Mark Scheme (Results)

Summer 2019

Pearson Edexcel GCE

In Mathematics (6665) Paper 1
Core Mathematics 3
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Summer 2019
Publications Code 6665_01_1906_MS
All the material in this publication is copyright
© Pearson Education Ltd 2019
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.
Question Scheme Marks
1 Way 1 Way 2
4x  6 4 x 2  14 x  10
x 2  0 x  4 4 x3  6 x 2  18 x  20 x  2 4 x  6 x 2  18 x  20
3

4 x3  0 x 2  16 x 4 x3  8 x 2 M1
 6 x 2  2 x  20  14 x 2  18 x  20
A1
 6 x 2  0 x  24  14 x 2  28 x
 2x  4 10 x  20
10 x  20
4x  6
x  2 4 x 2  14 x  20
4 x3  6 x 2  18 x  20
x2  4 4 x2  8 x2
M1
2 x  4  6 x  20
 4x  6 
( x  2)( x  2)  6 x  12
2
2
 4x  6  A1
( x  2)
(4)
Way 3  
4 x3  6 x 2  18 x  20   ax  b  x 2  4  c( x  2) o.e. 1st M1
Either substitutes/ and or equates coefficients to find a value for a, b or c 2nd M1
One of a  4, b  6, c  2 1st A1
All of a  4, b  6, c  2 2nd A1
(4)
(4 marks)
Notes
Way 1
M1 Divides 4 x3  6 x2 18 x  20 by x 2  4 to get a linear quotient and a linear remainder.
To award this look for a minimum of the following
4 x+ A
x 2 ( 0 x )  4 4 x3  6 x 2  18 x  20
4x 3 + 0 x 2  16 x
 Cx  + D
A1 Quotient = 4 x  6 and Remainder = 2 x  4
M1 Writes their expression in the appropriate form. Allow a slip with sign if the intention is clear.
 4 x3  6 x2  18 x  20  Their linear remainder
   Their Linear Quotient  and factorises x 2  4
 x 4
2
 x 4
2

Their linear remainder

into x 2 x 2 . This may be in one line ‘ Their Linear Quotient  ‘
 x  2  x  2 
Their linear remainder
This may be seen as Their Linear Quotient  followed by writing
x2  4
Their linear remainder
separately
 x  2  x  2 
Question Scheme Marks
2
A1 All values correct a  4, b  6 and c  2 or writes 4 x  6 
( x  2)
Way 2

M1 Divides 4 x3  6 x2 18 x  20 by ( x  2) to get a quadratic quotient and a constant remainder.

To award this look for a minimum of the following
4 x 2 + Ax + B
x  2 4 x  6 x 2  18 x  20
3

4x 3 + 0 x 2  16 x
D
A1 Quotient = 4 x 14 x  10 and Remainder = 0
2

M1 Divides their ‘ 4 x 2  14 x  10 ‘ by ( x  2) to get a linear quotient and a constant remainder.

To award this look for a minimum of the following
4x  E
x  2 4 x  14 x  20
2

4 x2  8 x2

F
2
A1 All values correct a  4, b  6 and c  2 or writes 4 x  6 
( x  2)
Way 3
1st M1 Forms the correct identity by muliplying through by x 2 4
2nd M1 Either equates coefficents and/or substitutes a value for x in an attempt to find a value for
either a, b or c
1st A1 At least one correct value a  4, b  6 or c  2
2
2nd A1 All values correct a  4, b  6 and c  2 or writes 4 x  6 
( x  2)
Question Scheme Marks

2(i)(a) (2 x  1)3 dy   3x  2  (2 x  1)  (2 x  1)

2 3

y   where   0 and   0
 3 x  2  dx  3x  2 
2

OR M1
dy
y  (2 x  1)  3x  2   (2 x  1) 2  3 x  2     3 x  2  (2 x  1) 3
1 1 2
3

dx
where   0 and   0
(2 x  1)3 dy  3x  2   6(2 x  1)  (2 x  1)  3
2 3

y  
 3 x  2  dx  3x  2 
2

OR A1
dy
y  (2 x  1)3  3x  2    3  2  (2 x  1) 2  3 x  2   3  3 x  2  (2 x  1) 3
1 1 2

dx
dy (2 x  1) 2 18 x  12  6 x  3
 M1
 3x  2 
2
dx
dy (2 x  1) 12 x  9  3(2 x  1) 2  4 x  3
2
or (2 x  1) 2 12 x  9  3x  2 
2
 or
 3x  2   3x  2 
2 2
dx A1
o.e
(4)
(i)(b) dy
0  either their 12 x  9  0 x or 2 x  1  0  x  ... M1
dx
1 3
x , x
2 4
(2)
Way 1
(ii) dy  sin 2 x M1
y  ln(1  cos 2 x)  
dx 1  cos 2 x
dy 2sin 2 x
 A1
dx 1  cos 2 x
dy 4sin x cos x
 M1
dx 2 cos 2 x
dy
 2 tan x A1
dx
(4)
Way 2 Careful with the order of the Method marks
y  ln(1  cos 2 x)  y  ln 2 cos 2 x  2nd M1

dy 4sin x cos x 1st M1

 
dx 2 cos 2 x 1st A1
dy
  2 tan x 2nd A1
dx
(4)

(10 marks)
Question Scheme Marks
Notes
(i)(a)

M1 Uses either quotient rule or product rule to the achieve the correct form.
dy
A1 Correct unsimplified or simplified expression for
dx
M1 Quotient Rule: Takes out a common factor of at least (2 x  1)2 from the numerator, allow
numerical slips but not algebraic slips, as long as the intention is clear.
Product Rule: Combines as a single fraction with a correct numerator allow numerical slips in
the denominator AND takes out a common factor of at least (2 x  1)2 from the numerator, allow
numerical slips as long as the intention is clear.
dy (2 x  1) 12 x  9 
2
dy
 (2 x  1) 2 12 x  9  3 x  2 
2
A1  o.e. such as
 3x  2 
2
dx dx

(i)(b)
dy dy dy
M1 Sets their 0 or  0 or  0 and proceeds correctly to find a critical value for their
dx dx dx
dy
 0 provided that their numerator is at least a quadratic. Condone setting equal to 0 and
dx
cancelling a common factor first as long as a value for x is achieved.
1 3
A1 x ,x
2 4

(ii)
Way 1
dy  sin 2 x
M1 Differentiates to a form 
dx 1  cos 2 x
dy 2sin 2 x
A1 A correct derivative 
dx 1  cos 2 x
M1 Uses the correct double angle identities sin2x = 2sinxcoss sin 2x  2sin x cos x and
cos 2 x  2 cos 2 x  1
If uses cos 2 x  cos 2 x  sin 2 x do not award this mark until either 1  sin 2 x becomes cos 2 x or
1  cos 2 x  sin 2 x becomes 2 cos 2 x in the denominator
dy
A1 Achieves  2 tan x with no incorrect work seen
dx
Way 2 Careful with the order of the Method marks

2nd M1 Replaces 1  cos 2x with 2 cos 2 x

dy f '  x 
1st M1 Differentiates y  ln  f  x     with their f ( x )
dx f  x 
dy 4sin x cos x
1st A1 Correct derivative 
dx 2 cos 2 x
dy
2nd A1 Achieves  2 tan x with no errors or omissions
dx
Question Scheme Marks

3.(a) R  65 B1
8
tan      awrt 82.87 M1A1
1
(3)
'R'
(b) 13   13.81 C  M1 A1
10
(2)
5
(c) cos 15t  82.87     M1
65
15t  82.87  128.33  t  3.03 A1

15t  82.87   360  128.33  t  ... 9.92 dM1

Both times correct 03: 02 and 09: 55 A1
(4)
(9 marks)
Notes
(a)
B1 Sight of R  65 . Condone R   65
(Do not allow decimals for this mark e.g. 8.06 but remember to isw after 65 )
8 1
M1 For sight of tan      =... or tan       ...
1 8
Condone either sin   8, cos   1  tan   8    ... or using  instead of 
8 1
If R is found first accept only sin    , cos       ...
R R
A1   awrt 82.87 Answer in radians (1.45) are A0
(b)
'R'
M1 For their 13 
10
A1 awrt 13.81 C

(c)
'R'
M1 Sets 13  cos 15t  '  '   12.5 with their values for R and  and rearranges to achieve
10
cos 15t  '  '  k where 1  k  1 . Condone starting from 13  ' R 'cos 15t  '  '  12.5

cos    '  '   12.5 leading to cos    '  '  k

'R'
Allow for 13 
10
A1 For one correct value of t. Accept either awrt t  3.03 or t  9.92
dM1 Dependent on the first method mark. For the correct method to find a second value for t.
Look for 15t  ''82.87''   360  ''128.33''  t  ..
A1 Both times correct 03: 02 and 09: 55

Note 1: Starting with 13  ' R 'cos 15t  '  '  12.5 can score a maximum of M1 A0 dM1 A0. It should
lead to t  0.712 and t  12.238

Note 2: Alpha in radians can score a maximum of M1 A0 dM1 A0. It should lead to t  0.012 and
t  0.213
Question Scheme Marks

 2x  5 
2 5
 x2 
4(a) ff(x)  M1
 2x  5 
 2
 x2 
2  2 x  5  5  x  2 
ff(x)  x dM1, A1
 2 x  5  2  x  2 
(3)
2 ln a  5
(b) Sets fg(a)  g(a)   ln a M1
ln a  2
  ln a   4 ln a  5  0
2
A1

  ln a  5 (ln a  1)  0  ln a  5, 1 dM1

 a  e5 , e 1 A1

(4)
(7 marks)

Notes
(a)
 2x  5 
2 5
 x2 
M1 Attempts ff(x)  condoning sign and numerical slips
 2x  5 
 2
 x2 
9 9
Alternatively writes f(x)  2  and hence ff(x)  2 
x2 9
2 2
x2
p
dM1 Correct processing to obtain a single fraction of the form . Achieved by either,
q
 multiplying all terms in both the numerator and the denominator by  x  2 
 attempting to write both the numerator and denominator as a single fraction followed by the
p
multiplication of the numerator by an inverted denominator to obtain a single fraction of the form
q
 attempting to write both the numerator and denominator as a single fraction followed by the
p
cancelling of the same denominators to obtain a single fraction of the form
q
A1 ff(x)  x
Question Scheme Marks

(b) Condone the use of x instead of a for the first 3 marks

2 ln a  5
M1 Sets fg(a)  g(a)   ln a
ln a  2
A1 Correct simplified quadratic equation in lna. Note ln 2 a  4 ln a  5  0 is fine
Condone poor notation ln a 2  4 ln a  5  0 if an attempt to solve as a quadratic equation for lna
dM1 Correct attempt to find a value for lna by solving 3TQ in lna
A1 a  e5 only, the solution a  e 1 must be rejected. A0 for x  e5
Question Scheme Marks

y
5(a)
V shape on the +ve x axis B1

 0, a   0, a  and 
a 
,0 B1
3 

x
O a 
 ,0
3 
(2)
(b) 1 1
Substitutes x  4 into 3x  a  x  2  3 4  a   2  2 M1
Way 1 2 2
Solves 12  a  4  a  8, 16 dM1 A1
(3)
Substitutes x = 4, squares both sides and forms a 3TQ in a
2

 3x  a    x  2   4a 2  96a  512  0
Way 2 2 1 M1
2 
Solves 3TQ to find values for a dM1
a  8, 16 A1
(3)
1
Sets   3x  a   x  2 and substitutes x  4 M1
2
Rearranges an equation to find a value for a dM1
a  8, 16 A1
(3)
1
(c) Chooses larger value of 'a’ solves 3x  a  x  2  x  ... M1
2
36
x or 7.2 A1
5
(7 marks)
Notes
(a)

B1 For a V shape on the positive x - axis in quadrants one and two. It must clearly pass through
the y - axis
a  a
B1 Points  0, a  and  , 0  both lie on the graph. Allow a on the y – axis and on the x - axis
3  3
Question Scheme Marks

(b)

Way 1
1
M1 Scored for setting 3x  a  x  2 and substituting in x  4
2
Implied by 12  a  4 , or 12  a  4 or 12  a  4

dM1 An acceptable method of finding one value of a

A1 Both a  8, 16

Way 2
M1 Substitutes x = 4 and squared both sides in either order to form a 3TQ in a

dM1 Solve their 3TQ to find a value for a

A1 Both a  8, 16

Way 3 (See if Way 1 is more relevant)

1 1 1
M1 Sets 3x  a  x  2 and either 3 x  a  x  2 or 3x  a   x  2 and substitutes in x  4
2 2 2

dM1 Rearranges an equation to find a value for a

Note: If they rearrange to find a = … then substitutes in x = 4 both M’s awarded at this point.

A1 Both a  8, 16

(c)
1
M1 Chooses the larger value of 'their a’ and solves 3x  a  x  2  x  ..
2
36
A1 x or 7.2
5
Note: If they use both values of their a then M1 and/or A1 is awarded when the largest value of x
following their values of a is selected.
Question Scheme Marks


d x 2  x  12   ln  x  3  d  ln  x  3  
6.(a) f ( x) 
dx dx
x 2
 x  12  M1


f ( x)  ln( x  3)  2 x  1  x 2  x  12   1
x3 A1
or f ( x)  ln( x  3)  2 x 1  x  4
(2)
1
(b) ln( x  3)  2 x  1   x  4)( x  3  0 M1
x3
 ln( x  3)  2 x 1   x  4  0

 2 x ln( x  3)  x  4  ln( x  3) dM1

ln( x  3)  4
 x  2ln( x  3)  1  4  ln( x  3)  x  * A1 *
2ln( x  3)  1
(3)
4  ln( x  3) 4  ln(4)
(c) Substitutes x0  1 in x   x1   awrt 1.428 M1, A1
2ln( x  3)  1 2ln(4)  1
x2  awrt 1.38  0 , x3  awrt 1.385 A1
(3)
(d) k   2  f (0)  k  24 ln 3 M1 A1

(2)
(10 marks)
Notes
(a)
M1  
Applies correctly the product rule to f ( x)  x 2  x  12 ln( x  3) . If they state
u  ...  u '  ... and v  ...  v '  ... follow through on their u '  ... v '  ... as long and u and v
are correct.
A1 
f ( x)  ln( x  3)  2 x  1  x 2  x  12  1
x3
correct un-simplified or simplified.
Award as soon as a correct version is seen, isw
Must have correct notation e. g. ln x  3 is A0

(b)
M1 Sets f ( x)  0 and attempts to factorise x 2

 x  12 which may have already been done in part
(a)
dM1 Rearranges to an equation of the form .. x ln( x  3)  .. x  ..  ..ln( x  3)
A1* Factorises and divides to form the given equation with no errors or omissions or poor notation
Question Scheme Marks

(c)
4  ln( x  3) 4  ln(4)
M1 Substitutes x0  1 in x  . Implied by x1  or awrt 1.4
2 ln( x  3)  1 2 ln(4)  1
A1 awrt 1.428
A1 x2  awrt 1.38  0 , x3  awrt 1.385

(d)
M1 k   2  f (0) This mark can be implied by seeing k = awrt 26.4 with no working seen
A1 k  24ln3
Question Scheme Marks

7(a) 2cos( A  30°)sec A  2  cos A cos30  sin A sin 30  sec A M1

Way 1
2  cos A cos 30  sin A sin 30 
 ...  tan x  k dM1
cos A
 tan A + 3 cso A1*
Way 2 2cos( A  30°) sec A  tan A + k  2cos( A  30°)  sin A  k cos A
 2  cos A cos30  sin A sin 30  sin A  k cos A M1
 3 cos A  sin A  sin A  k cos A dM1
Hence true and k  3 A1
(3)
(b)
2cos ( x  30°) = sec x and 2cos ( x  30°)sec x  tan A + 3
Way 1 For example
1) 2cos( x  30°)sec x = sec2 x  tan x+' 3' = sec2 x
OR
tan x  ' 3 '
2) = sec x  tan x+' 3' = sec2 x
sec x
OR
2cos ( x  30°)   tan A + ' 3 '  cos x  sec x   tan A + ' 3 '  cos x M1
3)
 tan x +' 3' = sec 2 x
OR
4) 2cos( x  30°) = sec x  2  cos x cos30  sin x sin 30  sec x
 2 cos 2 x cos 30  2sin x cos x sin 30  1
 cos2 x 5  sin x cos x  1  5  tan x  sec2 x
 tan 2 x  tan x+1  3 = 0 M1A1

1  1  4 1  3 
tan x   awrt 1.49,  0.49  x  ... M1
2
x  awrt 56.2,  26.2 A1
(5)
(8 marks)
Alt (b) 2cos( x  30°) = sec x  2  cos x cos30  sin x sin 30  sec x

 2 cos 2 x cos 30  2sin x cos x sin 30  1

 3  cos 2 A  1  1sin 2 A  2 M1

 2cos  2 A  30  2  3 M1A1

x  awrt 56.2,  26.2 M1, A1

(5)
 Question Scheme Marks
Notes
(a)
M1 Uses cos( A  30°)  cos A cos 30  sin A sin 30 (Condone a slip with a missing 2)
1
dM1 Uses sec A  (maybe implied) and divides with an intermediate line to reach tan A + k
cos A
A1 Achieves tan A + 3 with no errors or poor notation cso

(b)
M1 Uses part (a) and the equation correctly to reach the form tan x+'' 3'' = sec2 x
M1 Uses sec2 x  1  tan 2 x to form a quadratic in tan x , does not need to be collected onto one
side.
tan 2 x  tan x+1  3 = 0 or equivalent including tan x  tan x  awrt 0.73 = 0
2
A1
M1 Solves quadratic and proceeds to one value for x.
A1 x  awrt 56.2,  26.2
Question Scheme Marks

8 (a) Substitute N A  7200 in N A  3000  600e0.12t B1

e0.12t 7 M1 A1
ln 7
t 16.22 years M1, A1
0.12
(5)
dN A  dN A 0.12t 
(b) Differentiates to achieve  e0.12t  dt  600  0.12e  M1
dt  
N  3000 dN A
Substitutes e0.12t  A into  e0.12t
600 dt
dM1
OR
dN A
Substitutes 600e0.12t  N A  3000 into    600e0.12t
dt

 0.12  N A  3000  0.12 N A  360 or

dN A 3 A1
 N  360
dt 25 A
(3)
(c)(i) 200  Ce k and 500  Ce 2k M1
500 200
ek   k  ... or e-k   k  ... dM1
200 500
5
k  ln   * cso A1*
2
(ii) 80 B1
(4)
(12 marks)
Notes
(a)
B1 7200  3000  600e0.12t
M1 Moving from N A  3000  600e0.12t with a numerical N A and using the correct order of
operations and algebra to achieve an equation e0.12t λ where   0
A1 Achieves e0.12t 7 from correct work
ln 
dM1 Moving from e0.12t λ using lns to t  where   0
0.12
A1 cso 16.22 years

Note: If they work with 2 N A do not award any marks until a value for N A is substituted. If they
achieve the correct answer then all marks can be awarded. If they do not achieve the correct
answer B0 M1 A0 dM1 A0.

Special Case: A student who starts with N A  6000 and achieves t  13.41 years can score a
maximum B0, M1, A1 e0.12t 5 , M1, A0 for 3 out of 5
Question Scheme Marks

(b)
dN A
M1 Differentiates to achieves the correct form  e0.12t
dt
dM1 Rearranges the equation N A  3000  600e0.12t to e0.12t  ... and substitutes into
dN A
 e0.12t
dt
dN A
or rearranges to 600e0.12t  ... and substitutes into    600e0.12t
dt

 3000  600e0.12t  and substitutes

3 dN A
Alternatively writes  e0.12t to achieve
25 dt
  3000  600e0.12t   
dN A 3
dt 25
dN A 3
A1  0.12 N A  360 or N  360 o.e.
dt 25 A

(c)(i)
M1 States 200  Ce k and 500  Ce 2k
dM1 Uses a correct method to eliminate C from their equations to achieved either
500 200
e k  '' ''  k  ... or e-k   k  ...
200 500
5
A1* Achieves k  ln   with no errors or omissions or poor notation. cso
2

(c)(ii)
B1 80
Question Scheme Marks

1 dx 1 ...
9(a) x    ... OR OR ...(1  cot y)2 M1
Way 1 1  cot y dy 1  cot y  2
(1  cot y ) 2

1 dx 1 dx 1 cosec2 y
x     cosec 2
y OR  A1
1  cot y dy 1  cot y 2 dy 1  cot y 
2

1 1 x
cot y   1 or cot y  B1
x x
1
Uses cosec2 y  1  cot 2 y and cot y   1 to eliminate y and achieve
x
M1
dx
 f  x
dy
dx
 2 x2  2 x  1 * A1*
dy
(5)
1 1 1 x
Way 2 x  cot y   1 or cot y  B1
1  cot y x x
dy 1
Differentiates  ...   2 M1
dx x
dy 1
cosec 2 y   2 A1
dx x
1
Uses cosec2 y  1  cot 2 y and cot y   1 to eliminate y and achieve
x
M1
dx
 f  x
dy
dx
 2 x2  2 x  1* A1*
dy
(5)
1 1 x
Way 3 x  cot y   1  tan y  B1
1  cot y x 1 x
dy 1
Differentiates  ...   M1
dx 1  x 2
dy 1
sec 2 y   A1
dx 1  x 2
dx
Uses sec2 y  1  tan 2 y to eliminate y and achieve  f  x M1
dy
dx
 2 x2  2 x  1* A1*
dy
(5)
1 1
(b) x  B1
1 3 4
(1)
Question Scheme Marks

1 dx
(c) Subs their x  into their   2 x 2  2 x  1 and INVERTS M1
4 dy
dy 8
 1.6 or A1
dx 5
(2)
(8 marks)

(a)
Way 1
1
M1 Attempts to differentiate using the chain rule to reach a form on the rhs  ... or
(1  cot y)2
...
quotient rule to reach the form or ...(1  cot y)2
(1  cot y ) 2

dx 1 dx 1 cosec2 y
A1 Correct differentiation    cosec 2
y or 
dy 1  cot y 2 dy 1  cot y 
2

1 1 x
B1 States cot y   1 or cot y 
x x
1
M1 Attempts to use cosec2 y  1  cot 2 y with cot y   1 to eliminate y and find an expression for
x
2
1 
1    1
  2 
dx dx x
in terms of x only e.g.
dy dy 1
 
x
2 2
1  1  1
Note: cot 2 y    1 cosec 2 y  1    1 1  cot y 
x  x  x
dx
A1* Achieves the printed answer  2 x 2  2 x  1 with no errors or omissions
dy

Way 2
1 1 x
B1 States cot y   1 or cot y 
x x
dy 1
M1 Using implicit differentiation to reach  ...   2
dx x
dy 1
A1 Correct differentiation cosec 2 y   2
dx x
1
M1 Attempts to use cosec2 y  1  cot 2 y with cot y   1 to eliminate y and find an expression for
x
dx dx   1 2 
in terms of x only e.g.    1    1    x 2
dy dy  x  
 

dx
A1* Achieves the printed answer  2 x 2  2 x  1 with no errors or omissions
dy
Question Scheme Marks

Way 3

x
B1 tan y 
1 x
dy 1
M1 Using implicit differentiation to reach  ...  
dx 1  x 2
dy 1
A1 Correct differentiation sec2 y  
dx 1  x 2
x
M1 Attempts to use sec2 y  1  tan 2 y with tan y  to eliminate y and find an expression for
1 x
dx   x  
2
dx
 1  x 
2
in terms of x only  1    
dy dy   x  1  

dx
A1* Achieves the printed answer  2 x 2  2 x  1 with no errors or omissions
dy

(b)
1
B1 States x  o.e.
4
(c)
1 dx
M1 Subs their x  into   2 x 2  2 x  1 and INVERTS
4 dy
Note: If they do not show you the substitution you will need to check their value of x.

dy 8
A1  or 1.6 cso
dx 5

3 dy 8
Note: If part (b) x  this still leads to  this scores M1 A0
4 dx 5
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