Question 1:
1 | t− 3 x−
Attempts to get t in terms of x or y . E.g. x =  t =
t 3 x 1
Forms a Cartesian equation and makes progress to making y the subject
1 x − 1
E.g. y = + 2  y = + 2
t 3 x
7 x3 − 1
 y =
x | M1
dM1
A1
(3)
(3 marks)
2 (a)
(b) | 3 x) A B−
 +  V a l u e s f o r A a n d B
( ( ) 2 x − 1 x 2
2 x − 1 x − 2
One correct value, either A=−1 or B=2
− 1 2
Correct PF form +
2 x − 1 x − 2
 − 1 2 1 ( ) ( ) ( )
+ d x = − l n 2 x − 1 + 2 l n x − 2 + c
2 x − 1 x − 2 2
2 5 2 5
 3 x)  1 ( ) ( ) 
d x = − l n 2 x − 1 + 2 l n x − 2
( ( ) 2
2 x − 1 x − 2
5 5
 1   1 
=− ln49+2ln23−− ln9+2ln3
 2   2 
   
5 2 9
l n
2 1 | M1
A1
A1
(3)
M1, A1ft
dM1
A1
(4)
(7 marks)
3 (a)
(b) | ( ) ( )
Attempts 2 i − 3 j + 4 k − 8 i − 5 j + 3 k
R Q = − 6 i + 2 j + k
Attempts PQ RQ=2−6+−32+41
Full attempt to find c o s P Q R E.g.
2  − 6 + − 3  2 + 4  1 = 2 9 4 1 c o s P Q R
Angle PQR = 1 1 4  | M1
A1
(2)
M1
dM1
A1
(3)
(5 marks)
4(a)
(b)
(c) | 12
− 12
1  2  − ( )
= 4 − x = 4 1 − ...
4 − x 2
 12   32   12   32   52 
12 −  − −  −  −
− 2 3
 14 2   12   14 2   14 2   14 2 
1 − x = 1 + − − x + − x + − x
2 3 !
1 1 1 3 5
2 4 6
= + x + x + x
4 − x 2 2 1 6 2 5 6 2 0 4 8
x  2
Substitutes an appropriate value of x in both sides with LHS in terms of 3
2 0 4 8 3543
E.g. with x=1 3 = or
1 1 8 1 2048 | B1
M 1 , A 1
A1, A1
(5)
B1
(1)
M1
A1
(2)
(8 marks)
E.g. x=1
4
1
− 1 2
=
1
2
+
1
1
6
 1
2
+
2
3
5 6
 1
4
+
2
5
0 4 8
 1
6
Note that x = − 1 and x
2
= 1 all work with the same results
Alternatively correctly solves
4
1
− x 2
= 3 o.e. and substitutes this into their expansion
A1: With x = 1 3 =
2
1
0
1
4
8
8
1
or
3
2
5
0
4
4
3
8
. This can only be scored from a correct expansion
**There are lots of values of x that work, all more difficult than the one above.
E.g x 2 =
8
3

4
1
− x 2
=
2
3
and x 2 =
1 1
3

4
1
− x 2
= 3
In both these cases the fractions become much more difficult to find.
8 43
For x 2 =  3 = and for
3 27
x 2 =
1 1
3
 3 =
5
5
5
5
6
2
8
9
7
6
Alt I (a) By direct expansion
 1  3  1  3  5
1 = 

4−x 2

− 1 2 =4 − 1 2 +  − 1 4 − 3 2 

−x 2

1 +    − 2        − 2    4 − 5 2 

−x 2

2 +    − 2       − 2       − 2    4 − 7 2 

−x 2

3
 2 2 3!
4−x 2          
B1: For 4
−
12
+
2
M1: A correct attempt at the third or fourth terms condoning sign slips on the −x
A1: Correct and unsimplified expansion. See above
1 1 3 5
2 4 6
A1: Two correct and simplified terms of + x + x + x
2 16 256 2048
1 1 3 5
2 4 6
A1: + x + x + x . Accept as a list
2 16 256 2048
Alt II (a) By difference of two squares
1 1 1
1 = ( 2−x )− 2 ( 2+x )− 2 =4 − 2 +......
4−x 2
1
−
B1: For 4 2 +
M1: For correctly writing
4
1
− x 2
=
(
2 − x
) − 12

(
2 + x
) − 12
= ...

1 −
x2 
−
12


1 +
x2 
−
12
=
with a correct attempt at the third of fourth term in either expansion, followed by an attempt
to combine both the expansions.
2 4 6
A1: One correct term in x , x or x
A1: Two correct and simplified terms of
1
2
+
1
1
6
x
2
+
2
3
5 6
x
4
+
2
5
0 4 8
x
6
PMT
with no terms in
3 5
x, x or x
A1: Fully correct
5. | 2
k  
 1 2 x
States or implies Volume = π d x
12 ( 2 x 2 + 3 ) 1 .5
2
 
Attempts  1 2 x d x =  1 4 4 x d x = − 1 8 ( 2 x 2 + 3 ) − 2
( 2 x 2 + 3 ) 1 .5 ( 2 x 2 + 3 ) 3
1 8 1 8 7 1 3  
Sets − + = 
( 2 k 2 + 3 ) 2 ( 2  12 + 3 ) 2 6 4 8
1 8 9 7 1 3
 − + =  ( 2 k 2 + 3 ) 2 = 7 2 9  k 2 = . . .
( 2 k 2 + 3 ) 2 8 6 4 8
k = 2 3 | B1
M1A1
M1
ddM1
A1
(6 marks)
6 (a)
(b)
(c) | 
For correct parameter t = or x coordinate at P x = 4
4
d y
d y d t − 4 s i n 2 t
= =
d x d x 3 s e c 2 t
d t
2 ( ) 2 8
Equation of tangent y − 0 = − x − 4  y = − x +
3 3 3
k = 1 − 3
−1„ f „ 2 | B1
M1 A1
dM1 A1
(5)
B1
(1)
M1 A1
(2)
(8 marks)