= 0 and states ''hence there is a stationary point'' All aspects of the proof dx must be correct (c)

Transcription

1 Paper 1: Pure Matematics 1 Mark Sceme 1(a) (i) (ii) d d y 3 1x 4x x M1 A1 d y dx 1.1b 1.1b 36x 48x A1ft 1.1b Substitutes x = into teir dx (3) Sows d y 0 and states ''ence tere is a stationary point'' A1.1 dx Substitutes x = into teir d y dx () d y 48 0 and states ''ence te stationary point is a minimum'' A1ft.a dx (a)(i) M1: Differentiates to a cubic form 3 A1: 1x 4x dx (a)(ii) d y A1ft: Acieves a correct d 36x 48x x x for teir d M1: Substitutes x = into teir d y dx A1: Sows d y = 0 and states ''ence tere is a stationary point'' All aspects of te proof dx must be correct d y M1: Substitutes x = into teir dx Alternatively calculates te gradient of C eiter side of x A1ft: For a correct calculation, a valid reason and a correct conclusion. Follow troug on an incorrect d y dx () (7 marks) Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

2 (a) Uses sr 3 r 0.4 M1 1. Uses angle AOB 0.4 Uses area of sector OD 7.5 cm or uses radius is (1 7.5 ) cm M1 3.1a 1 1 (1 7.5) ( 0.4) r = 7.8cm A1ft 1.1b () (3) (5 marks) (a) M1: Attempts to use te correct formula s r wit s 3and 0.4 A1: OD = 7.5 cm (An answer of 7.5cm implies te use of a correct formula and scores bot marks) M1: AOB 0.4 may be implied by te use of AOB = awrt.74 or uses radius is (1 teir 7.5 ) M1: Follow troug on teir radius (1 teir OD) and teir angle A1ft: Allow awrt 7.8 cm. (Answer ). Follow troug on teir (1 teir 7.5 ) Note: Do not follow troug on a radius tat is negative. 38 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

3 3(a) Attempts x y 5... Centre (, 5) Sets k 5 0 M1.a () (a) k 9 A1ft 1.1b M1: Attempts to complete te square so allow x y 5... A1: States te centre is at (, 5). Also allow written separately x, y 5 (, 5) implies bot marks M1: Deduces tat te rigt and side of teir x y A1ft: k 9 Also allow is > 0 or 0 k Follow troug on teir rs of x y () (4 marks) Writes t 1 1 dt 1 dt and attempts to integrate M1.1 t t tln t c a a a a ln ln ln 7 7 a ln wit 7 k (4 marks) M1: Attempts to divide eac term by t or alternatively multiply eac term by t -1 1 M1: Integrates eac term and knows dt ln t. Te + c is not required for tis mark t M1: Substitutes in bot limits, subtracts and sets equal to ln7 A1: 7 7 Proceeds to a ln and states k or exact equivalent suc as 3.5 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

4 5 Attempts to substitute x 1 into Attempts to write as a single fraction y (x5)( x1) 6 ( x 1) x 1 6 y y 4 7 ( x 1) M1.1 M1.1 x 3x1 y a3, b1 x 1 M1: x 1 3 Score for an attempt at substituting t or equivalent into y 4t 7 + t M1: Award tis for an attempt at a single fraction wit a correct common denominator. x 1 Teir 4 7 term may be simplified first A1: Correct answer only y x 3x1 x 1 a3, b1 (3 marks) 40 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

5 6 (a)(i) barrels B1 3.4 (ii) Gives a valid limitation, for example Te model sows tat te daily volume of oil extracted would become negative as t increases, wic is impossible States wen t 10, V 1500 wic is impossible 64 States tat te model will only work for 0 t 7 B1 3.5b (i) Suggests a suitable exponential model, for example V Ae kt M Uses 0,16000 and 4,9000 in e k dm1 3.1b () 1 9 k ln awrt ln t 4 16 V 16000e or V 16000e 0.144t (ii) Uses teir exponential model wit t 3 V awrt barrels B1ft 3.4 (a)(i) B1: 10750barrels (a)(ii) B1: See sceme (i) M1: Suggests a suitable exponential model, for example V Ae kt t, V Ar or any oter suitable function suc as V Ae kt bwere te candidate cooses a value for b. dm1: Uses bot 0,16000 and 4,9000 in teir model. Wit V Ae kt candidates need to proceed to t Wit V Ar candidates need to proceed to Wit V Ae kt b e k r candidates need to proceed to 4 (5) (7 marks) b e k b were b is given as a positive constant and Ab M1: Uses a correct metod to find all constants in te model. A1: Gives a suitable equation for te model passing troug (or approximately troug in te case of decimal equivalents) bot values 0,16000 and 4,9000. Possible equations for te model could be for example V 0.144t 16000e V t (ii) B1ft: Follow troug on teir exponential model V 0.146t 15800e 00 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

6 7 Attempts AC AB BC i3jki9j3k3i6j4k M1 3.1a Attempts to find any one lengt using 3-d Pytagoras M1.1 Finds all of AB 14, AC 61, BC 91 A1ft 1.1b cos BAC M angle BAC * A1* 1.1b (5) (5 marks) M1: Attempts to find AC by using AC AB BC M1: Attempts to find any one lengt by use of Pytagoras' Teorem A1ft: Finds all tree lengts in te triangle. Follow troug on teir AC M1: Attempts to find BAC using cos BAC AB AC BC AB AC Allow tis to be scored for oter metods suc as cos BAC AB. AC AB AC A1*: Tis is a sow tat and all aspects must be correct. Angle BAC = Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

7 8 (a) f (3.5) = 4.8, f (4) = (+)3.1 Cange of sign and function continuous in interval [3.5, 4] Root * Attempts x f( x ) 0 1 x0 f( x0 ) A1*.4 () x x 1 = 3.81 y = ln(x 5) () y 30 x M1 3.1a Attempts to sketc bot y = ln(x 5) and y = 30 x States tat y = ln(x 5) meets y = 30 x in just one place, terefore y = ln(x 5) = 30 x as just one root f (x) = 0 as just one root A1.4 () (6 marks) (a) M1: Attempts f(x) at bot x = 3.5 and x = 4 wit at least one correct to 1 significant figure A1*: f (3.5) and f(4) correct to 1 sig figure (rounded or truncated) wit a correct reason and conclusion. A reason could be cange of sign, or f (3.5) f (4) 0 or similar wit f(x) being continuous in tis interval. A conclusion could be 'Hence root' or 'Terefore root in interval' f( x0 ) M1: Attempts x1 x0 evidenced by x1 4 f( x0 ) A1: Correct answer only x M1: For a valid attempt at sowing tat tere is only one root. Tis can be acieved by Sketcing graps of y = ln(x 5) and y = 30 x on te same axes Sowing tat f(x) = ln(x 5) + x 30 as no turning points Sketcing a grap of f(x) = ln(x 5) + x 30 A1: Scored for correct conclusion Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

8 9(a) sin cos tan cot cos sin M1.1 (a) sin cos sincos 1 1 sin States tancot 1sin AND no real solutions as 1 sin 1 M1: Writes sin tan cos and cos cot sin A1: sin cos Acieves a correct intermediate answer of sincos M1: Uses te double angle formula sin sincos A1*: Completes proof wit no errors. Tis is a given answer. M1.1 cosec * A1* 1.1b (4) B1.4 (1) (5 marks) Note: Tere are many alternative metods. For example 1 tan 1 sec 1 1 tan cot tan ten as te tan tan tan sin cos cossin cos main sceme. B1: Scored for sigt of sin and a reason as to wy tis equation as no real solutions. Possible reasons could be 1 sin 1...and terefore sin or sin arcsin wic as no answers as 1 sin 1 44 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

9 10 Use of sin( ) sin ( ) Uses te compound angle identity for sin( A B) wit A, B sin( ) sincos+ cossin Acieves B1.1 sin( ) sin sincos + cossin sin sin cos 1 cos sin Uses 0, sin 1 and cos 1 0 sin( ) sin Hence te limit0 cos and te gradient of ( ) te cord gradient of te curve cos * d B1: States or implies tat te gradient of te cord is M1.1 A1*.5 sin( ) sin or similar suc as sin( ) sin for a small or M1: Uses te compound angle identity for sin(a + B) wit A, B or A1: Obtains sincos + cossin sin or equivalent M1: Writes teir expression in terms of sin A1*: Uses correct language to explain tat d y d cos 1 and cos For tis metod tey sould use all of te given statements 0, sin 1, cos 1 sin( ) sin 0 meaning tat te limit cos 0 ( ) and terefore te gradient of te cord gradient of te curve cos d (5 marks) Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

10 10alt sin( ) sin Use of ( ) sin sin sin( ) sin Sets ( ) and uses te compound angle identity for sin(a + B) and sin(a B) wit A, B sin( ) sin Acieves sin cos + cos sin sin cos cos sin sin cos sin Uses 0, 0 ence 1and cos cos sin( ) sin Terefore te limit0 cos and te gradient of ( ) te cord gradient of te curve cos * d B1.1 M1.1 A1*.5 (5 marks) Additional notes: A1*: Uses correct language to explain tat d y cos d. For tis metod tey sould use te sin (adapted) given statement 0, 0 ence 1wit cos cos sin( ) sin meaning tat te limit0 cos and terefore te gradient of te ( ) cord gradient of te curve cos d 46 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

11 11(a) (a) Sets H d d M1 3.4 Solves using an appropriate metod, for example d 0.00 d Distance awrt 04 m only A1.a States te initial eigt of te arrow above te ground. B1 3.4 d d d d ( d 100) ( d 100) (d) (i).1 metres B1ft 3.4 (ii) 100 metres B1ft 3.4 M1: Sets H d0.00d 0 M1: Solves using formula, wic if stated must be correct, by completing square (look for d (3) (1) (3) () (9 marks) d..) or even allow answers coming from a grapical calculator A1: Awrt 04 m only B1: States it is te initial eigt of te arrow above te ground. Do not allow '' it is te eigt of te arcer'' M1: Score for taking out a common factor of 0.00 from at least te d and d terms M1: For completing te square for teir d 00d A1: ( d 100) or exact equivalent (d) B1ft: For teir ' ' =.1m B1ft: For teir 100m term Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

12 1 (a) b b N at log N log a log T M log10 N log10 a blog10 T so 10 m band c log a intercept Uses te grap to find eiter a or b a 10 or b = gradient M1 3.1b () Uses te grap to find bot a and b intercept a 10 and b = gradient Uses T 3in N b at wit teir a and b M1 3.1b Number of microbes 800 N log N 6 M (4) (d) We cannot extrapolate te grap and assume tat te model still olds States tat 'a' is te number of microbes 1 day after te start of te experiment A1 3.5b () B1 3.a (1) (9 marks) 48 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

13 Question 1 continued (a) M1: Takes logs of bot sides and sows te addition law M1: Uses te power law, writes log10 N log10 ablog10 T and states m band c log10 a M1: Uses te grap to find eiter a or b te sigt of 1.8 b.3 or a10 63 M1: Uses te grap to find bot a and b by te sigt of 1.8 b.3 and a10 63 intercept a 10 or b = gradient. Tis would be implied by intercept a 10 and b = gradient. Tis would be implied b M1: Uses T 3N at wit teir a and b. Tis is implied by an attempt at A1: Accept a number of microbes tat are approximately 800. Allow following correct work. Tere is an alternative to tis using a grapical approac. M1: Finds te value of log T from T =3. Accept as T 3 log T 0.48 M1: Ten using te line of best fit finds te value of log10 Accept log10 N.9 N from teir ''0.48'' '.9' M1: Finds te value of N from teir value of log N log N.9 N 10 A1: Accept a number of microbes tat are approximately 800. Allow following correct work M1 For using N = and stating tat log10 N 6 A1: Statement to te effect tat ''we only ave information for values of log N between 1.8 and 4.5 so we cannot be certain tat te relationsip still olds''. ''We cannot extrapolate wit any certainty, we could only interpolate'' Tere is an alternative approac tat uses te formula. M1: Use N = in teir N 63T log10 63 log10 T A1: Te reason would be similar to te main sceme as we only ave log10 T values from 0 to 1.. We cannot extrapolate te grap and assume tat te model still olds (d) B1: Allow a numerical explanation T 1 N a1 b N a giving a is te value of N at T =1 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

14 13(a) Attempts dt dx dx dt 3 sin t dx sin t 3 cost () Substitutes t 3 3 sin t M1.1 dx sin t in Uses gradient of normal = d 3 d M1.1 3 Coordinates of P = 1, B1 1.1b Correct form of normal y x M1.1 3 Completes proof x 3y1 0 * A1* 1.1b (5) Substitutes x = cos t and y = 3 cos t into x 3y1 0 M1 3.1a Uses te identity Substitutes teir cos t cos t 1 to produce a quadratic in cost M1 3.1a Finds 1cos t4cost cos t, M cost into x = cos t, y = 3 cos t, 6 Q 5 7, (6) (13 marks) 50 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

15 Question 13 continued (a) M1: Attempts dt dx dx and acieves a form sin t k Alternatively candidates may apply te sin t dt sin tcost double angle identity for cos t and acieve a form k sin t 3 sin t A1: Scored for a correct answer, eiter or 3 cost sin t M1: For substituting t 3 in teir d y d wic must be in terms of t x M1: Uses te gradient of te normal is te negative reciprocal of te value of d y. Tis may be dx seen in te equation of l. B1: States or uses (in teir tangent or normal) tat P = 1, M1: Uses teir numerical value of normal at P 1 wit teir d x 1, 3 3 to form an equation of te A1*: Tis is a proof and all aspects need to be correct. Correct answer only x 3y1 0 M1: For substituting x = cos t and y = 3 cos t into x 3y1 0 to produce an equation in t. Alternatively candidates could use y Ax B. M1: Uses te identity cos t cos t 1 cos t cos t 1 In te alternative metod it is for combining teir an equation in just one variable to set up an equation of te form to produce a quadratic equation in cost A1: For te correct quadratic equation 1cos t4cost5 0 Alternatively te equations in x and y are 3x x 5 0 y Ax B wit x 3y 1 0 to get 1 3y 4y M1: Solves te quadratic equation in cost (or x or y) and rejects te value corresponding to P. M1: 5 5 Substitutes teir cost or teir t arccos in x = cos t and y = cos t If a value of x or y as been found it is for finding te oter coordinate. A1: Q 5 7, Allow 5 7 x, y 3 but do not allow decimal equivalents Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

16 14(a) Uses or implies = 0.5 B1 1.1b For correct form of te trapezium rule = (3) Any valid statement reason, for example Increase te number of strips B1.4 Decrease te widt of te strips Use more trapezia (1) For integration by parts on ln xdx M1.1 3 x x ln x dx 3 3 All integration attempted and limits used Area of S = x 5d x x 5 x c B1 1.1b x3 x ln x x x 1 x5 dx ln x x 5x x1 M1.1 Uses correct ln laws, simplifies and writes in required form M1.1 Area of S = 8 ln 7 ( 8, 7, 7) 7 a b c (6) (10 marks) 5 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017

17 Question 14 continued (a) 0.5 B1: States or uses te strip widt = 0.5. Tis can be implied by te sigt of... in te trapezium rule M1: For te correct form of te bracket in te trapezium rule. Must be y values rater tan x values first y value last y value sum of oter y values A1: B1: See sceme M1: Uses integration by parts te rigt way around. Look for 3 x ln xdx Ax ln x Bx dx A1: 3 x x ln x dx 3 3 B1: Integrates te x 5 term correctly x 5x M1: All integration completed and limits used a M1: Simplifies using ln law(s) to a form ln c b A1: Correct answer only 8 ln 7 7 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited

18 15(a) Attempts to differentiate using te quotient rule or oterwise M1.1 (a) e 8cosx4sinx e x1 x1 x x1 f() e 1 Sets f() x 0and divides/ factorises out te e x terms M1.1 Proceeds via sin x 8 to tan cos x 4 x * A1* 1.1b (i) Solves tan 4x and attempts to find te nd solution M1 3.1a (4) x 1.0 (ii) Solves tan x and attempts to find te 1 st solution M1 3.1a x M1: Attempts to differentiate by using te quotient rule wit u 4sin x and v 1 x alternatively uses te product rule wit u 4sin x and v e A1: For acieving a correct f() x. For te product rule f ( x) e 8cos x4sin x e 1 x 1 x (4) 1 e x or (8 marks) M1: Tis is scored for cancelling/ factorising out te exponential term. Look for an equation in just cos x and sin x A1*: Proceeds to tan x. Tis is a given answer. (i) M1: Solves tan 4x attempts to find te nd solution. Look for x arctan 4 Alternatively finds te nd solution of tan x and attempts to divide by A1: Allow awrt x 1.0. Te correct answer, wit no incorrect working scores bot marks (ii) M1: Solves tan x attempts to find te 1 st solution. Look for x arctan A1: Allow awrt x Te correct answer, wit no incorrect working scores bot marks 54 Pearson Edexcel Level 3 Advanced GCE in Matematics Sample Assessment Materials Issue 1 April 017 Pearson Education Limited 017