Mathematics · Oct / Nov 2021 · Paper 21 · Question 8
YMA01
As printed in the paper
The question as text
8. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. y C M l R O x Figure 2 Figure 2 shows a sketch of part of the curve C with equation y = 4 3 x3 − 11x2 + kx where k is a constant The point M is the maximum turning point of C and is shown in Figure 2. Given that the x coordinate of M is 2
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Show the answer and mark scheme
8(a)[3]
y
=
x
3
−
11
x
2
+
kx
⇒
x =
4
x
2
−
22
x
+
k
3
d
d
y
uses
x
=
2,
=
0
⇒0
=
16
−
44
+
k
⇒
k
=
28
*
d
x
d
y
2
(
)(
)
8(b)[2]
=
4
x
−
22
x
+
28
=
0
⇒
2
x
−
4
2
x
−
7
=
0
⇒
x
=
..
.
d
x
7
x
<
2,
x
>
2
∫
4
1
11
x
3
−
11
x
2
+
28
x
d
x
⇒
x
4
−
x
3
+
14
x
2
8(c)[5]
3
3
3
68
Correct y coordinate of M
=
3
68
∫
2
4
Complete method to find R =
2 ''
×
''
−
3 x
3
−
11
x
2
+
28
x
d
x
3
0
68
1
11
4
3
2
=
2 ''
×
''
−
×
2
−
×
2
+
14 2
×
3
3
3
40
=
3
(a)
M1: Attempts d
y
x with one index correct. Must be seen in part (a).
d
into d
y
. The d
y
x =
2
=
0
ax
2
+
bx
+
dM1: Substitutes
x and sets
x must be of the form
d
d
the substitution must be clear. Going directly to 28
−
+
k
=
0
is M0. The “=0” may be
implied by an attempt to solve the equation for this mark.
A1*: Achieves
k =
28
via a correct intermediate line and no missing “=0”
(b)
M1: Attempts to find the critical values.
7
… Do not accept 7
7
A1:
x
<
2,
x
>
or
x
„
2,
x
<
x
<
2
Accept alternative set
2
2
2
notations.
(c)
M1: Attempts to integrate with one index correct
∫
4
1
11
x
3
−
11
x
2
+
28
x
d
x
⇒
x
4
−
x
3
+
14
x
2
A1:
Correct integration
need not be
3
3
3
simplified
68
B1:
Correct y coordinate of M
=
. Accept awrt 22.7 for this mark. May be seen anywhere
3
e.g. on the sketch
2
68
∫
4
2 ''
''
3 x
3
11
x
2
28
x
d
x
M1: Complete method to find R =
×
−
−
+
OR
3
0
∫
2
68
4
−
3 x
3
−
11
x
2
+
28
x
d
x
The lower limit may be implied. The integral
3
0
must be a changed function.
40
A1:
3 or exact equivalent
(If x and y are positive) (
)
2
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