Question 5:
--- 5(a) ---
5(a) | 6 6
( 2+ax)6 =26 + 25(ax)+ 24(ax)2+...
1 2 | M1
=64+192ax+240a2x2 +... | A1A1
(3)
(b) | 2
 1 6 1 3 3 1
3+  =9+ + or 9+ + +
 x x x2 x x x2 | B1
f ( x ) =  9 + 6 + 1  ( 6 4 + 1 9 2 a x + 2 4 0 a 2 x 2 + ) ... = ...
x x 2
Constant term is 9  6 4 + 6  1 9 2 a + 2 4 0 a 2 | M1
5 7 6 + 1 1 5 2 a + 2 4 0 a 2 = 5 7 6  1 1 5 2 a + 2 4 0 a 2 = 0
 1 1 5 2 + 2 4 0 a = 0  a = ... | dM1
24
a =−
5 | A1
(4)
Total 7
(b)
B1: Correct expansion of

3 +
1
x

2
unsimplified or simplified.
Note that this may be implied by the omission of the “9” as 9  6 4 = 5 7 6 and so the
“9” is not required.
Note that 3 2

1 + 2

3
1
x

+
( 3
1
x ) 2

is correct.
 1 2 6 12 6 1
Condone missing brackets if recovered e.g. 3+  =9+ + =9+ +
 x x x x x2
M1: Uses their expansion in part (a) and their expansion of

3 +
1
x

2
to extract the constant
 1 2  
term. This depends on having obtained 3+  =+ + oe with , , non-
 x x x2
 
zero and an attempt at "64"+ "192ax"+ "240a2x2"oe
x x2
An expression of this form is sufficient i.e. with the x’s still included.
Note that the 964 may not be seen as this cancels with the 576 so this mark may be
implied.
dM1: Sets their constant term = 576 and proceeds to obtain a non-zero value for a.
You do not need to be concerned about the processing for this mark.
This may be implied by e.g. 6  1 9 2 a + 2 4 0 a 2 = 0  a = ... as 9  6 4 = 5 7 6
PMT
May be implied by their value(s).
Condone x = … here.
Depends on the previous method mark.
48
A1: Correct value. Allow equivalents e.g. −4.8, −
10
The questions asks for the value of a so just look for the correct value e.g. “a = …” is
24
not required but x =− scores A0.
5
If a = 0 is also given and not rejected, score A0