Mathematics · May / June 2021 · Paper 41 · Question 2
YMA01
As printed in the paper
The question as text
2. Diagram NOT y drawn to scale 9 R O 6 x Figure 1 Figure 1 shows a sketch of part of the curve with equation 9 x > 3 y = ( 2 x − 3 1 25 ) . 2 The finite region R, shown shaded in Figure 1, is bounded by the curve, the line with equation y = 9 and the line with equation x = 6 This region is rotated through 2π radians about the x‑axis to form a solid of revolution. Find, by algebraic integration, the exact volume of the solid generated. (7)
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Show the answer and mark scheme
8
2
8
1
oe such as 2
A1:
k =
8 or 0.25
4
(a)(ii)
1
1
1
1
× −
× −
× −
2
2
2
2
M1: Correct attempt at 3rd or 4th term. Eg.
2
)2
or
Ax
=
×
(
kx
3
Bx
=
2!
3!
an attempt at substituting in their value for k to find a value for A or a value for B. Condone a missing
bracket around the kx terms so allow with k instead of k2 or k3 respectively. These may be simplified so
1
1
award for
)2
or
)3
again, condoning k instead of k2 or k3 respectively.
(
'' ''
(
'' ''
A
= −
×
k
B
=
×
k
8
16
1
A1:
A = −
o.e. There is no requirement to simplify the fraction
128
1
3
A1:
o.e You may occasionally see
which is fine.
B =
B =
1024
3072
(b)
M1: For an attempt to substitute
(
)
(
)
2
(
)
3
either
kx =
0.15
into an expansion of the form
1
±
p
×
kx
±
q
×
kx
± ×
r
kx
0.15
1
2
3
or
x
=
into their
1
+
8 x
+
''
A x
''
+
''
B x
''
'' ''
k
A1: 1.072398 Must be to 6 decimal places.
2
Achieves a lower limit of 2
α∫
2 d
∫
81
−
27
Attempts
y
x
=
α
d
x
=
α
×
(
)
(
)
2
x
−
3
2.5
2
x
−
3
1.5
Correct attempt at volume of solid generated under curve =
6
∫
6
k
27
27
2
=
π
y
d
x
=
=
−
−−
27
1
(
2
x
−
3
)
1.5
''2''
''2''
Volume
=
26
π
2
(
)
Correct attempt at volume of solid =
π
×
9
×
6
−
''2''
−
''26 ''
π
298
=
π
B1: Finds the lower limit. This may be awarded anywhere. Accept on the diagram or in the limits of an
integral
∫
...
...
M1: Integrates an expression of the form
d
x
and achieves
oe
(
)
2.5
(
)
1.5
2
x −
3
2
x −
3
α∫
∫
81
−
27
2 d
A1: Correct integration of
y
x
=
α
d
x
giving a solution
α
×
(
2
3
)
2.5
(
2
x
3
)
1.5
x −
−
following through on theirα . Typical values of α will be 1, π or 2π . No need to simplify here
dM1: Attempts to find the volume of the solid generated by rotating the area under the curve.
6
...
Look for
...
It is dependent upon the previous M1.
=
(
)
1.5
2
x
−
3
''2''
9
The top limit must be 6 and the bottom limit must be their solution to
=
9
(
)
1.25
2
x
−
3
A1: Correct exact volume for the solid generated by rotating the area under the curve. Volume
This may be the second part of a larger/complete expression. E.g. ....
−
26
π
ddM1: Correct method for the volume of the solid formed by rotating the area above the curve.
2
(
)
Accept
π
×
9
×
6
−
''2''
−
''26 ''
π
It is dependent both previous M's
A1: 298π
1
2
×
20
×
24
3
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