Mathematics · May / June 2022 · Paper 31 · Question 6
YMA01
As printed in the paper
The question as text
6. y C P O x Figure 3 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The function f is defined by 1 x – 9 2 – 2)(4x + 9) f(x) = 5(x 2 4
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[4]
1
1
1
(
)
−
f ( )
′
x
=
5(
x
2
−
2)
×
××
4 (4
x
+
9)
2
+
10 (4
x
x
+
9)
2
2
10(
x
2
−
2)
+
10 (4
x
x
+
9)
50
x
2
+
90
x
−
20
(
)
f ( )
′
x
=
=
1
1
(4
x
+
9)
(4
x
+
9)
2
2
10(5
x
2
+
9
x
−
2)
(
)
f ( )
′
x
=
1
(4
9)
x
+
2
1
6(b)[1]
Sets f ( )
′
x
=
0
to give
x = −
2, 5
6(c)[2]
(
)
(
)
1
(
)
2
(
)
y =
5 " 2"
−
−
2
4
×
" 2"
−
+
9
2
=
10
( 2, 10)
−
f(0)
30
6(d)[3]
= −
or Lowerbound
Upperbound
= ×
2 "10" 4
+
=
...
= ×−
2 " 30" 4
+
=
...
56
g( )
x
24
−
18
18
A1
k >
3.6
or eg
k >
but not
k
If
k <− is found then it must be rejected
2
5
5
(a)
1
1
−
M1
Attempts the product rule. Award for
±
A
(5
x
2
−
10)(4
x
+
9)
±
Bx
(4
x
+
9)
or equivalent where A, B
2
2
≠
0
The indices do not need to be processed for this mark.
uv
'
+
u v
'
and state the expressions for u, '
u , v and 'v then
If they quote the product rule eg
allow a miscopy when substituted. They must have
1
1
−
eg
u
=
5
x
2
−
10
→
u
'
=
Ax
and
v
=
(4
x
+
9)
2
→
v
'
=
B
(4
x
+
9)
2
1
1
May also attempt the product rule on
5
x
2
(4
x
+
9)
2
−
10(4
x
+
9)
2
so award for
1
1
1
−
−
≠
0
±
Ax
(4
x
+
9)
2
±
Bx
2
(4
x
+
9)
2
±
C
(4
x
+
9)
2
A, B, C
1
1
−
A1
Correct unsimplified expression for f ( )
′
x
eg
2(5
x
2
−
10)(4
x
+
9)
2
+
10 (4
x
x
+
9)
2
oe
dM1
Proceeds to a single fraction, multiplies out the brackets on the numerator and proceeds to
3TQ
1
−
form
or 3TQ
(4
x
+
9)
2
. They cannot proceed straight to the given 3TQ or a multiple of it
1
(4
x +
9)
2
without some working seen to score this mark.
3TQ
Alternatively, you may only see manipulation of the numerator and then
which is
1
(4
x +
9)
2
10(5
x
2
+
9
x
10(
2
2)
10 (4
9)
10(
2
2
4
2
9 )
10(5
2
9
2)
acceptable:
x
−
+
x
x
+
=
x
−
+
x
+
x
=
x
+
x
−
1
(4
x
+
9)
2
Condone arithmetical slips in their working but their method to proceeding to a single fraction
must be correct.
It is dependent on the previous method mark.
10(5
x
2
+
9
x
−
2)
A1
(
f ( )
′
x
=
)
with no errors seen in the main body of their solution.
1
(4
x
+
9)
2
10(5
x
2
+
9
x
−
2)
Accept
4
x
+
9
Condone invisible brackets as long as they are recovered, or their subsequent working
makes their
intention clear. They do not need to explicitly state k = 10 but if there is a
contradiction between their
expression and their stated value then the expression takes
precedence. Do not be concerned with what
they have before the =
(b) Mark (b) and (c) together
1
1
( 2,0)
,0
B1
Sets f ( )
′
x
=
0
to give
−
2, 5
Withhold if one is rejected. Condone
−
. May be
5
seen in (c)
(c)
1
M1
Substitutes their smaller root from solving
5
x
2
+
9
x
−
2
=
0
into
y
=
5(
x
2
−
2)(4
x
+
9)
2
and
proceeds to find a value for y. Condone bracket omissions, miscopying or arithmetical slips
long as it is clear that they are attempting to substitute their smaller root into f(x)
A1cao ( 2, 10)
−
only (Allow
x
= −
2,
y
=
10
)
(d)
B1
(
f(0)
)
30
seen (which may be implied by sight of 56
)
= −
−
M1
A correct attempt at the lower or upper bound.
Either attempts 2 "10" 4and identifies this as the upper bound. Follow through on their part (c)
×
+
y coordinate.
<
"24" or
"24"
is acceptable
" 56"
" 56"
Or attempts 2 " 30" 4×−
+
and identifies this as their lower bound.
>−
or
−
1
acceptable for this mark. It cannot be scored for using
2
×
f
+
4
as their lower bound.
5
Note that
g >−
56
or
g
−
56
will score B1M1.
Allow use of max or min for upper and lower bound for this mark.
oe eg “
A1
−
56
g( )
x
24
g x
( )
∈
: 56
−
g( )
x
∩
g( )
x
24
” or “ 56
−
g( ) AND g( )
x
x
[
]
g
∈−
56, 24
or
or other variations similar to these. Condone g for g(x) and 56
−
y
24
−
56
2f( )
x
+
4
24
but not 56
−
f
24
DO NOT ACCEPT “ 56
−
g( )
x
∪
g( )
x
24
” or
" 56
−
g( ) OR g( )
x
x
24"
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