CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 013 series 9709 MATHEMATICS 9709/11 Paper 1, maximum raw mark 75 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 11 Mark Scheme Notes Marks are of the following three types: M A B Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a correct result or statement independent of method marks. When a part of a question has two or more method steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously correct answers or results obtained from incorrect working. Note: B or A means that the candidate can earn or 0. B/1/0 means that the candidate can earn anything from 0 to. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10. Cambridge International Examinations 013
Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 11 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no follow through from a previous error is allowed) Correct Working Only often written by a fortuitous answer Ignore Subsequent Working Misread Premature Approximation (resulting in basically correct work that is insufficiently accurate) See Other Solution (the candidate makes a better attempt at the same question) Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR 1 PA 1 A penalty of MR 1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become follow through marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B marks in the case of premature approximation. The PA 1 penalty is usually discussed at the meeting. Cambridge International Examinations 013
Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 11 1 (i) 64 + 576x + 160x (ii) 576a ( x ) + 160( x ) = 0 160 a = oe (eg 576 15 ) or 3.75 4 [] Can score in (ii) Attempt integration 1 6 f x = x + 6 + c x 6 () 3 + c = 1 3 c = 3 ( ) ( ) ( ) [5] Accept unsimplified terms Sub.x = 3, y = 1. c must be present 3 (i) DB = 6i + 4j 3k cao DE = 3i +j 3k cao (ii) DB.DE = 18 + 8 + 9 = 35 DB = 61 or DE = 35 = 61 cosθ oe θ = 17. (0.300 rad) cao [] Use of x 1x + y1 y + z1z Correct method for moduli All connected correctly Use of e.g. BD. DE can score M marks (leads to obtuse angle) 4 (i) 4( 1 cos x ) + 8cos x 7 = 0 ( cos x 1)( cos x 3) 0 4c 8c + 3 = 0 = x = 60 or 300 (ii) 1 θ = 60 (or 300 ) θ = 10 only [] Use c + s = 1 Attempt to solve Allow 300 in addition 5 (i) x = ( ± ) y 1 OR y = x 1 (x/y interchange 1 st ) f 1 : x a x 1 for x > 1 (ii) ff ( x ) = ( x + 1) + 1 x + 1 = x = 3/ 1 = ( ± ) 13/ 4 Alt. (ii) f ( ) f ( 185/16) = 13/ 4 = f 1 ( 13/ 4) x x x = 3/ 4 = Or x + x ( 153/16) 0 Or x = 9 / 4, ( 17 / 4) www. Condone ± 3/ Alt.(ii) f(3/) = 13/4 f(13/4) = 185/16 x = 3/ SC.B answer 1.5 with no working Cambridge International Examinations 013
Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 11 6 (i) ( π α ) r + rα + r π r + rα + r ft for rα instead of rα or omission r SC1 for rα + 4r. (Plate = shaded part) 1 r + πr α (ii) ( ) α 3r α + πr 1 r Either can be scored in (iii) (iii) πr 1 r α = r α For equating their parts from (ii) α = π 5 [] 7 (i) mid-point = (3, 4) Grad. AB = ½ grad. of perp., = y 4 = ( x 3) y = x (ii) q = p p + q = 4 oe p ( p ) = 4 5p 8 = 0 + p {OR ¼ ( + ) + q = 4 5q + 4q 1 = 0 q } soi For use of 1/m soi ft on their (3, 4) and ft for 1 st eqn. Attempt substn (linear into quadratic) & simplify (0, ) and 8 6, 5 5 [5] 8 (i) (ii) A = xr + πr x + πr = 400 x = 00 πr A = 400r πr ( ) da = 400 πr dr = 0 00 r = oe π x = 0 no straight sections d A = π dr ( < 0 ) Max AG [5] Subst & simplify to AG (www) Differentiate Set to zero and attempt to find r Dep on reason π, or use of other valid Cambridge International Examinations 013
Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 11 10 a + d = oe 0 ( a + 19d ) = 1400 OR 10 ( a + 10d ) + 9d = d = 6 a = 13 9 (a) ( 9 ) 400 a (b) = 6 1 r [ ] 1000 ( r) 1 1 = 7 or 1 r 5 r = or 0.714 7 1 a = or 1.71(4) 7 a 1 r 1 r 1 r = = 7 1 7 [5] [5] a + 9d = 80 a + 19d = 140 or a + 9d = 00 Solve sim. eqns both from formulae Substitute or divide S n Ignore any other solns for r and a [ ] [ ] d y 1 dy At x =, = 4 1 y 8 = 4 x y = 4 x + 0 10 (i) = 3( 3 x) (ii) Area under curve = 81 8 ( 3 x) 4 4 + 1 Area under tangent = ( 4x 0) = 1x + 0x or 7 (from trap) D [5] OR OR 54 + 7x 4x B,1,0 3 4 7x + 1x B,1,0 7x x Limits 0 ½ applied to integral with intention of subtraction shown or area trap =½(0 + 8) ½ Could be implied 9 or 1.15 8 [6] Dep on both M marks Cambridge International Examinations 013
CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 013 series 9709 MATHEMATICS 9709/1 Paper 1, maximum raw mark 75 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 01 9709 1 Mark Scheme Notes Marks are of the following three types: M A B Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a correct result or statement independent of method marks. When a part of a question has two or more method steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously correct answers or results obtained from incorrect working. Note: B or A means that the candidate can earn or 0. B/1/0 means that the candidate can earn anything from 0 to. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10. Cambridge International Examinations 01
Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 01 9709 1 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no follow through from a previous error is allowed) Correct Working Only often written by a fortuitous answer Ignore Subsequent Working Misread Premature Approximation (resulting in basically correct work that is insufficiently accurate) See Other Solution (the candidate makes a better attempt at the same question) Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR 1 PA 1 A penalty of MR 1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become follow through marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B marks in the case of premature approximation. The PA 1 penalty is usually discussed at the meeting. Cambridge International Examinations 01
Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 1 1 (i) sinx = (1 p²) [1] Allow 1 p if following (1 p²) ± is B0. (ii) sin x tan x = = cos x 1 p p [1] for answer to (i) used. (iii) tan( 90 x) = p 1 p [1] for reciprocal of (ii) (i) slant length = 10 cm. circumference of base = 1π arc length = 10θ ( = 1π) θ = 1.π or 3.77 radians. (ii) ½r²θ = 188.5 cm ² or 60π. [] Use of rθ, θ calculated, not 6 or 8. Use of ½r²θ with radians and r = calculated 10, not 6 or 8. 3 y = 5x 6 dy 3 (i) = ½ (5x 6) 5 5 8 (ii) integral = 5x 6 1 5 Uses to 3.4 1.6 = 0.8 without 5. For 5 Use of uv or u/v ok. without 5. for 5 Use of limits in an integral. 4 OA = i + j and OB = 4 i + pk, (i) AB = b a = 3i j + 6k Unit vector = (3i j + 6k) 7 Must be AB = b a Divides by modulus. on vector AB. (ii) Scalar product = 4 = 5 (16 + p²) cos θ p = ±8 5 A (0, 8) B (4, 0) 8y + x= 33 m of AB = m of BC = ½ Eqn BC y 0 = ½(x 4) Sim eqns C (16, 6) Vector step method D (1, 14) (or AD y = ½x +8, CD y = x + 38) (or M = (8, 7) D = (1, 14) ) [7] Use of x 1 x + y 1 y + z 1 z For modulus. All linked correctly including correct use of cosθ=1/5. Use of m 1 m = 1for BC or AD Correct method for equation of BC Sim Eqns for BC, AC. valid method. Cambridge International Examinations 013
Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 1 6 y x (i) Sim triangles 16 y = 1 ¾x A = xy = 1x ¾x². (ii) 1 = (or trig) 16 da 6x = 1 4 = 0 when x = 8. A = 48. Trig, similarity or eqn of line (could also come from eqn of line) ag check working. Sets to 0 + solution. This is a Maximum. From ve quadratic or nd differential. Can be deduced without any working. Allow even if 48 incorrect. 7 (a) (i) a = 300, d = 1 540 = 300 + (n 1)1 n = 1 (ii) S 6 = 13 (600 + 5 1) = 11700 3 hours 15 minutes. (b) ar = 48 and ar² = 3 r = ⅔ a =7. S = 7 ⅓ = 16. 8 f : x a 3cosx for 0 Y x Y π. (i) 3cosx = 0 cos x = ⅔ x = 0.841 or 5.44 (ii) range is 5 Y f(x) Y 1 (iii) y 0 π x [] [] B,1 [], [] [1] Use of nth term. Ans 0 gets 0. Ignore incorrect units Correct use of s n formula. Needs ar and ar² + attempt at a and r. Correct S formula with r < 1 Makes cos subject, then cos 1 for π 1st answer. for [ 5. for Y 1. starts and ends at same point. Starts decreasing. One cycle only. for shape, not V or U. (iv) max value of k = π or 180º. (iv) g 1 + (x) = cos 1 x 3 [] Make x the subject, copes with cos. Needs to be in terms of x. Cambridge International Examinations 013
Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November 013 9709 1 8 9 y = + x x dy 8 (i) = x + ( 6 at A) dy dy dy = dt dt 0.4 Attempt at differentiation. algebraic unsimplified. Ignore notation needs product of 0.04 d y and his. (ii) y 64 = + 4x + 3 x 3 64 4x = ( + + 3x ) x 3 Limits to 5 used correctly 71.π or 85 (allow 71π or 851 to 85) 10 f : x a x 3x, g : x a 3x + k, (i) x 3x 9 > 0 x =3 or 1½ Set of x x > 3, or x < 1½ 3 (ii) x² 3x = ( x ) 4 3 9 Vertex (, ) 4 8 9 8 A3,,1 D [6] B3,,1 Use of integral of y² (ignore π) 3 terms 1 each error. Uses correct limits correctly. (omission of π loses last mark ) For solving quadratic. Ignore > or [ condone [ or Y x² in bracket is an error. on c and b. (iii) gf(x) = 6x² 9x + k = 0 Use of b² 4ac k = 8 7 oe. Used on a quadratic (even fg). Cambridge International Examinations 013
CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 013 series 9709 MATHEMATICS 9709/13 Paper 1, maximum raw mark 75 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper GCE A LEVEL October/November 013 9709 13 Mark Scheme Notes Marks are of the following three types: M A B Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a correct result or statement independent of method marks. When a part of a question has two or more method steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously correct answers or results obtained from incorrect working. Note: B or A means that the candidate can earn or 0. B/1/0 means that the candidate can earn anything from 0 to. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10. Cambridge International Examinations 013
Page 3 Mark Scheme Syllabus Paper GCE A LEVEL October/November 013 9709 13 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no follow through from a previous error is allowed) Correct Working Only often written by a fortuitous answer Ignore Subsequent Working Misread Premature Approximation (resulting in basically correct work that is insufficiently accurate) See Other Solution (the candidate makes a better attempt at the same question) Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR 1 PA 1 A penalty of MR 1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become follow through marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B marks in the case of premature approximation. The PA 1 penalty is usually discussed at the meeting. Cambridge International Examinations 013
Page 4 Mark Scheme Syllabus Paper GCE A LEVEL October/November 013 9709 13 1 (x + 1) (x ) or other valid method 1, x < 1, x > Attempt soln of eqn or other method Penalise, 1 f (x) = x + x (+c) 1 Attempt integ x or + x needed for M 5 = 1 + 4 + c Sub (4, 5). c must be present c = 3 (i) gradient of perpendicular = ½ soi y 1 = ½ (x 3) [] (ii) C = ( 9, 6) AC = [3 ( 9)] + [1 6] (ft on their C) AC = 13 soi in (i) or (ii) OR AB² = [3 ( 1)]² + [1 11]² AB = 6 AC = 13 4 (i) OD = 4i + 3j CD = 4i + 3j 10k [] for OD 10k (ii) OD.CD = 9 + 16 = 5 OD = 5 or CD = 15 5 = 5 15 cosθ oe ODC = 63.4 (or 1.11 rads) Use of x 1 x + y 1 y + z 1 z Correct method for moduli All connected correctly cao 5 (a) a = 8a 1( a) = 8( a)(1 r) 1 r oe [] (b) a + 4d = 197 10 [ a + 9d] = 040 d = 14 Or a + 9d = 408 Attempt to solve simultaneously 1 1 6 (i) sector areas are 11 α, 5 α 1 1 11 α 5 α k = 1 5 α 96 k = or 3.84 5 Sight of 11, 5 11 5 Or 5 Cambridge International Examinations 013
Page 5 Mark Scheme Syllabus Paper GCE A LEVEL October/November 013 9709 13 (ii) perimeter shaded region= 11α + 5α + 6 + 6 = 16α + 1 perimeter unshaded region = 5α + 5 + 5 = 5α + 10 16α + 1 = (5α + 10) α = 4/3 or 1.33 7 (a) x 1 = sin π 3 x = ± 1.366 for negative of 1 st answer (b) π 5π 13π θ + = or 3 6 6 π 11π θ = = or 6 π 11π θ =, 4 1 π or 6 1 correct angle on RHS is sufficient Isolating θ SC decimals 0.785 &.88 scores 8 (i) 81 (x 8 ) [1] (ii) 10 3 3 (x 8 ) soi leading to their answer 70 (x 8 ) for 10, 5C or 5C3. for 3 3. But must be multiplied. (iii) k (i) 405 soi + (ii) 675 (x 8 ) D k 1,0 dy x + = ± k = 9 = k ( x + ) + 1 0 x = ± k d y = k x + ( ) 3 When x = = k, When x = k, max d y = which is (> 0) min k d y = which is (< 0) k D [8] Attempt differentiation & set to zero Attempt to solve cao Attempt to differentiate again d y Sub their x value with k in it into Only 1 of bracketed items needed for each d y but and x need to be correct. Cambridge International Examinations 013
Page 6 Mark Scheme Syllabus Paper GCE A LEVEL October/November 013 9709 13 10 (i) Range is (y) c + 4c x + 4x = (x + ) 4 (Smallest value of c is) Allow > d y OR = x + 4 = 0 with no (wrong) working gets B (ii) 5a + b = 11 (a + b) + 4 (a + b) = 1 (11 5a + a) + 4 (11 5a + a) = 1 OR corresponding equation in b (8) (a 13a + 18) = (8) (a 9) (a ) = 0 OR (8) (b + 3) (b 1) = 0 9 a =, OR b = 3, 1 [6] for either a or b correct. Condone nd value. Spotted solution scores only B marks. Alt. (ii) Last 5 marks f 1 (x) = x + 4 g (1) = f 1 = (1) used Alt. (ii) Last 4 marks (a + b + 7) (a + b 3) = 0 a + b = 5 = 3 Solve a + b = 3, 5a + b = 11 a =, b = 1 (Ignore solution involving a + b = 7) Solve a + b = 3, 5a + b = 11 a =, b = 1 dy 1 1 3 11 (i) = ( x4 + 4x + 4) [ 4x + 4] dy 1 1 At x = 0, = 4 = (1) Equation is y = x 4 (ii) x + = x + 4x + 4 (x + ) = x4 + 4x + 4 x x 4 = 0 oe x = 0, ± 1 (iii) ( π ) 5 x 5 + x + 4x 1 ( π ) 0 + 4 5 B,1,0 D Sub x = 0 and attempt eqn of line following differentiation. AG www Attempt to integrate y 11π (6.91) oe 5 Apply limits 1 0 Cambridge International Examinations 013