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January 2019

Mark Scheme

Mock Paper (set1)

Pearson Edexcel GCE A Level Mathematics

Pure Mathematics 2 (9MA0/02)

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PEARSON EDEXCEL GCE MATHEMATICS

General Instructions for Marking

1. The total number of marks for the paper is 100

2. These 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

 o.e. – or equivalent (and appropriate)

 d or dep – dependent

 indep – independent

 dp decimal places

 sf significant figures

  The answer is printed on the paper or ag- answer given

4. All M marks are follow through.

A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft to

indicate that previous wrong working is to be followed through. After a misread

however, the subsequent A marks affected are treated as A ft, but answers

that don’t logically make sense e.g. if an answer given for a probability is >1 or

<0, should never be awarded A marks.

5. 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.

6. 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.

7. Ignore wrong working or incorrect statements following a correct answer.

8. 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. If no such alternative

answer is provided but the response is deemed to be valid, examiners

must escalate the response for a senior examiner to review.

Question Scheme Marks AOs

cos sin 2 tan

(a) 2

M1 1.

A1 1.1b

Question Scheme Marks AOs

x t y t

t

x

t y

x

M1 1.1b

A1 1.1b

x

y y y

x x x

x

y x

x

{ a 5, b  13, k 1}

A1 2.

Alt 1 4 4

t x

y y

M1 1.1b

A1 1.1b

x x y x xy y

y

x

x xy y x y x y x

x

A1 2.

(3 marks)

Question 2 Notes:

M1: An attempt to eliminate t

A1: Achieves a correct equation in x and y only which can be un-simplified or simplified

A1: Uses correct algebra to show

x

y x

x

Question Scheme Marks AOs

3 (a)

 

2 1 2

d

d

y

y x k x x x k x

x

 

M1 1.1b

A1 1.1b

2

d d 8

At 3, 0 2(3) 0

d d (3 5)

y y

x k

x x

dM1 1.1b

  k     k    k 

A1* 2.

(b)

2

2

2 3

d 16

d ( 5)

y

x

x x



When

2

2 3

d 16

d (3 5)

y

x

x

M1 1.1b

2

2

d

d

y

x

   {local} minimum {stationary point at P } A1 2.

(b)

Alt 1

E.g.

2

d

d

y

x

x



2

d

d

y

x

x



{local} minimum {stationary point at P }

M1 1.1b

A1 2.

(c)

Criteria 1 (Accept any one of the two following points)

2

2 3

d 16

At 7, 2 0

d (7 5)

y

x

x

2

3

2 3

d 16

d ( 5)

y

x x x

x x

Criteria 2 (Accept any one of the two following points)

2

2

d

At 6.9, 0.33... 0

d

y

x

x

   and

2

2

d

at 7.1, 0.27... 0

d

y

x

x

3

3 3

d 48

d ( 5)

y

x x

and

3

3 3

d 48

at 7, 6 0

d (7 5)

y

x

x

At least one of Criteria 1 or Criteria 2 M1 2.

Both Criteria 1 and Criteria 2 (with correct calculations)

and concludes the curve has a {non-stationary} point of inflection at

x  7

A1 2.

(d)

Sign change method is not valid because either

 the curve is not defined at

x  5

 the curve is not continuous over the interval (4.5, 5.5)

B1 2.

(9 marks)

Question 3 Notes:

Question Scheme Marks AOs

3 2

f ( ) x  x  6 x  7 x  2, x 

(a)

2

f ( ) x ( x  2)( x  4 x  1)

M1 2.2a

A1 1.1b

(b)

(i), (ii)

{Note:

Q

x    x 

is known and at P , R ,

2

( x  4 x  1)  0 }

2

( x  2)  4  1  0 or

x

M1 1.1b

P

 x   and 2 5

R

x  

A1 1.1b

(c)

3 2

sin   6sin   7sin  2 0,   „  „ 12  ,

Deduces that there are 14 real solutions for

B1 2.2a

Correct justification. E.g.

Both

sin   2 and sin   2  5  4.236...have no real solutions

and either

sin   2  5  0.236...

has 2 real solutions for each interval of

2 . So there are 12 real solutions in the interval

[0, 12  ]

and

2 real solutions in the interval

[  , 0]

sin   2  5  0.236...

has 2 real solutions for each interval of

2 . So there are 12 real solutions in the interval

[   , 11 ]

and

2 real solutions in the interval

[11  , 12 ]

sin   2  5  0.236...

has 2 real solutions for each interval of

2 . So there are 14 real solutions in the interval

[ 2  , 12 ]

and

no real solutions in the interval

[ 2  , ]

sin   2  5  0.236...

has two real solutions in each of

[   , 0], [ , 2  ], [3 , 4 ], [5 , 6  ], [7 , 8 ], [9 , 10 ]

and

[11  , 12 ]

B1ft 2.

(6 marks)

Question 4 Notes:

(a)

M1: Deduces

( x  2) is a factor of

f ( ) x and attempts to find a quadratic factor of

f ( ) x by either

equating coefficients or by algebraic long division

A1:

2

( x  2)( x  4 x 1)

(b)

(i), (ii)

M1: Correct method (i.e. completing the square or applying the quadratic formula) to solve a 3TQ.

Note: M1 can be given here for at least one of either 2  5 or 2  5 written down in part (b).

A1: Finds and identifies the correct exact x coordinate of P and the correct exact x coordinate of R

(c)

B1: Correct deduction of 14 (real solutions)

B1: See scheme

Question Scheme Marks AOs

4 2

4 4

kx kx kx

kx



 

(a)

For the

2

x

term:

2

2

k

k

M1 1.1b

A1 1.1b

2

2 2

k

k k k A

dM1 3.1a

A k A

M1 2.2a

or 0.

A  

A1 1.1b

(b)

f ( ) x is valid when

kx x

    x 

E.g.

 As

x  lies in the interval

x  the binomial expansion is valid

 As

, the binomial expansion is valid

B1ft 2.

(6 marks)

Question 6 Notes:

(a)

M1: For either

or

2

k  

or

2

kx  

or

or 10 as part of their

2

x

coefficient

A1: For

2

2

or

k

k

or equivalent as part of their

2

x

coefficient

dM1: dependent on the previous M mark

A complete strategy to find a value for k and use their k to find a value for A

M1: Deduces and applies

(their )

A  k or

(their )

A  k

A1:

or 0.

A  

(b)

B1ft: See scheme

Note: Allow follow through for applying either

their

x

k

or

their 1

k    

Question Scheme Marks AOs

2

f ( ) e 2 , , 0

x

x x x x

x

(a)

Evaluates both

f ( 1.5)

and

f ( 1)

M1 1.1b

f ( 1.5) 2.943536507... and

f ( 1)  0.3678794412...

Sign change and as

f ( ) x is continuous

lies between  1.5and  1

A1 2.

(b)

(i)

3

{ x }  1.

B1 1.1b

(ii)

{  }  1.06 (2 dp)

B1 2.2a

(c)

2

x

M1 1.1b

{ 2.829208695...} 2.83 (2 dp)

A1 1.1b

(d)

 Draws a tangent to the curve at x 1.5 and identifies (possibly by

writing

2

x )

where the tangent cuts the x -axis

M1 1.1b

and concludes either

 second approximation is not good because it is not in the

interval [1.5, 3]

2

x

(which is indicated on Figure 3) is nowhere near the root

A1 2.

(8 marks)

Question Scheme Marks AOs

8 (a)

Deduces that the radius of the circle is 6

B1 2.2a

2 2

( x  9)  ( y  6)  36

M1 1.1b

A1 1.1b

(b)

Let d be the distance from

to l

2 2 2 2

d  4  6  d ...

M1 3.1a

d  20 or 2 5 A1 1.1b

{ l : } y  6  2 5 , y  6  2 5

dM1 2.2a

A1 1.1b

(b)

Alt 1

Either

2 2

x y y

or

2 2

x y y

M1 3.1a

{ :} l y  6  2 5

A1 1.1b

{ l : } y  6  2 5

dM1 2.2a

A1 1.1b

(7 marks)

Question 8 Notes:

(a)

B1:

Deduces that the radius of the circle is 6. This can be achieved by either

 Stating that

r  6

or radius = 6 or

2

r  36

 Writing

2 2 2

( x   )  ( y  ) 36 or 6 ;  ,  0

M

2 2

( x 9)  ( y 6)  k ; k  0

A1:

2 2

( x  9)  ( y  6)  36 or

2 2 2

( x  9)  ( y  6)  6 o.e.

(b)

M1: Uses the circle property “the perpendicular from the centre to a chord bisects the chord” in a

complete strategy of writing an equation of the form

2

2 2

(their )

d r

and progresses as far as

2

d ...

A1: d  20 or 2 5

dM1: depends on the previous M mark

Deduces the horizontal line l is d units from the line

y  6

and so writes both

y  6  (their d ) and

y  6 (their d )

A1: For either:

y  6  2 5 and y  6  2 5

y  6  20 and y  6  20

(b)

Alt 1

M1: Uses the circle property “the perpendicular from the centre to a chord bisects the chord” in a

complete strategy of substituting either

x  13

or

x  5 into their circle equation and progresses as

far as

y ...

A1: For y  6  2 5 or y  6  20

dM1: depends on the previous M mark

Finds y in the form

y  6  (their d ),

deduces the other horizontal line l is d units below the line

y  6

and so writes

y  6 (their d )

A1: For either:

y  6  2 5 and y  6  2 5

y  6  20 and y  6  20

Question 9 Notes:

(a)

M1: Starting from ,

t

y  ab takes logs of both sides and uses the addition law of logarithms to

progress as far as

10 10 10

log log log

t

y  a  b

A1: Starting from ,

t

y  ab correctly shows that

10 10 10

log y  t log b  log a with no errors seen

Note: M1 (special case) can be given in part (a) for stating 10

c log a

(b)

M1: For either

a  10 or

b 10



A1:

a  170

and

b 0.

(c)(i)

B1ft: Correct practical interpretation of their a , where their a > 0

(c)(ii)

B1ft: Correct practical interpretation of their b , where their b :

0  b  1

(d)

M1: Substitutes

y  2. into the model (their )(their )

t

y  a b and rearranges their equation to give

t ...

A1: 9.9 (hours) (1 dp)

(d)

Alt 1

M1:

Substitutes

y  30, m  0.

and

c  2.

into the model

10

log y  mt  c and rearranges their

equation to give

t ...

A1: 9.9 (hours) (1 dp)

(e)

B1:

E.g. Estimate should be treated with caution because

t 9. is outside the range of times,

i.e.

0 „ t „ 5,

for which the model

t

y  is valid

Question Scheme Marks AOs

10 (a)

1

2

d d d

or

d d d

A A A

A k A k A

t t t

B1 3.1b

1

2

d A k d t

A

M1 1.1b

1

2

A d A k d t

  

 

1

2

1

2

A

 kt  c

or

1

2

2 A  kt { c } A1 1.1b

{ t 0, A  9  } 2 9  k (0) c M1 3.

1

2

 c  6  2 A  kt  6

{ t 6, A 56.25  } 2 56.25  k (6)  6

dM1 1.1b

  k   k   k 

1 1

2 2

2

A t A t A t

A1* 2.

(b)

(i), (ii)

Either

2

t A

t  18, A  272.25 { 271.19}

t  24, A  441 { 334.81}

t 30, A 650.25 {  337.33}

}

or

2

A t t

A  271.19  t 17.95713... { 18}

A  334.81  t 20.39709... { 24}

{ A 337.33  t  20.48873... { 30}}

M1 3.

Biologist’s model works well for

t  12 and

t  18 but appears to give an

overestimate for A (or does not work well) when t  24 and t  30

A1 3.5a

E.g.

 The biologist’s model appears to break down for large values of t.

This may be because the biologist’s model predicts values for A

which are greater than the total surface area of the piece of bread

used in the experiment.

 The biologist’s results indicate an upper limit for A , but the

biologist’s model does not give an upper limit for A.

B1 3.2a

(9 marks)