Mathematics · Oct / Nov 2023 · Paper 41 · Question 7
YMA01
As printed in the paper
The question as text
7. The number of goats on an island is being monitored. When monitoring began there were 3000 goats on the island. In a simple model, the number of goats, x, in thousands, is modelled by the equation k t 9 5 x t 4 3 where k is a constant and t is the number of years after monitoring began.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[2]
(
)
k
9
t
+
5
Substitutes x = 3 and t = 0 into
x
=
4
t
+
3
5
k
*
3
=
k
=
1.8
3
7(b)[1]
4050
7(c)[3]
3
A
B
+
(
)
x
9
−
2
x
x
9
−
2
x
1
2
Either
A =
or
B =
3
3
3
1
2
+
(
)
9
x
−
2
x
2
3
x
3 9
−
2
x
7(d)[5]
(
)
d
x
3
3
=
x
9
−
2
x
d
x
=
d
t
(
)
d
t
9
2
x
−
x
Or
d
x
(
)
1
1
3
x
9
2
x
d
x
d
t
=
−
=
(
)
d
t
3
x
9
−
2
x
1
1
(
)
(
)
1
1
(
)
1
(
)
or
ln
x
−
ln 9
−
2
x
=
t
+
c
ln
x
−
ln 9
−
2
x
=
t
+
c
3
3
9
9
3
Substitutes
t
=
0,
x
=
3
c
=
0
1
1
(
)
x
3
t
ln
x
−
ln 9
−
2
x
=
t
=
e
3
3
9
−
2
x
9
−
2
x
9
9
−
3
t
−
3
t
=
e
−
2
=
e
x
=
*
x
x
−
3
t
2
+
e
7(e)[1]
4500
(a)
M1: Substitutes x = 3 and t = 0 into the given equation
A1*: Shows that k =1.8 (oe e.g. 9
5 ) with no errors and with at least one correct line of the form ak
(
)
(
)
k
9 0 5
×+
k
0 5
+
5
k
e.g.
3
=
k
=
1.8, 3
=
k
=
1.8, 3
=
k
=
1.8
all score M1A1
4 0 3
×+
0 3
+
3
Alternative by verification:
M1: Substitutes k = 1.8 and t = 0 into the given equation
A1*: Shows that x = 3 with at least one correct intermediate line and a concluding statement that this is 3000
(
)
k
9 0
×+
5
1.
x
=
k
=
1.8
scores M1A0
4 0
×+
3
2. Substitutes x = 3000 and t = 0 into the given equation to obtain k = 1800 (hence k = 1.8) scores M1A0
(b)
B1: 4050 cao. Allow 4.05 thousand.
(c)
3
A
B
(
)
M1: Sets (
+
or equivalent e.g.
3
A
9
−
2
x
+
Bx
x
9
2
x
)
x
9
2
x
−
−
A1: One correct value for ‘A’ or ‘B’ or one correct fraction.
1
2
3
1
2
3
3
1
2
A1: Correct fractions e.g. (
+
oe e.g.
+
,
+
)
(
)
(
)
x
9
−
2
x
3
x
3 9
−
2
x
x
9
−
2
x
3
x
27
−
6
x
This mark is for the correct partial fractions not for the values of the constants.
Award once the correct fractions are seen and allow if seen in (d) if not seen here.
(d)
M1: Attempts to separate the variables and integrate to obtain a kt term and one of
α
ln
β
x
or
Condone missing brackets e.g. ln9
−
2x
The ln terms must come from a partial fraction of the form A
B
x or 9
.
−
2
x
1
1
(
)
(
)
B
(
)
A1ft:
ln
x
−
ln 9
−
2
x
=
t
+
c
but ft on their A and B so award for
A
ln
x
−
ln 9
−
2
x
=
3
3
2
There is no requirement for + c
Brackets must be present unless they are implied by later work.
1
1
(
)
(
)
Note that there are various correct alternatives here e.g.
ln3
x
−
ln 27
−
6
x
=
t
+
c
3
3
M1: Sets t = 0 and x = 3 in an attempt to find “c” having made some attempt to integrate at least one of their
partial fractions to obtain a ln term.
If this step is not attempted only the first two marks are available in this part.
They can “state” e.g. c = 0 or e.g. K = 1 provided it follows correct work and there was a “+ c”.
ddM1: Dependent upon both previous Method marks.
It is for using correct work to remove the ln’s having found a constant of integration.
A1*: Correct work to reach the given answer showing all necessary steps.
The scheme shows one such way with acceptable minimal working.
Note that some candidates may rearrange first before finding the constant of integration e.g.
d
(
)
3
x
3
=
x
9
−
2
x
d
x
=
d
t
(
)
d
t
x
9
2
x
−
1
1
(
)
ln
x
ln 9
2
x
t
c
−
−
= +
3
3
1 ln
x
x
x
3
t
= +
t
c
ln
=
3
t
+
d
=
K
e
3
9
−
2
x
9
−
2
x
9
−
2
x
t = 0 and x = 3
K
=
1
x
3
t
3
t
3
t
3
t
3
t
=
e
9e
−
2 e
x
=
x
x
2e
+
1
=
9e
9
−
2
x
3
t
9e
9
x
=
=
*
−
3
t
3
2
+
e
t
2e
+
1
In these cases, the first 2 marks are as already defined and then award M1 at the point t = 0 and x = 3 are used in
an attempt to find “c” and then ddM1 for using correct work to remove the ln’s and A1* for correct work to
reach the given answer showing all necessary steps.
As in the main scheme, if there is no “+ c” then only the first 2 marks are available in (d).
Note that in (d) it is possible to start again e.g.
d
x
(
)
1
1
3
9
2
d
d
=
x
−
x
x
=
t
(
)
d
t
3
x
9
−
2
x
1
A
B
1
2
+
+
(
)
(
)
x
9
−
2
x
x
9
−
2
x
9
x
9 9
−
2
x
etc.
(e)
B1: 4500 cao or 4.5 thousand
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