Mathematics · Oct / Nov 2021 · Paper 21 · Question 4
YMA01
As printed in the paper
The question as text
4. f (x) = (x2 − 2)(2x − 3) − 21
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[1]
States 21
−
(
)(
)
4(b)[2]
Attempts
f (3)
=
3
2
−
2
2 3
× −
3
−
21
⇒ (
)
Achieves f(3)
=
0
x −
3
is a factor of f ( )
x *
)
)(
3
(
(c) (i)
f ( )
x
=
2
x
−
3
x
2
−
4
x
−
15
=
x
−
3
2
x
2
....
±
5
)(
)
(
=
x
−
3
2
x
2
+
3
x
+
5
2
2
(ii)
Attempts
b
−
4
ac
for their
2
x
+
3
x
+
5
2
b
−
4
ac
<
0
x =
3
Achieves
and states that only root is
o.e. *
(a)
B1:
−
21
(b)
(
)(
)
M1: Attempts to substitute x = 3 into
f ( )
x
=
x
2
−
2
2
x
−
3
−
21
or its expanded form.
Condone slips but don't accept just f (3)
=
0
. Attempts via long division score M0.
and states that (
)
A1*: Achieves f (3)
=
0
x −
3
is a factor of f ( )
x . If this latter is given in a
preamble accept a minimal conclusion such as // or QED
(c)(i)
(
)(
)
B1:
Expands
f ( )
x
=
x
2
−
2
2
x
−
3
−
21
to reach correct 4 term cubic. Allow if the cubic
is seen anywhere.
M1: Correct attempt to find quadratic factor using either division by (
)
x −
3
or inspection.
To score the M mark look for e.g. correct first and last terms by inspection, or first two
terms by division. (Two correct terms will imply the mark.) Allow for work seen in (b) a
long as it is referred to in (c).
)(
)
A1: (
)(
f ( )
x
=
x
−
3
2
x
2
+
3
x
+
5
following a correct cubic. Must be seen together on one
line.
(c) (ii)
M1: Attempts to show their quadratic factor has no real roots. Factorisation attempts are M0.
Accept via
2
b
−
4
ac
an attempt at
for their quadratic factor
2
x
2
+
3
x
+
5
an attempt to solve their
using the quadratic formula or calculator
an attempt to complete the square
A1*: Requires correct factorisation, correct calculation, reason and conclusion
)(
)
(
2
For example after (c)(i)
f ( )
x
=
x
−
3
2
x
+
3
x
+
5
accept e.g.
23
2
2
x
+
3
x
+
5
has no roots as
−× × <
4 2 5
0
so f ( )
x =
0
only has root at
3
31
, (
)
2
x
2
+
3
x
+
5
=
0
⇒
x
= −
±
i
x
−
3
=
0
⇒
x
=
3
. So only one real root.
4
4
2
3
9
31
2
x
2
+
3
x
+
5
=
0
⇒
2
x
+
−
+
5
…
>
0
, so no roots hence 3 is the only
4
8
8
root
There must be some reference to the root either by stating it or indicating the linear term
has a root (e.g. writing “root” next to it), but do not accept incorrect statements such as
only real root is (
)
x −
3
Do not allow statements such as “Math error” without interpretation of what this means.
Attempts 480 13
15
4(c)[5]
States 21
−
(
)(
)
Attempts
f (3)
=
3
2
−
2
2 3
× −
3
−
21
⇒ (
)
Achieves f(3)
=
0
x −
3
is a factor of f ( )
x *
)
)(
3
(
(c) (i)
f ( )
x
=
2
x
−
3
x
2
−
4
x
−
15
=
x
−
3
2
x
2
....
±
5
)(
)
(
=
x
−
3
2
x
2
+
3
x
+
5
2
2
(ii)
Attempts
b
−
4
ac
for their
2
x
+
3
x
+
5
2
b
−
4
ac
<
0
x =
3
Achieves
and states that only root is
o.e. *
(a)
B1:
−
21
(b)
(
)(
)
M1: Attempts to substitute x = 3 into
f ( )
x
=
x
2
−
2
2
x
−
3
−
21
or its expanded form.
Condone slips but don't accept just f (3)
=
0
. Attempts via long division score M0.
and states that (
)
A1*: Achieves f (3)
=
0
x −
3
is a factor of f ( )
x . If this latter is given in a
preamble accept a minimal conclusion such as // or QED
(c)(i)
(
)(
)
B1:
Expands
f ( )
x
=
x
2
−
2
2
x
−
3
−
21
to reach correct 4 term cubic. Allow if the cubic
is seen anywhere.
M1: Correct attempt to find quadratic factor using either division by (
)
x −
3
or inspection.
To score the M mark look for e.g. correct first and last terms by inspection, or first two
terms by division. (Two correct terms will imply the mark.) Allow for work seen in (b) a
long as it is referred to in (c).
)(
)
A1: (
)(
f ( )
x
=
x
−
3
2
x
2
+
3
x
+
5
following a correct cubic. Must be seen together on one
line.
(c) (ii)
M1: Attempts to show their quadratic factor has no real roots. Factorisation attempts are M0.
Accept via
2
b
−
4
ac
an attempt at
for their quadratic factor
2
x
2
+
3
x
+
5
an attempt to solve their
using the quadratic formula or calculator
an attempt to complete the square
A1*: Requires correct factorisation, correct calculation, reason and conclusion
)(
)
(
2
For example after (c)(i)
f ( )
x
=
x
−
3
2
x
+
3
x
+
5
accept e.g.
23
2
2
x
+
3
x
+
5
has no roots as
−× × <
4 2 5
0
so f ( )
x =
0
only has root at
3
31
, (
)
2
x
2
+
3
x
+
5
=
0
⇒
x
= −
±
i
x
−
3
=
0
⇒
x
=
3
. So only one real root.
4
4
2
3
9
31
2
x
2
+
3
x
+
5
=
0
⇒
2
x
+
−
+
5
…
>
0
, so no roots hence 3 is the only
4
8
8
root
There must be some reference to the root either by stating it or indicating the linear term
has a root (e.g. writing “root” next to it), but do not accept incorrect statements such as
only real root is (
)
x −
3
Do not allow statements such as “Math error” without interpretation of what this means.
Attempts 480 13
15
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