Edexcel GCE Core Mathematics C2 Advanced Subsidiary

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1 Centre No. Candidate No. Paper Reference Paper Reference(s) 6664/01 Edecel GCE Core Mathematics C Advanced Subsidiary Thursday 4 May 01 Morning Time: 1 hour 0 minutes Materials required for eamination Mathematical Formulae (Pink) Surname Signature Items included with question papers Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation or symbolic differentiation/integration, or have retrievable mathematical formulae stored in them. Initial(s) Eaminer s use only Team Leader s use only Question Leave Number Blank Instructions to Candidates In the boes above, write your centre number, candidate number, your surname, initials and signature. Check that you have the correct question paper. Answer ALL the questions. You must write your answer for each question in the space following the question. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. The marks for individual questions and the parts of questions are shown in round brackets: e.g. (). There are 9 questions in this question paper. The total mark for this paper is 75. There are 8 pages in this question paper. Any blank pages are indicated. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You should show sufficient working to make your methods clear to the Eaminer. Answers without working may not gain full credit. This publication may be reproduced only in accordance with Pearson Education Ltd copyright policy. 01 Pearson Education Ltd. Printer s Log. No. P40685A W850/R6664/ /5/5/ *P40685A018* Total Turn over

2 1. Find the first terms, in ascending powers of, of the binomial epansion of Leave blank ( ) 5 giving each term in its simplest form. (4) Q1 (Total 4 marks) *P40685A08* Turn over

3 . Find the values of such that Leave blank log log ( ) = (5) 4 *P40685A048*

4 . y Leave blank C T r P Q O L Figure 1 The circle C with centre T and radius r has equation + y 0 16y + 19 = 0 (a) Find the coordinates of the centre of C. (b) Show that r = 5 () () The line L has equation = 1 and crosses C at the points P and Q as shown in Figure 1. (c) Find the y coordinate of P and the y coordinate of Q. () Given that, to decimal places, the angle PTQ is radians, (d) find the perimeter of the sector PTQ. () 6 *P40685A068*

5 4. f() = Leave blank (a) Use the factor theorem to show that ( + ) is a factor of f(). () (b) Factorise f() completely. (4) 10 *P40685A0108*

6 5. y Leave blank y = 10 8 A R O B y = 10 Figure Figure shows the line with equation y = 10 and the curve with equation y = 10 8 The line and the curve intersect at the points A and B, and O is the origin. (a) Calculate the coordinates of A and the coordinates of B. (5) The shaded area R is bounded by the line and the curve, as shown in Figure. (b) Calculate the eact area of R. (7) 1 *P40685A018*

7 6. (a) Show that the equation Leave blank tan = 5 sin can be written in the form (1 5 cos ) sin = 0 () (b) Hence solve, for 0 180, tan = 5 sin giving your answers to 1 decimal place where appropriate. You must show clearly how you obtained your answers. (5) 16 *P40685A0168*

8 7. y = ( + ) Leave blank (a) Complete the table below, giving the values of y to decimal places y () (b) Use the trapezium rule with all the values of y from your table to find an approimation for the value of 1 ( + ) d 0 You must show clearly how you obtained your answer. (4) 0 *P40685A008*

9 8. Leave blank h mm Figure mm A manufacturer produces pain relieving tablets. Each tablet is in the shape of a solid circular cylinder with base radius mm and height h mm, as shown in Figure. Given that the volume of each tablet has to be 60 mm, (a) epress h in terms of, (b) show that the surface area, A mm, of a tablet is given by A = + 10 (1) () The manufacturer needs to minimise the surface area A mm, of a tablet. (c) Use calculus to find the value of for which A is a minimum. (d) Calculate the minimum value of A, giving your answer to the nearest integer. (5) () (e) Show that this value of A is a minimum. () *P40685A08*

10 9. A geometric series is a + ar + ar +... Leave blank (a) Prove that the sum of the first n terms of this series is given by S n n a( 1 r ) = 1 r (4) The third and fifth terms of a geometric series are 5.4 and respectively and all the terms in the series are positive. For this series find, (b) the common ratio, (c) the first term, () () (d) the sum to infinity. () 6 *P40685A068*

11 Centre No. Candidate No. Paper Reference Paper Reference(s) 6664/01 Edecel GCE Core Mathematics C Advanced Subsidiary Friday 1 January 01 Morning Time: 1 hour 0 minutes Materials required for eamination Mathematical Formulae (Pink) Surname Signature Items included with question papers Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation or symbolic differentiation/integration, or have retrievable mathematical formulae stored in them. Initial(s) Eaminer s use only Team Leader s use only Question Leave Number Blank Instructions to Candidates In the boes above, write your centre number, candidate number, your surname, initials and signature. Check that you have the correct question paper. Answer ALL the questions. You must write your answer for each question in the space following the question. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. The marks for individual questions and the parts of questions are shown in round brackets: e.g. (). There are 9 questions in this question paper. The total mark for this paper is 75. There are 8 pages in this question paper. Any blank pages are indicated. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You should show sufficient working to make your methods clear to the Eaminer. Answers without working may not gain full credit. This publication may be reproduced only in accordance with Pearson Education Ltd copyright policy. 01 Pearson Education Ltd. Printer s Log. No. P4008A W850/R6664/ /4/4/4 *P4008A018* Total Turn over

12 Leave blank 1. A geometric series has first term a = 60 and common ratio r = 7 8 Giving your answers to significant figures where appropriate, find (a) the 0th term of the series, (b) the sum of the first 0 terms of the series, () () (c) the sum to infinity of the series. () *P4008A08*

13 Leave blank. A circle C has centre ( 1, 7 ) and passes through the point (, 0 0 ). Find an equation for C. (4) 4 *P4008A048*

14 . (a) Find the first 4 terms of the binomial epansion, in ascending powers of, of ( ) Leave blank giving each term in its simplest form. (4) (b) Use your epansion to estimate the value of (. 1 05), 8 giving your answer to 4 decimal places. () 6 *P4008A068*

15 4. Given that y =, Leave blank (a) show that log y = 1+ log () (b) Hence, or otherwise, solve the equation 1+ log = log ( 8 9) () 8 *P4008A088*

16 5. f( ) = + a + b +, where a and b are constants. Leave blank Given that when f( ) is divided by ( + ) the remainder is 7, (a) show that a b= 6 () Given also that when f( ) is divided by ( 1 ) the remainder is 4, (b) find the value of a and the value of b. (4) 1 *P4008A018*

17 6. y = 1 Leave blank y = O R 4 Figure 1 Figure 1 shows the graph of the curve with equation y = , > 0 The finite region R, bounded by the lines = 1, the -ais and the curve, is shown shaded in Figure 1. The curve crosses the -ais at the point ( 40, ). (a) Complete the table with the values of y corresponding to = and y () (b) Use the trapezium rule with all the values in the completed table to find an approimate value for the area of R, giving your answer to decimal places. (4) (c) Use integration to find the eact value for the area of R. (5) 14 *P4008A0148*

18 7. B Leave blank 6cm A 0.95 = Figure = D R C Figure shows ABC, a sector of a circle of radius 6 cm with centre A. Given that the size of angle BAC is 0.95 radians, find (a) the length of the arc BC, (b) the area of the sector ABC. () () The point D lies on the line AC and is such that AD = BD. The region R, shown shaded in Figure, is bounded by the lines CD, DB and the arc BC. (c) Show that the length of AD is 5.16 cm to significant figures. () Find (d) the perimeter of R, () (e) the area of R, giving your answer to significant figures. (4) 18 *P4008A0188*

19 8. Leave blank y Figure Figure shows a flowerbed. Its shape is a quarter of a circle of radius metres with two equal rectangles attached to it along its radii. Each rectangle has length equal to metres and width equal to y metres. Given that the area of the flowerbed is 4 m, (a) show that 16 ϖ y = 8 () (b) Hence show that the perimeter P metres of the flowerbed is given by the equation 8 P = + (c) Use calculus to find the minimum value of P. () (5) (d) Find the width of each rectangle when the perimeter is a minimum. Give your answer to the nearest centimetre. () *P4008A08*

20 o 1 9. (i) Find the solutions of the equation sin( 15 ) =, for which o (6) Leave blank (ii) y O P Q R Figure 4 Figure 4 shows part of the curve with equation y = sin( a b), where a> 0, 0< b< ϖ The curve cuts the -ais at the points P, Q and R as shown. Given that the coordinates of P, Q and R are ϖ ( 10 ) ( ), 0, ϖ, 0 and 11 5 ( 10 ), 0 find the values of a and b. ϖ respectively, (4) 6 *P4008A068*

21 Centre No. Candidate No. Paper Reference Paper Reference(s) 6664/01 Edecel GCE Core Mathematics C Advanced Subsidiary Thursday 6 May 011 Morning Time: 1 hour 0 minutes Materials required for eamination Mathematical Formulae (Pink) Surname Signature Items included with question papers Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation or symbolic differentiation/integration, or have retrievable mathematical formulae stored in them. Instructions to Candidates In the boes above, write your centre number, candidate number, your surname, initials and signature. Check that you have the correct question paper. Answer ALL the questions. You must write your answer for each question in the space following the question. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Initial(s) Eaminer s use only Team Leader s use only Question Number Leave Blank Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. The marks for individual questions and the parts of questions are shown in round brackets: e.g. (). There are 9 questions in this question paper. The total mark for this paper is 75. There are pages in this question paper. Any blank pages are indicated. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You should show sufficient working to make your methods clear to the Eaminer. Answers without working may not gain full credit. This publication may be reproduced only in accordance with Edecel Limited copyright policy. 011 Edecel Limited. Printer s Log. No. P8158A W850/R6664/ /5// *P8158A01* Total Turn over

22 1. f( ) = Leave blank (a) Find the remainder when f () is divided by ( 1). (b) Use the factor theorem to show that (+1) is a factor of f (). () () (c) Factorise f () completely. (4) *P8158A0*

23 . (a) Find the first terms, in ascending powers of, of the binomial epansion of Leave blank ( + b) 5 where b is a non-zero constant. Give each term in its simplest form. (4) Given that, in this epansion, the coefficient of is twice the coefficient of, (b) find the value of b. () 4 *P8158A04*

24 . Find, giving your answer to significant figures where appropriate, the value of for which Leave blank (a) 5 = 10, () (b) log ( ) = 1. () 6 *P8158A06*

25 4. The circle C has equation + y + 4 y 11 = 0 Leave blank Find (a) the coordinates of the centre of C, (b) the radius of C, () () (c) the coordinates of the points where C crosses the y-ais, giving your answers as simplified surds. (4) 8 *P8158A08*

26 5. O Leave blank 6cm π C A B Figure 1 The shape shown in Figure 1 is a pattern for a pendant. It consists of a sector OAB of a circle centre O, of radius 6 cm, and angle AOB = π. The circle C, inside the sector, touches the two straight edges, OA and OB, and the arc AB as shown. Find (a) the area of the sector OAB, (b) the radius of the circle C. () () The region outside the circle C and inside the sector OAB is shown shaded in Figure 1. (c) Find the area of the shaded region. () 1 *P8158A01*

27 6. The second and third terms of a geometric series are 19 and 144 respectively. Leave blank For this series, find (a) the common ratio, (b) the first term, (c) the sum to infinity, () () () (d) the smallest value of n for which the sum of the first n terms of the series eceeds (4) 16 *P8158A016*

28 7. (a) Solve for 0 60, giving your answers in degrees to 1 decimal place, Leave blank sin ( + 45 ) = (4) (b) Find, for 0, all the solutions of sin + = 7cos giving your answers in radians. You must show clearly how you obtained your answers. (6) 0 *P8158A00*

29 8. Leave blank Figure A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, cm, as shown in Figure. The volume of the cuboid is 81 cubic centimetres. (a) Show that the total length, L cm, of the twelve edges of the cuboid is given by L = (b) Use calculus to find the minimum value of L. () (6) (c) Justify, by further differentiation, that the value of L that you have found is a minimum. () 4 *P8158A04*

30 9. y y = Leave blank R B y = + 4 A O Figure The straight line with equation y = + 4 cuts the curve with equation y = at the points A and B, as shown in Figure. (a) Use algebra to find the coordinates of the points A and B. (4) The finite region R is bounded by the straight line and the curve and is shown shaded in Figure. (b) Use calculus to find the eact area of R. (7) 8 *P8158A08*

31 Centre No. Candidate No. Paper Reference(s) 6664/01 Edecel GCE Core Mathematics C Advanced Subsidiary Monday 10 January 011 Morning Time: 1 hour 0 minutes Materials required for eamination Mathematical Formulae (Pink) Paper Reference Surname Signature Items included with question papers Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them. Instructions to Candidates In the boes above, write your centre number, candidate number, your surname, initials and signature. Check that you have the correct question paper. Answer ALL the questions. You must write your answer to each question in the space following the question. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Initial(s) Eaminer s use only Team Leader s use only Question Number Leave Blank Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. The marks for individual questions and the parts of questions are shown in round brackets: e.g. (). There are 10 questions in this question paper. The total mark for this paper is 75. There are 8 pages in this question paper. Any blank pages are indicated. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You should show sufficient working to make your methods clear to the Eaminer. Answers without working may not gain full credit. This publication may be reproduced only in accordance with Edecel Limited copyright policy. 011 Edecel Limited. Printer s Log. No. H540A W850/R6664/ /5// *H540A018* Total Turn over

32 1. f() = a + b Leave blank where a and b are constants. When f() is divided by ( 1), the remainder is 7. (a) Show that a + b =. () When f() is divided by ( + ), the remainder is 8. (b) Find the value of a and the value of b. (5) *H540A08*

33 . In the triangle ABC, AB = 11 cm, BC = 7 cm and CA = 8 cm. Leave blank (a) Find the size of angle C, giving your answer in radians to significant figures. () (b) Find the area of triangle ABC, giving your answer in cm to significant figures. () 4 *H540A048*

34 . The second and fifth terms of a geometric series are 750 and 6 respectively. Leave blank Find (a) the common ratio of the series, (b) the first term of the series, () () (c) the sum to infinity of the series. () 6 *H540A068*

35 4. y Leave blank C A O R B Figure 1 Figure 1 shows a sketch of part of the curve C with equation y = ( + 1)( 5) The curve crosses the -ais at the points A and B. (a) Write down the -coordinates of A and B. (1) The finite region R, shown shaded in Figure 1, is bounded by C and the -ais. (b) Use integration to find the area of R. (6) 8 *H540A088*

36 40 40! 5. Given that =, 4 4! b! Leave blank (a) write down the value of b. (1) In the binomial epansion of (1 + ) 40, the coefficients of 4 and 5 are p and q respectively. (b) Find the value of q p. () 10 *H540A0108*

37 6. y = (a) Complete the table below, giving the values of y to decimal places. 5 Leave blank y (b) Use the trapezium rule, with all the values of y from your table, to find an 5 approimate value for d. () (4) y A S B O Figure 5 Figure shows a sketch of part of the curve with equation y =, > 1. At the points A and B on the curve, = and = respectively. The region S is bounded by the curve, the straight line through B and (, 0), and the line through A parallel to the y-ais. The region S is shown shaded in Figure. (c) Use your answer to part (b) to find an approimate value for the area of S. () 1 *H540A018*

38 7. (a) Show that the equation Leave blank sin + 7sin = cos 4 can be written in the form 4sin + 7sin + = 0 () (b) Hence solve, for 0 < 60, sin + 7sin = cos 4 giving your answers to 1 decimal place where appropriate. (5) 16 *H540A0168*

39 8. (a) Sketch the graph of y = 7,, showing the coordinates of any points at which the graph crosses the aes. () Leave blank (b) Solve the equation 7 4(7 ) + = 0 giving your answers to decimal places where appropriate. (6) 18 *H540A0188*

40 9. The points A and B have coordinates (, 11) and (8, 1) respectively. Leave blank Given that AB is a diameter of the circle C, (a) show that the centre of C has coordinates (, 6), (b) find an equation for C. (c) Verify that the point (10, 7) lies on C. (1) (4) (1) (d) Find an equation of the tangent to C at the point (10, 7), giving your answer in the form y = m + c, where m and c are constants. (4) *H540A08*

41 10. The volume V cm of a bo, of height cm, is given by Leave blank V = 4 (5 ), 0 5 (a) Find d V d. (b) Hence find the maimum volume of the bo. (4) (4) (c) Use calculus to justify that the volume that you found in part (b) is a maimum. () 6 *H540A068*

42 Mark Scheme (Results) Summer 01 GCE Core Mathematics C (6664) Paper 1

43 Summer Core Mathematics Mark Scheme Question number Scheme Marks ( ) =... + ( ) + ( ) +.., =, 40, + 70 B1, A1, A1 Notes Total 4 : The method mark is awarded for an attempt at Binomial to get the second and/or third term need correct binomial coefficient combined with correct power of. Ignore errors (or omissions) in powers of or or sign or bracket errors. Accept any notation for 5 C 1 and 5 C, 5 e.g. 1 and 5 (unsimplified) or 5 and 10 from Pascal s triangle This mark may be given if no working is shown, but either or both of the terms including is correct. Special Case Alternative Method Method 1: 5 B1: must be simplified to ( writing just is B0 ). must be the only constant term in the final answer- so is B0 but may be eligible for A0A0. A1: is cao and is for 40. (not +-40) The is required for this mark A1: is c.a.o and is for 70 (can follow omission of negative sign in working) A list of correct terms may be given credit i.e. series appearing on different lines 4 Ignore etra terms in and/or (isw) Special Case: Descending powers of would be ( ) + 5 ( ) + ( ) +.. i.e This is a misread but award as s.c. B1A0A0 if completely correct or B0A0A0 for correct binomial coefficient in any form with the correct power of ( ) = (1 + ( ) + ( ) +.. ) is B0A0A0 { The is 1 for the epression in the bracket and as in first method need correct binomial coefficient combined with correct power of. Ignore bracket errors or errors (or omissions) in powers of or or sign or bracket errors} answers must be simplified to =, 40, + 70 for full marks (awarded as before) ( ) = (1 + ( ) + ( ) +.. ) would also be awarded B0A0A0 1 Method : Multiplying out : B1 for and A1A1 for other terms with awarded if or ^ term is correct. Completely correct is 4/4

44 Question number Scheme Marks log = log B1 log log ( ) = log = 9 A1 o.e. Solves = 0 to give =... =, = 6 A1 Notes B1 for this correct use of power rule (may be implied) : for correct use of subtraction rule (or addition rule) for logs N.B. log log ( ) = log is M0 A1. for correct equation without logs (Allow any correct equivalent including Total 5 instead of 9.) for attempting to solve = 0 to give = (see notes on marking quadratics) A1 for these two correct answers Alternative Method log = + log ( ) is B1, Common Slips + log ( ) = needs to be followed by so ( ) = 9( ) for A1 Here is for complete method i.e.correct use of powers after logs are used correctly log log + log = may obtain B1 if log appears but the statement is M0 and so leads to no further marks log log ( ) = so log log ( ) = 1and log = 1 can earn for correct subtraction rule following error, but no other marks Special Case log = leading to = 9 and then to =, =6, usually earns B1M0A0, but may log( ) then earn A1 (special case) so /5 [ This recovery after uncorrected error is very common] Trial and error, Use of a table or just stating answer with both = and =6 should be awarded B0M0A0 then final A1 i.e. /5

45 Question number Scheme Marks Obtain ( ± 10) and ( y ± 8) (a) Obtain ( 10) and ( y 8) A1 (b) Centre is (10, 8). N.B. This may be indicated on diagram only as (10, 8) See ( ± 10) + ( y ± 8) = 5 (= r ) or A1 ( r = )"100" + "64" 19 () (c) (d) r = 5 * (this is a printed answer so need one of the above two reasons) A1 Use = 1 in either form of equation of circle and solve resulting quadratic to give y = ( ) ( y ) ( y ) e.g = = 5 8 = 16 or y y y y = = 0 so y= so y = y = 4 or 1 ( on EPEN mark one correct value as A1A0 and both correct as A1 A1) () A1, A1 () Use of rθ with r = 5 and θ = (may be implied by 9.75) Alternatives (a) OR (b) OR (c) Notes (a) (b) Perimeter PTQ = r + their arc PQ (Finding perimeter of triangle is M0 here) = or 19.8 or 19. Method : From + y + g + fy + c = 0 Centre is ( g, f ), and so centre is (10, 8). centre is ( ± g, ± f ) Method : Use any value of y to give two points (L and M) on circle. co-ordinate of mid point of LM is 10 and Use any value of to give two points (P and Q) on circle. y co-ordinate of mid point of PQ is 8 (Centre chord theorem). (10,8) is A1A1 Method : Using r = 5 * g + f c or ( r = )"100" + "64" 19 Method : Use point on circle with centre to find radius. Eg r = 5 * (1 10) + (1 8) Divide triangle PTQ and use Pythagoras with r (1 "10") = h, then evaluate "8 ± h" - (N.B. Could use,4,5 Triangle and 8 ± 4 ). Accuracy as before Mark (a) and (b) together as in scheme and can be implied by ( ± 10, ± 8). Correct centre (10, 8) implies A1A1 for a correct method leading to r =, or or for using equation of circle in A1 () 11 marks A1, A1 A1 A1 A1 A1 cao r = "100" + "64" 19 (not ) ( 10) ( y 8) k ± + ± = form to identify r= k = 5 or r = ) rd A1 r = 5 (NB This is a given answer so should follow Special case: if centre is given as (-10, -8) or (10, -8) or (-10, 8) allow A1 for r = 5 worked correctly as r = (d) Full marks available for calculation using major sector so Use of rθ with r = 5 and 4.48 θ = leading to perimeter of.14 for major sector () ()

46 Question number Scheme Marks 4 (a) ( ) ( ) ( ) ( ) f = = 0 so (+) is a factor A1 () (b) f ( ) = ( + )( ) A1 f ( ) = ( + )( )( 4) d A1 (4) 6 marks Notes (a) : Attempts f( ± ) (Long division is M0) A1 : is for =0 and conclusion Note: Stating hence factor or it is a factor or a (tick) or QED is fine for the conclusion. Note also that a conclusion can be implied from a preamble, eg: If f ( ) = 0, ( + ) is a factor. (Not just f(-)=0) (b) 1 st : Attempts long division by correct factor or other method leading to obtaining ( ± a ± b), a 0, b 0, even with a remainder. Working need not be seen as could be done by inspection. Or Alternative Method : 1 st : Use ( + )( a + b + c) = with epansion and comparison of coefficients to obtain a = and to obtain values for b and c 1 st A1: For seeing ( ). [Can be seen here in (b) after work done in (a)] nd : Factorises quadratic. (see rule for factorising a quadratic). This is dependent on the previous method mark being awarded and needs factors nd A1: is cao and needs all three factors together. Ignore subsequent work (such as a solution to a quadratic equation.) Note: Some candidates will go from {( + ) }( ) to { } list all three factors. Award these responses A1M0A0. =, =, 4, and not Finds = 4 and = 1.5 by factor theorem, formula or calculator and produces factors f ( ) = ( + )( )( 4) or f ( ) = ( + )( 1.5)( 4) o.e. is full marks f ( ) = ( + )( 1.5)( 4) loses last A1

47 Question number Method 1 5 (a) Puts 10 = 10 8 and rearranges to give three term quadratic Solves their " = 0" using acceptable method as in general principles to give = Obtains =, = 9 (may be on diagram or in part (b) in limits) Substitutes their into a given equation to give y = (may be on diagram) Scheme Or puts y = y y 10(10 ) (10 ) 8 and rearranges to give three term quadratic Solves their " y 9y + 8 = 0" using acceptable method as in general principles to give y = Obtains y = 8, y = 1 (may be on diagram) Substitutes their y into a given equation to give = (may be on diagram or in part (b)) Marks A1 (b) y = 8, y = 1 =, = 9 A1 (5) 10 (10 8) d = 8 { + c } A1 A = = 90 = 88 or 66 1 ( ) ( ) Area of trapezium = (8 + 1)(9 ) = 1.5 d B1 1 So area of R is = 57 or 4 A1 6 6 cao (7) 1 marks Notes (a) First : See scheme Second : See notes relating to solving quadratics Third : This may be awarded if one substitution is made Two correct Answers following tables of values, or from Graphical calculator are 5/5 Just one pair of correct coordinates no working or from table is M0M0A0A0 n n 1 (b) : + for any one term. 1 st A1: at least two out of three terms correct nd A1: All three correct d: Substitutes 9 and (or limits from part(a)) into an integrated function and subtracts, either way round 10 (NB: If candidate changes all signs to get ( ) d = { + c } This is A1 A1 Then uses limits d and trapezium is B1 Needs to change sign of value obtained from integration for final A1 so is M0A0 ) B1: Obtains 1.5 for area under line using any correct method (could be integration) or triangle minus triangle or rectangle plus triangle [may be implied by correct 57 1/6 ] : Their Area under curve Their Area under line (if integrate both need same limits) A1: Accept 57.16recurring but not PTO for Alternative method

48 Method for (b) Area of R 9 = (10 8) (10 ) d rd (in (b) ): Uses difference between two functions in integral. 9 n n 1 M: d + for any one term. A1 at least two out of these three A1 11 simplified terms = + 18 { + c} Correct integration. (Ignore + c). A = ( ) ( ) Substitutes 9 and (or limits from part(a)) into an integrated function and subtracts, either way round. d This mark is implied by final answer which rounds to 57. B1 See above working(allow bracketing errors) to decide to award rd mark for (b) here: ( 16 ) = 57 cao A1 6 (7) Special case of above method d 11 = + 18 { + c} = (...) (...) This mark is implied by final answer which rounds to 57. (not -57.) A1A1 D B1 Special Case Notes Difference of functions implied (see above epression) Integrates epression in y e.g ( 16 ) = 57 cao A1 y y + = : This can have first " 9 8 0" in part (b) and no other marks. (It is not a method for finding this area) Take away trapezium again having used Method loses last two marks Common Error: Integrates Integrates is likely to be A1A0dB0A0 11 is likely to e A0A0dB0A0 (7) Writing 9 (10 8) (10 ) d only earns final M mark

49 Question number Scheme Marks 6(a) States or uses sin tan = cos sin 5sin sin 5sin cos 0 sin (1 5cos ) 0 cos = = = A1 () (b) sin = 0 gives = 0, 180, 60 so = 0, 90, 180 B1 for two correct answers, second B1 for all three correct. Ecess in range lose last B1 cos = gives = (or 78.5 or 78.4) or = (or 81.6) 1 5 = 9. (or 9.), (or 141) B1, B1 A1, A1 (5) Notes 7 marks sinθ (a) : Statement that tanθ = or Replacement of tan (wherever it appears). Must be a correct cosθ statement but may involve θ instead of. A1: the answer is given so all steps should be given. 1 N.B. sin 5sin cos = 0 or 5sin cos + sin = 0 or sin ( 5) 0 cos = o.e. must be seen and be followed by printed answer for A1 mark sin = 5sin cos is not sufficient. (b) Statement of 0 and 180 with no working gets B1 B0 (bod) as it is two solutions : This mark for one of the two statements given (must relate to not just to ) A1, A1: first A1 for 9., second for Special case solving cos = 1/ 5 giving = or 58.5 is awarded A0A omitted would give A1A0 Allow answers which round to 9. or 9. and which round to and allow 141 Answers in radians lose last A1 awarded (These are 0, 0.68, 1.57,.46 and.14) Ecess answers in range lose last A1 Ignore ecess answers outside range. All 5 correct answers with no etras and no working gets full marks in part (b). The answers imply the method here

50 Question number Scheme Marks 7 (a) y B1, B1 () (b) 1 0.5, { (1 ) ( ) } o.e. B1,,A1 ft = A1 Notes (a) first B1 for and second B1 for ( is B0 ) Wrong accuracy e.g. 1.49, 1.74 is B1B0 6 marks (b) B1: Need ½ of 0.5 or 0.15 o.e. : requires first bracket to contain first plus last values and second bracket to include no additional values from the three in the table. If the only mistake is to omit one value from second bracket this may be regarded as a slip and M mark can be allowed ( An etra repeated term forfeits the M mark however) values: M0 if values used in brackets are values instead of y values A1ft follows their answers to part (a) and is for {correct epression} Final A1: Accept , or 1.50 only after correct work. (No follow through ecept one special case below following in table) Separate trapezia may be used : B1 for 0.15, for 1 h( a + b) used or 4 times (and A1ft if it is all correct ) e.g ( ) ( ) ( ) is A0 equivalent to missing one term in { } in main scheme Special Case: Bracketing mistake: i.e. 0.15(1+) +( ) scores B1 A0 A0 for 9.47 If the final answer implies that the calculation has been done correctly i.e (then full marks can be given). Need to see trapezium rule answer only (with no working) is 0/4 any doubts send to review Special Case; Uses to give or or or 1.50 gets, B1 B0 B1A1ft then A1 (lose 1 mark) NB Bracket is (4)

51 Question number Scheme Marks 8 60 B1 (a) ( h = ) or equivalent eact (not decimal) epression e.g. ( h = )60 π π (1) (b) ( A = )π + π h or ( A = )π r + π rh or ( A = )π r + π dh may not be simplified and may appear on separate lines (c) Either π ( A) = + 60 π π or As π h = then ( A = )π + 10 A = π + * A1 cso da 10 A1 ( ) = 4π or = 4π 10 d 10 4π = 0 implies B1 = (Use of > 0 or < 0 is M0 then M0A0) 10 = 4π or answers which round to.1 ( -.1 is A0) d A1 (d) 10, A1 A = π (.1) +, = 85 (only ft = or.1 both give 85).1 (e) d A 40 Or (method ) considers gradient to left and right Either = 4π + and sign d of their.1 (e.g at and.5) Notes considered ( May appear in (c) ) which is > 0 and therefore minimum (most substitute.1 but it is not essential to see a substitution ) (may appear in (c)) Or (method ) considers value of A either side Finds numerical values for gradients and observes gradients go from negative to zero to positive so concludes minimum (a) B1: This epression must be correct and in part (a) OR finds numerical values of A, observing greater than minimum value and draws conclusion 60 π r is B0 (b) B1: Accept any equivalent correct form may be on two or more lines. : substitute their epression for h in terms of into Area formula of the form A1: There should have been no errors in part (b) in obtaining this printed answer (c) : At least one power of decreased by 1 A1 accept any equivalent correct answer k + ch : Setting d A dy =0 and finding a value for ( = may be implied by answer). Allow = 0 d d d: Using cube root to find A1 : For any equivalent correct answer (need sf or more) Correct answer implies previous M mark (d) : Substitute the (+ve) value found in (c) into equation for A and evaluate. A1 is for 85 only A1 1 marks (e) : Complete method, usually one of the three listed in the scheme. For first method A ( ) must be attempted and sign considered A1: Clear statements and conclusion. (numerical substitution of is not necessary in first method shown, and must be correct. Must not see 85 substituted) or calculation could be wrong but A ( ) () (5) () ()

52 Question Scheme Marks 9 (a) n 1 ( S = ) a + ar + ( ar ) ar and (b) (c) (d) n n rsn ar ar ( ar )... ar n n n = S rs a ar = n S (1 r) = a(1 r ) d n And so result S n n a(1 r ) = (1 r) Divides one term by other (either way) to give r =... then square roots to give r = * A1 Or: ( Method ) Finds geometric mean i.e.4 and divides one term by.4 or.4 by one term r =, r = 0.6 (ignore 0.6) r = 0.6 (ignore 0.6) 5.4 Uses Uses 5.4 r or r a = 15 4, to give a = A1, A1ft () 15 S =, to obtain A1,A1 0.6 () 11 marks Notes (a) : Lists both of these sums ( S n =) may be omitted, r S n (or rs) must be stated 1 st n 1 two terms must be correct in each series. Last term must be ar n or ar in first series and the n n 1 corresponding ar or ar + in second series. Must be n and not a number. Reference made to other terms e.g. space or dots to indicate missing terms n : Subtracts series for rs from series for S (or other way round) to give RHS = ± ( a ar ). This may have been obtained by following a pattern. If wrong power stated on line 1 M0 here. (Ignore LHS)M0M0M0A0 d: Factorises both sides correctly must follow from a previous (It is possible to obtain M0A0 or M0A0) A1: completes the proof with no errors seen Special No errors seen: First line absolutely correct, omission of second line, third and fourth lines correct: Case M0A1 See net sheet of common errors. Refer any attempts involving sigma notation, or any proofs by induction to team leader. Also attempts which begin with the answer and work backwards. (b) : Deduces r by dividing either term by other and attempts square root A1: any correct equivalent for r e.g. /5 Answer only is / (Method ) Those who find fourth term must use ab and not 1 ( a b) + then must use it in a division with given term to obtain r = (c) : May be done in two steps or more e.g. 5.4 r then divided by r again A1ft: follow through their value of r. Just a = 15 with no wrong working implies A1 (4) () Common errors (d) : States sum to infinity formula with values of a and r found earlier, provided r <1 A1 : uses 15 and 0.6 (or /5) (This is not a ft mark) 5.4 (i) Fraction inverted in (b) r = and (ii) Uses r = 0.6: (b)m0a0 (c)a1ft (d) A0A0 i.e. /7 (iii) Uses ar 5 A1: 7.5 or eact equivalent r = 1, then correct ft gives A0 A1ft M0A0A0 i.e. /7 = 5.4, ar = Likely to have (b)a1 (c)m0a0 (d) A0A0 i.e./7

53 Mark Scheme (Results) January 01 GCE Mathematics Core Mathematics (6664)

54 January 01 C 6664 Mark Scheme Question number Scheme Marks 1 (a) 60, to obtain Uses ( ) 19 8, A1 () (b) Uses S = (1 ( 8) ) 7 1 8, or S = (( 8) 1) to obtain 680, A1 () (c) Uses S 60 = 1, to obtain , A1cao () 6 Notes Alternative method (a) : Correct use of formula with power = 19 A1: Accept 8.47, or or indeed (b) : Correct use of formula with n = 0 A1: Accept 681, 680.7, or or indeed (N.B or is A0) (c) : Correct use of formula A1: Accept 880 only Alternative to (a) Gives all 0 terms 15, 75.6(5), 41.17(1875), (1 st accurate) All correct and last term as above A1: Accept 8.5, 8.47, or or indeed Alternative to (b) Gives all 0 terms 15, 75.6(5), 41.17(1875), (1 st accurate) and adds Sum correct A1: Accept 680, 681, 680.7, or or indeed A1 A1

55 Question number The equation of the circle is Scheme Marks ( 1) ( y 7) ( r ) + + = A1 The radius of the circle is So ( 1) ( y 7) 50 ( 1) + 7 = 50 or 5 or r = + + = or equivalent A1 50 (4) 4 Notes is for this epression on left hand side allow errors in sign of 1 and 7. A1 correct signs (just LHS) Alternative method is for Pythagoras or substitution into equation of circle to give r or Giving this value as diameter is M0 A1, cao for cartesian equation with numerical values but allow ( 50) or (5 ) or any eact equivalent A correct answer implies a correct method so answer given with no working earns all four marks for this question. Equation of circle is + y ± ± 14y+ c = 0 Equation of circle is y y c = 0 A1 r Uses (0,0) to give c = 0, or finds r = ( 1) + 7 = 50 or 5 or r = 50 So + y + 14y = 0 or equivalent A1

56 Question number Scheme Marks (a). 8 (1 + ) = , B ( 4) + ( 4), = + + or = A1 A1 (4) (b) States or implies that = 0.1 B1 Substitutes their value of (provided it is <1) into series obtained in (a) Alternative for (b) Special case Notes i.e , = Starts again and epands ( ) to (0.05) + (0.05), ( Or 1 + 1/5 + 7/ /8000 = ) (a) B1 must be simplified = A1 cao () B1,,A1 7 The method mark () is awarded for an attempt at Binomial to get the third and/or fourth term need correct binomial coefficient combined with correct power of. Ignore bracket errors or 8 errors in powers of 4. Accept any notation for 8 C and 8 C, e.g. and 8 (unsimplified) or 8 and 56 from Pascal s triangle. (The terms may be listed without + signs) First A1 is for two completely correct unsimplified terms A1 needs the fully simplified and (b) B1 states or uses =0.1 or 4 = 40 for substituting their value of ( 0 < <1) into epansion (e.g. 0.1 (correct) or 0.01, or even 0.05 but not 1 nor 1.05 which would earn M0) A1 Should be answer printed cao (not answers which round to) and should follow correct work. Answer with no working at all is B0, M0, A0 States 0.1 then just writes down answer is B1 M0A0

57 Question number Scheme Marks 4. (a) log = log + log or log y log = log or log y log = log B1 log = log B1 Using log =1 B1 () (b) = 8 9 Solves 8 + 9= 0 1 to give = or = 9 A1 () 6 Notes (a) B1 for correct use of addition rule (or correct use of subtraction rule) B1: replacing log by log not log by log this is B0 These first two B marks are often earned in the first line of working 1 B1. for replacing log by 1 (or use of = ) If candidate has been awarded marks and their proof includes an error or omission of reference to logy withhold the last mark. So just B1 B1 B0 These marks must be awarded for work in part (a) only (b) Alternative to (b) using y for removing logs to get an equation in statement in scheme is sufficient. This needs to be accurate without any errors seen in part (b). for attempting to solve three term quadratic to give = (see notes on marking quadratics) A1 for the two correct answers this depends on second M mark only. Candidates often begin again in part (b) and do not use part (a). If such candidates make errors in log work in part (b) they score first M0. The second M and the A are earned as before. It is possible to get M0A1 or M0A0. Eliminates to give y 70y+ 4 = 0 with no errors is Solves quadratic to find y, then uses values to find A1 as before See etra sheet with eamples illustrating the scheme.

58 Question number Scheme Marks 5 (a) f ( ) = 8 + 4a b+ = 7 so a b = 6 * A1 () (b) f(1) = 1+ a+ b+ = 4 A1 Solve two linear equations to give a = and b = - A1 (4) 6 Notes (a) : Attempts f( ± ) = 7 or attempts long division as far as putting remainder equal to 7 (There may be sign slips) A1 is for correct equation with remainder = 7 and for the printed answer with no errors and no wrong working between the two (b) : Attempts f( ± 1) = 4 or attempts long division as far as putting remainder equal to 4 A1 is for correct equation with remainder = 4 and powers calculated correctly : Solving simultaneous equations (may be implied by correct answers). This mark may be awarded for attempts at elimination or substitution leading to values for both a and b. Errors are penalised in the accuracy mark. A1 is cao for values of a and b and eplicit values are needed. Special case: Misreads and puts remainder as 7 again in (b). This may earn A0A0 in part (b) and will result in a maimum mark of 4/6 Long Divisions a b a + ( ) + ( + 4) ( + ) + a + b ( a+ 1) + ( b+ a+ 1) ( 1) + a + b + A marks as before. and reach their - b + 4a - 8 = 7 and reach their + b + a + 1 = 4

59 Question number 6: (a) Scheme Marks y B1, B1 (b) () 1 0.5, { ( )} B1, A1ft = (or answers listed below in note) A1 (4) (c) d 1 16 = A1 A d =[ ] [ ] 4 1 = 11 or equivalent A1 4 Notes (a) B1 for 4 or any correct equivalent e.g B1 for.1 or.10 (b) B1: Need 0.5 or ½ of 0.5 : requires first bracket to contain first y value plus last y value (0 may be omitted Alternative Method for (b) or be at end) and second bracket to include no additional y values from those in the scheme. They may however omit one value as a slip. N.B. Special Case - Bracketing mistake 1 0.5( ) + ( ) scores B1 A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks ) A1ft: This should be correct but ft their 4 and.1 A1: Accept or or only (c) Attempt to integrate ie power increased by 1 or 1 becomes, A1 two correct terms, net A1 all three correct unsimplified (ignore +c) 1 (Allow or equivalent) d (This cannot be earned if previous M mark has not been awarded) Uses limits 4 and 1 in their integrated epression and subtracts (either way round) A or 11 ¼ or 45/4 or equivalent (penalise negative final answer here) Separate trapezia may be used : B1 for 0.5, for 1 ha ( + b) used 5 or 6 times ( and A1ft all correct for their 4 and.1 ) final A1 for etc. as before In part (b) Need to use trapezium rule answer only (with no working) is 0/4 -any doubts send to review In part (c) need to see integration (5) 11

60 Question number Scheme Marks 7 (a) rθ = , = 5.7 (cm), A1 () , 17.1 r θ = = ( cm ), A1 () (b) (c) 6 Let AD = then = so = 5.16 * sin 0.95 sin1.4 A1 () OR = / cos 0.95 OR so = / sin 0.6 so = 5.16 * OR = cos 0.95 leading to =, so = 5.16 * (d) Perimeter = = 11.7 or 6 + their 5.7 A1 ft () (e) Area of triangle ABD = sin0.95 = 1.6 or 1 6 tan0.95 = 1.6 (½ base height) or sin1.4 =1.6 So Area of R = = 4.5 Notes (a) : Needs θ in radians for this formula. Could convert to degrees and use degrees formula. A1: Does not need units (b) : Needs θ in radians for this formula. Could convert to degrees and use degrees formula. A1: Does not need units (c) : Needs complete correct trig method to achieve = May have worked in degrees, using 54.4 degrees and 71.1 degrees Using angles of triangle sum to 60degrees is not correct method so is M0 A1: accept answers which round to 5.16 (NB This is given answer) If the answer 5.16 is assumed and verified award A0 for correct work (d) : Accept answer only as implying method, or just A1 A1 (4) 1 Alternative For part (e) A1 : can be scored even following wrong answer to part (c) (e) : needs complete method for area of triangle ABD not ABC A1: Accept awrt 1.6 (If area of triangle is not evaluated or is given as 1.5 (truncated) this mark may be implied by 4.5 later) : Uses area of R = area of sector area of triangle ABD (not ABC) A1: Answers wrt 4.5 Finds area of segment and area of triangle BDC by correct methods Obtains.4585 and.0498 accept answers wrt.5,.1 A1 Uses area of segment + area of triangle BDC,to obtain 4.5 (not 4.6), A1 NB Just finding area of segment is M0

61 Question number Scheme Marks 8 (a) kr + cy = 4 or kr + c[( + y) y ] = π + = 4 y A π 16 π y = = * B1 cso 8 (b) P = + cy + k π r where c = or 4 and k = ¼ or ½ () π 16 π P= or P= o.e. A1 8 π 8 π 8 P = + + so P = + * A1 () dp 8 A1 (c) = + d 8 + = 0 =.. (d) 1 π 4 4 π and so = o.e. (ignore etra answer = -) A1 P = 4+ 4= 8 ( m ) B1 (5) 4 π y =, (and so width) = 1 (cm) 4, A1 Notes (a) : Putting sum of one or two y terms and one k r term equal to 4 (k and c may be wrong) A1: For any correct form of this equation with for radius (may be unsimplified) B1 : Making y the subject of their formula to give this printed answer with no errors (b) : Uses Perimeter formula of the form + cy + k π r where c = or 4 and k = ¼ or ½ A1: Correct unsimplified formula with y substituted as shown, 16 π π i.e. c = 4, k = ½, r = and y = or y = 8 A1: obtains printed answer with at least one line of correct simplification or epansion before giving printed answer or stating result has been shown or equivalent (c) : At least one power of decreased by 1 (Allow becomes ) A1: accept any equivalent correct answer : Setting d P =0 and finding a value for correct power of for candidate d A1 : For =. (This mark may be given for equivalent and may be implied by correct P) B1: 8 (cao) N.B. This may be awarded if seen in part (d) (d) : Substitute value found in (c) into equation for y from (a) ( or substitute and P into equation for P from (b)) and evaluate (may see and correct answer implies or need to see substitution if value was wrong.) A1 is for 1 or 1cm or 0.1m as this is to nearest cm () 1

62 Question number Scheme Marks 9 (i) 1 sin( 15) = so 15 = 0 ( α ) and = 15 A1 Need 15 = 180 α or 15 = 540 α Need 15 = 180 α and 15 = 60 + α and 15 = 540 α = 55 or 175 A1 = 55, 15, 175 A1 (6) Notes Correct order of operation: inverse sine then linear algebra - not just -15 = 0 (slips in linear algebra lose Accuracy mark) A1 Obtains first solution 15 Uses either 180 α or 540 α, uses all three 180 α and 60 + α and 540 α A1, for one further correct solution 55 or 175, (depends only on second ) A1 all further correct solutions If more than 4 solutions in range, lose last A1 Common slips: Just obtains 15 and 55, or 15 and 175 usually A1M0A1A0 Just obtains 15 and 15 is usually A1M0M0A0A0 (It is easy to get this erroneously) Obtains 5, 45, 15 and 165 usually A0A0A0 Obtains 5, 65, 145, (185) usually A0A0A0 π 11π Working in radians lose last A1 earned for 1, 6, π and 5 π or numerical 4 6 equivalents Mied radians and degrees is usually Method marks only Methods involving no working should be sent to Review 9 (ii) At least one of ( a π 10 b ) = 0 (or n π ) aπ ( 5 b) = π {or ( n+ 1) π} or in degrees 11 or ( a π 10 b ) = π {or ( n + ) π} If two of above equations used eliminates a or b to find one or both of these or uses period property of curve to find a or uses other valid method to find either a or b ( May see 5 π 10 a = π so a = ) Obtains a = A1 Obtains b = 5 π (must be in radians) A1 (4)

63 Notes a : Award for ( π a 10 b) = 0 or π a 10 = b BUT sin( π 10 b) = 0 is M0 a : As described above but solving ( π a 10 b) = 0 with ( π 5 b) = 0 is M0 (It gives a = b = 0) Special cases: Can obtain full marks here for both correct answers with no working A1A1 For a = only, with no working, award M0A1A0 For b = π 5 only with no working M0A0A1 Alternative Some use translations and stretches to give answers. If they achieve a= they earn second method and first accuracy. If they achieve correct value for b they earn first method and second accuracy. Common error is a = and b = π 10. This is usually M0A1A0 unless they have stated ( π b) = 0 earlier in which case they earn first. a 10

64 Mark Scheme (Results) June 011 GCE Core Mathematics C (6664) Paper 1

65 June 011 Core Mathematics C 6664 Mark Scheme Question Number Scheme Marks 1. f( ) = (a) Remainder = f (1) = = 6 Attempts f(1) or f( 1). = 6 6 A1 [] (b) Attempts f ( 1). f( 1) = ( 1) 7( 1) 5( 1) + 4 f( 1) = 0 with no sign or substitution and so ( + 1) is a factor. A1 [] errors and for conclusion. (c) f( ) = {( + 1) }( 9 + 4) A1 = ( + 1)( 1)( 4) d A1 (Note: Ignore the epen notation of (b) (should be (c)) for the final three marks in this part). [4] 8 (a) for attempting either f(1) or f( 1). Can be implied. Only one slip permitted. can also be given for an attempt (at least two subtracting processes) at long division to give a remainder which is independent of. A1 can be given also for 6 seen at the bottom of long division working. Award A0 for a candidate who finds 6 but then states that the remainder is 6. Award A1 for 6 without any working. (b) : attempting only f ( 1). A1: must correctly show f ( 1) = 0 and give a conclusion in part (b) only. Note: Stating hence factor or it is a factor or a tick or QED is fine for the conclusion. Note also that a conclusion can be implied from a preamble, eg: If f ( 1) = 0, ( + 1) is a factor. Note: Long division scores no marks in part (b). The factor theorem is required. (c) 1 st : Attempts long division or other method, to obtain ( ± a ± b), a 0, even with a remainder. Working need not be seen as this could be done by inspection. ( ± a ± b) must be seen in part (c) only. Award 1 st M0 if the quadratic factor is clearly found from dividing f ( ) by ( 1). Eg. Some candidates use their ( 5 10) in part (c) found from applying a long division method in part (a). 1 st A1: For seeing ( 9 + 4). nd d: Factorises a term quadratic. (see rule for factorising a quadratic). This is dependent on the previous method mark being awarded. This mark can also be awarded if the candidate applies the quadratic formula correctly. nd A1: is cao and needs all three factors on one line. Ignore following work (such as a solution to a quadratic equation.) 1 Note: Some candidates will go from {( + 1) }( 9 + 4) to { = 1, } =,4, and not list all three factors. Award these responses A1A0. Alternative: 1 st 1 : For finding either f (4) 0 f = 0. GCE Core Mathematics C (6664) June = or ( ) 1 st A1: A second correct factor of usually ( 4) or ( 1) found. Note that any one of the other correct factors found would imply the 1 st mark. nd d: For using two known factors to find the third factor, usually ( ± 1). nd A1 for correct answer of ( + 1)( 1)( 4). Alternative: (for the first two marks) 1 st : Epands ( + 1)( + a + b) {giving + ( a + ) + ( b + a) + b} then compare coefficients to find values for a and b. 1 st A1: a = 9, b = Not dealing with a factor of : ( + 1)( )( 4) or ( + 1)( )( 8) scores A1A0. Answer only, with one sign error: eg. ( + 1)( + 1)( 4) or ( + 1)( 1)( + 4) scores A1A0. (c) Award A1A1 for Listing all three correct factors with no working.

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