Mathematics · Oct / Nov 2024 · Paper 31 · Question 7

YMA01

As printed in the paper

The question as text

7. y C O x P Figure 3 The curve C has equation y = f (x) , where x  – 7 x + f (x) = x 3 4 7 4

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Show the answer and mark scheme

7(a)[4]
1
f( )
3
4
7
f ( )
′
3
2
4
7
2
3
(
4
7
)
−
x
=
x
x
+
x
=
x
x
+
+
x
x
+
2
(
)
2
3
3
2
4
+
7
+
2
3
x
x
x
x
2
f ( )
′
=
3
4
+
7
+
=
x
x
x
4
7
4
7
x
+
x
+
(
)
7
2
2
+
3
x
x
f ( )
′
=
x
4
7
x
+
7(b)[2]
3
3 4
Substitutes
into
7
...
x = −
x
x +
y
=
2
3
27
−
,
−
2
8
7(c)[2]
27
Attempts
−×−
4
''
''
8
27
y „
2
7(d)[2]
(
)
2, 7
(a)
1
M1: Attempts the product rule to achieve
f ( )
′
2
4
7
3
(
4
7
)
−
where P and Q are positive
x
=
Px
x
+
+
Qx
x
+
2
constants.
Other methods are possible so look at each one carefully.
1
(
)
1
−
7
7
For example:
f( )
=
3
4
+
7
f( )
=
4
+
7
6
f ( )
′
=
4
+
7
6
2
×
x
x
x
x
x
x
x
x
x
2
The same principles for the main mark scheme can be applied
1
(
)
A1:
f ( )
′
=
3
2
4
+
7
+
2
3
4
+
7
−
o.e. which may be left unsimplified
x
x
x
x
x
2
dM1: Dependent upon previous M1. It is for ''correctly'' producing a single fraction with a denominator of
4
x +
7
.
2
3
1
Px
(4
x
+
7)
+
Qx
(
)
Look for
f ( )
′
=
2
4
+
7
+
3
4
+
7
−
o.e
x
Px
x
Qx
x
2
4
x
+
7
7
2
(
2
3
)
x
x
+
A1:
f ( )
′
=
x
4
x
+
7
……………………………………………………………………………………………………
It is possible to attempt this by WMA14 methods using [
]
2
6
(
)
7
f( )
x
=
x
4
x
+
7
=
4
x
+
7
x
Score
6
5
M1, A1:
2f( )f ( )
′
=
28
+
42
x
x
x
x
5
5
2
14
(2
+
5)
14
(2
+
5)
7
(2
+
5)
x
x
x
x
x
x
dM1, A1:
f ( )
′
x
=
=
=
2f( )
x
3
4
+
7
2
4
+
7
x
x
x
……………………………………………………………………………………………………………………………
………………….
(b)
3
3 4
M1: Attempts to substitute
x = −
into
x +
7
to find the y value of the minimum point .
x
2
This may be implied by the correct y value.
It can be attempted even if part (a) is missing or incorrect
If they ignore the given form of f ( )
′
x
and use their own version, they must
1
2
3
(
)
−
Set their f ( )
′
x
which must be of the form
Px
4
x
+
7
+
Qx
4
x
+
7
=
0
2
1
1
Solve
2
4
7
3
(
4
7
)
−
0
by multiplying by (
4
7
)
o.e. to achieve a nonzero
Px
x
+
+
Qx
x
+
=
x +
2
2
value for x
3 4
Substitute this non zero value for x in into
7
to find y
x
x +
3
27
A1:
−
,
−
which may be awarded separately as, for example,
= −
1.5,
= −
3.375
x
y
2
8
Ignore (0, 0) if also given.
(c)
27
M1: Attempts
±×−
4
''
''
which may be seen within an inequality.
8
27
A1:
y „
which must be correctly simplified.
2
27
27
or (
Allow it to be written in other correct forms such as
g
„
,
g( )
x „
−∞
,13.5
2
2
27
The original function is f so
f ( )
x „
would be M1 A0
2
(d)
M1: For one correct coordinate. Look for (
2,... or (
)
)
. Allow for either
2
or
..., 7
x =
If the coordinates have been built up mark the final answer.
1
3
3
(
)
(
)
E.g.
,
→
2,
→
2,15
→−
6,15
would score M0 A0
2
8
8
A1: (
)
2, 7
. Allow
x =
2
and
y =
7
ISW after a correct answer

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