9MA0/01: Pure Mathematics Paper 1 Mark Scheme
9MA0/01: Pure Mathematics Paper 1 Mark schemeQuestion1 (a)SchemeMarksAOsB11.1bM11.1bA11.1b1Area(R ) 0.5 0.5 2 ( 0.6742 0.8284 0.9686 ) 1.0981 2 1 6.5405 1.635125 1.635 (3 dp) 4 (3)(b)Any valid reason, for example Increase the number of strips Decrease the width of the strips Use more trapezia between x 1 and x 3B12.4(1)(c)(i)(c)(ii) dx 5("1.635") 8.1751x 3 x 6 dx 6(2) ("1.635") 13.6351 1 x 1 B1ft2.2a B1ft2.2a35x(2)(6 marks)Question 1 Notes:(a)10.51 0.5 oror 0.25 or242B1:Outside bracketsM1:For structure of trapezium rule . .No errors are allowed, e.g. an omission of a y-ordinate or an extra y-ordinate or a repeatedy-ordinate.A1:Correct method leading to a correct answer only of 1.635(b)B1:See scheme(c)B1:8.175 or a value which is 5 their answer to part (a)Note: Allow B1ft for 8.176 (to 3 dp) which is found from 5(1.63125) 8.175625Note: Do not allow an answer of 8.1886 which is found directly from integration(d)B1:13.635 or a value which is 12 their answer to part (a)Note: Do not allow an answer of 13.6377 which is found directly from integrationA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme1
Question2 (a)Scheme112 1 1 2 15x 2 2 5 x 2 1 .2! 2 4 4 2 AOsB11.1bM11.1bA1ft1.1bA12.115x 25x 2 (4 5 x) ( 4 ) 1 2 1 4 4 12Marks 525 2x x .464(4)(b)(i)11 2x (4 5(0.1)) 10 4.5 32 or 1.5 2 or23543 2 2256So,(b)(ii)x 2 181or128orM11.1bM13.1aA11.1bB12.3332 or22235 1 25 1 2 . 2.121. 4 10 64 10 23543 2 2562 2 .25618114satisfies x (o.e.), so the approximation is valid.105(4)(8 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme2
Question 2 Notes:(a)11B1:Manipulates (4 5 x) 2 by taking out a factor of (4) 2 or 2M1:Expands (. x ) 2 to give at least 2 terms which can be simplified or un-simplified,1( 1 )( 12 ) 1 1 E.g. 1 ( x) or ( x) 2( x ) 2222! where is a numerical value and where 1 .A1ft:A1:1 . or( 12 )( 12 )( x ) 22!( 1 )( 12 ) 1 A correct simplified or un-simplified 1 ( x) 2( x) 2 expansion with consistent ( x)22! 525Fully correct solution leading to 2 x kx 2 , where k 464(b)(i)11or 0.1 into (4 5 x) 210M1:Attempts to substitute x M1:A complete method of finding an approximate value for substituting x 1or 0.1 into their part (a) binomial expansion and equating the result to10an expression of the form 2 or followed by re-arranging to giveA1:2 . E.g. 2; , 02 .181543362or any equivalent fraction, e.g.or384256128256Also allowor any equivalent fraction181(b)(ii)B1:Explains that the approximation is valid because x 41satisfies x 510A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme3
QuestionSchemeMarksAOsM11.1b 33(4.5) 3M12.2a 151.5A11.1ba1 3 , a2 0, a3 1.5 , a4 33 (a)100 arr 1(3)100 (b)99ar r 1 a (2)(151.5) 3 300B1ftr2.2ar 1(1)(4 marks)Question 3 Notes:(a)an 3, with a1 3 to generate values for a2 , a3 and a4an 2M1:Uses the formula an 1 M1:Finds a4 3 and deduces100 ar 33("3" "0" "1.5") "3"r 1A1:which leads to a correct answer of 151.5(b)B1ft:Follow through on their periodic function. Deduces that either100 99 a a (2)("151.5") 3 300rrr 1r 110099 a a "151.5" (33)("3" "0" "1.5")rr 1r 151.5 148.5 300r 1A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme4
QuestionSchemeMarksAOsM13.1aA11.1bOA i 7 j 2k , OB 4i 3 j 3k , OC 2i 10 j 9k4 (a)OD OC BA (2i 10 j 9k ) ( 3i 4 j 5k )or OD OA BC (i 7 j 2k ) ( 2i 7 j 6k )So(2)(b)AsM11.1bM13.1aA11.1bthenorSoonly(3)(5 marks)Question 4 Notes:(a)M1:A1:A complete method for finding the position vector of Dæ -1 öç -i 14j 4k or ç 14 ç 4 èø(b)M1:A complete attempt to findorM1:A complete process for finding the position vector of XA1:æ 7 öç 7i - j 8k or ç -1 ç 8 èøA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme5
QuestionSchemeMarksAOsM11.1bA1*1.1bM12.2aA11.1b5 (a)(i)f (-6) (-6)3 a(-6)2 - a(-6) 48 -216 36a 6a 48 0 Þ 42a 168 Þ a 4 *(a)(ii)Hence, f (x) (x 6)(x 2 - 2x 8)(4)(b)2log2 (x 2) log 2 x - log2 (x - 6) 3E.g. log 2 (x 2)2 log2 x - log2 (x - 6) 3 æ x ö2log 2 (x 2) log 2 ç 3è x - 6 øæ x(x 2)2 ölog 2 ç 3è (x - 6) øæ x(x 2)2 ö3çè (x - 6) ø 2()()é or log x(x 2)2 log 8(x - 6) ù22ëû{i.e. log a 3 Þ2}a 23 or 8M11.2M11.1bB11.1bA1 *2.1x(x 2)2 8(x - 6) Þ x(x 2 4x 4) 8x - 48Þ x 3 4x3 4x 8x - 48 Þ x 3 4x 3 - 4x 48 0 *(4)(c)2log2 (x 2) log 2 x - log2 (x - 6) 3 Þ x 3 4x 3 - 4x 48 0Þ (x 6)(x 2 - 2x 8) 0Reason 1: E.g. log 2 x is not defined when x - 6 log 2 (x - 6) is not defined when x - 6 x -6, but log 2 x is only defined for x 0Reason 2: b2 - 4ac - 28 0, so (x 2 - 2x 8) 0 has no (real) rootsAt least one of Reason 1 or Reason 2B12.4Both Reason 1 and Reason 2B12.1(2)(10 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme6
Question 5 Notes:(a)(i)M1:A1*:(a)(ii)M1:Applies f (-6)Applies f (-6) 0 to show that a 4Deduces (x 6) is a factor of f (x) and attempts to find a quadratic factor of f (x) by eitherequating coefficients or by algebraic long divisionA1:(x 6)(x 2 - 2x 8)(b)M1:Evidence of applying a correct law of logarithmsM1:Uses correct laws of logarithms to give either an expression of the form log 2 (h(x)) k , where k is a constant an expression of the form log 2 (g(x)) log 2 (h(x))B1:Evidence in their working of log 2 a 3 Þ a 23 or 8A1*:Correctly proves x 3 4x 3 - 4x 48 0 with no errors seen(c)B1:B1:See schemeSee schemeA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme7
Question6 (a)SchemeAttempts to use an appropriate model;e.g. y A(3- x)(3 x) or y A(9 - x 2 )MarksAOsM13.3M13.1bA11.1be.g. y A(9 - x 2 )Substitutes x 0, y 5 Þ 5 A(9 - 0) Þ A 5955y (9 - x 2 ) or y (3- x)(3 x),99(3)(b)Substitutes x 52.4into their y (9 - x 2 )925y (9 - x 2 ) 4.2 4.1 Þ Coach can enter the tunnel9M13.4A12.2b(2)(b)Alt 1æ9 2ö59 24.1 (9 - x 2 ) Þ x , so maximum width 2 ç 910è 10 øM13.4 2.545. 2.4 Þ Coach can enter the tunnelA12.2b(2)(c)E.g. Coach needs to enter through the centre of the tunnel. This willonly be possible if it is a one-way tunnelIn real-life the road may be cambered (and not horizontal)The quadratic curve BCA is modelled for the entrance to the tunnelbut we do not know if this curve is valid throughout the entirelength of the tunnelThere may be overhead lights in the tunnel which may block thepath of the coachB13.5b(1)(6 marks)Question 6 Notes:(a)M1:Translates the given situation into an appropriate quadratic model – see schemeM1:Applies the maximum height constraint in an attempt to find the equation of the model – see schemeA1:Finds a suitable equation – see scheme(b)M1:See schemeA1:Applies a fully correct argument to infer {by assuming that curve BCA is quadratic and the givenmeasurements are correct}, that is possible for the coach to enter the tunnel(c)B1:See schemeA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme8
Question7Scheme{ò{ò{òìïï u xxe 2x dx , íï dv e 2 xïî dx}}xe 2x dx 1 2xxe 22e2 x - xe 2x dx}du 1dx1Þ v e2 x2Þò 2e12xò 2e12x{dx}ö øæ11 ö e 2 x - ç xe 2 x - e 2 x 4 øè2Area(R) x}æ1 e 2 x - ç xe 2 x è2Marks2é5ù12e 2 x - xe 2 x dx ê e 2 x - xe 2x ú2ë4û0205æ5ö æ5ö 11 ç e 4 - e 4 - ç e 2(0) - (0)e0 e 4 42è4ø è4ø 4(5)7Alt 1{ò2e2 x - xe2x dx ò (2 - x)e 1(2 x)e 2 x 2 11(2 x)e 2 x e 2 x24ìíArea(R) î2xdx}ìduïï u 2 - x Þ dx - 1, íï dv e 2 x Þ v 1 e 2 xïî dx21 e 2 x dx 2òM13.1aM11.1bA11.1bM12.2aA12.12ü é11 ù(2 - x)e dx ý ê (2 - x)e 2x e2 x ú4 û00þ ë22üïïýïïþ2x5æ1 ö æ11 ö 1 ç 0 e 4 - ç (2)e0 e0 e 4 44 ø è24 ø 4è(5)(5 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme9
Question 7 Notes:M1:Attempts to solve the problem by recognising the need to apply a method of integration by parts oneither xe 2 x or (2 - x)e2x . Allow this mark for either{ } xe2x l xe2x ò m e2x dx (2 - x)e2x l (2 - x)e2x ò m e2x dx{ }where l , m ¹ 0 are constants.M1:For either 2e2 x - xe 2x e 2x 12x{dx}{ }12x(2 - x)e 2x (2 - x)e2xdx2Correct integration which can be simplified or un-simplified. E.g. A1:ò 2e1 ò 2e1 2xxe 2 æ11 ö2e2 x - xe 2 x e 2 x - ç xe 2 x - e2 x 4 øè2M1:1 2x 1 2xxe e2451 2e2 x - xe 2x e 2x - xe 2x4211 (2 - x)e 2x (2 - x)e 2x e 2x24Deduces that the upper limit is 2 and uses limits of 2 and 0 on their integrated functionA1:Correct proof leading to pe 4 q, where p 2e2 x - xe 2x e 2x -15, q 44A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme10
Question8 (a)Scheme2100(1 - (1.012)14 )Total amount 1 - 1.012or2100((1.012)14 -1)1.012 - 1 31806.9948 . 31800 (tonnes) (3 sf )MarksAOsM13.1bA11.1b(2)Total Cost 5.15(2000(14)) 6.45(31806.9948. - (2000)(14))M13.1bM11.1bA13.2a 5.15(28000) 6.45(3806.9948.) 144200 24555.116. 168755.116. 169000 (nearest 1000)(3)(5 marks)Question 8 Notes:(a)M1:Attempts to apply the correct geometric summation formula with either n 13 or n 14 ,a 2100 and r 1.012 (Condone r 1.12)A1:(b)Correct answer of 31800 (tonnes)M1:Fully correct method to find the total costM1:For eitherA1:{} 5.15(2000(14)) 144200 6.45("31806.9948." - (2000)(14)) 24555.116. 5.15(2000(13)) 133900 6.45("29354.73794." - (2000)(13)) 21638.059.{}{}{}Correct answer of 169000Note: Using rounded answer in part (a) gives 168710 which becomes 169000 (nearest 1000)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme11
Question9SchemeGradient of chord (2(x h)3 5) - (2x 3 5)x h-h(x h)3 x3 3x 2 h 3xh2 h3Gradient of chord MarksAOsB11.1bM12.1B11.1bA11.1bA12.2a(2(x 3 3x 2 h 3xh2 h3 ) 5) - (2x 3 5)1 h -1 2x 3 6x 2 h 6xh2 2h3 5 - 2x 3 - 51 h -1 6x 2 h 6xh2 2h3h 6x 2 6xh 2h2()dydylim 6x 2 6xh 2h2 6x 2 and so at P, 6(1)2 6h 0dxdx(5)9Alt 1Let a point Q have x coordinate 1 h, so yQ 2(1 h)3 5B11.1b2(1 h)3 5 - 71 h -1M12.1(1 h)3 1 3h 3h2 h3B11.1bA11.1bA12.2a{ P(1, 7), Q(1 h, 2(1 h) 3) Þ}3Gradient PQ 2(1 3h 3h2 h3 ) 5 - 71 h -1Gradient PQ 2 6h 6h2 2h3 5 - 7 1 h -16h 6h2 2h3h 6 6h 2h2()dylim 6 6h 2h2 6h 0dx(5)(5 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme12
Question 9 Notes:B1:2(x h)3 5, seen or impliedM1:Begins the proof by attempting to write the gradient of the chord in terms of x and hB1:(x h)3 x 3 3x 2 h 3xh2 h3 , by expanding brackets or by using a correct binomial expansionM1:Correct process to obtain the gradient of the chord as a x 2 b xh g h2 , a , b , g ¹ 0A1:Correctly shows that the gradient of the chord is 6x 2 6xh 2h2 and applies a limiting argument todeduce when y 2x 3 5,()dylim6x 2 6xh 2h2 6x 2 . Finally, deduces that 6x 2 . E.g.h 0dxdy 6.dxNote: d x can be used in place of hat the point P,Alt 1B1:Writes down the y coordinate of a point close to P.E.g. For a point Q with x 1 h,{ y } 2(1 h)Q3 5M1:Begins the proof by attempting to write the gradient of the chord PQ in terms of hB1:(1 h)3 1 3h 3h2 h3 , by expanding brackets or by using a correct binomial expansionM1:Correct process to obtain the gradient of the chord PQ as a b h g h2 , a , b , g ¹ 0A1:Correctly shows that the gradient of PQ is 6 6h 2h2 and applies a limiting argument to deducethat at the point P on y 2x 3 5,()dylim6 6h 2h2 6 6. E.g.h 0dxNote: For Alt 1, d x can be used in place of hA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme13
Question10 (a)Scheme3x - 5Þ y(x 1) 3x - 5 Þ xy y 3x - 5 Þ y 5 3x - xyx 1y Þ y 5 x(3 - y) ÞHence f -1 (x) y 5 x3- yx 5,3- xMarksAOsM11.1bM12.1A12.5(3)(b)æ 3x - 5 ö3ç-5è x 1 øff (x) æ 3x - 5 öçè x 1 ø 13(3x - 5) - 5(x 1)x 1 (3x - 5) (x 1)x 1 9x - 15 - 5x - 5 4x - 20 x - 5 3x - 5 x 14x - 4x -1(note that a - 5)M11.1aM11.1bA11.1bA12.1(4)(c)fg(2) f (4 - 6) f (- 2) 3(-2) - 5; 11-2 1M11.1bA11.1b(2)(d)g( x) x 3x ( x 1.5) 2.25 . Hence g min 2.2522Either g min 2.25 oror g(5) 25 15 10orM12.1B11.1bA11.1b(3)(e)E.g. the function g is many-onethe function g is not one-onethe inverse is one-manyB12.4g(0) g(3) 0(1)(13 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme14
Question 10 Notes:(a)M1:Attempts to find the inverse by cross-multiplying and an attempt to collect all the x-terms (orswapped y-terms) onto one sideA fully correct method to find the inverseA1:A correct f -1 (x) M1:x 5,3- x, expressed fully in function notation (including the domain)(b)3f (x) - 53x - 5intof (x) 1x 1M1:Attempts to substitute f (x) M1:Applies a method of “rationalising the denominator” for both their numerator and their denominator.A1:A1:3(3x - 5) - 5(x 1)x 1which can be simplified or un-simplified(3x - 5) (x 1)x 1Shows ff (x) x ax-5where a -5 or ff (x) , with no errors seen.x -1x -1(c)M1:Attempts to substitute the result of g(2) into fA1:Correctly obtains fg(2) 11(d)M1:Full method to establish the minimum of g.E.g. (x a )2 b leading to gmin b Finds the value of x for which g (x) 0 and inserts this value of x back into g(x) in orderto find to g minB1:For either finding the correct minimum value of g(Can be implied byor g(x) - 2.25 ) stating g(5) 25 15 10A1:States the correct range for g. E.g.(e)B1:See schemeorA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme15
Question11 (a)SchemeMarksAOsM11.1bB12.4A12.1f (x) k - 4x - 3x 2f (x) - 4 - 6x 0Criteria 1Eitherf (x) - 4 - 6x 0 Þ x 42Þ x -63oræ 2öæ 2öf ç - - 4 - 6 ç - 0è 3øè 3øCriteria 2Either f (-0.7) - 4 - 6(-0.7) 0.2 0f (-0.6) - 4 - 6(-0.6) -0.4 0oræ 2öf ç - - 6 ¹ 0è 3ø At least one of Criteria 1 or Criteria 2Both Criteria 1 and Criteria 2and concludes C has a point of inflection at x -23(3)(b)f (x) k - 4x - 3x 2 , AB 4 2{ }M11.1bA11.1bA12.2aM12.1A11.1bAB -1 k 1 - -1 - k 1 4 2 Þ k .M12.1So, 2 k 1 4 2 Þ k 7A11.1bf (x) k x - 2x 2 - x 3 cf (0) 0 or (0, 0) Þ c 0 Þ f (x) k x - 2x 2 - x3{f (x) 0 Þ} f (x) x(k - 2x - x ) 02{x2 2x - k 0} Þ (x 1)2Þ x -1 k 1() ({}Þ x 0, k - 2x - x 2 0- 1 - k 0 , x .)(7)(10 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme16
Question 11 Notes:(a)M1:E.g. æ 2öattempts to find f ç - è 3øfinds f (x) and sets the result equal to 0B1:See schemeA1:See scheme(b)M1:Integrates f (x) to give f (x) k x a x 2 b x 3 , a , b ¹ 0 with or without the constant ofintegrationA1:f (x) k x - 2x 2 - x 3 , with or without the constant of integrationA1:Finds f (x) k x - 2x 2 - x 3 c , and makes some reference to y f (x) passing through the originto deduce c 0. Proceeds to produce the result k - 2x - x 2 0 or x 2 2x - k 0M1:Uses a valid method to solve the quadratic equation to give x in terms of kA1Correct roots for x in terms of k. i.e. x -1 k 1M1:Applies AB 4 2 on x -1 k 1 in a complete method to find k .A1:Finds k 7 from correct solution onlyA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme17
M11.1bM11.1bA1*2.1sin 2qdq1 cosq20Attempts this question by applying the substitution u 1 cosqand progresses as far as achievingu 1 cosq Þò.(u - 1).udu - sinq and sin2q 2sinq cosqdqòòì sin 2qüdq ýíî 1 cosqþò2sinq cosqdq 1 cosqò- 2(u - 1)duuæ1ö- 2 ç 1 - du - 2(u - lnu)uøèìïíïîòp20ü1sin 2qïdq ý - 2 éë u - lnu ùû2 - 2((1- ln1) - (2 - ln 2))1 cosqïþ - 2(-1 ln2) 2 - 2ln2 *(7)12Alt 1Attempts this question by applying the substitution u cosq3.1aM11.1bA12.1M11.1bM11.1bü0sin 2qïdq ý - 2 éë u - ln(u 1) ùû1 - 2((0 - ln1) - (1- ln 2))1 cosqïþM11.1b - 2(-1 ln2) 2 - 2ln2 *A1*2.1u cosq Þòòò2sinq cosqdq 1 cosqì(u 1) - 1du - 2í - 2u 1îòp02.du - sinq and sin2q 2sinq cosqdqì sin 2qüdq ýíî 1 cosqþìïíïîòu.u 1M1and progresses as far as achievingò1-ò- 2uduu 1ü1du ý - 2(u - ln(u 1))u 1 þ(7)(7 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme18
Question 12 Notes:M1:See schemeM1:Attempts to differentiate u 1 cosq to giveA1:Applies u 1 cosq to show that the integral becomesdu . and applies sin2q 2sinq cosqdqò- 2(u - 1)duuM1:Achieves an expression in u that can be directly integrated (e.g. dividing each term by u or applyingpartial fractions) and integrates to give an expression in u of the form lu m lnu, l , m ¹ 0M1:For integration in u of the form 2(u - lnu)M1:Applies u-limits of 1 and 2 to an expression of the form lu m lnu, l , m ¹ 0 and subtracts eitherway roundA1*:Applies u-limits the right way round, i.e. òò12 12òò- 2(u - 1)du - 2u- 2(u - 1)du 2uand correctly provesòp2012211æ1öçè 1 - u ø du - 2 éë u - lnu ùû 2 - 2((1- ln1) - (2 - ln 2))2æ1öçè 1 - u ø du 2 éë u - lnu ùû1 2((2 - ln 2) - (1- ln1))sin 2qdq 2 - 2ln 2, with no errors seen1 cosqAlt 1M1:See schemeM1:Attempts to differentiate u cosq to giveA1:Applies u cosq to show that the integral becomesdu . and applies sin2q 2sinq cosqdqò- 2uduu 1M1:Achieves an expression in u that can be directly integrated (e.g. by applying partial fractions or asubstitution v u 1) and integrates to give an expression in u of the form lu m ln(u 1), l , m ¹ 0 or lv m lnv, l , m ¹ 0, where v u 1M1:For integration in u in the form 2(u - ln(u 1))M1:Either Applies u-limits of 0 and 1 to an expression of the form lu m ln(u 1), l , m ¹ 0 andsubtracts either way round Applies v-limits of 1 and 2 to an expression of the form l v m lnv, l , m ¹ 0, wherev u 1 and subtracts either way roundA1*:Applies u-limits the right way round, (o.e. in v ) i.e. òò01 01òò- 2udu - 2u 1- 2udu 2u 1and correctly provesòp0201100æ1 öçè 1 - u 1 ø du - 2 éë u - ln(u 1) ùû1 - 2((0 - ln1) - (1- ln 2))1æ1 öçè 1 - u 1 ø du 2 éë u - ln(u 1) ùû0 2((1- ln 2) - (0 - ln1))sin 2qdq 2 - 2ln 2, with no errors seen1 cosqA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme19
Question13 (a)SchemeR 2.5tan a 1.5o.e.2a 0.6435, so 2.5sin(q - 0.6435)MarksAOsB11.1bM11.1bA11.1b(3)(b)æ 4p (0) öæ 4p (0) öe.g. D 6 2sin ç- 1.5cos ç 4.5m è 25 øè 25 øæ 4p (0)öor D 6 2.5sin ç- 0.6435 4.5mè 25øB13.4(1)(c)Dmax 6 2.5 8.5 mB1ft3.4(1)(d)Sets4p t5pp- "0.6435" or2522Afternoon solution Þö4p t5p25 æ 5p- "0.6435" Þt "0.6435" ç2524p è 2øÞ t 16.9052. Þ Time 16 :54 or 4:54 pmM11.1bM13.1bA13.2a(3)(e)(i) An attempt to find the depth of water at 00:00 on 19th October2017 for at least one of either Tom’s model or Jolene’s model. At 00:00 on 19th October 2017,Tom: D 3.72. m and Jolene: H 4.5 mand e.g. As 4.5 ¹ 3.72 then Jolene’s model is not true (ii)M13.4A13.5aB13.3Jolene’s model is not continuous at 00:00 on 19th October 2017Jolene’s model does not continue on from where Tom’s model hasendedTo make the model continuous, e.g. (3)(11 marks)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme20
Question13 (d)Alt 1SchemeSets4p tp- "0.6435" 252æ 4p öPeriod 2p ç 12.5è 25 øö25 æ pAfternoon solution Þ t 12.5 "0.6435" ç4p è 2øÞ t 16.9052. Þ Time 16 :54 or 4:54 pmMarksAOsM11.1bM13.1bA13.2a(3)Question 13 Notes:(a)B1:R 2.5 Condone R 6.25M1:For either tan a A1:a awrt 0.64351.51.522or tan a or tan a or tan a 221.51.5(b)B1:Uses Tom’s model to find D 4.5 (m) at 00:00 on 18th October 2017(c)B1ft:(d)M1:Either 8.5 or follow through “6 their R ” (by using their R found in part (a))æ 4p t öæ 4p t öæ 4p töRealises that D 6 2sin ç- 1.5cos ç 6 "2.5"sin ç- "0.6435" and è 25 øè 25 øè 25øæ 4p tö4p tp5pso maximum depth occurs when sin ç- "0.6435" 1 Þ- "0.6435" orè 25ø2522M1:Uses the model to deduce that a p.m. solution occurs when4p t5pand rearranges- "0.6435" 252this equation to make t .A1:(d)Finds that maximum depth occurs in the afternoon at 16:54 or 4:54 pmAlt 1M1:æ 4p tö4p tpMaximum depth occurs when sin ç- "0.6435" 1 Þ- "0.6435" è 25ø252M1:æ 4p öRearranges to make t . and adds on the period, where period 2p ç 12.5è 25 øA1:Finds that maximum depth occurs in the afternoon at 16:54 or 4:54 pm{A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme}21
Question 13 Notes Continued:(e)(i)M1:A1:See schemeSee schemeNote: Allow Special Case M1 for a candidate who just states that Jolene’s model is not continuousat 00:00 on 19th October 2017 o.e.(e)(ii)B1:Uses the information to set up a new model for H. (See scheme)A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme22
Question14SchemeMarksAOsææpöæ p ööx y 4 ç cost cos ç - sint sin ç 2sintè 6øè 6øøèM13.1aM11.1bx y 2 3costA11.1bM13.1aA12.1æpöx 4cos ç t ,6øè x y 2 3y 2sint22 y 1 2 2(x y)2 y 1124(x y)2 3y 2 12(5)14Alt 1æöæpö(x y) ç 4cos ç t 2sint 6øèèø22æ æöæpöæ p öö ç 4 ç cost cos ç - sint sin ç 2sint è 6øè 6øøè èø( 2 3cost)22or 12cos2 tæ yöM13.1aM11.1bA11.1bM13.1aA12.12So, (x y)2 12(1- sin 2 t) 12 - 12sin 2 t 12 - 12 ç è 2ø(x y)2 3y 2 12(5)(5 marks)Question 14 Notes:M1:Looks ahead to the final result and uses the compound angle formula in a full attempt to write downan expression for x y which is in terms of t only.M1:Applies the compound angle formula on their term in x. E.g.ææpöæpöpöcos ç t cost cos ç sint sin ç 6øèè 6øè 6øA1:Uses correct algebra to find x y 2 3costM1:Complete strategy of applying cos2 t sin2 t 1 on a rearranged x y "2 3cost", y 2sintto achieve an equation in x and y onlyA1:Correctly proves (x y)2 ay 2 b with both a 3, b 12 , and no errors seenA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme23
Question 14 Notes Continued:Alt 1M1:Apply in the same way as in the main schemeM1:Apply in the same way as in the main schemeA1:Uses correct algebra to find x yM1:Complete strategy of applying cos2 t sin2 t 1 on x y()2( 2 3cost)2or(( x y))22( 12cos2 t)2 "2 3cost " to achieve anequation in x and y onlyA1:Correctly proves (x y)2 ay 2 b with both a 3, b 12 , and no errors seenA level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme24
A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme 1 9MA0/01: Pure Mathematics Paper 1 Mark scheme Question Scheme Marks AOs 1 (a) 1.09811 ( ) 2 R B1 1.1b M1 1.1b 1 5 4 (3 dp) A1 1.1b (3) (b) Any valid reason, for example Increas
Paper 1: Pure Mathematics 1 (*Paper code: 9MA0/01) Paper 2: Pure Mathematics 2 (*Paper code: 9MA0/02) Paper 3: Statistics and Mechanics (*Paper code: 9MA0/03) Each paper is: 2-hour written exam with calculator; 33.33% of the qualification; 100 marks. To get an E you need an average of
A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme 1 9MA0/02: Pure Mathematics Paper 2 Mark scheme Question Scheme Marks AOs 1 1 2) 2 r M1 1.1a A1 225 22. 24 r 1.1b length of minor arc ) dM1 3.1a 36 ab A1 1.1b (4) 1 Alt 1 2(4.8) 2 r M1 1.1
Pure Mathematics 1 9MA0 01 7 Oct 2020 PM 2h 00m Pure Mathematics 2 9MA0 02 14 Oct 2020 PM 2h 00m Statistics & Mechanics 9MA0 03 19 Oct 2020 PM 2h 00m . Autumn Series 2020 Written exams A-Level * Subject Component Title Compone
Pearson Edexcel Level 3 GCE Mathematics Advanced Level Paper 1 or 2: Pure Mathematics Practice Paper C Time: 2 hours Paper Reference(s) 9MA0/01 or 9MA0/02 You must have: Mathematical Formulae and Statistical Tables, calculator Candidates may use any calculator permitted by Pearson regulations.
Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) First teaching from September 2017 First certification from 2018 A Level Mathematics This draft qualification has not yet been accredited by Ofqual. It is published to enable teachers to have early sight of our proposed approach to Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0).
Edexcel Specification 9MA0 EXAM PAPER The Edexcel Maths A-level will be taught from a new linear syllabus. Students will start learning the material in September and will take three exams in May/June two years later. Exam 1 – Pure Mathematics, out of 100 marks. Counts for 33.33% of the qualification. Duration of 2 hours. Paper code: 9MA0/01.
CAPE Pure Mathematics Mark Schemes: Unit 1 Paper 01 103 Unit 1 Paper 02 105 Unit 1 Paper 032 118 Unit 2 Paper 01 123 Unit 2 Paper 02 125 Unit 2 Paper 032 138 CAPE Pure Mathematics Subject Reports: 2004 Subject Report 143 2005 Subject Report 181 2006 Subject Report 213 2007 Subject Report 243 2008 Subject Report Trinidad and Tobago 275
C is much more flexible than other high-level programming languages: C is a structured language. C is a relatively small language. C has very loose data typing. C easily supports low-level bit-wise data manipulation. C is sometimes referred to as a “high-level assembly language”. When compared to assembly language .
