January 2006
Question
Number
1
6675 Pure Mathematics P5/FP2 Mark Scheme
Scheme
Marks
x 2  2 x  17  ( x  1)2  16
4
I =

1
B1
dx = [ arsinh
( x - 1)
] or equiv.
4
=
arsinh ¾
( x  1) 2  16
[ M1 does not require limits; A1 f.t. on completing square, providing arsinh ]
1
3
Into ln form [ ln  
4

9
1  ;
16

=
M1 A1√
M1A1
ln2
[5]
[
If straight to ln form : B1,
ln ( x  1)  ( x  1)2  16 


Using limits correctly M1A1√ , ln2
2
(a)
A1 ]
Using b2  a 2 ( e2  1) ; [ 4 = 16 (e 2  1) ]
e =
5
or equiv. (1.12) M1A1
2
(2)
5
];
2
= 4
5
(b) Distance between foci = 2 a e [ 2 x 4 x
[
M1
M1A1√
(2)
A1√ dependent on both Ms ]
(c)
Ellipse, centred on origin
B1
Hyperbola, both branches
B1
Totally correct, touching, with
B1
correct intercepts
(3)
[7]
1
January 2006
3
6675 Pure Mathematics P5/FP2 Mark Scheme
dx
 1  cos t ,
dt
s 
dy
 sin t ,
dt
 (1  cos t )
2

 (sin t ) 2 dt ;
Use of “half-angle formula”
B1
(both)

=
[  4 cos t dt ] ;
2
Using limits correctly and surd form;
M1A1
2  2 cos t dt
 
(  
t  2

s = 4 sin 
2  ( 0)

= 2 √2 (allow
4
)
2
M1A1√
M1A1
[7]
4
(a) Using cosh x 
e x  e
2
x
and attempt to progress
 e3 x  3 e x  3e  x  e 3 x
Correct intermediate step as far as 4 
8

=
M1

 

  e x  e  x 

3 
2

 
e 3x  e 3x
 cosh 3 x
2
A1 (3)
M1
(b) Using part (a) to reduce to cosh2x = [ 2 ]
Correct method to form ln x or find e X or e 2 x
x = ln


2 1 ,
ln




2 1

A1
or equivalent

or ½ ln 3  2 2 , ½ ln 3  2 2 ,
M1
A1 A1√
(4)
( after finding e 2 x = ……)
[7]
2
Question
Number
5
Scheme
dy
1

. sec x tan x  tan x ;
dx sec x
s
2
 dy 
(1    dx 
 dx c

=

B1B1
  x
M1
(1  tan xc2 dx
 ln sec x  tan x
 sec x dx ;
s  0, x 
Method to find “c”:
Marks

3
 ( c)
 0  ln (2  3 )  c
ln (2  3 )
A1A1
M1
c =  ln (2  3 ) or  ln tan
 s  ln (sec  tan ) 
or equivalent
5
or – 1.32
12
or equivalent
A1
A1√
[8]
6
(a)
dy
 3 cosh 2 x sinh x
dx
B1
d2 y
 3 cosh 3 x  6 cosh x sinh 2 x
dx 2
or equivalent
Application of formula for radius of curvature [  =
M1A1
1  9 cosh

1  9 c

(c  1)
3c {c  2(c 2  1)}
(b) cosh x =
2
2
2

3
2
x sinh x
]
3
3 cosh x  6 cosh x sinh 2 x
Use of sinh2 x = cosh 2 x  1
4
4
M1
M1
3
2
=
9c

 9c  1
3c{ 3c 2  2)}
6
4
3
2
5
4
AG
A1*
(6)
B1
Using found value of cosh x in formula for  ;  = 4.8
M1A1
(3)
[9]
3
4
7
(a)
3
3
4

2 n
2
n 1
2
I n    x (4  x)   n  x (4  x) 2 dx
3 
3 0
 0
3
4
2
n x n  1 (4  x) 2 dx
3 0
=
A1√
1
4
1
4
2
2
n  4 x n  1 (4  x) 2 dx  n  x n (4  x) 2 dx
3 0
3 0
=
 In
[ 2n  3I
n
8
nI
3
 8n I
n 1
(b) Relating I 2 to I
I2 =
M1A1
16 8
.
7 5
I
0
n  1
]

2
nI
3
 I
n
n

8n
I
2n  3
using result from (a)
0
=
2048
105
M1A1
 53 
19

 105 
n  1
AG
A1*
(6)
M1
A1A1
(3)
[9]
4
8
1 

1 

1
1 
2

(a) ar tanh
 ln
2 1 1 
2


2 

=
1  2 1
ln 
x
2  2 1
=
1
ln
2

or equivalent
M1
2 1 
 or equivalent
2 1 
M1

2
2  1  ln


2 1
AG
A1*
(3)
Alternative Approach
 1 
ar tanh (sin x)  tanh y  

 2
and then use exponentials:
If using y 
Progression as far as e y  ....... or e2y  .......
Converting to ln form
Answer as given
Note:

1
ln 3  2 2
2

[ or cosh2 y  2 ]
M1
M1
A1*
can earn M1M1 but for A1* there must be a
convincing further step.
(b)
dy
1

. cos x ;
dx 1  sin 2 x

cos x
 sec x
cos 2 x
M1A1
(2)
Note: If tanh y  sin x is differentiated M1 requires
dy
 f ( x)
dx
(c) Attempt at by parts and use of result in (b)
=  cos x ar tanh (sin x) 
M1
 cos x sec x d dx
A1
M1
=  cos x ar tanh (sin x)  x
Using limits correctly :
= 


1

or exact equivalent
ln 1  2 
4
2
5
M1A1
(5)
[10 ]
9
(a) Correct method for finding
1
P
 
dy
dx
M1
Gradient of normal = – p
A1
y  2ap  ( p)( x  ap 2 )
Equation of normal:
y  px  2ap  ap3
M1
AG
A1*
(4)
(b) Using both equations and eliminating x or y
M1
( p  q) x  2a( p  q)  a ( p  q )
3
3
may be unsimplified
A1
x  2a  a ( p  pq  q )
2
2
A1
Finding the other coordinate
M1
y   apq ( p  q)
A1
(5)
(c) Using pq = 3 in both x and y
[ x  a ( p 2  q 2  5),
(in any form)
y   3 a ( p  q) ]
M1
Complete method for relating x and y, independent of p and q
M1
A correct equation, in any form
A1
[ e.g. :
 x  5a 

y 2  9 a 2 ( p 2  q 2  2 pq) = 9 a 2 
  6 ]
 a 

y 2  9 a ( x  a)
A1√
(4)
[13]
6