Question 8:
--- 8(a) ---
8(a) | 43
43  3  43
( 8 − 3 x ) = 1 6 1 − x but condone 1 6 ( 1 − k x ) , k  3
8 | M1
''Correct'' term 3 or term 4 in
4 1  2 
4 1   −

43 4 3 3 3 3 3
(1 − k x ) = 1  k x + (  k x ) 2 + (  k x ) 3 + . ..
3 2 6 | M1
x 2 x 3 3
Two terms correct of 1 6 − 8 x + + + ... following M1, M1 and k = 
2 2 4 8 | A1
x2 x3
=16−8x+ + +...
2 24 | A1
(4)
(b) | 1 5 x 2 x 2 x 3
Assume the curves meet so 8 x − + 8 = ''1 6 − 8 x + + '' (for some x)
2 2 2 4 | M1
x3 ( ) x3 ( )2
0=8−16x+8x2 + +...0=8 x2 −2x +8+ +...8 x−1 ..+..
24 24 | M1
x3
( )2
0=8 x−1 +
24 | A1
Requires
( ) 2 x 3
• Argument that 8 x − 1 +  0 .
2 4
( ) 2 x3 8 ( ) 2 x 3
E.g. 8 x − 1 0 and 0 (for 0 x ) so 8 x − 1 +  0
24 3 2 4
• Minimal statement (contradiction) and conclusion | A1
(4)
(8 marks)
Alt (a) by direct expansion. ( 8  3 x ) 43 = 8 43 + 4
3
 8 13  (  3 x ) +
4
3

2
1
3  8 − 23  (  3 x ) 2 +
4
3

1
3

6

−
2
3

 8 − 53  (  3 x ) 3 + ...
M1: For 8
43
or 16 oe seen as the constant term E.g. = 1 6  . . . x + . . .
M1: For the correct term 3 or term 4 of the expansion. Condone missing brackets on the (  3 x ) 2 , (  3 x ) 3
terms
A1: For two correct and simplified terms of 1 6 − 8 x +
x
2
2
+
x
2
3
4
+ ... Allow as a list
A1: Correct expansion from correct work. Allow as a list
(b)
M1: Sets up the contradiction hypothesis using the result of their (a) and setting equal to the quadratic
equation. There must be some words so look for a minimum of
Assume they meet (or simply let) 8 x −
1 5
2
x 2
+ 8 = ''1 6 − '' 8 x +
x
2
2
+
''
x
2
3
4 ''
M1: Rearranges terms and attempts to complete the square (or factorises as k (x-1) (x-1) ) on the
quadratic terms or other means to show the quadratic part is always positive (e.g. considers discriminant).
The work on the quadratic terms of the expression, which must be of the form 8 x 2 − 1 6 x + ( A − 8 ) should
take the form of one of
• 8 x 2 − 1 6 x + ( A − 8 ) = 8
(
x − 1
) 2
+ ...
• 8 x 2 − 1 6 x + ( A − 8 ) and attempts b2 −4ac=162 −48 ( A−8 )=...
A1: Correctly completes the square to form equation 0 = 8
(
x − 1
) 2
+
x
2
3
4
.
Via the discriminant you would need b2 −4ac=162 −488=0,
A1: Completes the argument by drawing together the contradiction.
States that 8
(
x − 1
) 2
0
x3
and 0(in the interval), hence
24
0  8
(
x − 1
) 2
+
x
2
3
4
so proven
( )2
x3
Stating that 8 x−1 + 0 is insufficient without some explanation. (See above)
24
( )2
Stating that both terms are positive (as x > 0) is not enough as 8 x−1 =0 at x = 1
For discriminant must mention b 2 − 4 a c = 0 , hence single root (at 1) and positive quadratic or
similar,
x
2
3
4
 0 hence 0  8
(
x − 1
) 2
+
x
2
3
4
so proven.
…………………………………………………………………………………………………….
There will be alternatives for the middle two marks in (b), so consider carefully what is attempted.
For example, candidates could work on the cubic 8 − 1 6 x + 8 x 2 +
x
2
3
4
. Condone use of calculus which
would involve some calculator work that proves that the function has a minimum value at x = awrt 0.99 y
= awrt 0.04 and hence 8 − 1 6 x + 8 x 2 +
x
2
3
4
 0 (for 0  x 
8
3
PMT
) proving that there is a contradiction. M1:
Differentiates, sets = 0 and solves to find x and y in the interval. A1: Correct with proof of minimum
x3
Note that merely solving 8−16x+8x2 + =0on a calculator to get x=−194and stating this is not in the
24
range is M0.
Question | Scheme | Marks