Mathematics · May / June 2025 · Paper 31 · Question 4
YMA01
As printed in the paper
The question as text
4. The function f is defined by x x 49 7 x x > 3 f ( ) x x x 2 4 12
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[3]
49
x
7
x
49
x
+
7
x x
−
3
+
=
(
)(
)
x
2
+
x
−
12
x
+
4
x
+
4
x
−
3
or e.g.
(
)
(
)
(
)
(
)
7
x x
2
+
x
−
12
49
x x
+
4
+
7
x x
2
+
x
−
12
49
x x
+
4
+
=
(
)(
)(
)
)(
)
)
(
(
x
2
+
x
−
12
x
+
4
x
+
4
x
2
+
x
−
12
x
+
4
x
2
+
x
−
12
(
)
49
x
+
7
x x
−
3
49
x
+
7
x
2
−
21
x
7
x
2
+
28
x
=
=
(
)(
)
(
)(
)
(
)(
)
x
+
4
x
−
3
x
+
4
x
−
3
x
+
4
x
−
3
(
)
7
x x +
4
7
x
=
=
*
(
)
(
)
(
)
x
−
3
x +
4
x
−
3
Do not allow a misread of the numerators of the fractions e.g. 7
x
7
49
x
as
or
as
x +
4
x +
4
x
2
+
x
−
12
M1: Factorises the denominator of the first term to obtain (
)(
)
x
+
4
x
±
3
and attempts to combine the two
...
...
(
3
)
49
x
7
x
x
+
x x
±
fractions using a common denominator to obtain
+
=
oe
x
2
+
x
−
12
x
+
4
(
x
+
4
)(
x
±
3
)
Alternatively multiplies numerator and denominator of both fractions by suitable expressions and
attempts to combine the two fractions e.g.
(
)
(
)
(
)
(
)
7
x x
2
+
x
−
12
...
x x
+
4
+
...
x x
2
+
x
−
12
49
x x
+
4
+
=
(
)(
)(
)
)(
)
)
(
(
x
2
+
x
−
12
x
+
4
x
+
4
x
2
+
x
−
12
x
+
4
x
2
+
x
−
12
A1:
For a correct unsimplified fraction.
Condone missing brackets if they are recovered.
A1*: Achieves the printed answer with sufficient working shown.
The factorisation required for the cancelling needs to be seen within their fraction so that e.g.
(
)
49
x
+
7
x x
−
3
2
49
x
7
x
7
x
+
28
x
7
x
+
=
=
=
scores M1A1A0*
(
)(
)
(
)(
)
(
)
x
2
+
x
−
12
x
+
4
x
+
4
x
−
3
x
+
4
x
−
3
x
−
3
(
)
7
x x +
4
Do not allow just
without the printed answer written down.
(
)
(
)
x +
4
x −
3
For reference if they take the more complicated method, the correct working to reach the printed could be as
follows:
(
)
(
)
49
x x
(
4
)
7
x x
2
x
12
7
x x
2
8
x
16
(
)(
)
+
+
+
−
+
+
7
x x
+
4
x
+
4
7
x
=
=
=
=
)(
)
(
(
)(
)(
)
(
)(
)(
)
(
)
x
+
4
x
2
+
x
−
12
x
+
4
x
+
4
x
−
3
x
+
4
x
+
4
x
−
3
x
−
3
4(b)[2]
(
x
−
3
)
×
7
−
7
x
×
1
7
x
( )
( )
f
x
=
f
′
x
=
(
)
x
−
3
(
x
−
3
)
2
7
x
21
( )
′
( )
f
x
=
f
x
= −
(
)
(
)
2
x
−
3
x
−
3
(b) ALT
7
x
21
21
f
( )
x
7
f
′
( )
x
=
=
+
= −
(
3
)
3
(
)
2
x
−
x
−
x
−
3
(b)
M1: Attempts to differentiate using the quotient rule or product rule.
(
)
α
x
−
3
−
β
x
( )
For the quotient rule it requires achieving
f
′
x
=
αβ
,
>
0
(
)
2
x
−
3
(
)
α
u
−
β
u
+
3
( )
If they use a substitution then similar conditions apply e.g.
u
=
x
−
3
f
′
u
=
u
2
( )
(
)
(
)
For the product rule it requires achieving
f
′
x
=
A x
−
3
−
1
−
Bx x
−
3
−
2
A B
,
>
0
( )
(
)
If they use a substitution then similar conditions apply e.g.
u
=
x
−
3
f
′
u
=
Au
−
1
−
B u
+
3
u
(
)
−
2
A1:
Correct simplified expression. Allow simplified equivalents e.g.
−
21
x
−
3
21
(
)
−
1
Allow
or e.g.
−
−
21
x
2
−
6
x
+
9
2
x
−
6
x
+
9
Isw once a correct simplified answer is seen.
Alternative:
7
x
β
...
β
M1: Expresses (
as
α+
and differentiates to obtain
or e.g.
α+
where u = x – 3
x −
3
)
(
x
3
)
(
)
u
−
x −
3
2
...
and differentiates to obtain
u
2
A1:
(
)
−
2
Correct simplified expression. Allow simplified equivalents e.g.
−
21
x
−
3
21
(
)
−
1
Allow
−
or e.g.
21
x
2
6
x
9
−
−
+
x
2
6
x
9
−
+
Isw once a correct simplified answer is seen.
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