EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
1.
(a)
Scheme
FINAL MARK SCHEME
Marks
f ( x)  2x3  6x2  7 x  4
f (4)  128  96  28  4  0
B1
[1]
(b)
f (4)  0  ( x  4) is a factor.
f ( x)  ( x  4)(2x2  2x  1)
So, x 
2 
M1 A1
4  4(2)(1)
M1
2(2)

1
  0   2   x    k 
2
2


 2  2i
 x  4,
4
 2   x 2
x
2
1
1
 k  0 x 
2
A1
[4]
2. (a)
1
 3 1 3

A
, B  1
4
5
5


0

1

2
1 
 31 0 3 23
 

 4 + 5 + 0 4  10  5 
M1
 4 2


9 9
A1
[2]
(b)
3 2
 5 2k 
C
, D  
 , where k is a constant,
8 6
4 k 
 3 2   5 2k 
 8 2k  2 
CD 

  

8 6  4 k 
 12 6  k 
M1
E does not have an inverse  det E  0.
8(6  k )  12(2k  2)
M1
8(6  k )  12(2k  2)  0
48  8k  24k  24
24  16k
k  32
M1
A1 oe
[4]
6 marks
1
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
3.
Scheme
f ( x)  x 2 
3
 3x  7 , x  0
4Цx
f ( x)  x 2 
3  12
x  3x  7
4
f ( x)  2 x 
3  32
x  3  0
8
f (4)   2.625  
or 42 
3
FINAL MARK SCHEME
Marks
M1A1
21
5
 2
8
8
B1
 3 4  7
4 4
f (4)  4.953125 
317
61
4
64
64
M1
 "  2.625" 

 "4.953125" 
2  4  
 4.529968454...
M1
 1436

 4 168

317 
317


 4.53 (2 dp)
A1 cao
[6]
6 marks
2
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
Scheme
n
4. (a)
r
3
FINAL MARK SCHEME
Marks
 6r  3
r 1

1 2
1
n (n  1) 2  6. n (n  1)  3n
4
2
M1A1B1
1 2
n (n  1)2  3n2
4
1
 n2  (n  1)2  12 
4


dM1
1 2 2
n  n  2n  13 (AG)
4
A1 *
[5]
(b)
Sn 

1
4
 r
3
 6r  3  S30  S15
 30
2
 2(30)  13 
30
r  16
 30 
2
1
4
15
2
15
2
 2(15)  13
 203850
M1
A1 cao
[2]
7 marks
3
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
5.
(a)
Scheme
C : y 2  8x 
a
8
4
FINAL MARK SCHEME
Marks
2
PQ  12  By symmetry yP 
12
2
6
B1
[1]
(b)
y 2  8x 
 x
36
8

62  8x
M1
9
2
(So P has coordinates
 92 , 6 
A1 oe
)
[2]
(c)
Focus S (2, 0)
Gradient PS 
B1
 6

60 
12 
 5 

9
2 
5
 2
2

M1
Either
y 0 
12
5
( x  2) or y  6 
12
5
( x  92 ) ;
M1
or y  125 x  c and
0
l:
12
5
 2  c
 c   245 ;
12 x  5 y  24  0
A1
[4]
7 marks
4
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
Scheme
 x
f ( x)  tan    3x  6,
2
6.
(a)
FINAL MARK SCHEME
Marks
  x  
f (1)   2.45369751...
f (2)  1.557407725...
M1
Sign change (and f ( x) is continuous) therefore a root
 is between x  1 and x  2.
A1
[2]
(b)
 1
"2.45369751..."

2
"1.557407725..."
or
M1
"2.45369751..."  "1.557407725" "2.45369751..."

1
 1


"2.45369751..."
1
 "1.557407725..."  "2.45369751..." 
 1 

6.464802745
4.011105235
 1.611726037...
A1
A1
[3]
5 marks
5
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
7. (a)
Scheme
arg z   tan 1
 
3
2
(b)

Marks
M1
  0.7137243789..   0.71 (2 dp)

z  2i 3 2i 3
2
FINAL MARK SCHEME
A1
[2]

M1
 4  2i 3  2i 3  3i 2


3  1  4i 3 
 2  i 3  4  4i 3  3
2i
M1A1
(Note: a  3, b   5.)
 3  5i 3
[3]
(c)
z7 2i 3 7

z 1 2  i 3 1



9  i 3   1  i 3 
1  i 3  1  i 3 
9  9i
3 i
dM1
3 3
13
12  8i
M1
3
4
 3  2i
(d)
M1
3
(Note: c  3, d  2.)
w    3i , and arg(4  5i  3w)  
A1
[4]

2
4

5
i

3
w

4

3


14i


So real part of (4  5i  3w) = 0 or 4  3   0
So,    43
M1
A1
[2]
11 marks
6
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
8.
(a)
Scheme
xy  c 2 at  ct ,
y
FINAL MARK SCHEME
Marks
c
t
.
dy
c2
c2
 c 2 x 1 
  c 2 x 2   2
x
dx
x
xy  c 2  x
dy
y0
dx
M1
dy dy dt
c 1
 .  2.
dx dt dx
t c
dy
dy
dy c 1
 c 2 x 2 or x  y  0 or

.
dx
dx
dx t 2 c
or equivalent expressions
A1
So, mT 
y
c
1
  2  x  ct 
t
t
( t 2 )
dy
1
 2
dx
t
M1
x  t2 y  2 c t
A1 *
(Allow t 2 y  x  2ct )
[4]
(b)
y  0  x  2 c t  A( 2 c t , 0).
B1
2ct
2c 

 B  0,
.
2
t 
t

B1
x0  y
Area OAB  36 
2c
1
 2 c t     36
2
 t 
 2c 2  36  c 2  18  c  3
M1
2
A1
[4]
8 marks
7
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
Scheme
det M  3( 5)  (4)(2)  15  8   23
9. (a)
FINAL MARK SCHEME
Marks
B1
[1]
(b)
4   2 a  7   25 
3
Therefore, 



 2  5   a  1    14 
M1
Either, 3(2a  7)  4(a  1)  25 or
2(2a  7)  5(a  1)  14
 3  2a  7   4  a  1 
or  2  2a  7   5  a  1  


A1
 25 


  14 
giving a  5
A1
[3]
1
 6 4 ;  12 (units)2
2
(c)
Area(ORS ) 
(d)
Area(OR S )   23  (12)
M1A1
[2]
M1
A1
[2]
(e)
(f)
Rotation; 90 anti-clockwise (or 270 clockwise) about  0, 0  .
B1;B1
M  BA
M1
A 1 
[2]
 0 1
 0 1
1

;  

(0)(0)  (1)(1)   1 0 
  1 0
A1
4   0 1
3
B


 2  5   1 0
M1
  4 3
B

 5 2
A1
[4]
14 marks
8
EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) – JUNE 2012
Question
Number
Scheme
10.
f (n)  22n 1  32n 1 is divisible by 5.
FINAL MARK SCHEME
Marks
f (1)  21  31  5,
B1
Assume that for n  k ,
f (k )  22k 1  32k 1 is divisible by 5 for k 

.
f (k  1)  f (k )  22( k  1)  1  32( k  1)  1   22 k  1  32 k  1 
M1A1
 22k  1  32k 1  22k 1  32k 1
 22k  1  2  32k 1  2  22k 1  32k 1
 4  22 k  1   9  32 k  1   22k  1  32k  1
M1
 3  22 k  1   8  32 k  1 
 3  22 k  1   3  32 k  1   5  32 k  1 
 3f (k )  5  32 k  1 
 f (k  1)  4f ( k )  5  32 k  1  or
A1
4(22 k 1  32 k 1 )  5  32 k  1 
If the result is true for n  k , then it is now true for n  k 1. As the result has shown to
be true for n  1, then the result is true for all n.
A1 cso
[6]
6 marks
9