EDEXCEL FURTHER PURE MATHEMATICS FP2 (R) (6668) – JUNE 2014
Question
Number
FINAL MARK SCHEME
Scheme
Marks
1.
(a)
2
4r  1
2

A
B

2r  1 2 r  1
2  A  2r 1  B  2r  1  A  1, B  1
2
4r  1
2

1
1

2r  1 2r  1
M1A1
(2)
n
1
1 
 1





2
2r  1 
r 1 4 r  1
r 1  2 r  1
n
(b)
(2)
1 1 1 1 1
1
1
1
 1       ... 

 1
3 3 5 5 7
2n  1 2n  1
2n  1

M1A1ft
2n  1  1
2n  1
n
 4r
r 1
1
2
1

n
2n  1
*
A1
(3)
[5]
2
3x  5 
2
0
x
3x 2  5 x  2
0
x
 3x  1 x  2   0
x
(or <)
or mult through by x2
(or <)
or x  3x  1 x  2  0
1
CVs x   , 2
3
x=0
1
x   , 0  x  2 or in set language (with curved brackets for A1)
3
Special case If „ used deduct final mark only.
1
M1
A1
B1
M1A1
(5)
[5]
EDEXCEL FURTHER PURE MATHEMATICS FP2 (R) (6668) – JUNE 2014
Question
Number
FINAL MARK SCHEME
Scheme
Marks
3.
(a)
dy
 2 y tan x  e 4 x cos 2 x
dx
2 tan x dx
e
 e 2ln sec x  sec2 x or
sec 2 x
1
cos2 x
M1A1
dy
 2 y tan x sec 2 x  e 4 x cos 2 x sec 2 x
dx
d
y sec 2 x   e 4 x

dx
1
y sec 2 x  e 4 x  c 
4
1

y   e 4 x  c  cos 2 x
4

dM1
B1ft( y sec2 x )
M1
oe
A1
(6)
(b)
1

y  1, x  0 1    c 
4

c
3
4
y
1 4x
e  3 cos 2 x oe

4
M1
A1
(2)
[8]
2
EDEXCEL FURTHER PURE MATHEMATICS FP2 (R) (6668) – JUNE 2014
Question
Number
4.
FINAL MARK SCHEME
Scheme
Marks
( y )r sin   2 cos 2 sin 
dy
 4sin 2 sin   2 cos 2 cos 
d
2sin 2 sin   cos 2 cos  0
4sin 2 cos   1  2sin 2   cos   0
 6sin
2
M1
M1A1
dM1
  1 cos   0
 cos  0
no solutions in range 
1
6
r sin   2cos 2 sin 
1
1 2
sin  
 cos 2  1  2sin 2   1  2  
6 3
6
2 1
4
Eqn. l: r sin   2  

3
6 3 6
2 6
r
co sec  oe  0     
Must be seen in exact form
9
 sin  
ddM1A1
M1
M1
A1
[9]
5.
2
(a)
d2 y
2  dy 
    2
2
dx
y  dx 
2
 dy  d y seen
  2
 dx  dx
B1
d3 y
4  dy  d 2 y 2  dy 


  
 
dx3
y  dx  dx 2 y 2  dx 
(b)
At x  0
M1 (  and
diff)
A1A1
(4)
3
2

d2 y 1 
9
1
  2     4   
2

dx
2 
4
2

 or  2.25
d3 y 1 
1
9
1  37
  5     2   
3
dx
2
2
4
2  16
2
3
1
 9  x  37  x
y  2  x         ...
2
 4  2!  16  3!
y  2
1.125
0.3854
2.325
A1
M1(2! or 2, 3!
or 6)
1
9
37 3
x  x2 
x  ...
2
8
96
0.5
 or
M1A1
A1
3 sf or better
3
(5)
[9]
EDEXCEL FURTHER PURE MATHEMATICS FP2 (R) (6668) – JUNE 2014
Question
Number
6. (a)
FINAL MARK SCHEME
Scheme
Marks
w  u  i 
x  iy
ix  y  1
M1
w  u  i 
1  y  ix 
x  iy

ix  y  1 1  y  ix 
M1
w  u  i 
1 
x  xy  xy  i  y  y 2  x 2 
1  y 
2
A1
 x2
y  y 2  x2
1  y 2   x2
M1
1  2y  y 2  x 2  y  y 2  x 2
(b)
y 1
A1
1
1
x i
x i
2
2
u  iv 

1
1 

i  x  i   1 ix 
2
2 

M1
1 1
x i
 xi
2 2
u  iv 
1 1
ix 
 xi
2 2
M1
1

x  i   x2 
4

u  iv 
1 2
x
4
A1
u
1 2
x
4
u v 
2
1 2
x
v 4
1 2
x
4
x
2
x 2   14  x 2 

1
4
 x2 
2
M1
2

u 2  v 2  1 , Centre O
(5)
1
16
 12 x 2  x 2  x 4

1
4
 x2 
2
*
1
M1
M1,A1
(6)
[11]
4
EDEXCEL FURTHER PURE MATHEMATICS FP2 (R) (6668) – JUNE 2014
Question
Number
7. (a)
FINAL MARK SCHEME
Scheme
 cos   i sin  
5
Marks
 cos5  i sin 5
B1
5 4 3
2
cos   i sin  
2!
5 4 3 2
5 4  3 2
3
4
5

cos   isin  
cos   i sin     i sin  
3!
4!
 cos5   5cos 4  i sin   
 cos5   5i cos 4  sin   10cos3  sin 2 
M1
A1
10i cos 2  sin 3   5cos  sin 4   i sin 5 
sin 5  5cos 4  sin   10cos 2  sin 3   sin 5 
 5 1  sin 2   sin   10 1  sin 2   sin 3   sin 5 
2
*
sin 5  16sin 5   20sin 3   5sin 
(b)
1
1
 sin 5  
2
2
5  210, 330, 570, 690, 930, 1050, 1290 (or in radians)
Or 210, 570, 930, 1290, 1650
Let x  sin 
16 x5  20 x3  5 x  
  42, 66, 114 , 138 , 186, 210, 258 (or in radians)
M1
A1
(5)
M1
A1, A1
dM1(at least 2
values)
Or 42, 114, 186, 258, 330
sin   0.669, 0.914,  0.105,  0.5,  0.978
(c)

4
0

4
0
1
  4sin   5sin   d  4  sin 5  5sin   d
5
3
A1
(5)
M1

1 1
4
   cos 5  5cos  
4 5
0
A1
1 1
5
  1 
 cos  5cos     5  

4 5
4
4  5 
1 1 1
5
4
  

4 
4 5 Ö 2 Ö 2
5
13 Ö 2 6


20
5
M1
A1
(4)
[14]
5
EDEXCEL FURTHER PURE MATHEMATICS FP2 (R) (6668) – JUNE 2014
Question
Number
8. (a)
Scheme
Marks
x  ez
dx
dz
 ez
dy
dy
M1
dy
dy
 e z
dx
dz
A1
d2 y
dy  z d 2 y dz 1  dy d 2 y 
 z dz


e

e
   

dx 2
dx dz
dz 2 dx x 2  dz dz 2 
M1A1A1
x2
d2 y
dy
 2 x  2 y  3ln x
2
dx
dx
 1 dy 1 d 2 y 
1 dy
x2   2
 2 2   2x 
 2 y  3z
x dz
 x dz x dz 
(b)
FINAL MARK SCHEME
d 2 y dy
  2 y  3z
dz 2 dz
Aux eqn: m 2  m  2  0
M1
A1
(7)
 m  2 m 1  0
m  2, 1
M1A1
y  Ae2 z  Be z
CF:
PI: Try
A1
y  az  b
dy
a
dz
d2 y
0
dz 2
a  2  az  b   3z
3
3
a , b
2
4
Complete soln:
(c)
M1
3
3
y  Ae 2 z  Be z  z 
2
4
3
3
y  Ax 2  Bx  ln x 
2
4
A1A1
(6)
B1 ft
(1)
[14]
6