3. (M1) 1. A A1 N2 (A1)(A1) A1 N4 - Free Download PDF

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–8–1.(a)(i)(ii)M13/5/MATME/SP1/ENG/TZ1/XX/M4 6correct expression for 2a b5, (5, 2) , 5i 2 jeg 22a correct substitution into length formulaeg52 2 2 , 5c 2M13/5/MATME/SP1/ENG/TZ1/XX/Mevidence of choosing product ruleeg uv′ vu′correct derivatives (must be seen in the product rule) cos x , 2xf ′ ( x) x cos x 2 x sin x2(M1)(A1)(A1)A1(A1)(b)A1valid approacheg c (2a b) , 5 x 0, 2 y 0(a)N252 222a b 29(b)3.(A1)A1–9–N2[4 marks](M1)A1substitutingegf′πinto their f ′( x)2ππ,222coseg(M1)πππ 2sin222correct values for both sinN2N4[4 marks]ππand cos seen in f ′( x)22(A1)π0 2 12[2 marks]Total [6 marks]2.(a)(b)x 1 , x 3 ( accept (1, 0), ( 3, 0) )A1A1N2[2 marks]METHOD 1attempt to find x-coordinate b1 3eg, x , f ′ ( x) 022acorrect value, x 1 (may be seen as a coordinate in the answer)attempt to find their y-coordinateegf ( 1) , 2 2 , y (M1)A1(M1) D4ay 4vertex ( 1, 4)A1N3[4 marks]METHOD 2attempt to complete the squareeg x 2 2 x 1 1 3(M1)attempt to put into vertex formeg ( x 1) 2 4 , ( x 1) 2 4(M1)vertex ( 1, 4)A1A1N3[4 marks]Total [6 marks]f′π π2A1N2[3 marks]Total [7 marks]

– 10 –4.(a)attempt to solve for Xeg XA C B , X B CA 1 , A 1 (C B ) , A 1 C BX (C B ) A(b)M13/5/MATME/SP1/ENG/TZ1/XX/M 1 1 1( CA BA )(M1)A1– 11 –5.(a)N2[2 marks]METHOD 1C B 1 2 4 2(seen anywhere)correct substitution into formula for 2 2 inverse 2 14 21, 3eg A 1 14 6 3 1 22attempt to multiply (C B ) and A 1 (in any order)X Note:A11 011 5(M1)A2N3[5 marks]METHOD 2attempt to multiply either BA 1 or CA 1 (in any order) 2 3 6 4 1 1 0 3 0 1,, two correct entrieseg3 3 922 4 6 2 2 22 2 2one correct multiplication51 1 3 1eg, 222 2 012 5X Note:1 011 5Award A1 for three correct elements.(M1)correct workingeg 4 y 5 , x 22 5(A1)A1interchanging x and y (seen anywhere)eg x y 5(M1)correct workingeg x 2 y 5 , y x 2 5(A1)f 1 (2) 9Award A1 for three correct elements.correct substitution into formula for 2 2 inverse 2 14 21, 3eg A 1 14 6 3 1 22attempt to set up equationeg 2 y 5 , 2 x 5N2[3 marks]METHOD 2 2 3 1 14 6 2 2, two correct elements,8 3 4 1 16 6 8 2egMETHOD 1f 1 (2) 9A1M13/5/MATME/SP1/ENG/TZ1/XX/M(b)A1recognizing g 1 (3) 30eg f (30)correct workingeg ( f g 1 ) (3) 30 5 ,(M1)(A1)25( f g 1 ) (3) 5A1N2[3 marks]A1N2Note: Award A0 for multiple values, eg 5 .[3 marks]Total [6 marks](M1)A1A2N3[5 marks]Total [7 marks]

– 12 –6.M13/5/MATME/SP1/ENG/TZ1/XX/Mattempt to integrate which involves lneg ln ( 2 x 5 ) , 12ln 2 x 5, ln 2 x(M1)(M1)C 6ln 3(A1) 6ln2x 53(a)(i)A2attempt to substitute (4, 0) into their integrated feg 0 6ln (2 4 5) , 0 6ln (8 5) C(accept 6ln ( 2 x 5 ) ln 36 )valid approach71 eg 4 2 , A B , AB AO OB 136 AB 24A1(ii)N5Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve ln.N2any correct equation in the form r a tb (accept any parameter for t)1 where a 2 and b is a scalar multiple of ABA2N23eg(a)evidence of correct formula40a, log, log8 log 5 log 5log a log b log5b(M1)A1Total [6 marks]7.M13/5/MATME/SP1/ENG/TZ1/XX/MSECTION B8.correct expression (accept absence of C)1eg 12ln ( 2 x 5 ) C , 6ln ( 2 x 5 )2f ( x) 6ln (2 x 5) 6ln 3– 13 –(M1)161 6tr 2 t 2 , ( x , y , z ) (1, 2, 3) t (3, 1, 2) , r 2 2t343 4tNote: Award A1 for a tb , A1 for L1 a tb , A0 for r b ta .eg[4 marks]Note: Ignore missing or incorrect base.(b)correct workingeg log 2 8 , 23 8(A1)log 2 40 log 2 5 3A1attempt to write 8 as a power of 2 (seen anywhere)eg (23 )log2 5 , 23 8 , 2a(M1)multiplying powerseg 23log2 5 , a log 2 5(M1)correct workingeg2log2 125 , log 2 53 ,8log2 5 125(2 )(b)N2[3 marks]correct calculation of scalar producteg 6(3) 2( 3) 4 p , 18 6 4 pR1(A1)correct workingeg 24 4 p 0 , 4 p 24A1p 6AGN0[3 marks]continued (A1)log 2 5 3A1recognizing that scalar product 0 (seen anywhere)N3[4 marks]Total [7 marks]

– 14 –M13/5/MATME/SP1/ENG/TZ1/XX/MQuestion 8 continued(c)9.setting lines equal16 13eg L1 L2 , 2 t 2 2 s 33415 6(M1)any two correct equations with different parameterseg 1 6t 1 3s , 2 2t 2 3s , 3 4t 15 6 sA1A1attempt to solve their simultaneous equations(M1)one correct parameter15eg t , s 23attempt to substitute parameter into vector equation1611eg 2 2 , 1 62234x 4 ( accept (4, 3, 5), ignore incorrect values for y and z )– 15 –(a)(i)attempt to find P (red) P (red)3 2 3 3 3 2eg , , 8 7 8 8 8 8P (none green) (ii)A1M13/5/MATME/SP1/ENG/TZ1/XX/M656 (M1)328A1attempt to find P (red) P (green)5 3 3 5 15eg , ,8 7 8 8 56(M1)recognizing two ways to get one red, one green5 3 3 5 3 52P( R ) P(G ) , , 28 7 8 7 8 8(M1)N2eg(M1)P (exactly one green) 3056 1528A1N2[5 marks](b)A1N3[7 marks]Total [14 marks]P (both green) 20(seen anywhere)56(A1)correct substitution into formula for E (X )63020 30 50eg 0 1 2 , 565656 64 64expected number of green marbles is7056A1 54A1N2[3 marks]continued

– 16 –M13/5/MATME/SP1/ENG/TZ1/XX/MQuestion 9 continued(c)– 17 –10.46(i)P ( jar B) (ii)P ( red jar B ) 6823A1 34A1(a)N1N1substitute 0 into feg ln (0 1) , ln1recognizing conditional probabilityP ( jar A and red )eg P ( A R ) ,, tree diagramP ( red )(M1)attempt to multiply along either branch (may be seen on diagram)11 3eg P ( jar A and red) 83 8(M1)(b)f ′ ( x) (M1)adding the probabilities of two mutually exclusive paths51 3 2 6eg P (red) 83 8 3 8(A1)(i)(ii)A1A1N3[6 marks]Total [16 marks]R1(M1)4 x3 0 , x3 0A1substituting x 1 into f ′′4(3 1) 4 2,eg(1 1) 24A1valid interpretation of point of inflexion (seen anywhere)eg no change of sign in f ′′ ( x) , no change in concavity,f ′ increasing both sides of zeroattempt to find f ′′ ( x) for x 0egf ′′ ( 1) ,4( 1) 2 ( 3 ( 1) 4 )( ( 1)4 1)N1[5 marks](A1)f ′′ (1) 2correct substitution151and A1 for 4x 3 .x4 1f increasing for x 0 (accept x 0 )(c)N2[2 marks]A1A1attempt to solve f ′ ( x) 0attempt to multiply along other branch2 61 P ( jar B and red) 23 8P ( jar A red ) 1 4 x 3 (seen anywhere)x4 1recognizing f increasing where f ′ ( x) 0 (seen anywhere)eg f ′ ( x) 0 , diagram of signsegegA1Note: Award A1 foreg1 31 3 8, 8P ( jar A red ) 1 3 2 6 5 3 8 3 8 8(M1)f (0) 0[2 marks](d)M13/5/MATME/SP1/ENG/TZ1/XX/M2N2R1(M1), diagram of signscorrect working leading to positive valueeg f ′′ ( 1) 2 , discussing signs of numerator and denominatorA1there is no point of inflexion at x 0AGN0[5 marks]continued

– 18 –M13/5/MATME/SP1/ENG/TZ1/XX/M-9-Question 10 continuedM09/5/MATME/SP1/ENG/TZ1/XX/M SECTION A(d)QUESTION 1(a)A1A1A1N3(b)(i)AB(ii)A14112321 1, B,541454A 1CA2N2A1N1[3 marks](M1)2614Total [15 marks]26( 4I )METHOD 1xyNotes:Award A1 for shape concave up left of POI and concave down right of POI.Only if this A1 is awarded, then award the following:A1 for curve through (0, 0) , A1 for increasing throughout.Sketch need not be drawn to scale. Only essential features need to be clear.4 00 4151232841454517xy84A1A1A1N3METHOD 25x y 8 , 6 x 2 y4for work towards solving their systemx5y17A1(M1)A1A1N3[7 marks]QUESTION 2111(a)P ( A)(b)P ( B A)(c)recognising that P( A210B) P ( A ) P( B A)correct valuese.g. P( AP( A1 211 102B)110A1N1A2N2(M1)(A1)B)A1N3[6 marks]

- 10 -M09/5/MATME/SP1/ENG/TZ1/XX/M QUESTION 3- 11 QUESTION 6evidence of choosing the product rulef ( x) e x ( sin x) cos x e x ( e x cos x e x sin x)(M1)A1A1substitutinge.g. f (e cos(M1)e sin , e ( 1 0) ,(a)(i)etaking negative reciprocal1e.g.f (1gradient ise(ii)A1interchanging x and y (seen anywhere)e.g. x e y 3M1correct manipulatione.g. ln x y 3 , ln yA1N3(b)0xcollecting like terms; using laws of logse.g. ln x lnx3 , ln x ln x 3 ; ln 1x1xsimplifyQUESTION 4e.g. ln x(a)Functiondisplacementaccelerationt3N4A2N2[6 marks](b)METHOD 1Finding coordinates of image on ge.g. 1 1 0, 1 2 2 , ( 1, 1) ( 1 1, 2 1) , (0, 2)P is (3, 0)(A1)(A1)A1A1METHOD 2h ( x) 2( x 4)2P is (3, 0)2(A1)(A1)A1A1aN4[6 marks]3 , ln x 23N20y 2dx(M1)A12x dx ,xcorrect integration1 2e.g. x dxx21 2a2equating their expression to 321 2a32e.g.22a 64a 8correct substitution VN4(A1)(A1)QUESTION 7e.g.N1N1N1[7 marks]integral expressionA1A1A1A1attempt to substitute into formula Vin any ordertranslated 1 unit to the rightstretched vertically by factor 2N0e3e3x e2A2A2AG(A1)3 2, x23GraphABQUESTION 5(a)x 3f 1 ( x) ln x 3(M1)[6 marks](b)M09/5/MATME/SP1/ENG/TZ1/XX/M (A1)(A1)M1A2N2[7 marks]

- 12 -M09/5/MATME/SP1/ENG/TZ1/XX/M SECTION B(b)(a)(i)x3cosA1N1(ii)y 3sinA1N1[2 marks]finding area(ii)13cos23cos , 8(ii)121A1N2PR224A1N1[3 marks]A1AG36cos 2A2for setting derivative equal to 0dA0,e.g. 36cos 20d(M1)2(A1)2A14(iii)PQ3sinA 18(2sin cos )A 18sin 2dAdvalid reason (seen anywhere)d2 A0 ; maximum when f ( x) 0e.g. at ,4 d 2finding second derivativeevidence of substituting72sin 2e.g.44,d2 Ad 272sin 2(b)N0[3 marks]N22,produces the maximum areaN2R1choosing correct vectors PQ and PR(A1)(A1)finding PQ PR, PQ , PRPQ PR(A1) (A1)(A1)2 4 4( 6)( 1)222 12226 , PR2242substituting into formula for angle between two vectors6ˆe.g. cos RPQ6241simplifying to expression clearly leading to2666e.g.,,6 2 6 144 121ˆcos RPQ224M1A1AGN0METHOD 2A1evidence of choosing cosine rule (seen anywhere)3QR0372AGMETHOD 1PQM1472sin(M1)PQ PO OQ , Q PA1e.g. A 4 3sin(i)evidence of approache.g.1bh2substituting(c)(i)(M1)e.g. A 2 x 2 y , A 8M09/5/MATME/SP1/ENG/TZ1/XX/M QUESTION 9QUESTION 8(a)- 13 -N0[8 marks]Total [13 marks]QRˆcos RPQˆcos RPQˆcos RPQ18 , PQ626 and PR242 66 24 18241222412241824(M1)A1(A1)(A1)(A1)2A1A1AGN0[7 marks]continued

- 14 -M09/5/MATME/SP1/ENG/TZ1/XX/M Question 9 continued(c)(i)(a)evidence of appropriate approachˆˆcos 2 RPQ1 , diagrame.g. using sin 2 RPQ(M1)substituting correctly(A1)ˆe.g. sin RPQ341122(M1)A1AGN0METHOD 232A1yf ( x) is reflection of y f ( x) in x axisand y f ( x) is reflection of y f ( x) in y axisN3(M1)sketch showing these are the same1, P̂ 602evidence of approache.g. drawing a right triangle, finding the missing side3sin Pˆ2since cos Pˆevidence of appropriate approach1e.g. attempt to substitute into ab sin C2correct substitution1e.g. area6242area 3 3METHOD 1evidence of substituting x for xa ( x)f ( x)( x) 2 1axf ( x)f ( x)x2 1METHOD 2(ii)M09/5/MATME/SP1/ENG/TZ1/XX/M QUESTION 10METHOD 1ˆsin RPQ- 15 -(A1)f ( x)f ( x)AG(A1)A1N3(M1)(b)evidence of appropriate approache.g. f ( x) 0(M1)to set the numerator equal to 0e.g. 2ax ( x2 3) 0; ( x 2 3) 0(A1)3,a 3,43,a 34(accept x 0, y0 etc.)A1A1A1A1A1N5[7 marks]A1A1N0[2 marks](0, 0),32axx2 1A1N2[6 marks]Total [16 marks]continued

- 16 -M09/5/MATME/SP1/ENG/TZ1/XX/M -9-Question 10 continued(c)(i)SECTION Acorrect expressionA27e.g.area(ii)aaln ( x 2 1) , ln 50223aln 52QUESTION 1aaln10 , (ln 50 ln10)22(a)A1A1N2METHOD 1(M1)recognizing that the factor of 2 doubles the area(M1)84842 f ( x 1) dx284f ( x 1) dx273A1x 32(M1)x 3 x 3,22accept yA1N2(A1)(M1)A1N3METHOD 1g (4) 5evidence of composition of functionsf (5) 25METHOD 2f ( x ) dxi.e. 2 their answer to (c)(i)2 f ( x 1)dx a ln 5for interchanging x and y (may be done later)e.g. x 2 y 3g 1 ( x)(b)recognizing the shift that does not change the area87ae.g.f ( x 1)dxf ( x)dx , ln 5432e.g.M09/5/MATME/SP1/ENG/TZ2/XX/M N3f g ( x) (2 x 3)2(M1)f g (4) (2 4 3)225(A1)A1METHOD 2changing variabledwlet w x 1 , so1dx2a2 f ( w)dwln ( w2 1) c2substituting correct limitse.g.848a ln[( x 1) 2 1] , a ln ( w2 1)42 f ( x 1)dx a ln 5N3[5 marks]QUESTION 2finding scalar product and magnitudes(M1)7, a ln 50 a ln103(M1)A1N3[7 marks]Total [16 marks]scalar product12 20 15 (magnitudes32(A1)(A1)(A1)23)42 52 , 42( 5)2( 3)2substitution into formula50M112 20 15e.g. cos3223(50cos50 ,425242( 5) 20.46)( 3) 2A2N4[6 marks]QUESTION 3(a)n 10(b)a(c)p , b 2q (or a10 5p (2q)552q , bp)A1N1A1A1N1N1A1A1A1N3[6 marks]

- 10 -M09/5/MATME/SP1/ENG/TZ2/XX/M QUESTION 45(b)METHOD 1A1328yN1(a)xlog 2 32 xlog 2 8 yx log2 32 y log2 8log2 8 3p 5, q3 (accept 5 x 3 y )(A1)(A1)(A1)A1(25 ) x(23 ) y25 x23 y25 x 3 ylog 2 (25 x 3 y ) 5 x 3 yp 5, q3 (accept 5 x 3 y )METHOD 1M(M 1)M1101(M1)2 01 55 01 2a bc dN31 00 15a b 1 , 2b(A1)M(A1)(b)(A1)A1A1A1N3METHOD 2METHOD 232 x8yM09/5/MATME/SP1/ENG/TZ2/XX/M QUESTION 5(a)log 2- 11 -N3[5 marks]0 , 5c d(M1)0 , 2d 10.200.1 0.5(A1)A1N3METHOD 1evidence of appropriate approache.g. X M 1Bxy5 01 217515(M1)A1A1N2METHOD 2evidence of appropriate approach0.20x1e.g.0.1 0.5 y70.2 x 1 ,xy5150.1x 0.5 y 7(M1)A1A1N2[6 marks]

- 12 -M09/5/MATME/SP1/ENG/TZ2/XX/M QUESTION 6(a)e2 xf ( x) 3( x 3)2A2N2METHOD 2(c)x3 9 x 2e3 sin x cos x2x(A1)0 not possible (seen anywhere)(M1)3 sin x cos xe.g.27 x 27f ( x) 3x 2 18 x 27A1N2EITHERf (3) 0 , f (3) 0A1N1tan xMETHOD 1f does not change sign at Pevidence for thisxR1R1(A1)f changes sign at P so P is a maximum/minimum (i.e. not inflexion)evidence for thisR1R1cos x ,sin xcos x1356ORN0sketch of 30 , 60 , 90 triangle with sides 1, 2, 35πwork leading to x65πverifyingsatisfies equation6METHOD 31( x 3) 4 c and sketching this function4indicating minimum at x 30, 3 sin xN0METHOD 2finding f ( x)0simplifyingattempt to expand ( x 3)3(b)M09/5/MATME/SP1/ENG/TZ2/XX/M QUESTION 7METHOD 1e.g. f ( x)- 13 -13A1A1A2N4A1A1A1N4[6 marks]R1R1N0[5 marks]

- 14 -M09/5/MATME/SP1/ENG/TZ2/XX/M - 15 -QUESTION 8(a)(b)SECTION B(i)3e3xA1N1(ii)cos xA1N13xsin xA13e3xcos xh3e3e3sin33e33cos34, 94, 104, 105, 95, 105, 10(b)12, 13, 14, 15 (accept 12, 13, 13, 13, 14, 14, 14, 15, 15)(c)P(12)A2N2[2 marks]A2N2[2 marks]3π3complete correct substitution of xe.g.3, 93, 103, 10(M1)correct expression3eQUESTION 9(a)evidence of choosing product rulee.g. uv vue.g.M09/5/MATME/SP1/ENG/TZ2/XX/M (A1)1, P(13)93, P(14)93, P(15)929A2N2[2 marks](d)3A1N3[6 marks]correct substitution into formula for E ( X )1332e.g. E ( S ) 121314159999123E (S )9A1A2N2[3 marks](e)METHOD 1correct expression for expected gain E ( A) for 1 game45e.g.50309950E ( A)9(A1)amount at end expected gain for 1 game 36200 (dollars)(M1)A1N2METHOD 2attempt to find expected number of wins and losses45e.g.36 ,3699(M1)attempt to find expected gain E (G)e.g. 16 50 30 20E (G) 200 (dollars)(M1)A1N2[3 marks]Total [12 marks]

- 16 -M09/5/MATME/SP1/ENG/TZ2/XX/M L1 : r810(c)0t 01A2(i)N2[2 marks](b)M09/5/MATME/SP1/ENG/TZ2/XX/M Question 10 continuedQUESTION 10(a)- 17 -evidence of equating62e.g. 24s91r and OAL1L2A1attempt to solvefinding s 3A1substitutingA1e.g. OB1 5sevidence of confirming for other two equations1 10e.g. 6 2 4, 2 4 2, 9so A lies on L 2M10t 01one correct equatione.g. 2 2s 8, 4 s 1, 1 5s t(M1)21 , A r5one correct equatione.g. 6 2 2s , 2 4 s , 9s 2evidence of approach228e.g. 4s 11150A1AGOBN0[4 marks](ii)(M1)A1M124123 158114AGevidence of appropriate approach(M1)e.g. AB AO OB , AB OB OA2AB15continued N0A1N2[7 marks](d)evidence of appropriate approach(M1)e.g. AB DCcorrect values2e.g. OD15OD029A12x1 , y4z215221 , 1452 x1 y4 zA1N2[3 marks]Total [16 marks]

- 18 -M09/5/MATME/SP1/ENG/TZ2/XX/M Note: In this question, do not penalize absence of units.(i)sss(c)(40 at )dt40t1 2at2(A1)(A1)c40t0 (c 100)(M1)1 2at2A1N5A1N1[6 marks](b)(i)(ii)stops at station, so v 040(seconds)ta(M1)evidence of choosing formula for s from (a) (ii)40substituting ta40 1402ae.g. 40a 2a2(M1)A1s , 50040a14021600a, 5002a2aa85(A1)A1M1A1R1AGN0METHOD 240104substituting into expression for s1e.g. s 40 104 1022s 200since 200 500,train stops before the stationA2M1A1R1AGN0METHOD 3a is deceleration845so stops in shorter timeso less distance travelledso stops before station800aevidence of simplification to an expression which obviously leads to ae.g. 500a 800, 5N2M140substituting into expression for s1s 40 104 1022s 200since 200 500 (allow FT on their s, if s 500 )train stops before the stationfrom (b) t(M1)setting up equatione.g. 500METHOD 1v 40 4t , stops when v 040 4t 0t 10(M1)substituting s 100 when t1 2s 40tat 1002(ii)M09/5/MATME/SP1/ENG/TZ2/XX/M Question 11 continuedQUESTION 11(a)- 19 -8A15A2A1(A1)R1AGN0[5 marks]8, 1000a 3200 1600aTotal [17 marks]AGN0[6 marks]continued

–9–M10/5/MATME/SP1/ENG/TZ1/XX/M SECTION A(a)evidence of setting function to zeroe.g. f (x) 0 , 8 x 2 x 2(M1)(b)2 (must be equation)(i)x(ii)substituting xy 82 into f ( x)evidence of expandinge.g. 24 4(23 ) x 6(22 ) x2A1(b)term is 25x4, )34do not accepte.g.1210e.g.S(A1)1012022A1A1N3[6 marks]A1N1(A1)(A1)A1A1correct substitution31 25725N3A124,52352A1N1[7 marks]2WPmultiplying WP by 2263x443, cos55correct substitution3 42e.g. sin 25 524sin 225cos 2(A1)N2(A1)(A1)sinNote: The first two steps may be done in any order.subtracting26A22e.g. cos 2(b)x4QUESTION 4N2[7 marks]Note: Award A1 for each correct element.x 4 , (4 4 x x 2 )(4 4 x x 2 )finding coefficients 24 and 1QUESTION 2(a)M14(2) x3(2 x)4 16 32 x 24 x 2 8x3evidence of correct working864e.g. 0 2 x (4 x) ,4x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0) , or xM10/5/MATME/SP1/ENG/TZ1/XX/M QUESTION 3QUESTION 1(a)– 10 –N2[6 marks]

– 11 –M10/5/MATME/SP1/ENG/TZ1/XX/M QUESTION 5(a)(b)(i)p 0.2A1N1(ii)q0.4A1N1(iii)r0.1A1N123A2N2P(A B )valid reason20.5, 0.35 0.3e.g.3R1thus, A and B are not independentAG2dx , 216dx 16 x , 4 x 2 dx20interchanging x and y (seen anywhere)e.g. x log y (accept any base)evidence of correct manipulation11e.g. 3xy , 3y x 2 , xlog 3 y , 2 y log3 x2f 1 ( x) 32 x0 , f 1 ( x) 0(b)y(c)METHOD 1attempt to substitute2 log 2e.g. ( f 1 g ) (2) 3 3attempt to set up integral expression16 4 x 2(a)finding g (2) log3 2 (seen anywhere)N0[6 marks]QUESTION 6e.g.M10/5/MATME/SP1/ENG/TZ1/XX/M QUESTION 70.2Note: Award A1 for an unfinished answer such as.0.3(c)– 12 –(16 4 x2 ) ,4x32dx3(seen anywhere)evidence of substituting limits into the integrand64323232e.g. 32, 64333128volume3A1A1(M1)A2g ) (2) 3(f1g )(2) 4AGN0A1N1A1(A1)log 3 2 2, 3A1N1METHOD 2M116 4 x 2log 3 41A1(M1)evidence of using log or index rulee.g. ( f(M1)N3[6 marks]attempt to form composite (in any order)2 log xe.g. ( f 1 g ) ( x) 3 3(M1)evidence of using log or index rule(A1)log 3 x 2, 3log3 x2e.g. ( f1g )( x) 3(f1g )( x)x2A1(f1g )(2) 4A1N1[7 marks]

– 13 –M10/5/MATME/SP1/ENG/TZ1/XX/M SECTION B(a)f ( x)x22x 3A1A1A1evidence of solving f ( x) 0e.g. x 2evidence of substituting their negative x-value into f ( x)111 3e.g. ( 1)3 ( 1)2 3( 1) ,335y35coordinates are 1,3ddsin x cos x ,cos xdxdxsin x (seen anywhere)evidence of using the quotient rulecorrect substitutionsin 2 x cos 2 xsin x ( sin x) cos x (cos x)e.g.,sin 2 xsin 2 x(sin 2 x cos 2 x)f ( x)sin 2 x1f ( x)sin 2 x(M1)2x 3 0evidence of correct working2 16e.g. ( x 1)( x 3) ,2x1 (ignore x 3 )M10/5/MATME/SP1/ENG/TZ1/XX/M QUESTION 9QUESTION 8(a)– 14 –A1(A1)(A1)(A1)M1A1A1AG(M1)(b)METHOD 1appropriate approache.g. f ( x)(sin x) 2A1(M1)f ( x) 2(sin 3 x)(cos x)N32cos xsin 3 xA1A1(i)( 3, 9)(ii)(1, 4)(iii)reflection gives (3, 9)stretch gives3,92A1N1A1A1N2Note: Award A1 for 2sin x , A1 for cos x .[3 marks]METHOD 2derivative of sin 2 x 2sin x cos x (seen anywhere)evidence of choosing quotient rulesin 2 x 0 ( 1) 2 sin x cos x1 , v sin 2 x , fe.g. u(sin 2 x) 22 cos x2sin x cos xf ( x)sin 3 x(sin 2 x) 2(A1)A1A1N33[8 marks](b)N0[5 marks]N3[6 marks]A1(M1)A1N3[3 marks]Total [14 marks](c)evidence of substitutingsinp(d)2cos1e.g.,2sin 321, q2M1220second derivative is zero, second derivative changes signA1A1N1N1[3 marks]R1R1N2[2 marks]Total [13 marks]

– 15 –M10/5/MATME/SP1/ENG/TZ1/XX/M any correct equation in the form r82e.g. r525a tb (accept any parameter)A2N2(d)(i)t 18Note: Award A1 for a tb , A1 for L a tb , A0 for rb ta .recognizing scalar product must be zero (seen anywhere)e.g. a b 0evidence of choosing direction vectors21 ,872kcorrect calculation of scalar producte.g. 2( 7) 1( 2) 8kk(c)503any two correct equationse.g. 3 2 p 5 7q , 1 p2q , 25 8 p 3 2qN0[5 marks]722A1A1attempting to solve equationsfinding one correct parameter ( p3, q4, 1)2)A1722N2A1evidence of finding magnitude(M1)qfinding AC(A1)(A1)AGevidence of equating vectors32e.g. L1 L 3 ,1p 1258the coordinates of A ar

–14 – M13/5/MATME/SP1/ENG/TZ1/XX/M Question 8 continued (c) setting lines equal (M1) eg LL 12= , 16 1 3 222 3 3 4 15 6 ts ...