Mathematics · May / June 2024 · Paper 21 · Question 9
YMA01
As printed in the paper
The question as text
9. y P C R1 k O x R2 Figure 1 Figure 1 is a sketch of the curve C with equation 3 2 (4 – x) x 0 y = 2x The point P is the stationary point of C.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[4]
(
)
y
=
2
x
2
4
−
x
=
8
x
2
−
2
x
2
d
1
3
y
x =
12
x
2
−
5
x
2
d
1
3
12
Stationary point
12
−
5
=
0
=
x
2
x
2
x
5
3
5
5
7
16
4
2
2
2
2
9(b)[4]
8
x
−
2
x
d
x
=
x
−
x
5
7
5
7
16
4
16 7
×
28
2
−
2
=
0
=
=
k
k
k
5
7
4 5
5
×
(a)
M1:
Complete attempt to differentiate.
Writes as a sum of two terms and reduces the power of at least one term by 1
1
3
Alternatively, by the product rule look for
Ax
2
(4
−
x
)
−
Bx
2
1
3
d
y
1
3
A1:
=
12
−
5
or
3
(4
−
)
−
2
. May be left unsimplified
x
2
x
2
x
2
x
x
2
d
x
dM1: Sets d
y
x =
0
and attempts to solve to find a nonzero value for x. Do not be concerned
d
about the method of solving for this mark.
12
A1cso:
x =
from correct work. Ignore reference to x = 0. Watch for incorrect methods of
5
2
1
3
12
solving, such as
3
, which score A0.
1
2
x
2
−
5
x
2
=
0
144
x
−
2
5
x
=
0
x
=
5
(b)
M1:
Complete attempt to integrate. Writes as a sum of two terms and increases the power
of at least one term by 1. For attempts at integrating by parts they must proceed
through the second integration to score the mark.
5
7
16
4
2
2
A1:
x
−
x
which may be unsimplified
5
7
5
7
16
4
dM1: Substitutes x = k in their
2
−
2
=
0
and solves. Alternatively, applies limits 0
x
x
5
7
1024
and 4 to find the area R1
=
and sets equal to the negative of the expression
35
from limits 4 to k (or reverses the limits and sets equal) and proceeds to reach a
value for k.
28
A1:
k =
(ignore references to k = 0 ) Accept as decimal 5.6.
5
Alt for the dM:
4
k
16
5
4
7
1024
16
5
4
7
1024
Area
R
=
x
2
−
x
2
=
Area
R
=
x
2
−
x
2
= −
k
=
1
5
7
3
5
2
5
7
35
0
4
4
k
16
5
4
7
1024
16
5
4
7
Note if
Area
R
= Area
R
=
x
2
−
x
2
=
=
x
2
−
x
2
k
=
...
1
2
5
7
35
5
7
0
4
this scores dM0. However, if a student realises the areas cancel and effectively achieves the
correct equation after stating areas equal then allow recovery for dM1A1.
6
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