Mathematics · May / June 2024 · Paper 41 · Question 8
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As printed in the paper
The question as text
8. f(x) = ( x x 8 3 ) 3 0 3
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Show the answer and mark scheme
8(a)[4]
4
3
3
4
(8
3 )
x
16 1
x
but condone
16(1
) ,
3
−
3
=
−
−
kx
3
k
≠
8
''Correct'' term 3 or term 4 in
4
1
2
4
1
×
×−
×
3
3
3
4
4
3
3
(1
−
kx
)
= ±
1
k
x
+
(
±
kx
)
2
+
(
±
k
x
)
3
+
...
3
3
2
6
3
x
2
x
3
Two terms correct of
16
−
8
x
+
+
+
...
following M1, M1 and
k = ±
2
24
8
x
2
x
3
16
8
x
...
=
−
+
+
+
2
24
15
x
2
x
2
x
3
8(b)[4]
Assume the curves meet so
8
x
−
+
8
=
''16
−
8
x
+
+
''
(for some x)
2
2
24
x
(
)
x
3
3
(
)
2
0
=
8
−
16
x
+
8
x
2
+
+
...
0
=
8
x
2
−
2
x
+
8
+
+
...
8
x
−
1
±
.. ..
+
24
2
4
(
)
x
3
2
0
=
8
x −
1
+
24
Requires
Argument that ()
x
3
2
8
1
0
x −
+
>
.
24
x >
8
x
E.g. (
)
3
) so (
)
3
2
2
8
x −
1
0
and
0
(for
0
<
x
<
8
x −
1
+
>
0
24
3
24
Minimal statement (contradiction) and conclusion(a)
4
4
4
M1: Attempts to takes out a factor of 16 or
38 . Look for
(8
−
3 )
x
=
16(1
−
kx
) ,
k
≠
3
3
3
4
4
4
or
(8
−
3 )
x
=
8 (1
−
kx
) ,
k
≠
3
Condone incorrect values of k other than 3
3
3
3
M1: Attempts the binomial expansion on (
)
4
1 kx
−
achieves at least a ''correct'' term 3 or term 4
3
(which may be unsimplified) for their k. Even allow with k =1
4
1
2
4
1
×
×−
×
3
3
3 (
Look for either
3
3 (
or
±
with the correct binomial coefficient and the correct
±
kx
)
2
kx
)
3
2
6
power of x. Condone missing brackets on the (
)
kx terms.
4
x
3
3
If they expand
without showing the unsimplified terms it is implied by
±
1
−
8 x
384
x
x
2
3
A1: For two correct and simplified terms of
16
−
8
x
+
+
+
...
following M1, M1 and
2
24
A1: Correct expansion from correct work. Terms may be as a list for both accuracy marks
4
1
2
4
1
×
×−
×
4
4
4
1
(
)
2
3
3
3
5
Alt (a) by direct expansion.
3
3
−
−
(8
±
3 )
x
=
8
+
8
±
3
x
+
8
( 3 )
±
x
2
+
8
( 3 )
±
3
3
3
3
3
3
2
6
4
M1: For
38 or 16 oe seen as the constant term E.g.
=
16
±
...
x
+
...
( 3 )x
M1: For the correct term 3 or term 4 of the expansion. Condone missing brackets on the
±
terms
x
2
x
3
A1: For two correct and simplified terms of
16
−
8
x
+
+
+
...
Allow as a list
2
24
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
15
x
x
x
2
2
3
Assume they meet (or simply let)
8
x
−
+
8
=
''16'' 8
−
x
+
+
2
2
''24''
M1: Rearranges terms and attempts to complete the square (or factorises as k (x1) (x1) ) on the
quadratic terms or other means to show the quadratic part is always positive (e.g. considers discriminant).
8
x
16
x
(
The work on the quadratic terms of the expression, which must be of the form
2
−
+
take the form of one of
(
)
2
8
x
2
−
16
x
+
(
A
−
8)
=
8
x
−
1
+
...
(
)
8
x
−
16
x
+
(
A
−
8)
and attempts
b
2
−
4
ac
=
16
2
−×
4 8
A
−
8
=
...
2
x
3
(
)
2
A1: Correctly completes the square to form equation
0
=
8 x
−
1
+
.
24
Via the discriminant you would need
b
2
−
4
ac
=
16
2
−××=
4 8 8
0
,
A1: Completes the argument by drawing together the contradiction.
States that (
)
x >
3
(
)
x
3
2
2
8
x −
1
0
and
0
(in the interval), hence
0
≠
8 x
−
1
+
so proven
24
24
x
3
Stating that (
)
2
8
x −
1
+
>
0
is insufficient without some explanation. (See above)
24
Stating that both terms are positive (as x > 0) is not enough as (
)
2
8
x −
1
=
0
at x = 1
For discriminant must mention
b
2
−
4
ac
=
0
, hence single root (at 1) and positive quadratic or
x >
3
(
)
x
3
2
similar,
0
hence
0
≠
8 x
−
1
+
so proven.
24
24
…………………………………………………………………………………………………….
There will be alternatives for the middle two marks in (b), so consider carefully what is attempted.
x
3
For example, candidates could work on the cubic
8 16
−
x
+
8
x
2
+
. Condone use of calculus which
2
4
would involve some calculator work that proves that the function has a minimum value at x = awrt 0.99 y
x
3
8
= awrt 0.04 and hence
8
16
8
0
(for
0
) proving that there is a contradiction. M1:
−
x
+
x
2
+
≠
<
x
<
24
3
Differentiates, sets = 0 and solves to find x and y in the interval. A1: Correct with proof of minimum
x
3
Note that merely solving
8
16
8
0
on a calculator to get
x = −
194
and stating this is not in the
−
x
+
x
2
+
=
24
range is M0.
3
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