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January 2019

Mark Scheme

Mock Paper (set1)

Pearson Edexcel GCE A Level Mathematics

Pure Mathematics 1 (9MA0/01)

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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

1 (a)

0.05 0.2 0.45 0.

Area( ) 0.5 1 2(e e e ) e

R        

B1 1.1b

M1 1.1b

Question Scheme Marks AOs

 

2 2 2

BC  7.5  8.5  2(7.5)(8.5) cos(   1.2)  BC 13.21743597...

M1 1.1b

A1 1.1b

Arc length AB  7.5(1.2)   Arc length AB  9 

B1 1.1a

Perimeter AOCBA

M1 3.1a

   38.21743597...  38.2 (cm) (1 dp)

A1 1.1b

(5 marks)

Question 2 Notes:

M1: Application of cosine rule for

2

BC or BC with any angle

A1:

Correct application of cosine rule for

2

BC

or BC using  1.

B1: Arc length AB = 7.5(1.2) or 9

M1: A complete strategy for finding the perimeter of the shape AOCBA

A1:

cao

Question Scheme Marks AOs

2

3 x  k  5 x  2

E.g.

2

3 x  5 x  k  2  0

or

2

 3 x  5 x  2  k  0

M1 1.1b

 

2

" b  4 ac "  0  25  4(3)( k  2)  0

M1 1.1b

25  12 k  24  0   12 k  49  0

Critical value obtained of

o.e.

B1 1.1b

o.e.

k 

A1 2.

(4 marks)

Question 3 Notes:

M1: Forms a one-sided quadratic equation or gathers all terms into a single quadratic expression

M1: Understands that the given equation has no real roots by applying

2

" b  4 ac "  0 to their one-sided

quadratic equation or to their one-sided quadratic expression

B1: See scheme

A1: Complete process leading to the correct answer, e.g.

k 

 k

k k

with no errors seen in their mathematical argument

4 (d)

Alt 2

Range of ff ( ) x is 0 „ ff ( ) x  3

M1 1.1b

As

, then

ff ( )

x 

has no solutions A1 2.

Question 4 Notes:

(a)

M1:

For one “end” fully correct; e.g. accept f ( ) x … 0 ( not x … 0 ) or f ( ) x  4 ( not x  4 );

or for both correct “end” values; e.g. accept 0  f ( ) x „4.

A1:

Correct range using correct notation.

Accept

0 „ f ( ) x  4, 0 „ y  4, [0, 4), f ( ) x …0 and f ( ) x  4

(b)

M1:

Attempts to find the inverse by cross-multiplying and an attempt to collect all the x- terms (or

swapped y -terms) onto one side.

M1: A fully correct method to find the inverse.

A1: A correct

1

f ( ) , 0 4

x

x x

x



„ , o.e. expressed fully in function notation, including the

domain, which may be correct or followed through from their part (a) answer for their range of f

Note: Writing

x

y

x

as

x

y y

x x

leads to a correct

1

f ( ) 4 , 0 4

x x

x



(c)

M1: Attempts to substitute

f ( )

x

x

x

into

12f ( )

3f ( ) 4

x

x 

M1: Applies a method of “rationalising the denominator” for their denominator.

A1:* Shows

ff ( )

x

x

x

with no errors seen.

Note: The domain of ff ( ) x is not required in this part.

(d)

M1: Sets

x

x 

to

and solves to find

x ...

A1: Finds

x  , rejects this solution as

ff ( ) x is valid for

x … 0 only

Concludes that

ff ( )

x  has no solutions.

Question 4 Notes Continued:

(d)

Alt 1

M1: Attempts to find

1

f



A1: Deduces

1

f ( ) f

x



and concludes

ff ( )

x  has no solutions because

f ( )

x  lies outside the range 0 „f ( ) x  4

(d)

Alt 2

M1: Evidence that the upper bound of ff ( ) x is 3

A1:

0 „ ff ( ) x  3 and concludes that

ff ( )

x  has no solutions because

Question 5 Notes:

B1: Writes down the y coordinate of a point close to P

E.g. For a point Q with x coordinate 2  h ,

2

Q

y   h   h

M1: Begins the proof by attempting to write the gradient of the chord PQ in terms of h

A1: Correct expression for the gradient of the chord PQ in terms of h

M1: Correct process to obtain the gradient of the chord PQ as  h   ;  ,  0

A1: Correctly shows that the gradient of PQ is 4 h  11 and applies a limiting argument to deduce that at

the point

P

on

2

d

d

y

y x x

x

   E.g.  

lim

h

h

Note:  x can be used in place of h

Alt 1

B1:

2

4( x  h )  5( x  h ), seen or implied

M1: Begins the proof by attempting to write the gradient of the chord in terms of x and h

A1: Correct expression for the gradient of the chord in terms of x and h

M1: Correct process to obtain the gradient of the chord as

 x   h   ;  ,  ,  0

A1:

Correctly shows that the gradient of the chord is 8 x  4 h  5 and applies a limiting argument to

deduce that when

2

d

d

y

y x x x

x

    E.g.  

lim

x h x

h

Finally, deduces that at the point P ,

d

d

y

x

Note: For Alt 1,  x can be used in place of h

Question Scheme Marks AOs

6 (a)

1

2 e or 2ln

x

u  x  u 

1

2

d 1

e

d 2

x u

x

or

d 1

d 2

u

u

x

or

d 2

d

x

u u

or

d x d u

u

or

2d u  u x d ,

etc. B1 1.1b

Criteria 1

1

2

(2) 0

x  0  a e and x  2  b e

a  1, b e or evidence of

and 2 e

Criteria 2 (dependent on the first B1 mark)

1

2

d. d d

(e 4)

x

x u u

u u u u

Either Criteria 1 or Criteria 2 B1 1.1b

Both Criteria 1 and Criteria 2

and correctly achieves the result

e

1

d

u

 u u 

B1 2.

(b)

A B

A u Bu

u u u u

u  0  A 3; u  4  B  3

M1 1.1b

A1 1.1b

d d 3ln 3ln( 4)

u u u u

u u u u

M1 3.1a

A1ft 1.1b

 

e

1

So, 3ln u  3ln( u 4)

 (3ln e  3ln(e  4))  (3ln1 3ln 5)

 3ln e  3ln(e  4) 3ln 5

5e

3ln

e 4

* A1* 2.

(8 marks)

Question Scheme Marks AOs

3sin   4cos   R sin(   ); R  0, 0   90 

(a)

tan

o.e. M1 1.1b

Either R  5 or  awrt 53.

B1 1.1b

5sin(   53.13 )

A1 1.1b

(b)(i) max

G  17  "5" 22 ( C ) 

B1ft 3.

(b)(ii)

G  17  3sin(15 ) t   4cos(15 ) ; 0 t  „ t „ 17

20  17  "5"sin(15 t  "53.13")

M1 3.

sin(15 "53.13")

t   or

sin( "53.13")

M1 1.1b

After midday solution  15 t  "53.13"  180 36.86989...

t

M1 3.1b

 t  13.0840...  Time 6 : 05 p.m. or 18 : 05

A1 3.2a

(8 marks)

Question 7 Notes:

(a)

M1: For either

tan

  or

tan

  or

tan

   or

tan

B1: At least one of either R  5 (condone R  25 ) or  awrt 53.

A1:

5sin(   53.13 )

(b)(i)

B1ft: Either 22 or follow through “17 + their R from part (a)”

(b)(ii)

M1:

Realisation that the model G  17  3sin(15 ) t   4cos(15 ) t  can be rewritten as

G  17  "5"sin(15 t  "53.13") and applies G  20

M1: Rearranges their equation to give either

sin(15 "53.13")

t   or

sin( "53.13")

Note: This mark can be implied by either

15 t  "53.13" 36.86989... or 143.1301...

   "53.13" 36.86989... or 143.1301...

M1: Uses the model in a complete strategy to find a value for t which is greater than 7

e.g. p.m. solution occurs when 15 t  "53.13"  180  36.86989... and so rearranges to give t ... ,

where t is greater than 7

A1:

Finds the p.m. solution of either

6 : 05 p.m. or 18 : 05 when the greenhouse temperature is predicted

by the model to be

20 C

Question 8 Notes:

(i)

M1: Attempts to

 complete the square or

 find the minimum by differentiation or

 draw a graph of

2

f ( y )  y  4 y  7

A1:

Completes the proof by showing

2

y  4 y  7 is positive for all real values of y with no errors seen in

their working.

(ii)

M1: See scheme

A1: See scheme

(ii)

Alt 1

M1: Assumes

3 2

e e

x x

… , takes logarithms and rearranges to make x the subject of their inequality

A1: See scheme

(iii)

M1: Begins the proof by negating Elsa’s claim and attempts to define n as an odd number

A1: Shows

2 2

n  4 k  4 k  1 , where n is correctly defined and gives a correct conclusion

(iv)

M1: See scheme

A1: See scheme

Question Scheme Marks AOs

9 (a)

sin 1 cos

1 cos sin

x x

x x

2 2

sin (1 cos )

(1 cos )sin

x x

x x

M1 2.

2 2

sin 1 2cos cos

(1 cos )sin

x x x

x x

A1 1.1b

1 1 2cos

(1 cos )sin

x

x x

M1 1.1b

2 2cos

(1 cos )sin

x

x x

2(1 cos )

(1 cos )sin

x

x x

2cosec

sin

x

x

{ k 2}

A1 2.

(b)

sin 1 cos

1 cos sin

x x

x x

2cosec x 1.6  cosec x 0.

As

cosec x is undefined for

 1  cosec x  1

then the given equation has no real solutions.

B1 2.

(b)

Alt 1

sin 1 cos

1 cos sin

x x

x x

2cosec x 1.6  sin x 1.

As

sin x is only defined for

 1 „ sin x „ 1

then the given equation has no real solutions.

B1 2.

(5 marks)

Question 9 Notes:

(a)

M1: Begins proof by applying a complete method of rationalising the denominator

Note:

2 2

sin (1 cos )

(1 cos )sin (1 cos )sin

x x

x x x x

is a valid attempt at rationalising the denominator

A1: Expands

2

(1  cos x ) to give the correct result

2 2

sin 1 2cos cos

(1 cos )sin

x x x

x x

M1: Evidence of applying the identity

2 2

sin x  cos x  1

A1: Uses

2 2

sin x  cos x  1 to show that

sin 1 cos

2cosec

1 cos sin

x x

x

x x

with no errors seen

(b)

B1: See scheme

(b)

Alt 1

B1: See scheme