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Mark Scheme (Results)
January 2016

Pearson Edexcel International A Level in Core Mathematics C34 (WMA02/01)

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Edexcel and BTEC Qualifications
Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at www.edexcel.com.
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Publications Code IA043148
All the material in this publication is copyright
© Pearson Education Ltd 2016

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

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EDEXCEL GCE MATHEMATICS
General Instructions for Marking
1. The total number of marks for the paper is 125.
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 and can be used if you are using the annotation facility on ePEN.


bod – benefit of doubt



ft – follow through



the symbol



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)



d… or dep – dependent



indep – independent



dp decimal places



sf significant figures



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



will be used for correct ft

or d… The second mark is dependent on gaining the first mark

4. All 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,

PhysicsAndMathsTutor.com the subsequent A marks affected are treated as A ft, but manifestly absurd answers 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. If you are using the annotation facility on ePEN, indicate this action by ‘MR’ in the body of the script.
6. If a candidate makes more than one attempt at any question:
 If all but one attempt is crossed out, mark the attempt which is NOT crossed out.
 If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt.
7. Ignore wrong working or incorrect statements following a correct answer.

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General Principles for Core 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 = …

(ax 2  bx  c)  (mx  p)(nx  q), where pq  c and mn  a , leading to x = …
2. Formula
Attempt to use correct formula (with values for a, b and c).
3. Completing the square
Solving x 2  bx  c  0 :

( x  b )2  q  c,
2

q  0,

leading to x = …

Method marks for differentiation and integration:
1. Differentiation
Power of at least one term decreased by 1. ( x n  x n 1 )
2. Integration
Power of at least one term increased by 1. ( x n  x n 1 )
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 mistakes 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.
Answers without working
The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done “in your head”, detailed working would not be required.

PhysicsAndMathsTutor.com
Question
Number

1

Scheme




(3  2 x )  4  3  4  1 



4

3  4 or

1
81

2

1 
 2    4   5   2 
  1   4    x  
 x   ... 


81 
2
 3 
 3 





B1

M1A1

1
8
40 2

x x  ...
81 243
729

Alternative: (3  2 x ) 4  3  4


2  x 3 

Marks

A1

(  4)(  5)
 (  4)(3) 5 (  2 x ) 
(3) 6 (  2 x ) 2  ...
2

1
8
40 2

x x  ...
81 243
729

(4 marks)
B1 M1 A1
A1
(4 marks)

B1

For taking out a factor of

3

 4

Evidence would be seeing either
M1

3  4 or

1 before the bracket.
81

For the form of the binomial expansion with n  4 and a term of  kx
To score M1 it is sufficient to see just the second and third term with the correct coefficient multiplied by the correct power of x. Condone sign slips. Look for ...   4  kx    4  5   kx 2 ...
2!

A1

Any (unsimplified) form of the binomial expansion. Ignore the factor before the bracket.
The bracketing must be correct but it is acceptable for them to recover from ''missing'' brackets for full marks. 2
 2    4  5  2 
8
40 2
Look for 1   4   x  
 x   or 1  x  x 

2
3
9
 3 
 3 

A1

cao 

1
8
40 2

x x  ... . Ignore any further terms.
81 243
729

Alternative

3  4 or

1 as the first term
81

B1

For seeing either

M1

It is sufficient to see the second and third term (unsimplified or simplified) condoning missing brackets. ie. Look for ...  (  4)(3)  5 ( kx )  (  4)(  5) (3)  6 ( kx ) 2
2

A1

Any (un simplified) form of the binomial expansion. ...  (  4)(3) 5 (  2 x )  (  4)(  5) (3) 6 (  2 x ) 2

A1

Must now be simplified cao

2

PhysicsAndMathsTutor.com
Question
Number

2 (a)
(b)

Scheme cosec 2 x  cosec x  12  0 or k = 12 cosec x  cosec x  12  0
2

so  cosecx  4 cosecx  3  0  cosecx  ....

1
1
or 
4
3
 x  14.5  or 165.5  or 340.5  or 199.5  s in x 

(a)
B1:

(b)
M1

dM1

Marks

B1
[1]
M1 dM1 dM1, A1 A1
[5]
(6 marks)

Accept cosec 2 x  cosec x  12  0 or k  12 . No working is required.
If they write cosec 2 x  cosec x  12  0 followed by k  12 allow isw

Solves quadratic in cosec x by any method – factorising, formula ( accept answers to 1 dp), completion of square. Correct answers (for cosec x of 4 and 3) imply this M mark.
Quadratic equations that have ‘imaginary’ roots please put into review.
Uses sin x 

1 by taking the reciprocal of at least one of their previous answers cosecx This is dependent upon having scored the first M1 dM1 For using arcsin to produce one answer inside the range 0 to 360 from their values.
Implied by any of 14.5 or 165.5  or 340.5 or 199.5 following (cosec x -4) (cosec x+3) =0
A1
Two correct answers inside the range 0 to 360
A1
All four answers in the range, x  awrt 14.5  165.5  340.5  199.5 
Any extra solutions in the range withhold the last A mark.
Ignore any solutions outside the range 0  x  360
..............................................................................................................................................................................
Radian solutions will be unlikely, but could be worth dM1 for one solution and dM1A1 A0 for all four solutions (maximum penalty is 1 mark) but accuracy marks are awarded for solutions to 3dp
FYI: Solutions awrt are 0.253, 2.889, 3.481, 5.943
.....................................................................................................................................................................
The first two M marks may be achieved 'the other way around' if a candidate uses cosec x 

1 in line 1 and sin x

produces a quadratic in sin x .
Award M1 for using cosec x 

1
(twice) and producing a quadratic in sin x and dM1 for solving as above. sin x

PhysicsAndMathsTutor.com
Question
Number

3

Scheme

Differentiates wrt x

3x ln 3  6

Marks

dy 3 2 dy  2 y  3xy dx dx

B1 B1,M1, A1

Substitutes (2, 3) AND rearranges to get dy

dx
 9  ln 729 d y 27 dy d y 9 ln 3  27 6 ln 3  9
2
 9 ln 3  6

 18



, dx 2 dx dx
12
8
8

M1 A1, A1

(7)
(7 marks)

B1
B1
M1

Differentiates 3 x  3 x ln 3 or e x ln 3  e x ln 3 ln 3 dy Differentiates 6 y  6 dx Uses the product rule to differentiate 3 xy 2 . Evidence could be sight of
2

3
2

y2  kxy

dy dx If the rule is quoted it must be correct. It could be implied by u=.., u'=.., v =.., v'=.. followed by their dy vu'+uv'. For this M to be scored y 2 must differentiate to ky , it cannot differentiate to 2y. dx A1

A completely correct differential of 3 xy 2 . It need not be simplified.

M1

Substitutes x  2, y  3 into their expression containing a derivative to find a ‘numerical’ value for dy

2

dx

The candidate may well have attempted to change the subject. Do not penalise accuracy errors on this method mark

A1

27
Any correct numerical answer in the form p ln q  r where p, q, r and s are constants e.g. 9 ln 3  2

A1

Exact answer. Accept either

s

12

 9  ln 729 or ln 729  9
8
8

.......................................................................................................................................................................
Note: There may be candidates who multiply by 2 first and start with

23x 12y 3xy2

This is perfectly acceptable and the mark scheme can be applied in a similar way.
......................................................................................................................................................................

PhysicsAndMathsTutor.com
Question
Number

Scheme

Marks

2
3

V   

4(a)


1

( )



4 dx (4  3x)2

B1

4
 4

dx  ( )   (4  3x)1 
2
(4  3x)
 3


M1A1

2

4
 4
3
 4

 ( )   (4  3 x ) 1   ( )   (4  2) 1   (4  3) 1 
3
 3
 1
 3

10


9

(b)

M1
A1
[5]

Length scale factor is 9 so volume scale factor is 9 3
So volume  9 3  10   810 or 2545 (cm 3 )

M1 A1

9

[2]
(7 marks)

(a)
2
3

B1

Need a correct statement including





2

 2  and correct limits and dx . Allow V    
 dx
 4  3x 
1

2

Allow if candidate initially writes down V   





2
 d x attempts to integrate and later uses the
(4  3 x ) 

2
3

correct limits either way around. Also allow if the
M1
A1

 is later multiplied by their

Uses substitution or reverse chain rule to do integral achieving  k (4  3 x )
For  4 (4  3 x ) 1 They do not need
3

1





1

2

 2 

 dx
 4  3x 

 or the limits

M1

Substitutes the correct limits in a changed/integrated function and subtracts (either way around)

A1

This answer or equivalent fraction. Accept answer with recurring decimals ie 1.1

.

(b)
M1

Attempts to multiply their answer in (a) by 729. May be implied by 10   810 (missing the
9


This may be implied by ( a )   15

1 2
3


A1







3

Any correct equivalent awrt 2540 or 2550 or 810



)

PhysicsAndMathsTutor.com

Question
Number
5. (a)

Scheme f (1)   3 ,

Marks

f (2)  2

Sign change (and as f ( x ) is continuous) therefore a root interval 1, 2 

M1


lies in the
A1
[2]

(b)

f( x )   x  4 x  6  0  x (4  x )  6
3

2

M1

2

 6 
 6  *
 x2  
 and so x  

4 x 

 4 x 
(c)

 6  x2  

 4  1.5  x 2  aw rt 1.5492

A1*
[2]
M1

,

x 3  aw rt 1.5647 ,

A1 and x 4  a w rt 1 .5 6 9 6 / 1 .5 6 9 7

A1
[3]

(d)

f (1.5715)   0.00254665... , f (1.5725)  0.0026157969

Sign change (and as f ( x ) is continuous) therefore a root  lies in the interval 1 .5 7 1 5, 1 .5 7 2 5     1.572 (3 dp)

M1A1
[2]
(9 marks)

(a)
M1
A1

(b)
M1

Attempts to evaluate both f (1) and f (2) and achieves at least one of f (1)   3 or f (2)  2
If a smaller interval is chosen, eg 1.57 and 1.58, the candidate must refer back to the region 1 to 2
Requires (i) both f (1)   3 and f (2)  2 correct,
(ii) sign change stated or equivalent Eg f (1)  f (2)  0 and
(iii) some form of conclusion which may be : or “so result shown” or qed or tick or equivalent
Must either state f(x) = 0 or set  x 3  4 x 2  6  0 before writing down at least the line equivalent to

 x2 (x  4)  6
A1* Completely correct with all signs correct. There is no requirement to show  6 
4 x

 6 
 6 
 and x  

 4 x 
 x4

2
Expect to see a minimum of the equivalent to x  

Alternative working backwards

M1

Starts with answer and squares, multiplies across and expands
6
 6 
2
x 
 x 2 (4  x )  6  4 x 2  x 3  6
x 
4 x
4 x


A1

Completely correct  x 3  4 x 2  6  0 and states ''therefore f(x) = 0'' or similar

6 x4 PhysicsAndMathsTutor.com

(c)
M1
A1
A1
(d)
M1

A1

Ignore any reference to labelling. Mark as the first, second and third values given.
An attempt to substitute x0  1.5 into the iterative formula. Eg. Sight of

 6  or x 2  aw rt 1.55


 4  1.5 

x 2  a w rt 1 .5 4 9 2

Both x 3  a w rt 1 .5 6 4 7 and x 4  a w rt 1 .5 6 9 6 o r 1 .5 6 9 7
Choose suitable interval for x, e.g. 1 .5 7 1 5, 1 .5 7 2 5  and at least one attempt to evaluate f(x) not the iterative formula. A minority of candidate may choose a tighter range which should include1.57199
(alpha to 5dp). This would be acceptable for both marks, provided the conditions for the A mark are met. Continued iteration is M0
Needs
(i) both evaluations correct to 1 sf, (either rounded or truncated)
Eg f (1.5715)   0.003 rounded f (1.5715)   0.002 truncated
(ii) sign change stated or equivalent Eg f ( a )  f ( b )  0 and
(iii)some form of conclusion which may be : or “so result shown” or qed or tick or equivalent x f(x)

1

‐3

1.1

‐2.491

1.2

‐1.968

1.3

‐1.437

1.4

‐0.904

1.5

‐0.375

1.6

0.144

1.7

0.647

1.8

1.128

1.9

1.581

2

2

x

f(x)

1.5715

‐0.002546651

1.5716

‐0.002030342

1.5717

‐0.001514047

1.5718

‐0.000997766

1.5719

‐0.0004815

1.572

3.4752E‐05

1.5721

0.00055099

1.5722

0.001067213

1.5723

0.001583422

1.5724

0.002099617

1.5725

0.002615797

PhysicsAndMathsTutor.com

Question
Number

Scheme

Marks

320  C 

6(a)

B1
[1]

T = 1 8 0  3 0 0 e  0 .0 4 t  1 6 0 ,  e  0 .0 4 t

(b)

t

160

 a w rt 0 .5 3 
300

1
1
 160 
 300  ln  ln 
 or

0.04  300 
0.04  160 

dM1
A1cso

15.7 (minutes) cao

[4]

dT
 (  0.04)  300e  0.04 t = (  0 .0 4 )  ( T  2 0 ) dt = 20  T *
25

(c)

M1, A1

M1 A1
A1*
[3]
(8 marks)

Alt (b)

Puts T = 180 so

180 = 300e–0.04t + 20 and 300e  0.04 t  160

ln300  0.04 t  ln 160  t  ..,

ln300  ln 160
0.04

dM1, A1
A1cso

15.7 (minutes) cao
(a)
B1
(b)
M1
A1

M1

[4]

320 cao - do not need  C
Substitutes T= 180 and proceeds to a form Ae 0.04 t  B or C e 0.04 t  D
Condone slips on the power for this mark. For example condone A e  0.4 t  B
For e 0.04t  160 or e 0.04 t  300 or exact equivalent such as e  0.04 t  8
300

160

15

Accept decimals here e 0.04 t  0.53.. or e 0.04 t  1.875 dM1 A1
(c)
M1

Dependent upon having scored the first M1, it is for moving from e kt  c , c  0  t  ln c k 15.7 correct answer and correct solution only. Do not accept awrt
Differentiates to give d T  k e  0.04 t . Condone d T  ke 0.4 t following T = 3 00 e  0.4 t  20 dt dt

This can be achieved from T=300e0.04t  20  t 
A1

1 dt k
 T  20  ln 

for M1

0.04  300  dT T  20

Correct derivative and correctly eliminates t to achieve
If candidate changes the subject it is for d t  dT  25

T  20 

dT
 (  0.04)  T  20  oe dt oe

Alternatively obtains the correct derivative, substitutes T in
A1*

dT
20  T dT 

  12e  0.04 t and dt 25 dt compares. To score the A1* under this method there must be a statement.
Obtains printed answer correctly – no errors

PhysicsAndMathsTutor.com

Question
Number

Scheme

dy

dx

7.(a)

(b)

( x2  1) 

Marks

6 x 1  ln( x 2  1) 

6x
 6 x  ln( x2  1) x 1
Or
( x2  1)2
2

M1

ln(x2 1)  1 so x  e 1

3 e M1A1 ddM1A1 [5]

3 ln 2 or 1.0397
2

(c)

[2]

( x 2  1) 2

dy
2x
 0  (x2 1) 3 2
 3ln(x2 1)(2x)  0 dx (x 1)

y

M1A1

B1
[1]

(d)
1
 1   ...........................
2

1
3
3
3

ln10  2  ln 2  ln 5  
1   0 
2
10
5
2





1
2

B1 oe

M1

(0.6907755279..  4.010767..) 

A1

= 2.351 (awrt 4 sf)

[3]
(11 marks)

(a)
M1

3ln( x 2  1)
( x 2  1)
If the formula is quoted it must be correct. There must have been some attempt to differentiate both terms.
If the rule is not quoted nor implied by their working, meaning that terms are written out

Applies the Quotient rule, a form of which appears in the formula book, to

u  3ln  x 2  1 , v  x 2  1, u '  .., v '  ... followed by their dy  dx ( x 2  1) A

vu ' uv '
, then only accept answers of the form v2 x
 Bx ln( x 2  1)
Ax  Bx ln( x 2  1) x 1
. Condone invisible brackets for the M. or ( x 2  1) 2
( x 2  1) 2
2

Alternatively applies the product rule with u  3ln(x 1), v   x2 1
If the formula is quoted it must be correct. There must have been some attempt to differentiate both terms.
If the rule is not quoted nor implied by their working, meaning that terms are written out
2

1

u  3ln( x2  1), v   x2  1 , u '  .., v '  ... followed by their vu ' uv ' , then only accept answers of the form
1

1
 x 2  1  A

A1

x
2
  x 2  1  Bx ln( x 2  1) . x 1
2

Condone invisible brackets for the M.
Any fully correct (unsimplified) form of f '( x ) . Remember to isw.

PhysicsAndMathsTutor.com

Using quotient rule look for variations of

dy

dx

( x2  1) 

6x
 6 x  ln( x2  1) x 1
( x2  1)2
2

6
Using the product rule look for d y   x 2  11  2 x   x 2  12  2 x  3 ln( x 2  1) dx (b)
M1

M1
A1

x 1

Setting their numerator (with more than one term) of their f '( x )  0 and proceeds to a form that does not include fractional terms.
If the product rule has been applied in (a) they also need an equation without fractions to score this.
Allow all marks in part (b) if denominator was incorrect in (a), for example v rather than v2 in their quotient rule.
Proceeds using correct work to ln(x2 1)  A  x  .. x  e 1 achieved from a  correct numerator. Ie condone it arising from   v u ' u v '

dM1

Dependent upon both M's having been scored. It is for substituting in their value of x
(which may be decimal) and finding a value of y from the correct function

A1

Correct solution only y 

3 and no other solution for x > 0. Ignore solutions x  0 e (c)
B1

3
1
ln2 or 1.0397 or exact equivalent such as ln8
2
2

(d)
1
1 outside the (main) bracket
 1 or
2
2

B1

for h = 1. This is implied by

M1

For inside the brackets: 0  3 ln 10  2  3 ln 2  3 ln 5  You can follow through on their 3 ln2


10
2
5
2
The decimal equivalent is



0  0.691  2 1.040  0.966 

Allow if you have an invisible bracket. That is you see 1  0  3 ln 10  2  3 ln 2  3 ln 5 


2
10
2
5


A1

awrt 2.351



PhysicsAndMathsTutor.com

Question
Number

8(a)

Scheme

Either

Marks

f ( )  9 cos 2   sin 2   9 cos 2   1  cos 2 
(cos 2  1)
 8cos2   1  8
1
2

[3]

Or f ( )  9 (cos 2  1)  1 (1  cos 2 )

M1 M1

2

A1

 5  4cos 2

[3]




2

(b)

M1
A1

 5  4 cos 2
2

M1

Either :Way1 splits as

 a

2

2

0

d   b 2 cos 2 d
0


2

 b
0

2

cos 2 d = ... 2 sin 2   ... sin 2 d

M1

=  .. 2 sin 2  .. cos 2   ..cos 2 d

dM1

Integral = 2 sin 2  2 cos 2  sin 2  +



5

2

   3

5 

  2      0
Use limits to give
 3






3

θ

3

 5 3

 24   



=

A B1ft
1

ddM1 A1
[6]
(9 marks)



1st 4 marks 

2

2

2
  f ( )d =   ( a  b cos 2 )d  =

Or: Way 2

2

0

0

  (..  ..sin 2 )   .. (..  ..sin 2 )d
2

  (...  ..sin 2 )  .. ..  ...cos 2    . ..  ...cos 2  d
2
2

︵

 θ

1st 4 marks 5

2

θ  2 s in 2  ) 

2



θ 


5

2

θ2

2

 
 c o s 2   
 

5

θ3

3


 s in 2  


M1 dM1 A1 B1ft

Or: Way 3 Way 2 that goes back to Way One

2



2



2
( a  b cos 2 )d    (..  ..sin 2 )  .. (..  ..sin 2 )d

0

  2 (..  ..sin 2 ) 
  2 (..  ..sin 2 ) 

  ...



  ...

2



d    .. sin 2 d 

d   .. cos 2   ..cos 2 d 
1 0
5
2
  ( θ  2 sin 2 )  3 θ 3  2 cos 2  sin 2
2

M1

dM1
A1 B1ft

PhysicsAndMathsTutor.com

(a)
M1
M1

Uses sin 2   1  cos 2  or cos 2   1  sin 2  to reach an expression in either sin 2  or cos 2 
Attempts to use the double angle formula cos 2   1  2 sin 2  or cos 2   2 cos 2   1 to convert their expression in sin 2  or cos 2  to form an expression a  b cos 2
A1 cao  5  4cos 2
......................................................................................................................................................................
Alternative
M1 One attempted application of double angle formula on either sin 2  or cos 2  . See above for rules
M1 A second attempted applications of double angle formula to form an expression a  b cos 2
A1
cao  5  4cos 2
(b) Note: On e pen this is marked up M1 M1 A1 M1 M1 A1. We are scoring it M1 M1 A1 B1 M1 A1
M1 An attempt at using integration by parts the correct way around.
IF THE CANDIDATE DOES NOT STATE OR IMPLY AN INCORRECT FORMULA ACCEPT
In Way One look for  b 2 cos 2 d  .. 2 sin 2   .. sin 2 d

In Way Two look for



2

(a  b cos 2 )d   2 (..  ..sin 2 )    .. (..  ..sin 2 )d



dM1 Dependent upon M1 having been scored, it is for an attempted use of integration by parts the correct way around for a second time.
In Way One look for
 .. 2 sin 2  .. cos 2   ..cos 2 d
In Way Two look for
  2 (..  ..sin 2 )   .. (.. 2  ..cos 2 )   (.. 2  ..cos 2 )d



Way 3 : You may see a candidate multiplying out their second integral and reverting to a type one integral.   2 (..  ..sin 2 )   .. 3  ... cos 2   ..cos 2 d



A1

B1ft ddM1 A1

cao  2 2 sin 2  2 cos 2  sin 2 


2
 a d  a

3
3

Accept in any unsimplified form

It is scored for the term independent of the trigonometrical terms.

Dependent upon both previous M's. For using both limits although you may not see the 0.
A decimal answer of 3.318 following correct working implies this mark cso. Note that a correct answer does not necessarily imply a correct solution

PhysicsAndMathsTutor.com

Question
Number

9(a)

Marks

Scheme

3x 2  4
A B
C
  2
2
x (3x  2) x x 3x  2

2 6
,
(B = 2 , C = 6) x2 3x  2
3x2 4  Ax(3x 2)  B(3x 2) Cx2 A..

B1, B1,
M1

3
(A = 3 ) is one of the fractions x A1
[4]

 3x  4
1
dx
 dy   2
 y
 x (3 x  2)

B1

lny =  A  B2 


M1

2

(b)

C dx  x x
3x  2
B C
= A ln x   ln(3 x  2) x 3

ye

A ln x 

B C
 ln( 3 x  2 )  D x 3

y  K x 3 (3 x  2 ) 2 e

2

x

(+k) or y  D e
3

or



2 x M1A1ft
A ln x 

B C
 ln(3 x  2 ) x 3

k

3



2 x Kx e e xe or oe
2
(3 x  2)
(3 x  2) 2

M1
A1cso
[6]
(10 marks)

PhysicsAndMathsTutor.com

(a)

6 being one of the ''partial'' fractions
3x  2 x For two of the partial fractions being  22 and  6
3x  2 x For either  22 or

B1
B1
M1

Need three terms in pfs and correct method either compares coefficients or substitutes a value to obtain A
Look for 3 x 2  4  A x (3 x  2)  B (3 x  2)  C x 2  A  ..
3
x

A1
(b)
B1:
M1
M1
A1ft
M1

Separates variables correctly. No need for integral signs
Integrates left hand side to give lny and uses their partial fractions from part (a) (may only have two pf 's)
Obtains two ln terms and one reciprocal term on rhs (need not have constant of integration for this mark) (must have 3 pf 's here). Condone a missing bracket on the ln(3x-2)
Correct (unsimplified) answer for rhs for their A, B and C (do not need constant of integration at this stage)
For undoing the logs correctly to get y = …. now need constant of integration.
A ln x 

B C
 ln(3 x  2)  d x 3

B C
A ln x   ln(3 x  2) x 3

A ln x 

B C
 ln(3 x  2) x 3

D
Accept y  e
OR y  D e
BUT NOT y  e
A1
cso One of the forms of the answer given in the scheme o.e.
.............................................................................................................................................................................
Special case: For students who use two partial fractions
Very common incorrect solutions using two partial fractions are
3x 2  4
A
B
2
3
 2
 2+ using substitution and comparing terms in x2
2
x (3x  2) x 3x  2 x 3 x - 2 or 3x 2  4
A
B
2
-6
 2
 2+ using substitution
2
x (3x  2) x 3x  2 x 3 x - 2

Both of these will scoring B1B1M0A0 in SC in (a)
In part (b) this could score B1, M1 M0 A0 M1 A0 for a total of 5 out of 10.


...

 ... ln ( 3 x  2 )  D



...

 ... ln ( 3 x  2 )

For the final M1 they must have the correct form y  e x or y  D e x or equivalent
............................................................................................................................................................................

PhysicsAndMathsTutor.com

Question
Number
10. (a)

Scheme
R 

Marks
B1

34

tan 

5
3

M1

   1.03

awrt 1.03 A1

(b)

sin  2 x  "1.03"  

[3]

3 4 s in  2 x  1 .0 3   4

3 s in 2 x + 5 c o s 2 x = 4 

4
" 34 "

One solution in range Eg.



0.68599... 

 4 
2 x  ''1.03 ''  2  arcsin 
  x  ...
 " 34 " 

M1
M1

Either x = awrt 3.0 or awrt 0.68

A1

 4 
Second solution in range Eg 2x  ''1.03''    arcsin 
  x  ...
 " 34" 

M1

Both x = awrt 2sf 3.0 and 0.68
(c)

Greatest value is 4



34



2

 3 = 139

Least value is 4  0  3 =3

A1
[5]
M1 A1
. M1 A1
[4]
(12 marks)

PhysicsAndMathsTutor.com

(a)
B1

R  34 Condone  34

M1 For tan   

5
3

or tan   

3
This may be implied by awrt 1.0 rads or awrt 59 degrees
5

If R is used to find α only accept cos   

3 their R

or sin   

5 their R

A1 accept   awrt 1.03; also accept 3 4 s in  2 x  1 .0 3  .
If the question is done in degrees only the first accuracy mark is withheld. The answer in degrees (59.04) is A0

(b) On epen this is marked up M1M1M1A1A1. We are scoring it M1M1A1M1A1
M1 For reaching s in  2 x  th e ir   

4 th e ir R

(Uses part (a) to solve equation)

It may be implied by  2 x  th eir    arcsin  4  = 0.75 rads


 th eir R 
M1 For an attempt at one solution in the range. It is acceptable to find the negative solution, -0.14 and add
 4 
Look for 2x  their   2  arcsin 
  x  .. (correct order of operations)
 their R 

 4 
Alternatively 2x  their    arcsin 
  x  ..
 their R 
A1 Awrt 3.0 or awrt 0.68. Condone 3 for 3.0. In degrees accept awrt 38.8 or 172.1
 4 
M1 For an attempt at a second solution in the range. This can be scored from their ''arcsin 
 ''
 their R 
 4 
2 x  their    their arcsin 
  x  .. (correct order of operations)
 their R 
 4 
Or 2 x  their   2  their arcsin 
  x  ..
 their R 
Look for

A1 Awrt 3.0 AND awrt 0.68 in radians or awrt 38.8 and awrt 172.1 in degrees. Condone 3 for 3.0

(c) (i)
M1 Attempts to find 4  R   3
2

A1 139 cao

(c)(ii)
M1 Uses 0 for minimum value. Accept 4  0   3
2

A1

3



PhysicsAndMathsTutor.com
Question
Number

Marks

Scheme

Shape y4 11(a)

 0,3

Asymptote y  4 y intercept (0, 3)

 1
 3

 1

  ln 4, 0 
 3





Touches x axis at   ln 4,0 

B1

B1
B1

B1
[4]

Shape B1
Asymptote y   2

(b)

B1

Passes though origin B1 y  2

[3]
(c)
(d)

B1

f(x) >  4 y = e 3 x  4

[1]
 e 3 x  y  4

M1 dM1   3 x  ln( y  4) and x =

1
1
f 1 ( x )   ln( x  4) or ln
1 ,
3
( x  4) 3

(x > 4) cao

A1
[3]

(e)

 1 
3 ln 

 x2 

fg(x) = e
= (x 

M1

4

2)3  4

,

= x 3  6 x 2  12 x  4

dM1, A1
[3]
(14 marks)

PhysicsAndMathsTutor.com

(a)
B1

B1
B1

B1
(b)
B1
B1
B1
(c)
B1

For a correct shape. The curve must lie completely in first and second quadrants with a cusp on – ve x axis and approaching an asymptote as x becomes large and positive. The curvature must be correct on both sections. The lh section does not need to extend above the asymptote. On the rh section do not allow if the curve drops below the y intercept.
See practice items for clarification
The asymptote of the curve is given as y = 4. Do not award if a second asymptote is given or the curve does not appear to have an asymptote at y = 4 y intercept is (0, 3). Allow if given in the body of their answer say as A= .
3 sufficient if given on the y-axis. Condone (3,0) being marked on the correct axis.
Do not award if there are two intercepts x intercept is (–1/3 ln 4, 0). Allow if given in the body of their answer say as B= .
–1/3 ln 4 is sufficient if given on the x axis. Do not award if there are two intercepts shape – similar to original graph ( do not try to judge the stretch)
Equation of the asymptote given as y = 2 Do not award if a second asymptote is given or the curve does not appear to have an asymptote here.
Curve passes through O only. cao f(x) >  4 . Accept alternatives such as  4, 
Note that f(x) ≥  4 is B0. Accept range or y for f(x)

(d)
M1
dM1

A1

For an attempt to make x (or a switched y) the subject of the formula. For this to be scored they must make e  3 x or e3x as subject. Allow numerical slips.
This is dependent upon the first M being scored. It is for undoing the exp correctly by using ln’s.
Condone imaginary brackets for this mark. Accept x being given as a function of y and involving ln's.
If the rhs is written ln(4x) this implies taking the ln of each term which is dM0
This is cao. Accept any correct equivalent. The domain is not required for this mark but the bracket is.
Accept y =..

(e)
M1
dM1

3ln

1

Correct order of operations. Allow for e ( x2)  4
Dependent Method Mark. Simplifies this expression using firstly the power law and then the fact that e ln ..  .. to reach fg(x) as a function of x.
 1 
3 ln

 x2 

You may condone sign errors here so tolerate e
A1

Correct expansion to give this answer.

4 

1

 x  2

3

4

PhysicsAndMathsTutor.com
Question
Number

12 (a)

Scheme

12  5  2
 12 
 5  2
 0
 
   
 
 4  4  2  6 any two of these
 4     4    2     6  
 5
 2   0
 3
5  2  3
 
   
 
Full method to find either  or 
(1)    2
Sub   2 into (2) to give   1 (need both)
3
Check values in 3rd equation

52(2)  3(1) and make statement eg. True
3

 12   5 
 2   0  2 
   
     
Position vector of intersection is  4   2  4  OR  2   1  6  =  4 
3
 5   2
 0  3  1 
   
     

Marks

M1
M1
A1
B1
M1,A1
[6]

(b)

 5  0
   
 4  .  6 
 2  3
18
   

 0.4 cos  
2
2
2
2
2
45
5  ( 4)  2 6  3
So acute angle is 66.4 degrees

M1 A1

A1
[3]

(c)

 7
 
When λ= 1 this gives  0  so B lies on l1
 3
 

B1

(d)

Way 1:

M1 A1

A B 

45

h 45  sin 66.4 h = 6.15

Way 2:

5 





XB :    2  6 
 3  3 



0
  



 1
Find XB.  6   0       followed by calculation of XB by Pythag
 15 
 3
  h = 6.15

[1]

M1
A1
[4]
M1A1

M1
A1
[4]
(14 marks)

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(a)
M1
M1

For writing down any two equations that give the coordinates of the point of intersection.
Accept two of 12  5   2 ,  4  4   2  6  , 5  2   3  condoning slips.
A full method to find either  or  .

A1

Both values correct   1 and   2
3

B1

M1

A1

(b)
M1

The correct values must be substituted into both sides of the third equation. There must be some minimal statement (a tick will suffice) that the values are the same. This can also be scored via the substitution of   1   2 into both of the equations of the lines but there must be the same
3
minimal statement.
Substitutes their value of  into l1 to find the coordinates or position vector of the point of intersection. It is dependent upon having scored second method mark. Alternatively substitutes their value of  into l2 to find the coordinates or position vector of the point of intersection.
Correct answer only. Accept as a vector or a coordinate. Accept (2, 4, 1) (A correct answer here implies previous M mark). Note that the correct answer can be achieved by solving just the first equation. A clear attempt to use the correct formula for a.b  a b cos (where a and b are the gradient vectors)
2
2
2
2
2
Expect to see 50 46  23  5  (4)  2  6 3 cos allowing for slips.

A1
A1
(c)
B1
(d)
M1

For cos  0.4 . This may be implied by 66.4 or 113.6.Also accept cos  0  24  6 oe
45 45

cao for awrt 66.4
States or uses λ= 1 and checks all 3 coordinates
Way 1:
Finds distance AB using a correct method
Using Pythagoras look for

 AB 

 7  '' 2 '' 

45 or

2

  0  '' 4 ''    3  ''1'' 

 AB  3

2

2

or 'one' gradient

52  (4)2  22

5 or (AB) = awrt 6.71

A1

Correct answer

M1

Reaches h 45sin66.4 with their values for AB and sin  to find h

A1

awrt 6.15 ( allow also if it follows 113.6). The exact answer of

3
105 is fine
5

Way 2: Setting up a point X on l2
M1




Finds distance XB2 or vector X B using a correct method. For this to be scored X   2,2  6,3 
B  (7, 0, 3) and there must be an attempt at differences

A1

 2



2
2
Correct answer h  XB  45  6  38 or XB    5, 2  6 ,3  3

M1

Find minimum value of h by completion of square or differentiation giving h =

0
  

1

Alternatively by the vector method uses XB.  6   0    .. followed by substitution of this     
15 

 3
 

A1

 5 

 into l2 to find length of BX    1.6  using pythagoras’ theorem.
 3.2 

 awrt 6.15

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Question
Number

Scheme dy dy d t  2 cos t

dx dx
 12 sin 2 t dt 2 cos t

 24 sin t cos t

13 (a)



(b)

Marks

dy

When t =  , dx 3

M1 dM1 2 cost
1
=  12 cosec t
24sin t cos t

1
12  3

2

M1 A1
[4]

 3
  
 18 



M1 A1

So Normal has gradient  1  6 3

M1

m

When t =  , x  3 and y  3

B1

3

Equation of normal is
(c)

3  6

so

3 (x  3)

y  6

3x  19

3

x  6(1 2sin t )  x  f ( y)
2

So
(d)

y 

x  6  3y2

or

M1 A1
[6]
M1 dM1 A1

f ( y)  6  3 y2

[3]
B1

-2 < y < 2 or k = 2

[1]
(14 marks)

Alt (a)

Via cartesian must start with

x  A  B y2

or

y CDx

1

2

dx dy x

 ky or
 b  2   then as before dy dx
3
 followed by correct (double angle) substitution
Alt (b)

M1 dM1 Must start with

x ABy or
2

y CD x

dx dy 1 x  6 y or
  2  dy dx
6
3
For substituting their y  3 into a



1
2

dx of the form Py dy 1st M1

Or alternatively substituting their x  3 into a dy of the form P  Q  Rx 

1

2

dx

For using the 'correct numerical' grad of the normal either 

dx
1
or dy  dy dx 2nd M1

PhysicsAndMathsTutor.com

(a)

dy

M1

dM1

dt   A cos t dx  B sin 2t dt They may use any double angle formula for cos first. Condone sign slips in this formula dy dt   A cos t
Eg cos 2 t   2 cos 2 t  1 to get dx  B sin t cos t dt Correct double angle formula used sin 2t  2sin t cos t
In the alternative method the correct double angle formula must have been used

Differentiates both x and y wrt t and establishes

M1

Cancels cost and replaces 1/sint by cosect correctly achieving a form dy   cosect

A1

cao d y   1 cosect

dx

dx

12

(b)



M1

Substitute t 

A1

dy
1
 dx 12  3

3

into their dy   cosect dx or exact equivalent. It may be implied by normal gradient of 6 3

2

Accept decimals here.

dy
  0.096 or implied by normal gradient of 10.4 dx M1

Use of negative reciprocal in finding the gradient of the normal.

B1
M1

for x = -3, y  3
Correct method for line equation using their normal gradient and their (-3, one of their coordinates.
Look for y  y1  

3)

allowing a sign slip on

dy dx  x  x1  or x  x1   dx   y  y1  dy t   t 3

3

If the candidate uses y  mx  c they must proceed as far as c =.. for this mark
A1
(c)
M1
dM1

cao y  6 3x 19 3

Attempts to use the double angle formula cos 2 t   1  2 sin 2 t leading to an equation linking x and y
If cos 2 t  cos 2 t  sin 2 t is initially used there must be an attempt to replace the cos 2 t by 1  sin 2 t
Uses correct cos 2t  1  2 sin 2 t and attempts to replace sint by y and cos2t by x
2

6

Condone poor bracketing in cases such as cos 2t  1  2 sin 2 t  x  1  2 y
6

x  6  3y

or

f ( y)  6  3 y

A1

Correct equation. Accept

(d)
B1

States k  2 or writes the range of y as -2 < y < 2

2

2

2

2

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