June 2006
6675 Further Pure Mathematics FP2
Mark Scheme
Question
number
1.
Scheme Marks
 e x  ex   e x  ex
  2
5
2
2

 
2x
x
3e  22e  7  0
(3e x  1)(e x  7)  0
x  ln
1
or  ln 3
3

  11

M: Simplify to form quadratic in e x
1
e x  , e x  7 M: Solve 3 term quadratic.
3
B1
M1 A1
M1 A1
A1
x  ln 7
(6)
6 Marks
2.
(a)
Using b2  a 2 (1  e2 ) or equiv. to find e or ae: (a = 2 and b = 1)
M1 A1
3
2
Using y 2  4(ae) x
M1 A1 (4)
e
(b)
y 2  4 3 x (M requires values for a and e)
x 3
B1ft
(1)
5 Marks
3.

ds
d
s = 0 at  = 0:
s   e sin cos d  e sin
 k 
M1 A1
(M mark may be scored by a full substitution method)
M1 A1cso
s  esin  1
k = –1
(4)
(M mark requires a k, or use of limits)
4 Marks
Question
number
4.
Scheme Marks
dy
2x

dx 1  ( x 2 ) 2
M:
d 2 y 2(1  x 4 )  2 x.4 x3

dx 2
(1  x 4 )2
dy

dy
2x


or
2 2
dx 1  x 2
dx 1  ( x )
 2  6x4 
 

4 2 
 (1  x ) 
M1
M1 A1cso
3
  dy  2  2
1    
3
  dx  
  (1  1) 2

2
d y
1
2
dx
A1cso (6)
3
= –22
M1 A1
(or 22) (or exact equivalent, e.g. 8, 2 2 )
6 Marks
2nd M: Quotient or product rule attempt. (Chain rule, if used, must be
‘good’.)
3rd M: Attempt  with derivative values
A: Correct derivative values (1 and 1) seen or implied by working.
Alternative: (involving implicit differentiation).
dy
sec 2 y
 2x
[M1 A1] (allow one ‘slip’ for M1)
dx
d 2 y dy
dy
2
sec y 2   2 sec y (sec y tan y )
2
dx
dx
dx
(or alternative
dy
d2 y
dy
 2 x cos 2 y , so
 2 cos 2 y  2 x  2 cos y ( sin y )
2
dx
dx
dx
2
Then marks as in main scheme (n.b. tan y  1, sec y  2 ).
e.g.
)
[M1]
Question
number
5.
(a)
Scheme Marks
dy
 4 sec h 2 4 x  1
dx
dy
0
cosh 2 4 x  4
cosh 4 x  2
Put
dx
4 x  ln( 2  3 ) or 8 x  ln( 7  4 3 ) or e 4 x  2  3 or e 4 x  7  4 3
( or +)
1
1
x  ln( 2  3 ) or x  ln( 7  4 3 )
(or equiv.)
4
8
(b)
1
y   ln( 2  3 )  tanh(....)
(Substitute for x)
4
1
3
sec h 4 x   1  tanh 2 4 x ,
tanh 4 x 
2
2
3 1
1
y
 ln( 2  3 )  2 3  ln 2  3
(*)
2 4
4





B1
M1
A1
A1
(4)
M1
M1
A1
(3)
7 Marks
(a) ‘Second solution’, if seen, must be rejected to score the final mark.
(b) 2nd M requires an expression in terms of 3 without hyperbolics,
exponentials and logarithms.
Question
number
6.
(a)
Scheme Marks
dx
1
 1
dt
t
B1 B1
1

dy
 2t 2
dt
2
2
1
2 1
1
t 1
 1    2 
 1   2  1  or
1     2t  ,
t
t t
t
 t 

4
 1
4
Length  1   dt  t  ln t 1  4  ln 4  1  3  ln 4 (*)
t
1 

(b) Surface area =

2
2 
4
1
2
1
 1    2 
4 t 1     2t  dt
 t 

M1 M1 A1
(7)
M1
1
1


 
4
  8  t 2  t 2  dt 



1



 

4
 32
1
 16
2t
 2
 160

 (8 )
 2t 2   (8 )  4     2  
 3

3
 3

 3

 1
M1, A1
 1 
 53  
3 

M1 M1 A1
(4)
11 Marks
Note dependence of M’s.
(a) 2nd M: Complete integration attempt. 3rd M: Subs. correct limits and
subtract.
(b) 2nd M: Complete integration attempt. 3rd M: Subs. correct limits and
subtract.
In (b), missing 2 or  or 2 throughout could score M0 M1 M1 A0.
If 2 is there initially, then lost, could score M1 M1 M1 A0.
Question
number
7.
Scheme Marks

M1 A1 A1

x3
x3
arsinh x 
dx
3
3 x2  1
x 2 arsinh xdx 
3
 x3

 arsinh x = 9arsinh3
3
0
du
 2x
Let u  x 2  1
dx
or 9 ln 3 
10

B1
 2
2
u  x  1
1
1
 
1
x3
1 u 1
1  2
2
d
u

d
u

u

u
1
1

 du
3
6
6


u 2 .2 x
u2



 32
1
1  2u
 
 2u 2 
6 3


2u
du

 2 x
dx

M1

M1

1

2
 3  u  1 du 
M1
 1 u 3
 
   u  
 3  3
 
When x = 0, u = 1 and when x = 3, u = 10
......u 
10

M1 A1
10
 3
1

 2
1  2u 2
1  20 10
2

    2 

2
u


2
10

 3

6 3
6  3



1
1  14 10 4 
1
  9 ln 3  10  7 10  2
Area = 9arsinh 3  
(*)
6 3
3
9

 

A1cso (10)
10 Marks
Dependent M marks:
M: Choose an appropriate substitution & find
du
or ‘Set up’ integration
dx
by parts.
M: Get all in terms of ‘u’
or Use integration by parts.
M: Sound integration.
M: Substitute both limits (for the correct variable) and subtract.
Question
number
7.
Scheme Marks
Alternative solution:
dx
 cosh 
Let x  sinh 
d

M1

M1
x 2ar sinh x dx   sinh 2  cosh  d
sinh  
1

sinh  cosh 2   1 d

3
 3 


3
M1 A1 A1

B1
arsinh3
 sinh 3  


3

0
= 9arsinh3

1
1  cosh 3 
2
sinh  cosh   1 d  
 cosh  
3
3 3

arsinh
3
3
  1 

1  cosh 
1
 10 10

 cosh  
 
 10     1

3 3
3
0
  3 
 3



M1


 
1  7 10 2 
1
  9 ln 3  10  7 10  2
Area = 9arsinh 3  
3 3
3
9
M1 A1

(*)
A1cso
(10)
10 Marks
Question
number
7.
Scheme Marks
A few alternatives for:
(i)

x3
x2 1
dx .
du
 2x
dx
Let u  x 2
3

u2
1 u
1

1
2
du 
1
u
du

2 1 u
2u
No marks yet… needs another substitution, or parts, or perhaps…
u
1
 1 u 
1 u
1 u
1
1
1
1  u du  
du

2
2 1 u
3
1
1
1  u  2  1  u  2
3
Limits (0 to 9)
(ii)
Let x  sinh 
sinh 3 
2
 cosh   cosh  d   sinh  cosh   1 d
Then, as in the alternative solution,

1
1  cosh 3 
sinh  cosh 2   1 d  
 cosh  
3
3 3

Limits (0 to arsinh3)



M1
M1
M1
M1
dx
 cosh 
d

M1

M1
M1
M1
Question
number
7.
(iii)
Scheme Marks
M1
du
 sec 2 
d
Let u  tan 
tan 3 
2
2
 sec   sec  d   tan  sec  sec   1 d
sec 3 
  sec 2  sec  tan   d   sec  tan   d 
 sec 
3
Limits sec   1 to sec   10
(iv) (By parts… must be the ‘right way round’, not integrating x 2 )
du
dv
x
u  x2 ,
 2x

, v  1 x2
2
dx
dx
1 x




x 2 1  x 2   2 x 1  x 2 dx
M1
M1
M1
M1

M1
Limits
(v) (By parts)
du
dv
1
u  x3 ,
 3x 2

, v  arsinh x
dx
dx
1 x2
(vi)
x3
x( x 2  1)  x x( x 2  1)
x



2
2
2
1 x
1 x
1 x
1 x2
x
2
 x 1  x dx   1  x 2 dx
3
1
1
 1 x2 2  1 x2 2
3
Limits
M1
x2 1 x2 


M1
3
2 2
x 1 2
3
 

No progress
M0
M1
M1
M1
M1
Question
number
8.
(a)
Scheme Marks
 x cosh x dx  x sinh x   nx
n
n
n 1
M1 A1
sinh x dx

M1
 x n sinh x  nx n 1 cosh x  n(n  1) x n  2 cosh x dx
I n  x sinh x  nx
n
(b)
n 1
cosh x  n(n  1) I n  2
A1
(*)
I 4  x 4 sinh x  4 x 3 cosh x  12I 2
I 4  x 4 sinh x  4 x 3 cosh x  12x 2 sinh x  2 x cosh x  2 I 0 
(This M may also be scored by finding I 2 by integration.)




I 4  x 4  12 x 2  24 sinh x,   4 x 3  24 x cosh x
(c)
x




 C 

A1, A1 (5)
1
 12 x 2  24 sinh x   4 x 3  24 x cosh x 0
= 37sinh 1 – 28cosh 1
M: x = 1 substituted throughout (at some
stage)
 e  e 1 
 e  e 1 
  28

 37
 2 
 2 
M: Use of exp. Definitions (can be in terms of x)
1
 9e  65e 1
2
4
M1
M1
B1
I 0  cosh x dx  sinh x  k

(4)

M1
M1
A1
(3)
12 Marks
(b) Integration constant missing throughout loses the B mark
.
Question
number
9.
(a)
(b)
Scheme Marks
x 2 mx  c 

1
b 2 x 2  a 2 (mx  c) 2  a 2 b 2
a2
b2
b 2  a 2 m 2 x 2  2a 2 mcx  a 2 c 2  b 2  0
M1
2


2a mc  4b
2
 



 a 2m2 a 2 c 2  b2
4a 4 m 2 c 2  4a 2 b 2 c 2  b 4  a 2 m 2 c 2  a 2 m 2 b 2
2
2

(*)

c2  b2  a2m2
(*)
(c) Find height and base of triangle (perhaps in terms of c).
c 
b 2  a 2 m 2 
OB  c  b 2  a 2 m 2
and
AO =


m 
m

2
2
2
2
c
b a m
Area of triangle OAB =
M: Find area and subs. for c.

2m
2m
(d)
b 2  a 2 m 2 b 2 1 a 2


m 
m
2m
2
2
b
d
b2
a2
b2
m
  m 2 
0
 a2
2
a
dm
2
2
m
2
2
 b  a   a  b 
(*)
          ab
 2  b   2  a 
(e)
 a 2 mc
Root of quadratic: x  2
(Should be correct if quoted
b  a 2m2
directly)
b
a
Using m 
and c  b2  a 2m2 : x  
a
2
nd
(The 2 M is dependent on using the quadratic equation).


A1
M1
(2)
A1
(2)
M1
A1
M1 A1 (4)
M1 A1
A1
M1
M1 A1 (3)
14 Marks
(d) Alternative: b 2  a 2 m 2  2bam (since (b  am) 2  0 )
b a m
 ab
2m
2
2
[M1]
2
[A1]
Conclusion [A1]
(e)
Alternative:
Begin with full eqn. b 2  a 2 m 2 x 2  2a 2 mcx  a 2 c 2  b 2  0 .
b
In the eqn., use conditions m  and c  b 2  a 2 m 2 ( b 2 )
a
a
Simplify and solve eqn., e.g. 2 x 2  2a 2 x  a 2  0 x  
2



(3)

[M1]