2010 Mathematics Higher Finalised Marking Instructions

(a) (i) Show that ( x 1) is a factor of f ( x) 2 x 3 x 2 8 x 5.225(ii) Hence factorise f ( x) fully.(b) Solve 2 x3 x2 8 x 5 0.1Generic SchemeIllustrative Scheme22 (a)Method 1 : Using synthetic division 1ssknow to use x 1 1121 85 2iccomplete evaluation 212123 83 55 502 3ic 3state conclusionpdfind quadratic factor 5pdfactorise completely 5 ( x 1) is a factor44see note 222 x 3x 5stated, or implied by 5( x 1)( x 1)(2 x 5)stated explicitlyMethod 2 : Using substitution and inspection 1know to use x 1 22 1 8 5 0 3( x 1) is a factor 4( x 1)(2 x 3 x 5) 5( x 1)( x 1)(2 x 5)see note 22stated explicitlyNotes1. Communication at 3 must be consistent with working at 2 .i.e. candidate’s working must arrive legitimately at zero before 3 is awarded.If the remainder is not 0 then an appropriate statement would be '( x 1) is not a factor ' .2For 3 , minimum acceptable statement is ‘factor’.Unacceptable statements : x 1 is a factor, ( x 1) is a factor, x 1 is a root, ( x 1) is a root etc.3. At 5 the expression may be written as ( x 1)2 (2 x 5) .22 (b) 6icstate solutions 6x 1 and x 52or 2 5 or 2(5212These may appear inthe working at (a).Notes4.From (a) ( x 1)( x 1)(2 x 5) leading to x 1, x 52)then ( 1, 0 ) and , 0 gains 6 .(52)However, ( x 1)( x 1)(2 x 5) leading to ( 1, 0 ) and , 0 only does not gain 6 .5.From (a) ( x 1)(2 x 5) only leading to x 1, x 52( x 1)( x 1)(2 x 5) leading to x 1, x 1 and x Page 1052does not gain 6 as equation solved is not a cubic, butgains 6 as follow through from a cubic equation.

Detailed Marking Instructions : Higher Mathematics 2010 Vs 12(c) The line with equation y 2 x 3 is a tangent to the curve with equation22y 2 x 3 x 2 6 x 2 at the point G.5Find the coordinates of G.(d) This tangent meets the curve again at the point H.Write down the coordinates of H.1Generic SchemeIllustrative Scheme22 (c)Method 1 : Equating curve and lineMethod 1 : Equating curve and line 7 7ssset yCURVE yLINE2x3 x2 6x 2 2x 3 82x x 8x 5compare with (a) or factorise 9( x 1)( x 1)(2 x 5) 10 icidentify xG 10 x 1 11 pdevaluate yG 11 y 1 8 9icssexpress in standard form3stated explicitly2 0see note 6Method 2 : DifferentiationMethod 2 : Differentiation 7ssknow to and differentiate curve 76x2 2x 6 8icset derivative to gradient of line 86x2 2x 6 2 9pdsolve quadratic equation 9x ssprocess to identify xG complete to yCURVE yLINE 11 y 1 from both curve and line10 11 ic1043and 1at x 1 evaluate yCURVE and yLINENotesIn method 1:6. 8 is only available if ‘ 0’ appears at either the 8 or 9 stage.7. 9 , 10 and 11 are only available via the working from 7 and 8 .8.If ( x 1)( x 1)(2 x 5) does not appear at 9 stage, it can be implied by 5 and 10 .9.At 9 a quadratic used from (a) may gain 9 , 11 and 12 but a quadratic from 8 may gain 11 and 12 only.10. If G and H are interchanged then 10 is lost but 11 and 12 are still available.11. Candidates who obtain three distinct factors at 9 can gain 11 for evaluating all y values, but lose 10and 12.12. A repeated factor at 5 or 9 stage is required for 10 to be awarded without justification.In both methods:13. All marks in (c) are available as a result of differentiating 2 x 3 x 2 6 x 2 and solving this equal to 2 (frommethod 2).Only marks 7 and 8 (from method 1) are available to those candidates who choose to differentiate2 x 3 x 2 8 x 5 and solve this equal to 0.14. Candidates may choose a combination of making equations equal and differentiation.22 (d) 12 pdstate solution 12( 52, 8)may appear in (c)Notes15. Method 2 from (c) would not yield a value for H and so 12 is not available.Page 11

23y(a) Diagram 1 shows a right angled triangle,where the line OA has equation 3 x 2 y 0.A3(i) Show that tan a 2 .a(ii) Find the value of sin a.(b) A second right angled triangle is addedas shown in Diagram 2.4xODiagram 1yAThe line OB has equation 3 x 4 y 0.B4Find the values of sin b and cos b.a bxODiagram 2Generic SchemeIllustrative Scheme23 (a)323 2 1ss write in slope/intercept form 1y x or y 1 5 x 2icconnect gradient and tan a 2m 3pd calculate hypotenuse 313 4ic 43state value of sine ratio13and tan a stated explicitly32or m tan a and tan a 32stated, or implied by 4or3 1313may not appear until (c)Notes1. 4 is only available if 1 sin a 1.2.Only numerical answers are acceptable for 3 and 4 .Regularly occurring responsesResponse 13x 2 y 0 3133 (2) 2 (3) 03tan a a2sin a 32313 1 2 42 marks out of 423 (b) 5ss determine tan b 5tan b 6ss know to complete triangle 6right angled triangle with 3 and 4 correctly shown77 pd determine hypotenuse 5 8ic 8sin b state values of sine and cosine ratios34stated, or implied by 6stated, or implied by 835and cos b Notes3. 8 is only available if 1 sin b 1 and 1 cos b 1 .4.sin b 5.Only numerical answers are acceptable for 7 and 8 .Page 1235and cos b 45without working is awarded 3 marks only.45may not appear until (c)

Detailed Marking Instructions : Higher Mathematics 2010 Vs 1223(c) (i) Find the value of sin( a b).(ii) State the value of sin(b a).4Generic SchemeIllustrative Scheme23 (c) 9know to use addition formula 9 10 icsubstitute into expansion 10 11 pdevaluate sine of compound angle 11 12 ssuse sin( x) sin xss12 sin a cos b cos a sin b345 1362 13355 136 5 13Notes6. sin(A B) sin A cos B cos A sin B, or just sin A cos B cos A sin B, with no further working does not gain 9 .7. Candidates should not be penalised further at 10 , 11 and 12 for values of sine and cosine outside the range 1 to 1.8. Candidates who use sin( a b) sin a sin b lose 9 , 10 and 11 but can gain 12 , as follow through, only fora non-zero answer which is obtained from the result sin( x) sin x.9. Treat sin31345cos cos23sin35as bad form only if ‘sin’ and ‘cos’ subsequently disappear.10. It is acceptable to work through the whole expansion again for 12 .Regularly occurring responsesResponse 19sin( a b) sin a sin b x 10 6 6 x Response 2sin a 3 cos a 2Marks lost in (a) or (b)sin b 3 cos b 4sin( a b) sin a cos b cos a sin b 910 3 4 2 3 11 6 Eased - not dealing with11 0 x sin(b a) 0 x 12fraction containing a surd.sin(b a) 6 0 marks out of 4123 marks out of 4Response 3Response 4From (a) and (b) sin a sin b (c) (i)2323cos a cos b sin( a b) sin a cos b cos a sin b 2 33 541513 25(i) sin( a b) sin a sin b cos a cos b x 91335 10 93 35 1315 131(ii) sin( b a) 5 132 1345 10 11 12 11(ii) sin(b a) sin b cos a cos b sin a251335 41523 123 marks out of 43 marks out of 4Here the working was not necessary; the answerwould gain 12 , provided it is non zero.Page 13

Paper 21 The diagram shows a cuboid OPQR,STUV relative to the coordinate axes.zP is the point (4, 0, 0) ,VQ is (4, 2, 0) and U is (4, 2, 3) .U (4, 2, 3)yM is the midpoint of OR.S13R NTQ (4, 2, 0)M MN is the point on UQ such that UN UQ.O(a) State the coordinates of M and N. (b) Express the vectors VM and VN in component form.P (4, 0, 0)x22Treat as bad form, coordinates written as components and vice versa, throughout this question.Generic SchemeIllustrative Scheme1 (a) 1 2 3icinterpret midpoint for M 1(0, 1, 0)2(4, 2, 2)interpret ratio for N icintepret diagram 3pdprocess vectors 4ic1 (b) 4 0 VM 1 3 4 VN 0 1 Using evidence from (a)or may have been takendirectly from diagram.Notes1.V is the point (0, 2, 3) , which may or may not appear in the working to (b).Regularly occurring responsesResponse 1Response 212(a) M(2, 0, 0) x N(4, 2, 1) x (b)0 marks out of 2(b) 2 VM 2 3 3 Consistent with (a) 4 4VN 0 1 From diagramIncorrectV statedV(0, 3, 2) 0 VM 2 x 3 2 4 4VN 1 0 1 mark out of 2Response 31(a) M(0, 2, 0) x N(4, 2, 2)1 mark out of 2(b) 4 VM 0 x 3 3 0 4VN 0 x 1 V (4, 2, 3)used inboth butnot stated0 marks out of 22 marks out of 2Page 14 2

Detailed Marking Instructions : Higher Mathematics 2010 Vs 121 The diagram shows a cuboid OPQR,STUV relative to the coordinate axes.zP is the point (4, 0, 0) ,VQ is (4, 2, 0) and U is (4, 2, 3) .M is the midpoint of OR.S13N is the point on UQ such that UN UQ.(c)U (4, 2, 3)y NTRMM Q (4, 2, 0)OP (4, 0, 0)Calculate the size of angle MVN.x5Treat as bad form, coordinates written as components and vice versa, throughout this question.Generic SchemeIllustrative Scheme1 (c)Method 1 : Vector Approach 5ssknow to use scalar product 6pdfind scalar product 7pdfind magnitude of a vector 8pdfind magnitude of a vectorMethod 1 : Vector Approach VMVN5ˆ . cos MVN VM VN 6 VM . VN 3 7 VM 10 8 VN 17 9pdevaluate angle 976 7 or 1 339 rads or 85 2 gradsMethod 2 : Cosine Rule ApproachMethod 2 : Cosine Rule Approach 5ssknow to use cosine rule 5 6pdfind magnitude of a side 6222ˆ VM VN MNcos MVN2 VM VNVM 10 7pdfind magnitude of a side 7VN 17find magnitude of a side 8MN 21 976 7 or 1 339 rads or 85 2 grads 8 9pdpdevaluate anglestated, or implied by 9stated, or implied by 9Notes2. 5 is not available to candidates who choose to evaluate an incorrect angle.3.For candidates who do not attempt 9 , then 5 is only available if the formula quoted relates to the labellingin the question.4. 9 should be awarded for any answer that rounds to 77 or 1 3 rads or 85 grads (i.e. correct to twosignificant figures.)Regularly occurring responsesResponse 1 OM . ONˆWrong anglecos MON 5OM ON OM . ON 2 6 Eased because only oneOM 1 7non-zero component. 8ON 24 65 9 or 1 150 rads or 73 2 grads 93 marks out of 5Response 2 VM . VNˆcos MVN VM VN VM . VN 0 6 VM 17 7 8VN 2 590 or equivalent 9Goingdirectly to90 from 6would lose 7 and 8 .4 marks out of 5Page 15

2(a) 12 cos x 5 sin x can be expressed in the form k cos( x a) , where k 0 and 0 a 360.Calculate the values of k and a.4Generic SchemeIllustrative Scheme(a) 1ssuse addition formula 1 k cos x cos a k sin x sin a or k (cos x cos a sin x cos a )stated explicitly 2iccompare coefficients 2 k cos a 12 and k sin a 5 or k sin a 5stated explicitly 3pdprocess k4pdprocess a 3 13no justification required, but do not accept4 22 6169accept any answer which rounds to 23Notes1. Do not penalise the omission of the degree symbol.2. Treat k cos x cos a sin x sin a as bad form only if the equations at the 2 stage both contain k.3.13 cos x cos a 13 sin x sin a or 13 (cos x cos a sin x sin a ) is acceptable for 1 and 3 .4. 2 is not available for k cos x 12 and k sin x 5 or k sin x 5 , however , 4 is still available.5. 4 is lost to candidates who give a in radians only.6. 4 may be gained only as a consequence of using evidence at 2 stage.7.Candidates may use any form of the wave equation for 1 , 2 and 3 , however 4 is only available if thevalue of a is interpreted for the form k cos( x a) .Regularly occurring responsesResponse 1Ak(cos x cos a sin x sin a ) sin a 5 x Response 1Bk cos x cos a k sin x sin a 1k 13 23 Response 2k cos( x a) 1 13 cos x cos a 13 sin x sin a 32 k cos x cos a k sin x sin a5124a 22 6 cos a 12tan a 1tan a 51213 cos a 12then a 22 6 x 4a 22 6 413 cos( x 22 6) 3orResponse 3Ak sin a 5 1 2Response 4k cos x cos a sin x sin a k sin a 12 1 x 2k cos x cos a k sin x sin a k cos a 12k cos a 12k cos a 5k sin a 5412k 13tan a x 35 a 67 412k 13tan a 35 4a 67 4 k 13 33 marks out of 4Page 16See note 73 marks out of 4Response 3Bk cos x cos a sin x sin a See note 64a 337 4 x 2 marks out of 42 marks out of 4 213 sin a 53 marks out of 4 1 x 2See note 6512a 337 4tan a 43 marks out of 4

Detailed Marking Instructions : Higher Mathematics 2010 Vs 122(b) (i) Hence state the maximum and minimum values of 12 cos x 5 sin x .(ii) Determine the values of x, in the interval 0 x 360, at which these maximum andminimum values occur.Generic Scheme3Illustrative Scheme(b) 5ssstate maximum and minimum 513, 13 6icfind x corresponding to max. value 6maximum at 337 4 and no otherspd find x corresponding to min. value77 orminimum at 157 4 and no others 6337 4 and 157 4 and no others7 maximum at 337 4 and minimum at 157 4Notes8. 5 is available for169 and 169 only if169 has been penalised at 3 .9. Accept answers which round to 337 and 157 for 6 and 7 .10. Candidates who continue to work in radian measure should not be penalised further.11. Extra solutions, correct or incorrect, should be penalised at 6 or 7 but not both.12. 6 and 7 are not available to candidates who work with 13 cos( x 22 6) 0 or 13cos( x 22 6) 1.13. Candidates who use 13 cos( x 22 6) from a correct (a) lose 6 but 7 is still available.Regularly occurring responsesResponse 1Response 2From (a) a 67 4From (a)5max/ min 13 max at 292 6 6min at 112 6 7169 cos( x 22 6) max 169 min 1696max at 22 6 x min at 202 6 7 52 marks out of 33 marks out of 3169 alreadypenalised at (a)Response 3A13 cos( x 22 6) 1322 66max at 22 6 x 7min at 202 6 Insufficient evidence for1 mark out of 3 5Response 3BN.B. Candidates who usedifferentiation in (b) can gain22 613 cos( x 22 6) max at 22 6 x 6min at 202 6 71 mark out of 3 5 only, as a direct result oftheir response in (a).This question is in degrees andso calculus is not appropriatefor 6 and 7 .Page 17

3 (a) (i) Show that the line with equation y 3 x is a tangent to the circle withequation x 2 y 2 14 x 4 y 19 0 .(ii) Find the coordinates of the point of contact, P.5Generic SchemeIllustrative Scheme(a) 1sssubstitute 1x 2 (3 x)2 14 x 4(3 x) 19 0Method 1 : Factorisingpdexpress in standard form 22 x2 4x 2 3icstart proof 32( x 1)( x 1) 4iccomplete proof 4equal roots so line is a tangent 2 0see note 1Method 2 : Discriminant 2 5pdcoordinates of P2 x2 4x 2 0stated explicitly 34 4 2 2 4b2 4ac 0 so line is a tangent 52x 1, y 4NotesFor method 1 :1. 2 is only available if “ 0” appears at either 2 or 3 stage.2. Alternative wording for 4 could be e.g. ‘repeated roots’, ‘repeated factor’, ‘ only one solution’, ‘only onepoint of contact’ along with ‘line is a tangent’.For both methods :3. Candidates must work with a quadratic equation at the 3 and 4 stages.4. Simply stating the tangency condition without supporting working cannot gain 4 .5. For candidates who obtain two distinct roots, 4 is still available for ' not equal roots so not a tangent ' or' b2 4 ac 0 so line is not a tangent ' , but 5 is not available.Page 18

Detailed Marking Instructions : Higher Mathematics 2010 Vs 123 (b) Relative to a suitable set of coordinate axes, the diagram below shows thecircle from (a) and a second smaller circle with centre C.C PThe line y 3 x is a common tangent at the point P.The radius of the larger circle is three times the radius of the smaller circle.6Find the equation of the smaller circle.x 2 y 2 14 x 4 y 19 0Generic SchemeIllustrative Scheme(b)Method 1 : via centre and radius6icstate centre of larger circleMethod 1 : via centre and radius 6 ( 7, 2)see note 117 ssfind radius of larger circle 772see note 6 8pdfind radius of smaller circle 88see note 79ssstrategy for finding centre10icinterpret centre of smaller circle11icstate equation ss(1, 6)11( x 1)2 ( y 6)2 8 6x 2 y 2 2 x 12 y 29 0orx 2 y 2 2 x 12 y 29 0see note 11e.g. " Stepping out "8(1, 6) strategy for finding centre( 7, 2)7 icstate centre of smaller circle 9ssstrategy for finding radius 910pdfind radius of smaller circle 1011icstate equation orMethod 2 : via ratios8 e.g " Stepping out"10 Method 2 : via ratios 6 icstate centre of larger circle79stated, or implied by 811 2 2 228see note 102( x 1) ( y 6)2 8NotesFor method 1:6. Acceptable alternatives for 7 are 6 2 or decimal equivalent which rounds to 8 5 i.e. to two significant figures.7. Acceptable alternatives for 8 are8.723or 2 2 or decimal equivalent which rounds to 2 8.(1, 6) without working gains 10 but loses 9 .For method 2:9.(1, 6) without working gains 8 but loses 7 .10. Acceptable alternatives for 10 are 2 2 or decimal equivalent which rounds to 2 8.In both methods:11. If m 1 is used in a ‘stepping out’ method the centre of the larger circle need not be stated explicitly for 6 .12. For the smaller circle, candidates who ‘guess’ values for either the centre or radius cannot be awarded 11 .13. At 11 e.g.28 , 2 8 2 are unacceptable, but any decimal which rounds to 7 8 is acceptable.14. 11 is not available to candidates who divide the coordinates of the centre of the larger circle by 3.Page 19

4 Solve 2 cos 2 x 5 cos x 4 0 for 0 x 2 π.5Generic SchemeIllustrative Scheme4Method 1 : Using factorisation1 ssknow to use double angle formula 12 (2 cos 2 x 1 ). . . 2icexpress as quadratic in cos x 24 cos 2 x 5 cos x 6 0 must appear at either of(4 cos x 3)(cos x 2)these lines to gain 2 . 3ss start to solve3Method 2 : Using quadratic formula 12 (2 cos 2 x 1) . . . 24 cos 2 x 5 cos x 6 0 ( 5) ( 5)2 4 4 ( 6)2 4In both methods : 4 5pd reduce to equations in cos x onlypd complete solutions to include onlyone where cos x k with k 1 3cos x 4cos x 345 2 419, 3 864or 4 5cos x 2cos x 3 4andcos x 2and no solutionandno solutionand2 419, 3 864Notes1. 1 is not available for simply stating that cos 2A 2 cos 2 A 1 with no further working.2. Substituting cos 2A 2 cos 2 A 1 or cos 2a 2 cos 2 a 1 etc. should be treated as bad form throughout.3. In the event of cos 2 x sin 2 x or 1 2 sin 2 x being substituted for cos 2x , 1 cannot be given until theequation reduces to a quadratic in cos x.4. Candidates may express the quadratic equation obtained at the 2 stage in the form 4c 2 5c 6 0,4 x 2 5 x 6 0 etc. For candidates who do not solve

Detailed Marking Instructions : Higher Mathematics 2010 Vs 12 Page 5 9 Normally, do not penalise: Working subsequent to a correct answer; Omission of units; Legitimate variations in numerical answers; Bad form; Correct working in the wrong part of a question; unless specifically mentioned in the marking scheme. 10