Mathematics · Oct / Nov 2020 · Paper 41 · Question 3
YMA01
As printed in the paper
The question as text
3. y y = e 0.5x −2 O x R Figure 1 Figure 1 shows a sketch of part of the curve with equation y = e 0.5x −2 The region R, shown shaded in Figure 1, is bounded by the curve, the x-axis and the y-axis. The region R is rotated 360° about the x-axis to form a solid of revolution. Show that the volume of this solid can be written in the form a ln 2 + b, where a and b are constants to be found. (6)
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Show the answer and mark scheme
Calculates the upper limit as 2ln 2 or ln 4
∫
(
)
2
∫
(
)
e
0.5
x
−
2 d
x
=
e
x
−
4e
0.5
x
+
4 d
x
=
e
x
−
8e
0.5
x
+
4
x
''2ln2''
x
0.5
x
Volume =
π
''e
−
8e
+
4 ''
x
=
...
0
Volume = 8 ln 2
π
−
5
π
B1: Calculates the upper limit as 2ln 2 or ln 4 only. Do not accept 1.386, ln 2
0.5 or anything else.
Recovery from ln 2
0.5 is allowed so if the candidate progresses to 8 ln 2
π
−
5
π
do not withhold this mark.
M1: Attempts to square (
)
∫
x −
kx x →
e
0.5
2
with
e d
e
kx
seen at least once. Ignore any factor of π
This may be awarded when candidates fails to square
e
0.5
x correctly
∫
(
)
2
So for example
e
0.5
x
−
4e
0.5
x
+4 d
x →
.... ...e
±
0.5
x
is sufficient
A1: Two terms correct
e
x
8e
0.5
x
4
x
Ignore any factor of π
−
+
A1: All three terms correct
e
x
−
8e
0.5
x
+
4
x
Ignore any factor of π
dM1: A full and complete method of finding the volume of the solid generated.
'' ''
a
(
)
∫
2
with a being 2ln 2, ln 4 , 1.38.. or ln 2
0.5
Look for an attempt to find
π
e
x
−
2 d
x
2
0
It is dependent upon the previous M so some correct integration must be seen as well as use of
both limits. There must be some evidence for the 0. Those terms cannot just ''disappear'' or be set to 0.
Condone (
)
A1: cao 8 ln 2
π
−
5
π
or 5
−
π
+
8 ln 2
π
π
8ln 2
−
5
(Note that a requirement of the question is to write in the form ln 2
a
+ )
b
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