9MA0/02: Pure Mathematics Paper 2 Mark Scheme
9MA0/02: Pure Mathematics Paper 2 Mark schemeQuestion1SchemeMarksAOs1 2r (4.8)2M11.1a1 2225r (4.8) 135 r 2 r 7.5 o.e.24A11.1bdM13.1aA11.1blength of minor arc 7.5(2 4.8) 15 36 a 15, b 36 (4)1Alt1 2r (4.8)2M11.1a1 2225r (4.8) 135 r 2 r 7.5 o.e.24A11.1bdM13.1aA11.1blength of major arc 7.5(4.8) 36 length of minor arc 2 (7.5) 36 15 36 a 15, b 36 (4)(4 marks)Question 1 Notes:M1:Applies formula for the area of a sector with 4.8; i.e.Note: Allow M1 for considering ratios. E.g.A1:dM1:135 4.8 r 2 2 1 Uses a correct equation e.g. r 2 (4.8) 135 to obtain a radius of 7.52 Depends on the previous M mark.A complete process for finding the length of the minor arc AB, by either (their r ) (2 4.8) A1:1 2r with 4.822 (their r ) (their r )(4.8)Correct exact answer in its simplest form, e.g. 15 36 or 36 15 A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme1
QuestionScheme1Attempts to substitute cos 1 2 into either 1 4cos or 3cos 2 22(a)1 1 1 4cos 3cos 1 4 1 2 3 1 2 2 2 MarksAOsM11.1bM11.1bA1*2.1221 1 1 4 1 2 3 1 2 4 2 4 3 1 4 2 2 3 3 2 44 8 5 2 *(3)E.g.(b)(i) (b)(ii)Adele is working in degrees and not radians5 Adele should substitute and not 5 into the180approximation 5 8 5 awrt 7.962 , so 5 gives a good approximation. 180 B12.3B12.42(2)(5 marks)Question 2 Notes:(a)(i)M1:M1:See scheme1Substitutes cos 1 2 into 1 4cos 3cos 2 and attempts to apply21 2 1 2 2Note: It is not a requirement for this mark to write or refer to the term in 4A1*:(a)(ii)B1:Correct proof with no errors seen in working.Note: It is not a requirement for this mark to write or refer to the term in 4See scheme(b)(i)B1:See scheme(b)(ii)B1:Substitutes 5 orinto 8 5 2 to give awrt 7.962 and an appropriate conclusion.18036A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme2
Question3 (a)Scheme t 0, 75 75 25 A A 50 25 50e 0.03tMarksAOsB13.3(1)(b) 60 60 25 "50"e 0.03t e 0.03t t ln(0.7) 11.8891648 11.9 minutes (1 dp) 0.0360 25"50"M13.4A11.1b(2)(c)A valid evaluation of the model, which relates to the large values of t.E.g. As 20.3 25 then the model is not true for large values of t 20.3 25 0.094 does not have any solutions and so"50"the model predicts that tea in the room will never be 20.3 C. Sothe model does not work for large values of te 0.03t B13.5at 120θ 25 50e 0.03(120) 26.36 which is notapproximately equal to 20.3, so the model is not true for largevalues of t(1)(4 marks)Question 3 Notes:(a)B1:Applies t 0, 75 to give the complete model 25 50e 0.03t(b)M1:Applies 60 and their value of A to the model and rearranges to make e 0.03t the subject.Note: Later working can imply this mark.A1Obtains 11.9 (minutes) with no errors in manipulation seen.(c)B1See schemeA level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme3
QuestionSchemeMarksAOsB11.1bB11.1by4(a)Correct graph inquadrant 1 and quadrant 2with V on the x-axis 5 States (0, 5) and , 0 2 55marked in the correct position2on the x-axisand 5 marked in the correct positionon the y-axisorOx52(2)2x 5 7(b)2 x 5 7 x . and (2 x 5) 7 x .M11.1b{critical values are x 6, 1 } x 1 or x 6A11.1b(2)(c)2x 5 x 52E.g. 55to give x 2255and solves (2 x 5) x to also give x 225Sketches graphs of y 2 x 5 and y x .2Solves 2 x 5 x M13.1aA12.5 5 Indicates that these graphs meet at the point , 0 2 Hence using set notation, e.g. 5 x : x 2 5 x , x 2 5 x : x 2 5 2 (2)(6 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme4
Question 4 Notes:(a)B1:See schemeB1:See scheme(b)M1:A1:See schemeCorrect answer, e.g. x 1 or x 6 x 1 x :x 6x 1 x :x 6 (c)M1:A complete process of finding that y 2 x 5 and y x 5meet at only one point.2This can be achieved either algebraically or graphically.A1:See scheme.Note: Final answer must be expressed using set notation.A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme5
1bo.e.A11.1b3x 2 y k intersects y 2 x 2 5 at two distinct pointsEliminate y and forms quadratic equation 0 or quadratic expression 0 3x 2(2 x 5) k 4 x 3x 10 k 0 "b 4ac " 0 3 4( 4)(10 k ) 022229 16(10 k ) 0 169 16k 0Critical value obtained ofk 16916(5)5Eliminate y and forms quadratic equation 0 or quadratic expression 0 M13.1aA11.1bdM12.116916B11.1bo.e.A11.1b22 1 y 2 (k 2 y ) 5 y (k 2 4ky 4 y 2 ) 59 3 Alt 18 y 2 (8k 9) y 2k 2 45 0 "b2 4ac " 0 (8k 9) 2 4(8)(2k 2 45) 064k 2 144k 81 64k 2 1440 0 144k 1521 0Critical value obtained ofk 16916(5)5Alt 22M13.1aA11.1bdM12.116916B11.1bo.e.A11.1bdy333151 3 4 x , ml 4 x x . So y 2 5 dx22832 8 3 151 k 3 2 k . 8 32 Critical value obtained ofk 16916(5)(5 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme6
Question 5 Notes:M1:Complete strategy of eliminating x or y and manipulating the resulting equation to form a quadraticequation 0 or a quadratic expression 0 A1:Correct algebra leading to either 4 x 2 3x 10 k 0 or 4 x 2 3x 10 k 0or a one-sided quadratic of either 4 x 2 3x 10 k or 4 x 2 3x 10 k 8 y 2 (8k 9) y 2k 2 45 0or a one-sided quadratic of e.g. 8 y 2 (8k 9) y 2k 2 45dM1:Depends on the previous M mark.Interprets 3x 2 y k intersecting y 2 x 2 5 at two distinct points by applying"b 2 4ac " 0 to their quadratic equation or one-sided quadratic.B1:See schemeA1:Correct answer, e.g.169 k 16 Alt 2M1:169 k : k 16 Complete strategy of using differentiation to find the values of x and y where 3x 2 y k is atangent to y 2 x 2 5A1:3151Correct algebra leading to x , y 832dM1:Depends on the previous M mark.3151Full method of substituting their x , y into l and attempting to find the value for k.832B1:A1:See schemeDeduces correct answer, e.g.169 k 16 169 k : k 16 A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme7
Question6(a)SchemeMarksAOsB11.1bf ( x) (8 x)ln x , x 0Crosses x-axis f ( x) 0 (8 x)ln x 0x coordinates are 1 and 8(1)(b)Complete strategy of setting f ( x) 0 and rearranges to make x . u (8 x) du dx 1M13.1aM11.1bA11.1bA1*2.1v ln x dv 1 dx x f ( x) ln x 8 xx8 x8 0 ln x 1 0xx88* 1 ln x x x1 ln x ln x (4)(c)Evaluates both f (3.5) and f (3.6)M11.1bA12.4f (3.5) 0.032951317. and f (3.6) 0.058711623.Sign change and as f ( x) is continuous, the x coordinate of Q lies betweenx 3.5 and x 3.6(2)(d)(i) x5 3.5340B11.1b(d)(ii) x 3.54 (2 dp)B12.2aQ(2)(9 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme8
Question 6 Notes:(a)B1:Either 1 and 8 on Figure 2, marks 1 next to A and 8 next to B(b)M1:Recognises that Q is a stationary point (and not a root) and applies a complete strategy of settingf ( x) 0 and rearranges to make x .M1:Applies vu uv , where u 8 x , v ln xNote: This mark can be recovered for work in part (c)A1:8 x, or equivalentxNote: This mark can be recovered for work in part (c)A1*:Correct proof with no errors seen in working.(c)M1:Evaluates both f (3.5) and f (3.6)A1:f (3.5) awrt 0.03 and f (3.6) awrt 0.06 or f (3.6) 0.05 (truncated)(8 x)ln x ln x and a correct conclusion(d)(i)B1:(d)(ii)B1:See schemeDeduces (e.g. by the use of further iterations) that the x coordinate of Q is 3.54 accurate to 2 dpNote: 3.5 3.55119 3.52845 3.53848 3.53404 3.53600 3.53514 ( 3.535518.)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme9
Question7(a)SchemeMarksAOsdpdp p kpdtdtB13.3 k dtM11.1bln p kt c A11.1bA1 *2.11dp pln p kt c p ekt c ekt ec p aekt *(4)(b)p ae ln p ln a kt and evidence of understanding that eitherkt gradient k or " M " k vertical intercept ln a or "C " ln agradient k 0.14vertical intercept ln a 3.95 a e3.95 51.935 52 (2 sf )M12.1A11.1bA11.1b(3)(c)e.g. p aekt p a(ek )t abt , p 52e0.14 t p 52(eB12.2aB11.1b0.14 t)b 1.15 which can be implied by p 52(1.15)t(2)2(d)(i)Initial area (i.e. "52" mm ) of bacterial culture that was first placed onto thecircular dish.(d)(ii)E.g. Rate of increase per hour of the area of bacterial cultureThe area of bacterial culture increases by “15%” each hourB13.4B13.4(2)(e)The model predicts that the area of the bacteria culture will increaseindefinitely, but the size of the circular dish will be a constraint on this area.B13.5b(1)(12 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme10
Question 7 Notes:(a)M1:Translates the scientist’s statement regarding proportionality into a differential equation, whichdpdp p kpinvolves a constant of proportionality. e.g.dtdtCorrect method of separating the variables p and t in their differential equationA1:ln p kt , with or without a constant of integrationA1*:Correct proof with no errors seen in working.B1:(b)M1:See schemeA1:Correctly finds k 0.14A1:Correctly finds a 52(c)B1:Uses algebra to correctly deduce either p abt from p aekt p "52"(e"0.14" )t from p "52"e"0.14"tB1:(d)(i)See schemeB1:(d)(ii)B1:See scheme(e)B1:See schemeGives a correct long-term limitation of the model for p. (See scheme).A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme11
Question8(a)SchemeMarksAOsM11.1bA11.1bdhdV dV dh 160 50 h h 2 dtdt dh dt M13.1adVdV160 dh 160 When h 10, 2 dtdh dt 50 (10) (10) 400 dM13.4A11.1bdV11 160 , V h 2 (75 h) 25 h 2 h3dt33dV 50 h h 2dh()dh 0.4 (cm s -1 )dt(5)(b)dh300 dt50 (20) (20) 2M13.4dh 0.5 (cms -1 )dtA11.1b(2)(7 marks)Question 8 Notes:(a)M1:A1:Differentiates V with respect to h to give h h 2 , 0, 050 h h 2M1:dV dh Attempts to solve the problem by applying a complete method of their 160 dh dt M1:Depends on the previous M mark.Substitutes h 10 into their model fordh160 which is in the formdV dt their dh A1:(b)Obtains the correct answer 0.4M1:Realises that rate for of 160 cm3 s 1 for 0 „ h „ 12 has no effect when the rate is increased todh300 cm3 s 1 for 12 h „ 24 and so substitutes h 20 into their model forwhich is in thedt300 formdV their dh A1:Obtains the correct answer 0.5A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme12
Question9(a)Scheme 9 15 8 10 E.g. midpoint PQ , 22 ( 3, 1) , which is the centre point A,so PQ is the diameter of the circle.MarksAOsM11.1bA12.1(2)(a)mPQ Alt 1 10 833 PQ : y 8 ( x 9)15 9443355PQ : y x . So x 3 y (3) 14444so PQ is the diameter of the circle.M11.1bA12.1(2)(a)Alt 2PQ ( 9 15)2 (8 10)2 and either 900 30 AP (3 9)2 ( 1 8)2 AQ (3 15)2 ( 1 10)2 225 15 M11.1bA12.1 225 15e.g. as PQ 2 AP , then PQ is the diameter of the circle.(2)Uses Pythagoras in a correct method to findeither the radius or diameter of the circle.(b)(( x 3)2 ( y 1)2 225 or (15)2)M11.1bM11.1bA11.1b(3)(c)Distance ("15")2 (10)2 125 5or 1(2("15")) 2 (2(10)) 225M13.1aA11.1b(2)(d)ˆ ) sin( ARQ20ˆ 90 cos 1 10 or ARQ 2("15") "15" ˆ 41.8103. 41.8 (to 0.1 of a degree)ARQM13.1aA11.1b(2)(9 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme13
Question 9 Notes:(a)M1:Uses a correct method to find the midpoint of the line segment PQA1:Completes proof by obtaining (3, 1) and gives a correct conclusion.(a)Alt 1M1:Full attempt to find the equation of the line PQA1:Completes proof by showing that (3, 1) lies on PQ and gives a correct conclusion.(a)Alt 2M1:A1:Attempts to find distance PQ and either one of distance AP or distance AQCorrectly shows either PQ 2 AP , supported by PQ 30, AP 15 and gives a correct conclusion (b)M1:PQ 2 AQ , supported by PQ 30, AQ 15 and gives a correct conclusionEither uses Pythagoras correctly in order to find the radius. Must clearly be identified as theradius. E.g. r 2 ( 9 3) ( 8 1)2r 2 (15 3) ( 10 1) or r 222or r (15 3)2( 9 3) ( 10 1)2 ( 8 1) or22or uses Pythagoras correctly in order to find the diameter. Must clearly be identified as thediameter. E.g. d 2 (15 9 ) ( 10 8)22or d (15 9 )2 ( 10 8 )2Note: This mark can be implied by just 30 clearly seen as the diameter or 15 clearly seen as theradius (may be seen or implied in their circle equation)M1:Writes down a circle equation in the form ( x "3") 2 ( y " 1") 2 (their r ) 2A1:( x 3) 2 ( y 1) 2 225 or ( x 3)2 ( y 1) 2 15 2 or x 2 6 x y 2 2 y 215 0(c)M1:Attempts to solve the problem by using the circle property “the perpendicular from the centre to achord bisects the chord” and so applies Pythagoras to write down an expression of the form(their "15") 2 (10) 2 .A1:5 5 by correct solution only(d)M1:Attempts to solve the problem by e.g. using the circle property “the angle in a semi-circle is a right 1020ˆ 90 cos 1 ˆ ) angle” and writes down either sin( ARQor ARQ 2(their "15") their "15" ˆ ) Note: Also allow cos( ARQA1:( ( )) 152 (15 ) ( 2 ( 5 5 ) )152 2 5 522 5 3 41.8 by correct solution onlyA level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme14
Question10 (a)Scheme 4 x ln 3 MarksAOsB12.2a(1)(b)Attempts to apply ydxdtdt dx y dt dt 1 1 dt t 1 t 2 1AB 1 A(t 2) B(t 1)(t 1)(t 2) (t 1) (t 2)11 A 1, B 1 gives(t 1) (t 2) 1 1 dt ln(t 1) ln(t 2) (t 1) (t 2) Area( R) ln(t 1) ln(t 2) 02 (ln 3 ln 4) (ln1 ln 2) (3)(2) 6 ln 3 ln 4 ln 2 ln ln 4 4 3 ln * 2 )(b)Alt 1Attempts to apply ydx 1dx e 2 1x 1dx ,e 1xM13.1aA11.1b1AB 1 A(u 1) Buu (u 1) u (u 1)M13.1a11 u (u 1)A11.1bM11.1bA11.1bM12.2aA1 *2.1with a substitution of u e 1x ydx A 1, B 1 1 1 du u u 1 gives 1 1 du ln u ln(u 1) u (u 1) Area( R) ln u ln(u 1) 13 (ln 3 ln 4) (ln1 ln 2) (3)(2) 6 ln 3 ln 4 ln 2 ln ln 4 4 3 ln * 2 (8)(9 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme15
Question10 (b)Alt 2SchemeMarksAOsM13.1aA11.1b1AB 1 Av B(v 1)(v 1)v (v 1) vM13.1a11 (v 1) vA11.1bM11.1bA11.1bM12.2aA1 *2.1Attempts to apply ydx with a substitution of v e ydx A 1, B 1 1dx e 2 1x 1dx ,e 1xx 1 1 dv v 1 v gives 1 1 dv ln(v 1) ln v (v 1) v Area( R) ln(v 1) ln v 24 (ln 3 ln 4) (ln1 ln 2) (3)(2) 6 ln 3 ln 4 ln 2 ln ln 4 4 3 ln * 2 (8)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme16
Question 10 Notes:(a)B1:Uses x ln(t 2) with t 2 4 to deduce the correct domain, x ln 3 3 (b)M1:Attempts to solve the problem by either a parametric process or A1:Obtains M1:a Cartesian process with a substitution of either u e x 1 or v e x t 1 t 2 dt from a parametric approach 1 1 u u 1 du from a Cartesian approach with u e 1 1 1 v 1 v dv from a Cartesian approach with v e 1 1 xxApplies a strategy of attempting to express either111,or(t 1)(t 2) u (u 1)(v 1)vas partial fractionsA1:M1:Correct partial fractions for their methodIntegrates to give either ln(t 1) ln(t 2) ln u ln(u 1) ; , 0, where u e x 1 ln(v 1) ln v ; , 0, where v e xA1:Correct integration for their methodM1:Either Parametric approach: Deduces and applies limits of 2 and 0 in t and subtracts the correctway round Cartesian approach: Deduces and applies limits of 3 and 1 in u, where u e x 1,and subtracts the correct way round Cartesian approach: Deduces and applies limits of 4 and 2 in v, where v e x ,and subtracts the correct way roundA1*: 3 Correctly shows that the area of R is ln , with no errors seen in their working 2 A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme17
Question11SchemeMarksAOs(5k 10) (2k ) (7k 14) (5k 10) k .M12.1 3k 10A11.1bM12.2an( 2(4) (n 1)(8) )2M11.1bn( 8 8n 8 ) 4n 2 (2n) 2 which is a square number2A12.1Arithmetic sequence, T2 2k , T3 5k 10, T4 7k 14 k 6Sn 2k 4 k 6 T2 12 , T3 20, T4 28 . So d 8, a 4(5)(5 marks)Question 11 Notes:M1:Complete method to find the value of kA1:Uses a correct method to find k 6M1:Uses their value of k to deduce the common difference and the first term ( T2 ) of the arithmeticseries.M1:Applies S n A1:Correctly shows that the sum of the series is (2n) 2 and makes an appropriate conclusion.n( 2a (n 1)d ) with their a T2 and their d.2A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme18
Question12SchemeMarksAOsComplete process to find at least one set of coordinates for P.The process must include evidence of differentiatingdy 0 to find x . settingdx substituting x . into sin x cos y 0.5 to find y .M13.1a dy dx B11.1bM12.2aA11.1bM11.1bA11.1bB12.1Appliescos x sin ydy 0dxdycos x 0 (e.g. cos x 0 or 0 cos x 0) x .dxsin ygiving at least one of either x x 2or x 212 2 sin cos y 0.5 cos y y or 2233 22 2 So in specified range, ( x , y ) , and , 2323 , by cso sin cos y 0.5 cos y 1.5 has no solutions, 2 and so there are exactly 2 possible points P.x 2(7)(7 marks)Question 12 Notes:M1:See schemeB1:Correct differentiated equation. E.g. cos x sin yM1:Uses the information “the tangent to C at the point P is parallel to the x-axis”dy 0 and finds x .to deduce and applydxA1:See schemeM1:For substituting one of their values fromA1:Selects coordinates for P on C satisfyingdy 0dxdy 0 into sin x cos y 0.5 and so finds x ., y .dxdy 3 , y 0 and „ x 22dx2 2 i.e. finds , and , and no other points by correct solution only3 2 3 2B1:Complete argument to show that there are exactly 2 possible points P.A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme19
Question13(a)SchemeMarksAOs1cos 2 x sin 2 x sin 2 xM11.2 1 cos 2 xsin 2 xM11.1b 1 2cos 2 x 12cos 2 x 2sin x cos x2sin x cos xM12.1A11.1b cos x cot x *sin xA1*2.1cosec2 x cot 2 x cot x , x 90n , n cosec2 x cot 2 x (5)(b)cosec(4 10 ) cot (4 10 ) 3 ; 0 „ 180 ,cot (2 . ) 3M12.2a2 . 30 12.5 M11.1bA11.1b2 5 180 PV . M12.1 102.5 A11.1b(5)(10 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme20
Question 13 Notes:(a)M1:cos 2 x1and cot2 x sin 2 xsin 2 xCombines into a single fraction with a common denominatorM1:Applies sin 2 x 2sin x cos x to the denominator and applies eitherM1:Writes cosec2 x cos 2 x 2cos 2 x 1 cos 2 x 1 2sin 2 x and sin 2 x cos 2 x 1 cos 2 x cos 2 x sin 2 x and sin 2 x cos 2 x 1to the numerator and manipulates to give a one term numerator expressionA1:A1*:2cos 2 xor equivalent.2sin x cos xCorrect proof with correct notation and no errors seen in workingCorrect algebra leading to(b)M1:Uses the result in part (a) in an attempt to deduce either 2 x 4 10 or x 2 . and usesx 2 . to write down or imply cot (2q . ) 3M1:Applies arccot( 3 ) 30 or arctan 1 30 3 and attempts to solve 2 . 30 to give .A1:Uses a correct method to obtain 12.5 M1:Uses 2 5 180 their PV in a complete method to find the second solution, .A1:Uses a correct method to obtain 102.5 , with no extra solutions given either inside or outsidethe required range 0 „ 180 A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme21
Question14 (i)SchemeMarksAOsM12.3A12.4For an explanation or statement to show when the claim 3x 2 x failsThis could be e.g.1111 or is not greater than or equal to3223 when x 1, when x 0, 3x 2 x or 3x is not greater than or equal to 2xfollowed by an explanation or statement to show when the claim 3x 2 x istrue. This could be e.g. x 2, 9 4 or 9 is greater than or equal to 4 when x 0, 3x 2 xand a correct conclusion. E.g. so the claim 3x 2 x is sometimes true(2)(ii)Assume that3 is a rational numberM12.1 p 3 q p2 3 q2M11.1b p 2 is divisible by 3 and so p is divisible by 3A12.2aFrom earlier, p 2 3 q 2 (3c)2 3q 2M12.1 q 2 3c 2 q 2 is divisible by 3 and so q is divisible by 3A11.1bA12.4So3 p, where p and q integers, q 0, and the HCF of p and q is 1qSo p 3c , where c is an integerAs both p and q are both divisible by 3 then the HCF of p and q is not 1This contradiction implies that3 is an irrational number(6)(8 marks)A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme22
Question 14 Notes:(i)M1:A1:See schemeSee scheme(ii)M1:Uses a method of proof by contradiction by initially assuming that3 in the form3 3 is rational and expressesp, where p and q are correctly defined.qpand rearranges to make p 2 the subjectqM1:WritesA1:Uses a logical argument to prove that p is divisible by 3M1:Uses the result that p is divisible by 3, (to construct the initial stage of proving that q is alsodivisible by 3), by substituting p 3c into their expression for p 2A1:Hence uses a correct argument, in the same way as before, to deduce that q is also divisible by 3A1:Completes the argument (as detailed on the scheme) that3 is irrational.Note: All the previous 5 marks need to be scored in order to obtain the final A mark.A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme23
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
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 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
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
us88685733 agma 1012-f 1990 us88685736 agma 2003-b 1997 us88685805 agma 6110-f 1997 us88685810 agma 9004-a 1999 us88685815 agma 900-e 1995 de88686925 tgl 18790/01 1972-09 de88686928 tgl 18791/01 1982-06 de88686929 tgl 18791/02 1983-07 us88687101 a-a-20079 2002-08-20 us88687113 a-a-50800 1981-04-23 us88687199 a-a-59173 1998-03-04 us88687222 a-a-55106 1992-07-15 us88687243 a-a-20155 1992-11-16 .
