Mathematics · Oct / Nov 2021 · Paper 31 · Question 1

YMA01

As printed in the paper

The question as text

1. The function f is defined by x x 5 5 x = + x > 0 f ( ) x + x + x + 2 7 12 4 x 5

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

1(a)[3]
(
)
5
x
5
x
5
x
+
5
x x
+
3
+
=
(
)(
)
x
2
+
7
x
+
12
x
+
4
x
+
3
x
+
4
2
5
x
+
20
x
5 (
x x
+
4)
5
x
=
=
=
*
(
)(
)
(
)
x
+
3
x
+
4
x
+
3 (
x
+
4)
x
+
3
1(b)[3]
5
x
y
=
⇒
xy
+
3
y
=
5
x
⇒
5
x
−
xy
=
3
y
(
)
x
+
3
3
y
3
x
−
1
( )
f
x
=
⇒
x
=
So
5
−
y
5
−
x
Domain 0
<
x
<
5
(c) (i)
) (
)
5
x
5
x
+
3
−
5
x
15
f ( )
(
f ( )
′
x
=
⇒
x
=
=
(
)
x
+
3
(
)
2
(
)
2
x
+
3
x
+
3
(ii)
States (f is an) increasing function with a suitable reason
E.g. Since (
)
x +
3
2
is positive
(a)
M1
Attempts to combine the two fractions using a common denominator.
Allow errors on the numerator but at least one of the terms must have been adapted.
Usual rules apply for factorising the quadratic denominator.
Condone invisible brackets and slips on the numerator when combining the two
fractions.
Allow the two fractions to be written separately with the same denominator.
A1
For a correct unsimplified fraction with a quadratic numerator and denominator (which
may be implied)
(
)
(
)
2
(
)
5
x x
+
4
5
x x
+
7
x
+
12
3
2
x
5
x
+
20 (
x
+
4)
5
x
5
x
5
x
+
40
x
+
80
x
+
=
+
=
=
(
)(
(
)
x
2
7
x
12
...
...
2
2
+
+
x
+
4
x
+
7
x
+
12
)
x
+
7
x
+
12
x
+
4
(
x
+
4)
Correctly achieves the given answer of 5
x
A1*
showing intermediate steps (cso). Expect
x +
3
to see the two fractions combined and then both the numerator and denominator
factorised before cancelling terms to achieve full marks. In the case of forming a cubic
numerator and denominator you must see the terms collected before factorising. Bracket
errors at some point in their working is A0*
(b)
5
x
5
y
M1
Attempts to change the subject on
y
=
(or
x
=
). Look for cross
x
+
3
y
+
3
multiplication with an attempt to collect terms. Do not follow through on their answer
to part (a)
3
x
−
3
x
−
1
( )
−
1
( )
f
x
=
f
x
=
A1
or
Must be in terms of x. Condone
f
−=
1
...
(or
5
−
x
x
−
5
f
−=
1
y
=
...
) but do not allow just
y =
...
or
f
−
1
:
y
=
...
Correct domain 0
<
x
<
5
A1
(c) Mark (i) and (ii) together
M1
Attempts to use the quotient rule or product rule. Look for an expression of the form
A x
(
3
)
5
x
+
−
or
5 (
3)
1
(
3)
1
5 (
3)
2
where A and B are non
x x
+
−
→
B x
+
−
±
x x
+
−
(
)
2
x
+
3
zero constants.
15
15
5
5
x
(
)
A1
f ( )
′
x
=
or
15(
x
+
3)
−
2
or
Do not allow
−
for
(
x
3
)
2
x
2
+
6
x
+
9
x
+
3
(
x
+
3)
2
+
this mark.
15
They achieve (
)
f ( )
′
x
=
A1
and states
(
)
2
x
+
3
(that f is an) increasing function
since (
3
)
2
is positive
x +
Alt methods in (a)
Taking out common factors
(
)
(
)
1
+
x
+
3
1
+
x
+
3
5
x
5
x
5
x
+
=
5
x
=
M1A1:
or
(
)(
)
(
)
(
)
2
7
12
x
+
3
x
+
4
x
+
4
x
+
3
x
+
x
+
x
+
4
Methods where candidates ''split up'' fractions.
5
x
5
M1
Attempts
OR
by partial fractions
x
2
+
7
x
+
12
x
2
+
7
x
+
12
5
x
5
x
5
x
A1
−
+
x
+
3
x
+
4
x
+
4
5
x
You would expect to see 5
x
5
x
5
x
A1*
−
+
before proceeding to the
x +
3
x
+
3
x
+
4
x
+
4
given answer.
Alt method in (b)
5
x
k
y
=
y
=
5
±
M1
Writes
as
and then attempts to make x the subject.
x
+
3
x
+
3
15
−
1
( )
f
x
=
−
3
A1
or equivalent
5
−
x
0
<
x
<
5
A1
Alt method in (c)
5
x
k
C
5
M1
Writes
y
=
as
y
=
±
and attempts the chain rule to get
±
x
+
3
x
+
3
(
)
x
+
3
2
15
(
)
A1
f ( )
′
x
=
or exact equivalent
(
)
2
x
+
3
States that since (
)
2
A1
x +
3
is positive so f is increasing function
1(c)[3]
(
)
5
x
5
x
5
x
+
5
x x
+
3
+
=
(
)(
)
x
2
+
7
x
+
12
x
+
4
x
+
3
x
+
4
2
5
x
+
20
x
5 (
x x
+
4)
5
x
=
=
=
*
(
)(
)
(
)
x
+
3
x
+
4
x
+
3 (
x
+
4)
x
+
3
5
x
y
=
⇒
xy
+
3
y
=
5
x
⇒
5
x
−
xy
=
3
y
(
)
x
+
3
3
y
3
x
−
1
( )
f
x
=
⇒
x
=
So
5
−
y
5
−
x
Domain 0
<
x
<
5
(c) (i)
) (
)
5
x
5
x
+
3
−
5
x
15
f ( )
(
f ( )
′
x
=
⇒
x
=
=
(
)
x
+
3
(
)
2
(
)
2
x
+
3
x
+
3
(ii)
States (f is an) increasing function with a suitable reason
E.g. Since (
)
x +
3
2
is positive
(a)
M1
Attempts to combine the two fractions using a common denominator.
Allow errors on the numerator but at least one of the terms must have been adapted.
Usual rules apply for factorising the quadratic denominator.
Condone invisible brackets and slips on the numerator when combining the two
fractions.
Allow the two fractions to be written separately with the same denominator.
A1
For a correct unsimplified fraction with a quadratic numerator and denominator (which
may be implied)
(
)
(
)
2
(
)
5
x x
+
4
5
x x
+
7
x
+
12
3
2
x
5
x
+
20 (
x
+
4)
5
x
5
x
5
x
+
40
x
+
80
x
+
=
+
=
=
(
)(
(
)
x
2
7
x
12
...
...
2
2
+
+
x
+
4
x
+
7
x
+
12
)
x
+
7
x
+
12
x
+
4
(
x
+
4)
Correctly achieves the given answer of 5
x
A1*
showing intermediate steps (cso). Expect
x +
3
to see the two fractions combined and then both the numerator and denominator
factorised before cancelling terms to achieve full marks. In the case of forming a cubic
numerator and denominator you must see the terms collected before factorising. Bracket
errors at some point in their working is A0*
(b)
5
x
5
y
M1
Attempts to change the subject on
y
=
(or
x
=
). Look for cross
x
+
3
y
+
3
multiplication with an attempt to collect terms. Do not follow through on their answer
to part (a)
3
x
−
3
x
−
1
( )
−
1
( )
f
x
=
f
x
=
A1
or
Must be in terms of x. Condone
f
−=
1
...
(or
5
−
x
x
−
5
f
−=
1
y
=
...
) but do not allow just
y =
...
or
f
−
1
:
y
=
...
Correct domain 0
<
x
<
5
A1
(c) Mark (i) and (ii) together
M1
Attempts to use the quotient rule or product rule. Look for an expression of the form
A x
(
3
)
5
x
+
−
or
5 (
3)
1
(
3)
1
5 (
3)
2
where A and B are non
x x
+
−
→
B x
+
−
±
x x
+
−
(
)
2
x
+
3
zero constants.
15
15
5
5
x
(
)
A1
f ( )
′
x
=
or
15(
x
+
3)
−
2
or
Do not allow
−
for
(
x
3
)
2
x
2
+
6
x
+
9
x
+
3
(
x
+
3)
2
+
this mark.
15
They achieve (
)
f ( )
′
x
=
A1
and states
(
)
2
x
+
3
(that f is an) increasing function
since (
3
)
2
is positive
x +
Alt methods in (a)
Taking out common factors
(
)
(
)
1
+
x
+
3
1
+
x
+
3
5
x
5
x
5
x
+
=
5
x
=
M1A1:
or
(
)(
)
(
)
(
)
2
7
12
x
+
3
x
+
4
x
+
4
x
+
3
x
+
x
+
x
+
4
Methods where candidates ''split up'' fractions.
5
x
5
M1
Attempts
OR
by partial fractions
x
2
+
7
x
+
12
x
2
+
7
x
+
12
5
x
5
x
5
x
A1
−
+
x
+
3
x
+
4
x
+
4
5
x
You would expect to see 5
x
5
x
5
x
A1*
−
+
before proceeding to the
x +
3
x
+
3
x
+
4
x
+
4
given answer.
Alt method in (b)
5
x
k
y
=
y
=
5
±
M1
Writes
as
and then attempts to make x the subject.
x
+
3
x
+
3
15
−
1
( )
f
x
=
−
3
A1
or equivalent
5
−
x
0
<
x
<
5
A1
Alt method in (c)
5
x
k
C
5
M1
Writes
y
=
as
y
=
±
and attempts the chain rule to get
±
x
+
3
x
+
3
(
)
x
+
3
2
15
(
)
A1
f ( )
′
x
=
or exact equivalent
(
)
2
x
+
3
States that since (
)
2
A1
x +
3
is positive so f is increasing function

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