Mathematics · Oct / Nov 2020 · Paper 21 · Question 6
YMA01
As printed in the paper
The question as text
6. y C1 B R A C2 O x Figure 1 Figure 1 shows a sketch of part of the curves C1 and C2 with equations C1 : y = x3 – 6x + 9 x 0 C2 : y = –2x2 + 7x – 1 x 0 The curves C1 and C2 intersect at the points A and B as shown in Figure 1. The point A has coordinates (1, 4). Using algebra and showing all steps of your working,
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Show the answer and mark scheme
6(a)[4]
x
3
−
6
x
+
9
=−
2
x
2
+
7
x
−
1
M1
Sets C1 = C2, and collects terms
⇒
...
(
)
Correct cubic equation. The “= 0” may
⇒±
x
3
+
2
x
2
−
13
x
+
10
=
0
A1
be implied by their attempt to solve.
Examples:
)(
)
(
x
3
+
2
x
2
−
13
x
+
10
=
x
−
1
x
2
+
...
x
+
...
=
(
x
−
1)(
x
+
...)(
x
+
..)
⇒
x
=
...
Attempts to factorise using (x – 1) as a factor or uses long division by (x – 1) to obtain a
quadratic factor and proceeds to solve quadratic or factorise and solve
)(
)
(
NB
x
3
+
2
x
2
−
13
x
+
10
=
x
−
1
x
2
+
3
x
−
10
M1
or
x
3
+
2
x
2
−
13
x
+
10
=
(
x
−
1)(
x
+
...)(
x
+
..)
⇒
x
=
...
Attempts 3 factors directly (by considering roots)
or
x
3
+
2
x
2
−
13
x
+
10
=
0
⇒
x
=
...
Solves (using calculator) to obtain 3 roots (may need to check if cubic incorrect)
(
)
x
=
2,
y
=
5 or
2, 5
Correct values from a correct cubic.
A1
Allow as a coordinate pair or written separately.
If there are any errors in the algebra e.g. wrong factors, wrong working etc. this mark should
be withheld even if they have (2, 5) and score as M1A1B1(Second M on EPEN)A0
Special Case
If you see:
x
3
−
6
x
+
9
=−
2
x
2
+
7
x
−
1
⇒
x
3
+
2
x
2
−
13
x
+
10
=
0
(
)
⇒
x
=
2,
y
=
5 or
2, 5
Score M1A1B1(Second M on EPEN)A0
6(b)[5]
For increasing any power of x by 1 for
x
n
→
x +
n
1
M1
C1 or C2 or for ± (C1 – C2)
∫
{
(
)
}
∫
(
)
±
−
2
x
2
+
7
x
−−
1
x
3
−
6
x
+
9
d
x
= ±
−
x
3
−
2
x
2
+
13
x
−
10 d
x
x
4
2
x
3
13
x
2
= ± −
−
+
−
10
x
4
3
2
or
{
}
∫
(
)
∫
(
)
±
−
2
x
2
+
7
x
−
1 d
x
−
x
3
−
6
x
+
9 d
x
2
x
3
7
x
2
x
4
6
x
2
dM1A1
= ± −
+
−
x
−
−
+
9
x
3
2
4
2
or
∫
(
)
2
x
3
7
x
2
∫
(
)
x
4
6
x
2
−
2
x
2
+
7
x
−
1 d
x
= −
+
−
x
,
x
3
−
6
x
+
9 d
x
=
−
+
9
x
3
2
4
2
dM1: For correct integration of 1 term for C1 and one term for C2
or for correct integration for 2 terms of their ± (C1 – C2)
A1: Fully correct integration of both C1 and C2 or for ± (C1 – C2). Award this mark as soon
as fully correct integration is seen and ignore subsequent work.
2 2
( )
3
13 2
( )
2
2 1
( )
3
13 1
( )
2
2
4
1
4
( )
( )
= −
−
+
−
10 2
−−
−
+
−
10 1
4
3
2
4
3
2
ddM1
Fully correct strategy for the area. Depends on both previous M marks.
Uses the limits “2” and 1 in their “changed” expression(s) and subtracts either way round.
13
=
12
If the attempt is correct apart from subtracting the wrong way round (for limits or functions)
A1
13
and
is obtained, allow recovery if they then make their answer positive.
−
12
Some values for reference:
2
2
−
2
x
3
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