Pure Maths October 2020 mark scheme PDF

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AS Pure Maths October 2020 mark scheme...

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Mark Scheme (Result)

October 2020 Pearson Edexcel GCE In AS Level Mathematics 8MA0 Paper 1 Pure Mathematics

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October 2020 Publications Code 8MA0_01_2010_MS All the material in this publication is copyright © Pearson Education Ltd 2020

General Marking Guidance

•

All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last.

•

Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.

•

Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.

•

There is no ceiling on achievement. All marks on the mark scheme should be used appropriately.

•

All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved,

i.e.

if

the

answer

matches

the

mark

scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme. •

Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited.

•

When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted.

•

Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

EDEXCEL GCE MATHEMATICS General Instructions for Marking 1. The total number of marks for the paper is 100. 2. The Edexcel Mathematics mark schemes use the following types of marks: • M marks: method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated. • A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. • B marks are unconditional accuracy marks (independent of M marks) • Marks should not be subdivided. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. • bod – benefit of doubt • ft – follow through • the symbol will be used for correct ft • cao – correct answer only • cso - correct solution only. There must be no errors in this part of the question to obtain this mark • isw – ignore subsequent working • awrt – answers which round to • SC: special case • oe – or equivalent (and appropriate) • dep – dependent • indep – independent • dp decimal places • sf significant figures •  The answer is printed on the paper • The second mark is dependent on gaining the first mark 4. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 5. Where a candidate has made multiple responses and indicates which response they wish to submit, examiners should mark this response. If there are several attempts at a question which have not been crossed out, examiners should mark the final answer which is the answer that is the most complete. 6. Ignore wrong working or incorrect statements following a correct answer. 7. Mark schemes will firstly show the solution judged to be the most common response expected from candidates. Where appropriate, alternatives answers are provided in the notes. If examiners are not sure if an answer is acceptable, they will check the mark scheme to see if an alternative answer is given for the method used.

General Principles for Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles)

Method mark for solving 3 term quadratic: 1. Factorisation

( x 2 + bx + c) = ( x + p)( x + q), where pq = c , leading to x = ... (ax2 + bx + c) = (mx + p)(nx + q), where pq = c and mn = a , leading to x = ... 2. Formula Attempt to use the correct formula (with values for a, b and c)

3. Completing the square



2 Solving x + bx + c = 0 :  x 



2

b  q  c = 0, q  0 , leading to x = ... 2 

Method marks for differentiation and integration: 1. Differentiation −1

n n Power of at least one term decreased by 1. ( x → x )

2. Integration n n1 Power of at least one term increased by 1. ( x → x + )

Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners’ reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are small errors in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values but may be lost if there is any mistake in the working.

Exact answers Examiners’ reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.

Question 1

Scheme Attempts to differentiate x n → x n−1 seen once 3

y = 2 x − 4x + 5 

2 dy = 6x − 4 dx

2 dy = 6x − 4 dx For a correct method of finding a tangent at P (2,13). Score for y − 13 = ''20'' ( x − 2 )

For substituting x = 2 into their

y = 20 x − 27

Marks

AOs

M1

1.1b

A1

1.1b

dM1

1.1b

ddM1

1.1b

A1

1.1b

(5) (5 marks) Notes 3

2

M1:

Attempts to differentiate x n → x n −1 seen once. Score for x → x or 4 x → 4 or +5 → 0

A1:

2  dy  2  dx =  6 x − 4 which may be unsimplified 6 x − 4 + C is A0  

dy . The first M must have been awarded. dx dy Score for sight of embedded values, or sight of '' at x = 2 is'' or a correct follow through. dx Note that 20 on its own is not enough as this can be done on a calculator.

dM1: Substitutes x = 2 into their

ddM1: For a correct method of finding a tangent at P (2,13).Score for y − 13 = ''20''( x − 2) It is dependent upon both previous M's. If the form y = mx + c is used they must proceed as far as c = ... A1:

Completely correct y = 20 x − 27 (and in this form)

Question

Scheme

Marks

AOs

7 3

M1

1.1b

A full attempt to find the bearing Eg 180  + ''67 ''

dM1

3.1b

A1

1.1b

2(a)

√20 63.4° 49.8°

√34

√58

66.8° 3

23.2° 7

Attempts to find an ''allowable'' angle Eg tan  =

Bearing = awrt 246.8

(3) (b) Attempts to find the distance travelled =

( 4 − −3) + (− 2 + 5 ) = 2

2

(

58

Attempts to find the speed =

)

58 2.75

= awrt 2.77 km h −1

M1

1.1b

dM1

3.1b

A1

1.1b

(3) (6 marks) Notes: Score these two parts together. (a) M1:

Attempts an allowable angle. (Either the “66.8” , “23.2” or (“49.8” and “63.4” )) 7 3 −2 − −5 tan  =  , tan  =  , tan  =  etc 3 7 4 − −3 There must be an attempt to subtract the coordinates (seen or applied at least once) 7 7 , cos =  , etc If part (b) is attempted first, look for example for sin =  '' 58 '' '' 58 '' They may use the cosine rule and trigonometry to find the two angles in the scheme. See 4 "58"+ "20"− "34" and tan  =  or equivalent. above. Eg award for cos = 2 2 " 58 " " 20 " 7 3 dM1: A full attempt to find the bearing. 180 + arctan , 270 − arctan , 3 7 360  − "49.8 "− "63.4 " . It is dependent on the previous method mark.

A1:

Bearing = awrt 246.8 oe. Allow S 66.8  W

(b) M1:

2

Attempts to find the distance travelled. Allow for d = ( 4 − −3) + ( −2 + 5) 2

2

You may see this on a diagram and allow if they attempt to find the magnitude from their “resultant vector” found in part (a). dM1: Attempts to find the speed. There must have been an attempt to find the distance using the coordinates and then divide it by 2.75. Alternatively they could find the speed in km min − 1 and then multiply by 60 A1:

awrt 2.77 km h −1

Question 3 (i)

Scheme

x 2 − 18 = x  x

(

)

18 2− 1

2 − 1 = 18  x = 18 2+1  2 −1 2 +1

x=

18

x=

(

) = 6+3

2+ 1 1

2

Marks

AOs

M1

1.1b

dM1

3.1a

A1

1.1b

(3) (ii)

3x −2

4

=

1 2 2

2

6x −4

=2

−

3 2

3 6 x − 4 = −  x = ... 2 5 x= 12

M1

2.5

dM1

1.1b

A1

1.1b

(3) (6 marks) Notes (i) M1:

Combines the terms in x, factorises and divides to find x. Condone sign slips and ignore any attempts to simplify 18 2

Alternatively squares both sides x 2 − 18 = x  2 x − 12 x + 18 = x

2

dM1: Scored for a complete method to find x. In the main scheme it is for making x the subject and then multiplying both numerator and denominator by 2 + 1 In the alternative it is for squaring both sides to produce a 3TQ and then factorising their quadratic equation to find x. (usual rules apply for solving quadratics) A1:

x = 6 + 3 2 only following a correct intermediate line. Allow

6+ 3 2 as an intermediate 1

line. In the alternative method the 6 − 3 2 must be discarded. (ii) M1:

Uses correct mathematical notation and attempts to set both sides as powers of 2 or 4. ax + b

dx + e

c

f

= 2 or 4 = 4 is sufficient for this mark. Eg 2 Alternatively uses logs (base 2 or 4) to get a linear equation in x. 3x −2 3x −2 1 1 1 . 4 =  log 2 4 = log 2  2 (3 x − 2 ) = log 2 2 2 2 2 2 2 3x − 2 1 1 =  3x − 2 = log 4 Or 4 2 2 2 2

Or 4

3x − 2

=

1 2 2

3x

 4 = 4 2  3 x = log 4 4 2

dM1: Scored for a complete method to find x. Scored for setting the indices of 2 or 4 equal to each other and then solving to find x. There must be an attempt on both sides. 1 −1 + 3x − 2 2 3x − 2 1 2 =2 =2 You can condone slips for this mark Eg bracketing errors 4 or 2 2 In the alternative method candidates cannot just write down the answer to the rhs. 3 − 1 3 2 = log 2 2 = − So expect some justification. E.g. log 2 2 2 2

A1:

−

3 2

3 1 = −  condoning slips as per main scheme 2 2 2 2 1 or 3 x = log4 4 2  3 x = 1 + 4 5 x= with correct intermediate work 12

or log 4

1

= log 4 2

Question 4 (a)

Scheme Attempts A = mn + c with either ( 0,190) or(8,169 ) 190 − 169 Or attempts gradient eg m =  (= − 2.625 ) 8 Full method to find a linear equation linking A with n E.g. Solves 190 = 0n + c and 169 = 8n + c simultaneously

A = −2.625n + 190

Marks

AOs

M1

3.3

dM1

3.1b

A1

1.1b

(3) (b)

Attempts A = −2.625 19 + 190 = ...

A = 140.125 g km- 1 It is predicting a much higher value and so is not suitable

M1

3.4

A1

1.1b

B1ft

3.5a

(3) (6 marks) Notes (a) M1:

Attempts A = mn + c with either ( 0,190) or (8,169 ) considered. Eg Accept sight of 190 = 0n + c or 169 = 8m + c or A −169 = m( n − 8) or A = 190 + mn where m could be a value. 21 190 − 169 or sight of  2.625 or  oe Also accept an attempt to find the gradient  8 8

dM1: A full method to find both constants of a linear equation Method 1: Solves 190 = 0n + c and 169 = 8n + c simultaneously 190 − 169 Method 2: Uses gradient and a point Eg m =  (= − 2.625) and c = 190 8 Condone different variables for this mark. Eg. y in terms of x. A1:

A = −2.625n + 190 or A = −

21 n + 190 oe 8

(b) M1:

Attempts to substitute '' n '' = 19 into their linear model to find A. They may call it x = 19 Alternatively substitutes A = 120 into their linear model to find n.

A1:

A = 140.125 from n = 19 Allow A = 140 or n = 26 / 27 following A = 120

B1ft: Requires a correct calculation for their model, a correct statement and a conclusion E.g For correct (a) A = 140 is (much) higher than 120 so the model is not suitable/appropriate. Follow through on a correct statement for their equation. As a guide allow anything within [114,126] to be regarded as suitable. Anything less than 108 or more than 132 should be justified as unsuitable. Note B0 Recorded value is not the same as/does not equal/does not match the value predicted

Question 5 (a)

Scheme States

Marks

AOs

M1

1.1b

A1

1.1b

A1

1.1b

sin  sin 27 = 12 7

Finds  = awrt 51 or awrt 129 = awrt 128.9

(3) (b)

Attempts to find part or all of AD 2

2

2

Eg AD = 7 +12 − 2 12  7cos101.9 = ( AD = 15.09) Eg ( AC) =7 +12 −2 12 7cos( 180 −''128.9 '' − 27) 2

2

2

M1

1.1b

dM1

3.1a

A1

3.2a

Eg 12 cos 27 or 7 cos"51" Full method for the total length = 12 + 7 + 7 + ''15.09 '' = = 42 m

(3) (6 marks) Notes (a) M1:

States

sin  sin 27 oe with the sides and angles in the correct positions = 12 7

Alternatively they may use the cosine rule on  ACB and then solve the subsequent quadratic to find AC and then use the cosine rule again A1:

awrt 51 or awrt 129

A1:

Awrt 128.9 only (must be seen in part a))

(b) M1:

Attempts a ''correct'' method of finding either AD or a part of AD eg (AC or CD or forming a perpendicular to split the triangle into two right angled triangles to find AX or XD) which may be seen in (a). You should condone incorrect labelling of the side. Look for attempted application of the cosine rule 2

2

2

( AD) =7 +12 −2 12 7cos ( ''128.9 ''− 27) or ( AC) =7 +12 −2 12 7cos( 180 −''128.9 '' − 27) 2

2

Or an attempted application of the sine rule Or

2

( AD) 7 = sin 27 sin ( ''128.9 ''− 27 )

( AC) 7 = sin 27 sin (180 − ''128.9''− 27)

Or an attempt using trigonometry on a right-angled triangle to find part of AD 12 cos 27 or 7 cos"51" This method can be implied by sight of awrt 15.1 or awrt 6.3 or awrt 8.8 or awrt 10.7 or awrt 4.4 dM1: A complete method of finding the TOTAL length. There must have been an attempt to use the correct combination of angles and sides. Expect to see 7 + 7 +12 + '' AD '' found using a correct method. This is scored by either 7 + 7 +12 + '' AD '' if ACB = 128.9 in a) or 7 + 7 +12 + awrt15.1 by candidates who may have assumed ACB = 51.1 in a) A1:

Rounds correct 41.09 m (or correct expression) up to 42 m to find steel bought Candidates who assumed ACB = 51.1(acute) in (a): Full marks can still be achieved as candidates may have restarted in (b) or not used the acute angle in their calculation which is often unclear. We are condoning any reference to AC = 15.1 so ignore any labelling of the lengths they are finding.

Diagram of the correct triangle with lengths and angles:

Diagram using the incorrect acute angle:

Question 6 (a)

Scheme

10  1  10  2  10  3  ( kx ) +   (kx ) +   (kx ) ... 1 2 3      

(1 + kx )10 = 1 + 

=1 +10 kx + 45k2 x2 +120 k3 x3 ...

Marks

AOs

M1 A1

1.1b 1.1b

A1

1.1b

(3) (b)

3

Sets 120k = 3 10k 2

4k = 1  k = ... 1 k = 2

B1

1.2

M1

1.1b

A1

1.1b

(3) (6 marks) (a) M1:

A1:

An attempt at the binomial expansion. This may be awarded for either the second or third  10  10 9 8 term or fourth term. The coefficients may be of the form 10C1 ,   etc or eg 3! 2 A correct unsimplified binomial expansion. The coefficients must be numerical so cannot  10  10  9  8 10 be of the form C1 ,   . Coefficients of the form are acceptable for this mark. 3! 2

( kx ) 2 but allow recovery 2 3 1 +10 kx + 45k2 x2 +120 k3 x3 ... or 1 +10 ( kx) + 45 ( kx) +120( kx) ... The bracketing must be correct on

A1:

Allow if written as a list. (b) B1:

3

Sets their 120 k = 3  their 10k (Seen or implied) For candidates who haven't cubed allow 120k = 3  their 10k 3 3

If they write 120 k x = 3  their 10kx only allow recovery of this mark if x disappears afterwards. M1:

A1:

3

Solves a cubic of the form Ak = Bk by factorising out/cancelling the k and proceeding B correctly to at least one value for k. Usually k = A 1 k =  o.e ignoring any reference to 0 2

Question 7 (a)

Scheme

Marks

AOs

x n → x n+1

M1

1.1b



A1

1.1b

dM1

1.1b

A1*

2.1

 5  + 3  dx = 5 x + 3x  2 x  k

5 x + 3 x  = 4  5 k + 3k − 8 = 4 1 3k + 5 k −12 = 0 *

(

(b)

3k + 5 k − 12 = 0  3 k − 4

)(

k=

k+3 = 0

)

M1

3.1a

4 , ( − 3) 3

A1

1.1b

dM1

1.1b

A1

2.3

k = ...  k = ... oe k=

(4)<...