Mathematics · May / June 2021 · Paper 31 · Question 8

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The question as text

8. A scientist is studying a population of fish in a lake. The number of fish, N, in the population, t years after the start of the study, is modelled by the equation N = 600 e 0 3 . t t  0 2 + e 0 3 . t Use the equation of the model to answer parts (a), (b), (c), (d) and (e).

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8(a)[1]
200
8(b)[1]
600
8(c)[4]
600e
0.3
t
500
=
⇒
100e
0.3
t
=
1000,
⇒
e
0.3
t
=
10
2
+
e
0.3
t
ln10
= 7 years 8 months
⇒=
t
0.3
(
)
8(d)[3]
2
+
e
0.3
t
×
180e
0.3
t
−
180e
0.3
t
×
e
0.3
t
600e
0.3
t
d
N
N
=
⇒
=
2
+
e
0.3
t
d
t
(
)
2
2
+
e
0.3
t
d
N
360e
0.3
t
⇒
=
(
)
d
t
2
2
+
e
0.3
t
8(e)[4]
360e
0.3
t
(
)
(
)
2
8
=
⇒
8 e
0.3
t
−
328 e
0.3
t
+
32
=
0
(
)
2
2
+
e
0.3
t
41
+
1665
e
0.3
t
=
2
41
+
1665
ln
2
(
t
)
=
0.3
(
)awrt 12.4
T =
(a)
B1: 200
(b)
B1: 600
(c)
M1: Sets
N =
500
and proceeds to
or equivalent
C
e
0.3
t
=
D
t =
A1:
. If ln’s are taken earlier it would be for ln100
+
0.3
t
=
ln1000
e
0.3
10
dM1: Full method to find a value for t using correct log work.
A1: 7 years 8 months. Accept 7 years 9 months following a correct value of t.
(d)
M1: Uses the quotient rule to obtain an expression of the form (
)
2
+
e
0.3
t
×
a
e
0.3
t
−
b
e
0.3
t
×
e
0.3
t
(
)
2
2
+
e
0.3
t
In general condone missing brackets for the M mark.
vu
′
−
uv
′
A correct rule may be implied by their u, v, u′, v′ followed by applying
etc.
v
2
E.g. if they quote u = 600e0.3t and v = 2 + e0.3t and do not make the differentiation easier, they can be awarded
this mark for applying the correct rule. Also allow this mark if they quote the correct quotient rule but only
have v rather than v2 in the denominator.
(
)
2
+
e
0.3
t
×
180e
0.3
t
−
180e
0.3
t
×
e
0.3
t
d
N
A1: Correct derivative in any form e.g.
=
(
)
d
t
2
2
+
e
0.3
t
d
360e
0.3
t
N
=
A1: Correctly obtains
d
t
(
)
2
2
e
0.3
t
+
t or
Withhold this mark if you see
written as
e
0.3
2
t
e
0.3
t
×
e
0.3
t
e
0.09
..................................................................................................................................................................................
Alt (d)
600e
0.3
t
1200
d
N
1200 0.3e
×
0.3
t
M1 A1 :
N
=
⇒
N
=
600
−
⇒
=
(
)
2
+
e
0.3
t
2
+
e
0.3
t
d
t
2
2
+
e
0.3
t
600e
0.3
t
B
d
N
k
e
0.3
t
Score M1 for splitting
into
A±
leading to
t =
2
+
e
0.3
t
2
+
e
0.3
t
d
(
)
2
2
+
e
0.3
t
and A1 for correct derivative in any form.
d
360e
0.3
t
N
=
A1:
(following fully correct work)
d
t
(
)
2
2
e
0.3
t
+
.............................................................................................................................................................................
May also see product rule in (d):
600e
0.3
t
d
N
(
)
1
(
)
1
(
−
−
N
=
=
600e
0.3
t
2
+
e
0.3
t
⇒
=
180e
0.3
t
2
+
e
0.3
t
−
180e
0.3
t
×
e
0.3
t
2
+
e
0.3
2
+
e
0.3
t
d
t
180e
0.3
t
180e
0.6
t
360e
0.3
t
+
180e
0.6
t
−
180e
0.6
t
360e
0.3
t
=
−
=
=
2
e
0.3
t
(
)
2
(
)
2
(
)
2
+
2
e
0.3
t
2
e
0.3
t
2
e
0.3
t
+
+
+
(
)
−
1
(
)
−
2
a
e
0.3
t
2
+
e
0.3
t
−
b
e
0.3
t
×
e
0.3
t
2
+
e
0.3
t
Score M1 for an expression of the form
and A1 A1 as above.
A correct rule may be implied by their u, v, u′, v′ followed by applying vu
′
+
uv
′
etc.
(e)
M1: Sets d
N
t = and proceeds to a quadratic in
e
0.3
t
8
d
t . Condone both roots to be found
dM1: Correct attempt to solve the quadratic in
e
0.3
t = k where k > 0 using correct log work
ddM1: Correct attempt to find the value of t for
e
0.3
A1: Achieves (
)awrt 12.4
T =

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