Mathematics · May / June 2022 · Paper 41 · Question 3
YMA01
As printed in the paper
The question as text
3. x 3x Figure 1 A tablet is dissolving in water. The tablet is modelled as a cylinder, shown in Figure 1. At t seconds after the tablet is dropped into the water, the radius of the tablet is x mm and the length of the tablet is 3x mm. 2 s –1 The cross-sectional area of the tablet is decreasing at a constant rate of 0.5 mm
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[4]
3
3
2
k
the multiple outside. B0 if brackets on
missing unless implied by recovery.
3
M1: Sets their coefficient of
2x equal to 3 times their coefficient of x to produce a two term quadratic equation
in terms of k.
3−. May be
2
A1*: Achieves given answer from a correct equation, but condone if B and C both missed the
scored if A was incorrect.
(c)(i)
k = − only. The k = 0 solution must be rejected.
6
B1:
(ii)
M1: Substitutes their nonzero value for k into a correct expression for the coefficient of
3x . Must include the
3−
2
A1: 32
9 oe
2(a)
1
A
B
=
+
1
=
A
(1
−
x
)
+
B
(1 3 )
+
x
(1 3 )(1
+
x
−
x
)
1 3
+
x
1
−
x
1
4
when
x =
1
=
1
4
B
B
=
...
or when
x = −
=
1
3 A
A
=
...
3
3
1
+
4(1
+
3 )
x
4(1
−
x
)
cot
y
d
y
=
... d
x
"lnsin "
y
=
... d
x
3
1
...
=
"
" "
+
" d
x
=
...ln(1
+
3 )
x
±
...ln(1
−
x
)
(
+
c
)
4(1
+
3 )
x
4(1
−
x
)
1
1
lnsin
y
=
ln(1
+
3 )
x
−
ln(1
−
x
)
(
+
c
)
oe
4
4
π
1
1
1
1
1
lnsin
=
ln 1
+ ×
3
−
ln 1
−
+
c
c
=
...
= −
ln5
2
4
2
4
2
4
k
lnsin
y
m
ln(...)
si
n
k
y
...
m
=
=
or
k
lnsin
y
=
...
sin
k
y
=
exp(...)
1
+
3
x
sin
4
y
=
5(1
)
−
x
Notes
(a)
B1: For a correct suitable identity without fractions, such as 1
=
A
(1
−
x
)
+
B
(1 3 ),
+
x
seen or implied.
M1: Attempts to find one of the constants by either substitution or equating coefficients. May be implied by a
correct value for A or B via cover up rule.
3
1
+
oe allow values for A and B to be stated following a correct partial fraction form, or if
A1:
4(1
+
3 )
x
4(1
−
x
)
correct partial fractions see in (b).
(b)
d
y
(oe for cot y) and integrate cot y . Accept any
M1: Attempts to separate variables to form
c t
o
y x
=
g
(
x
)
d
changed function for the attempt but must be attempting to integrate cot y (oe).
...
M1: Attempts to integrate their partial fractions from (a) so award for
x →
...ln(1
+
3 ) or ...ln 4
x
(1
3 )
+
...
(
)
and
x →
...ln(1
−
x
) or ...ln 4
−
4
x
oe
(1
−
)
A1ft: Correct expression (any equivalent) (both sides). Follow through on their constants for the partial
fractions. Condone the absence of the constant of integration.
dM1: Depends on second M, and must have attempted to integrate both sides. Uses the initial conditions in an
equation with a constant of integration. May integrate between limits to achieve this. (Accept if a value for
c cannot be reached from their equation.)
s n
i
M1: Attempts to rearrange their equation by correctly using log work to reach the required form
Must have had kln sin y = … (k may be 1). Not dependent may be gained before finding the constant if ln A
is used, and allow if the constant is missing.
1
3
+
x
A1:
sin
3(b)[4]
cot
y
d
y
=
... d
x
"lnsin "
y
=
... d
x
3
1
...
=
"
" "
+
" d
x
=
...ln(1
+
3 )
x
±
...ln(1
−
x
)
(
+
c
)
4(1
+
3 )
x
4(1
−
x
)
1
1
lnsin
y
=
ln(1
+
3 )
x
−
ln(1
−
x
)
(
+
c
)
oe
4
4
π
1
1
1
1
1
lnsin
=
ln 1
+ ×
3
−
ln 1
−
+
c
c
=
...
= −
ln5
2
4
2
4
2
4
k
lnsin
y
m
ln(...)
si
n
k
y
...
m
=
=
or
k
lnsin
y
=
...
sin
k
y
=
exp(...)
1
+
3
x
sin
4
y
=
5(1
)
−
x
Notes
(a)
B1: For a correct suitable identity without fractions, such as 1
=
A
(1
−
x
)
+
B
(1 3 ),
+
x
seen or implied.
M1: Attempts to find one of the constants by either substitution or equating coefficients. May be implied by a
correct value for A or B via cover up rule.
3
1
+
oe allow values for A and B to be stated following a correct partial fraction form, or if
A1:
4(1
+
3 )
x
4(1
−
x
)
correct partial fractions see in (b).
(b)
d
y
(oe for cot y) and integrate cot y . Accept any
M1: Attempts to separate variables to form
c t
o
y x
=
g
(
x
)
d
changed function for the attempt but must be attempting to integrate cot y (oe).
...
M1: Attempts to integrate their partial fractions from (a) so award for
x →
...ln(1
+
3 ) or ...ln 4
x
(1
3 )
+
...
(
)
and
x →
...ln(1
−
x
) or ...ln 4
−
4
x
oe
(1
−
)
A1ft: Correct expression (any equivalent) (both sides). Follow through on their constants for the partial
fractions. Condone the absence of the constant of integration.
dM1: Depends on second M, and must have attempted to integrate both sides. Uses the initial conditions in an
equation with a constant of integration. May integrate between limits to achieve this. (Accept if a value for
c cannot be reached from their equation.)
s n
i
M1: Attempts to rearrange their equation by correctly using log work to reach the required form
Must have had kln sin y = … (k may be 1). Not dependent may be gained before finding the constant if ln A
is used, and allow if the constant is missing.
1
3
+
x
A1:
sin
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