Mathematics · May / June 2021 · Paper 11 · Question 2

YMA01

As printed in the paper

The question as text

2. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. f(x) = ax3 + (6a + 8)x2 – a2x where a is a positive constant. Given f(–1) = 32

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Show the answer and mark scheme

2(a)[5]
→
... ,...
x
1
x
→
...
x
,...
x
−
1
→
...
x
−
2
, 5
→
0
2
2
3
−
2
One of 2
8
8
−
A1
3 x ,
−
2x
3 x−
2
or exact simplified equivalent terms eg
or
2
,
−
or
−
(
)
3
2
3x
x
−
2.6x−
&
2
3
2
−
Condone
3 x and condone double signs such as
1
+ −
2x
The terms do not need to be on
2
the same line for this mark.
Two of 2
3
−
2
8
8
−
A1
3 x ,
−
2x
,
−
3 x−
2
or exact simplified equivalent terms eg
or
−
.
2
(
)
3
3x
2
x
2
3
−
Condone
3 x and condone double signs such as
1
+ −
2x
2
. The terms do not need to be on
the same line for this mark.
3
2
2
8
or exact simplified equivalent all on one line eg 2
8
−
A1
3 x
−
2x
3 x−
2
3 x
−
2
−
−
(
)
3
3x
2
x
2
3
8
−
Do not allow
3 x
1
−
2x
−
3 x−
2
or double signs and do not isw in this part (including
2
rounding decimals).
(
a
+
8)(
a
−
3)
=
0
or
a = − and chooses
with reason *
a =
3
8
a =
3
3
x
3
+
26
x
2
−
9
x
=
0
⇒
x
(3
x
2
+
26
x
−
9)
=
0
(ii)
x
(3
x
−
1)(
x
+
9)
1
(
)
x =
0,
, 9
−
3
(b)(i)
(
) 0
y =
1
1
3
1
1
=
y
=
" "or
y
= −
" 9"
⇒
y
=
...
(or
or
...
)
3
3
( 9)
−
3
=
...
3
3
1
(
)
y =
,
−
729
27
(b)(ii)
1
9
z
=
" "
→
z
=
...
3
1
(
)
z =
−
only
2
(a)(i)
M1
Substitutes in
x
= ±
1,
y
=
32
and proceeds to a 3TQ in terms of a with all terms on one side
of the equation. Condone the lack of = 0 and condone slips in their rearrangement.
dM1
Attempts to solve their quadratic equation by either factorising, completing the square or the
quadratic formula. They cannot just state the roots. It is dependent on the first method mark.
In all cases they must show their working to score this mark so:
Solving by factorising requires the factorised form of their
a
2
5
a
24
+
−
=
(
)
stated before proceeding to the roots i.e
(
a
+
"8")(
a
−
"3")
=
0
Solving by using the quadratic formula requires the values for a, b and c to be
stated in the formula before proceeding to the roots. They cannot just state the
values of a, b and c.
2
to find the roots.
A1*
a =
2(b)[5]
→
... ,...
x
1
x
→
...
x
,...
x
−
1
→
...
x
−
2
, 5
→
0
2
2
3
−
2
One of 2
8
8
−
A1
3 x ,
−
2x
3 x−
2
or exact simplified equivalent terms eg
or
2
,
−
or
−
(
)
3
2
3x
x
−
2.6x−
&
2
3
2
−
Condone
3 x and condone double signs such as
1
+ −
2x
The terms do not need to be on
2
the same line for this mark.
Two of 2
3
−
2
8
8
−
A1
3 x ,
−
2x
,
−
3 x−
2
or exact simplified equivalent terms eg
or
−
.
2
(
)
3
3x
2
x
2
3
−
Condone
3 x and condone double signs such as
1
+ −
2x
2
. The terms do not need to be on
the same line for this mark.
3
2
2
8
or exact simplified equivalent all on one line eg 2
8
−
A1
3 x
−
2x
3 x−
2
3 x
−
2
−
−
(
)
3
3x
2
x
2
3
8
−
Do not allow
3 x
1
−
2x
−
3 x−
2
or double signs and do not isw in this part (including
2
rounding decimals).
(
a
+
8)(
a
−
3)
=
0
or
a = − and chooses
with reason *
a =
3
8
a =
3
3
x
3
+
26
x
2
−
9
x
=
0
⇒
x
(3
x
2
+
26
x
−
9)
=
0
(ii)
x
(3
x
−
1)(
x
+
9)
1
(
)
x =
0,
, 9
−
3
(b)(i)
(
) 0
y =
1
1
3
1
1
=
y
=
" "or
y
= −
" 9"
⇒
y
=
...
(or
or
...
)
3
3
( 9)
−
3
=
...
3
3
1
(
)
y =
,
−
729
27
(b)(ii)
1
9
z
=
" "
→
z
=
...
3
1
(
)
z =
−
only
2
(a)(i)
M1
Substitutes in
x
= ±
1,
y
=
32
and proceeds to a 3TQ in terms of a with all terms on one side
of the equation. Condone the lack of = 0 and condone slips in their rearrangement.
dM1
Attempts to solve their quadratic equation by either factorising, completing the square or the
quadratic formula. They cannot just state the roots. It is dependent on the first method mark.
In all cases they must show their working to score this mark so:
Solving by factorising requires the factorised form of their
a
2
5
a
24
+
−
=
(
)
stated before proceeding to the roots i.e
(
a
+
"8")(
a
−
"3")
=
0
Solving by using the quadratic formula requires the values for a, b and c to be
stated in the formula before proceeding to the roots. They cannot just state the
values of a, b and c.
2
to find the roots.
A1*
a =

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