 9p(p ) 7 The diagram shows a rectangle. Write down an epression, in terms of, for the area of the rectangle. Show that the area of the rectangle can be written as 5 8 Epand and simplify. a ( + 4)( + 5) b ( + 7)( + ) c ( + 7)( ) d ( + 5)( 5) e ( + )( ) f ( )( + ) g (5 )( 5) h ( )(7 + 4) i ( + 4y)(5y + 6) j ( + 5) k ( 7) l (4 y) Etend 9 Epand and simplify ( + )² + ( 4)² 0 Epand and simplify. a b DBO/GAT 08 8

Epand and simplify. a ( + 4)( + )( + ) b ( + 7)( + )( ) c ( + 7)( )( 9) d ( - 5)( 4)( ) e ( + )( )( ) f ( )( + )( ) g ( ) h ( ) 4 DBO/GAT 08 9

Surds and rationalising the denominator A LEVEL LINKS Scheme of work: a. Algebraic epressions basic algebraic manipulation, indices and surds Key points A surd is the square root of a number that is not a square number, for eample,, 5, etc. Surds can be used to give the eact value for an answer. ab a b a a b b To rationalise the denominator means to remove the surd from the denominator of a fraction. To rationalise a b you multiply the numerator and denominator by the surd b To rationalise a b c you multiply the numerator and denominator by b c Eamples Eample Simplify 50 50 5 5 5 5 Choose two numbers that are factors of 50. One of the factors must be a square number Use the rule ab a b Use 5 5 Eample Simplify 47 47 49 4 49 4 7 7 4 Simplify 47 and. Choose two numbers that are factors of 47 and two numbers that are factors of. One of each pair of factors must be a square number Use the rule ab a b Use 49 7 and 4 4 Collect like terms DBO/GAT 08 0

Eample Simplify 7 7 7 7 = 49 7 7 4 = 7 = 5 Epand the brackets. A common mistake here is to write 7 49 Collect like terms: 7 7 7 7 0 Eample 4 Rationalise = = 9 Multiply the numerator and denominator by Use 9 = Eample 5 Rationalise and simplify = Multiply the numerator and denominator by = 4 Simplify in the numerator. Choose two numbers that are factors of. One of the factors must be a square number = = 6 Use the rule ab a b 4 Use 4 5 Simplify the fraction: simplifies to 6 DBO/GAT 08

Eample 6 Rationalise and simplify 5 5 = = 5 5 5 5 5 5 Multiply the numerator and denominator by 5 Epand the brackets = 6 5 4 5 5 5 Simplify the fraction = 6 5 = 5 6 4 Divide the numerator by Remember to change the sign of all terms when dividing by Practice Simplify. a 45 b 5 c 48 d 75 e 00 f 8 g 7 h 6 Hint One of the two numbers you choose at the start must be a square number. Simplify. a c e 7 6 b 45 5 50 8 d 75 48 8 8 f 7 Watch out! Check you have chosen the highest square number at the start. Epand and simplify. a ( )( ) b ( )(5 ) c (4 5)( 45 ) d (5 )(6 8) DBO/GAT 08

4 Rationalise and simplify, if possible. a 5 c 7 e b d f 8 5 5 g 8 4 h 5 45 5 Rationalise and simplify. a 5 b 4 c 6 5 Etend 6 Epand and simplify y y 7 Rationalise and simplify, if possible. a 9 8 b y DBO/GAT 08

Rules of indices A LEVEL LINKS Scheme of work: a. Algebraic epressions basic algebraic manipulation, indices and surds Key points a m a n = a m + n m a mn a n a (a m ) n = a mn a 0 = a n n a i.e. the nth root of a m n n m n a a a a m a m m The square root of a number produces two solutions, e.g. 6 4. Eamples Eample Evaluate 0 0 0 0 = Any value raised to the power of zero is equal to Eample Evaluate 9 9 9 = Use the rule a n n a Eample Evaluate 7 7 7 = = 9 m n n Use the rule a a Use 7 m DBO/GAT 08 4

Eample 4 Evaluate 4 4 4 6 Use the rule Use 4 6 a m a m Eample 5 Simplify 6 5 6 5 = 6 = and use the rule give 5 5 a a m n mn a to Eample 6 Simplify 5 4 5 5 8 4 4 4 = 8 4 = 4 Use the rule Use the rule a a m n m n m n a a a a mn Eample 7 Write as a single power of Use the rule a m a m, note that the fraction remains unchanged Eample 8 Write 4 as a single power of 4 4 4 Use the rule Use the rule a n n a m a m a DBO/GAT 08 5

Practice Evaluate. a 4 0 b 0 c 5 0 d 0 Evaluate. a 49 b 64 c 5 d 4 6 Evaluate. a 5 b 5 8 c 49 d 4 6 4 Evaluate. a 5 b 4 c 5 d 6 5 Simplify. a b 0 5 c y e y g y 4 0 d f h 7y 5 4y c c c Watch out! Remember that any value raised to the power of zero is. This is the rule a 0 =. 6 Evaluate. a d 4 b 4 6 e 7 c 9 6 f 9 7 64 7 Write the following as a single power of. a d 5 b 7 e c f 4 DBO/GAT 08 6

8 Write the following without negative or fractional powers. a d b 0 c 5 e f 5 4 9 Write the following in the form a n. a 5 b d e 4 c 4 f Etend 0 Write as sums of powers of. a 5 b c 4 DBO/GAT 08 7

Factorising epressions A LEVEL LINKS Scheme of work: b. Quadratic functions factorising, solving, graphs and the discriminants Key points Factorising an epression is the opposite of epanding the brackets. A quadratic epression is in the form a + b + c, where a 0. To factorise a quadratic equation find two numbers whose sum is b and whose product is ac. An epression in the form y is called the difference of two squares. It factorises to ( y)( + y). Eamples Eample Factorise 5 y + 9 4 y 5 y + 9 4 y = y(5y + ) The highest common factor is y. So take y outside the brackets and then divide each term by y to find the terms in the brackets Eample Factorise 4 5y 4 5y = ( + 5y)( 5y) This is the difference of two squares as the two terms can be written as () and (5y) Eample Factorise + 0 b =, ac = 0 So + 0 = + 5 0 = ( + 5) ( + 5) = ( + 5)( ) Work out the two factors of ac = 0 which add to give b = (5 and ) Rewrite the b term () using these two factors Factorise the first two terms and the last two terms 4 ( + 5) is a factor of both terms DBO/GAT 08 8

Eample 4 Factorise 6 0 b =, ac = 60 So 6 0 = 6 5 + 4 0 = ( 5) + ( 5) = ( 5)( + ) Work out the two factors of ac = 60 which add to give b = ( 5 and 4) Rewrite the b term ( ) using these two factors Factorise the first two terms and the last two terms 4 ( 5) is a factor of both terms Eample 5 Simplify 4 99 4 99 Factorise the numerator and the denominator For the numerator: b = 4, ac = So 4 = 7 + = ( 7) + ( 7) = ( 7)( + ) For the denominator: b = 9, ac = 8 So + 9 + 9 = + 6 + + 9 = ( + ) + ( + ) = ( + )( + ) So 4 ( 7)( ) 99 ( )() 7 = Work out the two factors of ac = which add to give b = 4 ( 7 and ) Rewrite the b term ( 4) using these two factors 4 Factorise the first two terms and the last two terms 5 ( 7) is a factor of both terms 6 Work out the two factors of ac = 8 which add to give b = 9 (6 and ) 7 Rewrite the b term (9) using these two factors 8 Factorise the first two terms and the last two terms 9 ( + ) is a factor of both terms 0 ( + ) is a factor of both the numerator and denominator so cancels out as a value divided by itself is DBO/GAT 08 9

Practice Factorise. a 6 4 y 0 y 4 b a b 5 + 5a 5 b c 5 y 0 y + 5 y Factorise a + 7 + b + 5 4 c + 0 d 5 4 e 7 8 f + 0 g 40 h + 8 Hint Take the highest common factor outside the bracket. Factorise a 6 49y b 4 8y c 8a 00b c 4 Factorise a + b 6 + 7 + 5 c + 7 + d 9 5 + 4 e 0 + + 9 f 8 + 0 5 Simplify the algebraic fractions. a c e 4 8 4 4 b d f 5 5 4 470 6 Simplify a c Etend 9 6 78 4 5 0 6 b d 75 70 6 74 7 Simplify 0 5 8 Simplify ( ) ( ) 4 DBO/GAT 08 0

Completing the square A LEVEL LINKS Scheme of work: b. Quadratic functions factorising, solving, graphs and the discriminants Key points Completing the square for a quadratic rearranges a + b + c into the form p( + q) + r If a, then factorise using a as a common factor. Eamples Eample Complete the square for the quadratic epression + 6 + 6 = ( + ) 9 = ( + ) Write + b + c in the form b b c Simplify Eample Write 5 + in the form p( + q) + r 5 + = 5 5 = 4 4 5 Before completing the square write a + b + c in the form b a c a Now complete the square by writing 5 in the form b b = = 5 5 4 8 5 7 4 8 Epand the square brackets don t 5 forget to multiply by the 4 factor of 4 Simplify DBO/GAT 08

Practice Write the following quadratic epressions in the form ( + p) + q a + 4 + b 0 c 8 d + 6 e + 7 f + Write the following quadratic epressions in the form p( + q) + r a 8 6 b 4 8 6 c + 9 d + 6 8 Complete the square. a + + 6 b c 5 + d + 5 + Etend 4 Write (5 + 0 + ) in the form (a + b) + c. DBO/GAT 08

Solving quadratic equations by factorisation A LEVEL LINKS Scheme of work: b. Quadratic functions factorising, solving, graphs and the discriminants Key points A quadratic equation is an equation in the form a + b + c = 0 where a 0. To factorise a quadratic equation find two numbers whose sum is b and whose products is ac. When the product of two numbers is 0, then at least one of the numbers must be 0. If a quadratic can be solved it will have two solutions (these may be equal). Eamples Eample Solve 5 = 5 5 = 5 5 5 = 0 5( ) = 0 So 5 = 0 or ( ) = 0 Therefore = 0 or = Rearrange the equation so that all of the terms are on one side of the equation and it is equal to zero. Do not divide both sides by as this would lose the solution = 0. Factorise the quadratic equation. 5 is a common factor. When two values multiply to make zero, at least one of the values must be zero. 4 Solve these two equations. Eample Solve + 7 + = 0 + 7 + = 0 b = 7, ac = + 4 + + = 0 ( + 4) + ( + 4) = 0 ( + 4)( + ) = 0 So ( + 4) = 0 or ( + ) = 0 Therefore = 4 or = Factorise the quadratic equation. Work out the two factors of ac = which add to give you b = 7. (4 and ) Rewrite the b term (7) using these two factors. Factorise the first two terms and the last two terms. 4 ( + 4) is a factor of both terms. 5 When two values multiply to make zero, at least one of the values must be zero. 6 Solve these two equations. DBO/GAT 08

Eample Solve 9 6 = 0 9 6 = 0 ( + 4)( 4) = 0 So ( + 4) = 0 or ( 4) = 0 4 or 4 Factorise the quadratic equation. This is the difference of two squares as the two terms are () and (4). When two values multiply to make zero, at least one of the values must be zero. Solve these two equations. Eample 4 Solve 5 = 0 b = 5, ac = 4 So 8 + = 0 ( 4) + ( 4) = 0 ( 4)( + ) = 0 So ( 4) = 0 or ( +) = 0 4 or Factorise the quadratic equation. Work out the two factors of ac = 4 which add to give you b = 5. ( 8 and ) Rewrite the b term ( 5) using these two factors. Factorise the first two terms and the last two terms. 4 ( 4) is a factor of both terms. 5 When two values multiply to make zero, at least one of the values must be zero. 6 Solve these two equations. Practice Solve a 6 + 4 = 0 b 8 = 0 c + 7 + 0 = 0 d 5 + 6 = 0 e 4 = 0 f + 0 = 0 g 0 + 4 = 0 h 6 = 0 i + 8 = 0 j 6 + 9 = 0 k 7 4 = 0 l 0 = 0 Solve a = 0 b = c + 5 = 4 d 4 = e ( + ) = + 5 f 0 = g ( + ) = + 5 h ( ) = ( + ) Hint Get all terms onto one side of the equation. DBO/GAT 08 4

Solving quadratic equations by completing the square A LEVEL LINKS Scheme of work: b. Quadratic functions factorising, solving, graphs and the discriminants Key points Completing the square lets you write a quadratic equation in the form p( + q) + r = 0. Eamples Eample 5 Solve + 6 + 4 = 0. Give your solutions in surd form. + 6 + 4 = 0 ( + ) 9 + 4 = 0 ( + ) 5 = 0 ( + ) = 5 + = 5 = 5 So = 5 or = 5 Write + b + c = 0 in the form b b c 0 Simplify. Rearrange the equation to work out. First, add 5 to both sides. 4 Square root both sides. Remember that the square root of a value gives two answers. 5 Subtract from both sides to solve the equation. 6 Write down both solutions. Eample 6 Solve 7 + 4 = 0. Give your solutions in surd form. 7 + 4 = 0 4 7 = 0 Before completing the square write a + b + c in the form b a c a 7 7 4 4 4 7 49 4 = 0 4 8 7 7 = 0 4 8 = 0 Now complete the square by writing 7 in the form b b a a Epand the square brackets. 4 Simplify. (continued on net page) 5 Rearrange the equation to work out DBO/GAT 08 5

7 7 4 8 7 7 4 6 7 7 4 4 7 7 4 4 So 7 7 or 4 4 7 7 4 4. First, add 7 8 6 Divide both sides by. to both sides. 7 Square root both sides. Remember that the square root of a value gives two answers. 8 Add 7 4 to both sides. 9 Write down both the solutions. Practice Solve by completing the square. a 4 = 0 b 0 + 4 = 0 c + 8 5 = 0 d 6 = 0 e + 8 5 = 0 f 5 + 4 = 0 Solve by completing the square. a ( 4)( + ) = 5 b + 6 7 = 0 c 5 + = 0 Hint Get all terms onto one side of the equation. DBO/GAT 08 6

Solving quadratic equations by using the formula A LEVEL LINKS Scheme of work: b. Quadratic functions factorising, solving, graphs and the discriminants Key points Any quadratic equation of the form a + b + c = 0 can be solved using the formula b b 4ac a If b 4ac is negative then the quadratic equation does not have any real solutions. It is useful to write down the formula before substituting the values for a, b and c. Eamples Eample 7 Solve + 6 + 4 = 0. Give your solutions in surd form. a =, b = 6, c = 4 b b 4ac a 6 6 4()(4) () 6 0 6 5 5 So 5 or 5 Identify a, b and c and write down the formula. Remember that b b 4ac is all over a, not just part of it. Substitute a =, b = 6, c = 4 into the formula. Simplify. The denominator is, but this is only because a =. The denominator will not always be. 4 Simplify 0. 0 45 4 5 5 5 Simplify by dividing numerator and denominator by. 6 Write down both the solutions. DBO/GAT 08 7

Eample 8 Solve 7 = 0. Give your solutions in surd form. a =, b = 7, c = b b 4ac a ( 7) ( 7) 4()( ) () Identify a, b and c, making sure you get the signs right and write down the formula. Remember that b b 4ac is all over a, not just part of it. Substitute a =, b = 7, c = into the formula. 7 7 6 7 7 So or 6 7 7 6 Simplify. The denominator is 6 when a =. A common mistake is to always write a denominator of. 4 Write down both the solutions. Practice Solve, giving your solutions in surd form. a + 6 + = 0 b 4 7 = 0 Solve the equation 7 + = 0 Give your solutions in the form a c Solve 0 + + = 5 Give your solution in surd form. b, where a, b and c are integers. Hint Get all terms onto one side of the equation. Etend 4 Choose an appropriate method to solve each quadratic equation, giving your answer in surd form when necessary. a 4( ) = b 0 = ( + ) c ( ) = 0 DBO/GAT 08 8

Sketching quadratic graphs A LEVEL LINKS Scheme of work: b. Quadratic functions factorising, solving, graphs and the discriminants Key points The graph of the quadratic function y = a + b + c, where a 0, is a curve called a parabola. Parabolas have a line of symmetry and for a > 0 for a < 0 a shape as shown. To sketch the graph of a function, find the points where the graph intersects the aes. To find where the curve intersects the y-ais substitute = 0 into the function. To find where the curve intersects the -ais substitute y = 0 into the function. At the turning points of a graph the gradient of the curve is 0 and any tangents to the curve at these points are horizontal. To find the coordinates of the maimum or minimum point (turning points) of a quadratic curve (parabola) you can use the completed square form of the function. Eamples Eample Sketch the graph of y =. The graph of y = is a parabola. When = 0, y = 0. a = which is greater than zero, so the graph has the shape: Eample Sketch the graph of y = 6. When = 0, y = 0 0 6 = 6 So the graph intersects the y-ais at (0, 6) When y = 0, 6 = 0 ( + )( ) = 0 = or = So, the graph intersects the -ais at (, 0) and (, 0) Find where the graph intersects the y-ais by substituting = 0. Find where the graph intersects the -ais by substituting y = 0. Solve the equation by factorising. 4 Solve ( + ) = 0 and ( ) = 0. 5 a = which is greater than zero, so the graph has the shape: DBO/GAT 08 6 = 6 4 (continued on net page) 6 To find the turning point, complete the square. 9

5 = 4 When 0, and 5 y, so the turning point is at the 4 5 point, 4 7 The turning point is the minimum value for this epression and occurs when the term in the bracket is equal to zero. Practice Sketch the graph of y =. Sketch each graph, labelling where the curve crosses the aes. a y = ( + )( ) b y = ( ) c y = ( + )( + 5) Sketch each graph, labelling where the curve crosses the aes. a y = 6 b y = 5 + 4 c y = 4 d y = + 4 e y = 9 f y = + 4 Sketch the graph of y = + 5, labelling where the curve crosses the aes. Etend 5 Sketch each graph. Label where the curve crosses the aes and write down the coordinates of the turning point. a y = 5 + 6 b y = + 7 c y = + 4 6 Sketch the graph of y = + +. Label where the curve crosses the aes and write down the equation of the line of symmetry. DBO/GAT 08 0

Solving linear simultaneous equations using the elimination method A LEVEL LINKS Scheme of work: c. Equations quadratic/linear simultaneous Key points Two equations are simultaneous when they are both true at the same time. Solving simultaneous linear equations in two unknowns involves finding the value of each unknown which works for both equations. Make sure that the coefficient of one of the unknowns is the same in both equations. Eliminate this equal unknown by either subtracting or adding the two equations. Eamples Eample Solve the simultaneous equations + y = 5 and + y = + y = 5 + y = = 4 So = Using + y = + y = So y = Check: equation : + ( ) = 5 YES equation : + ( ) = YES Subtract the second equation from the first equation to eliminate the y term. To find the value of y, substitute = into one of the original equations. Substitute the values of and y into both equations to check your answers. Eample Solve + y = and 5 y = 5 simultaneously. + y = + 5 y = 5 6 = 8 So = Using + y = + y = So y = 5 Check: equation : + 5 = YES equation : 5 5 = 5 YES Add the two equations together to eliminate the y term. To find the value of y, substitute = into one of the original equations. Substitute the values of and y into both equations to check your answers. DBO/GAT 08

Eample Solve + y = and 5 + 4y = simultaneously. ( + y = ) 4 8 + y = 8 (5 + 4y = ) 5 + y = 6 7 = 8 So = 4 Using + y = 4 + y = So y = Check: equation : 4 + ( ) = YES equation : 5 4 + 4 ( ) = YES Multiply the first equation by 4 and the second equation by to make the coefficient of y the same for both equations. Then subtract the first equation from the second equation to eliminate the y term. To find the value of y, substitute = 4 into one of the original equations. Substitute the values of and y into both equations to check your answers. Practice Solve these simultaneous equations. 4 + y = 8 + y = 7 + y = 5 + y = 5 4 + y = 4 + 4y = 7 y = 4y = 5 5 + y = 6 + y = y = 9 + y = 4 DBO/GAT 08

Solving linear simultaneous equations using the substitution method A LEVEL LINKS Scheme of work: c. Equations quadratic/linear simultaneous Tetbook: Pure Year,. Linear simultaneous equations Key points The subsitution method is the method most commonly used for A level. This is because it is the method used to solve linear and quadratic simultaneous equations. Eamples Eample 4 Solve the simultaneous equations y = + and 5 + y = 4 5 + ( + ) = 4 5 + 6 + = 4 + = 4 = So = Using y = + y = + So y = Check: equation : = + YES equation : 5 + = 4 YES Substitute + for y into the second equation. Epand the brackets and simplify. Work out the value of. 4 To find the value of y, substitute = into one of the original equations. 5 Substitute the values of and y into both equations to check your answers. Eample 5 Solve y = 6 and 4 + y = simultaneously. y = 6 4 + ( 6) = 4 + 6 48 = 0 48 = 0 = 45 So = 4 Using y = 6 y = 4 6 So y = 7 Check: equation : 4 ( 7) = 6 YES equation : 4 4 + ( 7) = YES Rearrange the first equation. Substitute 6 for y into the second equation. Epand the brackets and simplify. 4 Work out the value of. 5 To find the value of y, substitute = 4 into one of the original equations. 6 Substitute the values of and y into both equations to check your answers. DBO/GAT 08

Practice Solve these simultaneous equations. y = 4 y = + 5y = 4 5 y = y = 4 + 5 4 = y 9 + 5y = 8 5y = 5 + 4y = 8 6 y = 4 7 y = y = 4 7 = y 8 + y + = 0 y = 4y = 8 Etend 9 Solve the simultaneous equations + 5y 0 = 0 and ( y ) ( y). 4 DBO/GAT 08 4

Solving linear and quadratic simultaneous equations A LEVEL LINKS Scheme of work: c. Equations quadratic/linear simultaneous Key points Make one of the unknowns the subject of the linear equation (rearranging where necessary). Use the linear equation to substitute into the quadratic equation. There are usually two pairs of solutions. Eamples Eample Solve the simultaneous equations y = + and + y = + ( + ) = + + + + = + + = + = 0 ( 4)( + ) = 0 So = or = Using y = + When =, y = + = When =, y = + = Substitute + for y into the second equation. Epand the brackets and simplify. Factorise the quadratic equation. 4 Work out the values of. 5 To find the value of y, substitute both values of into one of the original equations. So the solutions are =, y = and =, y = Check: equation : = + YES and = + YES equation : + = YES and ( ) + ( ) = YES 6 Substitute both pairs of values of and y into both equations to check your answers. DBO/GAT 08 5

Eample Solve + y = 5 and y + y = simultaneously. 5 y 5 y y y 5y y y 4y 5yy 4 y 5y 4 0 (y + 8)(y ) = 0 So y = 8 or y = Using + y = 5 When y = 8, + ( 8) = 5, = 4.5 When y =, + = 5, = Rearrange the first equation. Substitute 5 y for into the second equation. Notice how it is easier to substitute for than for y. Epand the brackets and simplify. 4 Factorise the quadratic equation. 5 Work out the values of y. 6 To find the value of, substitute both values of y into one of the original equations. So the solutions are = 4.5, y = 8 and =, y = Check: equation : 4.5 + ( 8) = 5 YES and ( ) + = 5 YES equation : ( 8) + 4.5 ( 8) = YES and () + ( ) = YES 7 Substitute both pairs of values of and y into both equations to check your answers. Practice Solve these simultaneous equations. y = + y = 6 + y = 0 + y = 0 y = 4 y = 9 + y = 5 + y = 7 5 y = 5 6 y = 5 y = + y = 5 7 y = + 5 8 y = + y = 5 + y = 4 9 y = 0 + y = y y = 8 y = 5 Etend y = y = + y = + y = DBO/GAT 08 6

Solving simultaneous equations graphically A LEVEL LINKS Scheme of work: c. Equations quadratic/linear simultaneous Key points You can solve any pair of simultaneous equations by drawing the graph of both equations and finding the point/points of intersection. Eamples Eample Solve the simultaneous equations y = 5 + and + y = 5 graphically. y = 5 y = 5 has gradient and y-intercept 5. y = 5 + has gradient 5 and y-intercept. Rearrange the equation + y = 5 to make y the subject. Plot both graphs on the same grid using the gradients and y-intercepts. Lines intersect at = 0.5, y = 4.5 Check: First equation y = 5 + : 4.5 = 5 0.5 + YES Second equation + y = 5: 0.5 + 4.5 = 5 YES The solutions of the simultaneous equations are the point of intersection. 4 Check your solutions by substituting the values into both equations. DBO/GAT 08 7

Eample Solve the simultaneous equations y = 4 and y = 4 + graphically. 0 4 y Construct a table of values and calculate the points for the quadratic equation. Plot the graph. Plot the linear graph on the same grid using the gradient and y-intercept. y = 4 has gradient and y-intercept 4. The line and curve intersect at =, y = and =, y = Check: First equation y = 4: = 4 YES = 4 YES Second equation y = 4 + : = 4 + = 4 + YES YES 4 The solutions of the simultaneous equations are the points of intersection. 5 Check your solutions by substituting the values into both equations. Practice Solve these pairs of simultaneous equations graphically. a y = and y = + b y = 5 and y = 7 5 c y = + 4 and y = Solve these pairs of simultaneous equations graphically. a + y = 0 and y = + 6 b 4 + y = and y = c + y + 4 = 0 and y = Hint Rearrange the equation to make y the subject. Solve these pairs of simultaneous equations graphically. a y = and y = 4 + DBO/GAT 08 8

b y = and y = c y = and y = + + 5 4 Solve the simultaneous equations + y = and + y = 5 graphically. Etend 5 a Solve the simultaneous equations + y = and + y = 4 i graphically ii algebraically to decimal places. b Which method gives the more accurate solutions? Eplain your answer. DBO/GAT 08 9

Linear inequalities A LEVEL LINKS Scheme of work: d. Inequalities linear and quadratic (including graphical solutions) Key points Solving linear inequalities uses similar methods to those for solving linear equations. When you multiply or divide an inequality by a negative number you need to reverse the inequality sign, e.g. < becomes >. Eamples Eample Solve 8 4 < 6 8 4 < 6 < 4 Divide all three terms by 4. Eample Solve 4 5 < 0 4 5 < 0 4 5 < Divide all three terms by 5. Eample Solve 5 < 7 5 < 7 < < 6 Add 5 to both sides. Divide both sides by. Eample 4 Solve 5 8 5 8 5 0 Subtract from both sides. Divide both sides by 5. Remember to reverse the inequality when dividing by a negative number. Eample 5 Solve 4( ) > (9 ) 4( ) > (9 ) 4 8 > 7 7 8 > 7 7 > 5 > 5 Epand the brackets. Add to both sides. Add 8 to both sides. 4 Divide both sides by 7. DBO/GAT 08 40

Practice Solve these inequalities. a 4 > 6 b 5 7 c + 4 d 5 < e 5 f 8 < Solve these inequalities. a 4 b 0 + c 7 > 5 5 Solve a 4 8 b 7 + 0 < 45 c 6 4 d 4 + 7 < e 4 5 < f 4 4 4 Solve these inequalities. a t + < t + 6 b (n ) n + 5 5 Solve. a ( ) > (4 ) + 4 b 5(4 ) > (5 ) + Etend 6 Find the set of values of for which + > and 4 > 6. DBO/GAT 08 4

Quadratic inequalities A LEVEL LINKS Scheme of work: d. Inequalities linear and quadratic (including graphical solutions) Key points First replace the inequality sign by = and solve the quadratic equation. Sketch the graph of the quadratic function. Use the graph to find the values which satisfy the quadratic inequality. Eamples Eample Find the set of values of which satisfy + 5 + 6 > 0 + 5 + 6 = 0 ( + )( + ) = 0 = or = Solve the quadratic equation by factorising. Sketch the graph of y = ( + )( + ) Identify on the graph where + 5 + 6 > 0, i.e. where y > 0 < or > 4 Write down the values which satisfy the inequality + 5 + 6 > 0 Eample Find the set of values of which satisfy 5 0 5 = 0 ( 5) = 0 = 0 or = 5 Solve the quadratic equation by factorising. Sketch the graph of y = ( 5) Identify on the graph where 5 0, i.e. where y 0 0 5 4 Write down the values which satisfy the inequality 5 0 DBO/GAT 08 4

Eample Find the set of values of which satisfy + 0 0 + 0 = 0 ( + )( + 5) = 0 = or = 5 y Solve the quadratic equation by factorising. Sketch the graph of y = ( + )( + 5) = 0 Identify on the graph where + 0 0, i.e. where y 0 5 O 5 Write down the values which satisfy the inequality + 0 0 Practice Find the set of values of for which ( + 7)( 4) 0 Find the set of values of for which 4 0 Find the set of values of for which 7 + < 0 4 Find the set of values of for which 4 + 4 > 0 5 Find the set of values of for which + 0 Etend Find the set of values which satisfy the following inequalities. 6 + 6 7 ( 9) < 0 8 6 5 + DBO/GAT 08 4

Sketching cubic and reciprocal graphs A LEVEL LINKS Scheme of work: e. Graphs cubic, quartic and reciprocal Key points The graph of a cubic function, which can be written in the form y = a + b + c + d, where a 0, has one of the shapes shown here. The graph of a reciprocal a function of the form y has one of the shapes shown here. To sketch the graph of a function, find the points where the graph intersects the aes. To find where the curve intersects the y-ais substitute = 0 into the function. To find where the curve intersects the -ais substitute y = 0 into the function. Where appropriate, mark and label the asymptotes on the graph. Asymptotes are lines (usually horizontal or vertical) which the curve gets closer to but never touches or crosses. Asymptotes usually occur with reciprocal functions. For eample, the a asymptotes for the graph of y are the two aes (the lines y = 0 and = 0). At the turning points of a graph the gradient of the curve is 0 and any tangents to the curve at these points are horizontal. A double root is when two of the solutions are equal. For eample ( ) ( + ) has a double root at =. When there is a double root, this is one of the turning points of a cubic function. DBO/GAT 08 44

Eamples Eample Sketch the graph of y = ( )( )( + ) To sketch a cubic curve find intersects with both aes and use the key points above for the correct shape. When = 0, y = (0 )(0 )(0 + ) = ( ) ( ) = 6 The graph intersects the y-ais at (0, 6) When y = 0, ( )( )( + ) = 0 So =, = or = The graph intersects the -ais at (, 0), (, 0) and (, 0) Find where the graph intersects the aes by substituting = 0 and y = 0. Make sure you get the coordinates the right way around, (, y). Solve the equation by solving = 0, = 0 and + = 0 Sketch the graph. a = > 0 so the graph has the shape: Eample Sketch the graph of y = ( + ) ( ) To sketch a cubic curve find intersects with both aes and use the key points above for the correct shape. When = 0, y = (0 + ) (0 ) = ( ) = 4 The graph intersects the y-ais at (0, 4) When y = 0, ( + ) ( ) = 0 So = or = Find where the graph intersects the aes by substituting = 0 and y = 0. Solve the equation by solving + = 0 and = 0 (, 0) is a turning point as = is a double root. The graph crosses the -ais at (, 0) a = > 0 so the graph has the shape: DBO/GAT 08 45

Practice Here are si equations. A 5 y B y = + 0 C y = + D y = E y = F + y = 5 Here are si graphs. i ii iii Hint Find where each of the cubic equations cross the y-ais. iv v vi a Match each graph to its equation. b Copy the graphs ii, iv and vi and draw the tangent and normal each at point P. Sketch the following graphs y = y = ( )( + ) 4 y = ( + )( + 4)( ) 5 y = ( + )( )( ) 6 y = ( ) ( + ) 7 y = ( ) ( ) 8 y = Etend 0 Sketch the graph of Hint: Look at the shape of y = a in the second key point. y Sketch the graph of 9 y = y DBO/GAT 08 46

Straight line graphs A LEVEL LINKS Scheme of work: a. Straight-line graphs, parallel/perpendicular, length and area problems Key points A straight line has the equation y = m + c, where m is the gradient and c is the y-intercept (where = 0). The equation of a straight line can be written in the form a + by + c = 0, where a, b and c are integers. When given the coordinates (, y ) and (, y ) of two points on a line the gradient is calculated using the y y formula m Eamples Eample A straight line has gradient and y-intercept. Write the equation of the line in the form a + by + c = 0. m = and c = So y = + + y = 0 + y 6 = 0 A straight line has equation y = m + c. Substitute the gradient and y-intercept given in the question into this equation. Rearrange the equation so all the terms are on one side and 0 is on the other side. Multiply both sides by to eliminate the denominator. Eample Find the gradient and the y-intercept of the line with the equation y + 4 = 0. y + 4 = 0 y = 4 4 y Gradient = m = 4 y-intercept = c = Make y the subject of the equation. Divide all the terms by three to get the equation in the form y = In the form y = m + c, the gradient is m and the y-intercept is c. DBO/GAT 08 47

Eample Find the equation of the line which passes through the point (5, ) and has gradient. m = y = + c = 5 + c = 5 + c c = y = Substitute the gradient given in the question into the equation of a straight line y = m + c. Substitute the coordinates = 5 and y = into the equation. Simplify and solve the equation. 4 Substitute c = into the equation y = + c Eample 4 Find the equation of the line passing through the points with coordinates (, 4) and (8, 7)., 8, y 4 and y 7 y y 7 4 m 8 6 y c 4 c c = y Substitute the coordinates into the y y equation m to work out the gradient of the line. Substitute the gradient into the equation of a straight line y = m + c. Substitute the coordinates of either point into the equation. 4 Simplify and solve the equation. 5 Substitute c = into the equation y c Practice Find the gradient and the y-intercept of the following equations. a y = + 5 b y = 7 c y = 4 d + y = 5 e y 7 = 0 f 5 + y 4 = 0 Copy and complete the table, giving the equation of the line in the form y = m + c. Gradient y-intercept Equation of the line 5 0 4 7 Hint Rearrange the equations to the form y = m + c DBO/GAT 08 48

Find, in the form a + by + c = 0 where a, b and c are integers, an equation for each of the lines with the following gradients and y-intercepts. a gradient, y-intercept 7 b gradient, y-intercept 0 c gradient, y-intercept 4 d gradient., y-intercept 4 Write an equation for the line which passes though the point (, 5) and has gradient 4. 5 Write an equation for the line which passes through the point (6, ) and has gradient 6 Write an equation for the line passing through each of the following pairs of points. a (4, 5), (0, 7) b (0, 6), ( 4, 8) c (, 7), (5, ) d (, 0), (4, 7) Etend 7 The equation of a line is y + 6 = 0. Write as much information as possible about this line. DBO/GAT 08 49

Parallel and perpendicular lines A LEVEL LINKS Scheme of work: a. Straight-line graphs, parallel/perpendicular, length and area problems Key points When lines are parallel they have the same gradient. A line perpendicular to the line with equation y = m + c has gradient. m Eamples Eample Find the equation of the line parallel to y = + 4 which passes through the point (4, 9). y = + 4 m = y = + c 9 = 4 + c 9 = 8 + c c = y = + As the lines are parallel they have the same gradient. Substitute m = into the equation of a straight line y = m + c. Substitute the coordinates into the equation y = + c 4 Simplify and solve the equation. 5 Substitute c = into the equation y = + c Eample Find the equation of the line perpendicular to y = which passes through the point (, 5). y = m = m y c 5 ( ) 5 = + c c = 4 y 4 c As the lines are perpendicular, the gradient of the perpendicular line is. m Substitute m = into y = m + c. Substitute the coordinates (, 5) into the equation y c 4 Simplify and solve the equation. 5 Substitute c = 4 into y c. DBO/GAT 08 50

Eample A line passes through the points (0, 5) and (9, ). Find the equation of the line which is perpendicular to the line and passes through its midpoint. 0, 9, y 5 and y y y 5 m 90 m y c 6 9 0 9 5 ( ) 9 Midpoint =,, 9 c 9 c 4 9 y 4 Substitute the coordinates into the y y equation m to work out the gradient of the line. As the lines are perpendicular, the gradient of the perpendicular line is. m Substitute the gradient into the equation y = m + c. 4 Work out the coordinates of the midpoint of the line. 5 Substitute the coordinates of the midpoint into the equation. 6 Simplify and solve the equation. 9 7 Substitute c into the equation 4 y c. Practice Find the equation of the line parallel to each of the given lines and which passes through each of the given points. a y = + (, ) b y = (, ) c + 4y + = 0 (6, ) d y + = 0 (8, 0) Find the equation of the line perpendicular to y = which passes through the point ( 5, ). Hint If m = a then the negative b reciprocal b m a Find the equation of the line perpendicular to each of the given lines and which passes through each of the given points. a y = 6 (4, 0) b y = + (, ) c 4y 4 = 0 (5, 5) d 5y + 5 = 0 (6, 7) 4 In each case find an equation for the line passing through the origin which is also perpendicular to the line joining the two points given. a (4, ), (, 9) b (0, ), ( 0, 8) DBO/GAT 08 5

Etend 5 Work out whether these pairs of lines are parallel, perpendicular or neither. a y = + b y = c y = 4 y = 7 + y = 0 4y + = d y + 5 = 0 e + 5y = 0 f y = 6 + y = y = + 7 6 y + = 0 6 The straight line L passes through the points A and B with coordinates ( 4, 4) and (, ), respectively. a Find the equation of L in the form a + by + c = 0 The line L is parallel to the line L and passes through the point C with coordinates ( 8, ). b Find the equation of L in the form a + by + c = 0 The line L is perpendicular to the line L and passes through the origin. c Find an equation of L DBO/GAT 08 5

The cosine rule A LEVEL LINKS Scheme of work: 4a. Trigonometric ratios and graphs Tetbook: Pure Year, 9. The cosine rule Key points a is the side opposite angle A. b is the side opposite angle B. c is the side opposite angle C. You can use the cosine rule to find the length of a side when two sides and the included angle are given. To calculate an unknown side use the formula a b c bc cos A. Alternatively, you can use the cosine rule to find an unknown angle if the lengths of all three sides are given. b c a To calculate an unknown angle use the formula cos A. bc Eamples Eample 4 Work out the length of side w. Give your answer correct to significant figures. Always start by labelling the angles and sides. a b c bc cos A w 8 7 8 7 cos 45 w =.804 040 5... w =.8040405 w = 5.8 cm Write the cosine rule to find the side. Substitute the values a, b and A into the formula. 4 Use a calculator to find w and then w. 5 Round your final answer to significant figures and write the units in your answer. DBO/GAT 08 5

Eample 5 Work out the size of angle θ. Give your answer correct to decimal place. Always start by labelling the angles and sides. b c a cos A bc 0 7 5 cos 07 76 cos 40 θ =.878 49... θ =.9 Write the cosine rule to find the angle. Substitute the values a, b and c into the formula. 4 Use cos to find the angle. 5 Use your calculator to work out cos ( 76 40). 6 Round your answer to decimal place and write the units in your answer. Practice Work out the length of the unknown side in each triangle. Give your answers correct to significant figures. a b c d DBO/GAT 08 54

Calculate the angles labelled θ in each triangle. Give your answer correct to decimal place. a b c d a Work out the length of WY. Give your answer correct to significant figures. b Work out the size of angle WXY. Give your answer correct to decimal place. DBO/GAT 08 55

The sine rule A LEVEL LINKS Scheme of work: 4a. Trigonometric ratios and graphs Tetbook: Pure Year, 9. The sine rule Key points a is the side opposite angle A. b is the side opposite angle B. c is the side opposite angle C. You can use the sine rule to find the length of a side when its opposite angle and another opposite side and angle are given. a b c To calculate an unknown side use the formula. sin A sin B sin C Alternatively, you can use the sine rule to find an unknown angle if the opposite side and another opposite side and angle are given. sin sin sin To calculate an unknown angle use the formula A B C. a b c Eamples Eample 6 Work out the length of side. Give your answer correct to significant figures. Always start by labelling the angles and sides. a b sin A sin B 0 sin6 sin75 0 sin 6 sin75 = 6.09 cm Write the sine rule to find the side. Substitute the values a, b, A and B into the formula. 4 Rearrange to make the subject. 5 Round your answer to significant figures and write the units in your answer. DBO/GAT 08 56

Eample 7 Work out the size of angle θ. Give your answer correct to decimal place. Always start by labelling the angles and sides. sin A sin B a b sin sin7 8 4 8sin7 sin 4 θ = 7. Write the sine rule to find the angle. Substitute the values a, b, A and B into the formula. 4 Rearrange to make sin θ the subject. 5 Use sin to find the angle. Round your answer to decimal place and write the units in your answer. Practice Find the length of the unknown side in each triangle. Give your answers correct to significant figures. a b c d DBO/GAT 08 57

Calculate the angles labelled θ in each triangle. Give your answer correct to decimal place. a b c d a Work out the length of QS. Give your answer correct to significant figures. b Work out the size of angle RQS. Give your answer correct to decimal place. DBO/GAT 08 58

Areas of triangles A LEVEL LINKS Scheme of work: 4a. Trigonometric ratios and graphs Tetbook: Pure Year, 9. Areas of triangles Key points a is the side opposite angle A. b is the side opposite angle B. c is the side opposite angle C. The area of the triangle is sin ab C. Eamples Eample 8 Find the area of the triangle. Always start by labelling the sides and angles of the triangle. Area = sin ab C Area = 8 5 sin8 Area = 9.805 6... Area = 9.8 cm State the formula for the area of a triangle. Substitute the values of a, b and C into the formula for the area of a triangle. 4 Use a calculator to find the area. 5 Round your answer to significant figures and write the units in your answer. DBO/GAT 08 59

Practice Work out the area of each triangle. Give your answers correct to significant figures. a b c The area of triangle XYZ is. cm. Work out the length of XZ. Hint: Rearrange the formula to make a side the subject. Etend Find the size of each lettered angle or side. Give your answers correct to significant figures. a b Hint: For each one, decide whether to use the cosine or sine rule. c d DBO/GAT 08 60

4 The area of triangle ABC is 86.7 cm. Work out the length of BC. Give your answer correct to significant figures. DBO/GAT 08 6

Rearranging equations A LEVEL LINKS Scheme of work: 6a. Definition, differentiating polynomials, second derivatives Tetbook: Pure Year,. Gradients of curves Key points To change the subject of a formula, get the terms containing the subject on one side and everything else on the other side. You may need to factorise the terms containing the new subject. Eamples Eample Make t the subject of the formula v = u + at. v = u + at v u = at v u t a Get the terms containing t on one side and everything else on the other side. Divide throughout by a. Eample Make t the subject of the formula r = t πt. r = t πt r = t( π) r t All the terms containing t are already on one side and everything else is on the other side. Factorise as t is a common factor. Divide throughout by π. Eample Make t the subject of the formula t r t. 5 t r t 5 t + r = 5t r = t r t Remove the fractions first by multiplying throughout by 0. Get the terms containing t on one side and everything else on the other side and simplify. Divide throughout by. DBO/GAT 08 6

Eample 4 Make t the subject of the formula t 5 r t r(t ) = t + 5 rt r = t + 5 rt t = 5 + r t(r ) = 5 + r 5 r t r t 5 r. t Remove the fraction first by multiplying throughout by t. Epand the brackets. Get the terms containing t on one side and everything else on the other side. 4 Factorise the LHS as t is a common factor. 5 Divide throughout by r. Practice Change the subject of each formula to the letter given in the brackets. C = πd [d] P = l + w [w] D = S T [T] 4 7 0 q r p t [t] 5 u = at t [t] 6 V = a + 4 [] y 7 7 y a b c [y] 8 [a] 9 [d] a d 7g 9 h [g] e(9 + ) = e + [e] y [] g 4 Make r the subject of the following formulae. 4 a A = πr b V r c P = πr + r d V r h 4 Make the subject of the following formulae. y ab 4 c z a b z cd d py a b 5 Make sin B the subject of the formula sin A sin B 6 Make cos B the subject of the formula b = a + c ac cos B. Etend 7 Make the subject of the following equations. p p p a ( s t ) b ( a y) ( y) q q q DBO/GAT 08 6

Area under a graph A LEVEL LINKS Scheme of work: 7b. Definite integrals and areas under curves Key points To estimate the area under a curve, draw a chord between the two points you are finding the area between and straight lines down to the horizontal ais to create a trapezium. The area of the trapezium is an approimation for the area under a curve. The area of a trapezium = ( ) h a b Eamples Eample Estimate the area of the region between the curve y = ( )( + ) and the -ais from = 0 to =. Use three strips of width unit. 0 y = ( )( + ) 6 6 4 0 Trapezium : a 6 0 6, b 6 0 6 Trapezium : a 6 0 6, b 4 0 4 Trapezium : a 4 0 4, a 0 0 0 Use a table to record the value of y on the curve for each value of. Work out the dimensions of each trapezium. The distances between the y-values on the curve and the -ais give the values for a. (continued on net page) DBO/GAT 08 64

h( a b) (6 6) 6 h( a b) (6 4) 5 h( a b) (4 0) Area = 6 + 5 + = units Work out the area of each trapezium. h = since the width of each trapezium is unit. 4 Work out the total area. Remember to give units with your answer. Eample Estimate the shaded area. Use three strips of width units. 4 6 8 0 y 7 4 4 6 8 0 y 7 6 5 4 Trapezium : a 7 7 0, b 6 6 Trapezium : a 6 6, b 5 8 Trapezium : a 5 8, a 4 4 0 h( a b) (0 6) 6 h( a b) (6 8) 4 h( a b) (8 0) 8 Area = 6 + 4 + 8 = 8 units Use a table to record y on the curve for each value of. Use a table to record y on the straight line for each value of. Work out the dimensions of each trapezium. The distances between the y-values on the curve and the y-values on the straight line give the values for a. 4 Work out the area of each trapezium. h = since the width of each trapezium is units. 5 Work out the total area. Remember to give units with your answer. DBO/GAT 08 65

Practice Estimate the area of the region between the curve y = (5 )( + ) and the -ais from = to = 5. Use four strips of width unit. Hint: For a full answer, remember to include units. Estimate the shaded area shown on the aes. Use si strips of width unit. Estimate the area of the region between the curve y = 8 + 8 and the -ais from = to = 6. Use four strips of width unit. 4 Estimate the shaded area. Use si strips of width unit. DBO/GAT 08 66

5 Estimate the area of the region between the curve y = 4 + 5 and the -ais from = 5 to =. Use si strips of width unit. 6 Estimate the shaded area. Use four strips of equal width. 7 Estimate the area of the region between the curve y = + + 5 and the -ais from = to = 5. Use si strips of equal width. 8 Estimate the shaded area. Use seven strips of equal width. DBO/GAT 08 67