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BITSAT 2006 Solved Question Paper
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© examsnet.com
Question : 36
Total: 150
A population
p
(
t
)
of 1000 bacteria introduced into nutrient medium grows according to the relation
p
(
t
)
=
1000
+
1000
t
100
+
t
2
. The maximum size of this bacterial population is:
1100
1250
1050
5250
Validate
Solution:
p
(
t
)
=
1000
+
1000
t
100
+
t
2
...(i)
On differentiating both side w.r.t.t.
p
′
(
t
)
=
0
+
(
100
+
t
2
)
(
1000
)
−
1000
t
(
2
t
)
(
100
+
t
2
)
2
=
1000
(
100
−
t
2
)
(
100
+
t
2
)
2
...(ii)
Put
p
′
(
t
)
=
0
for maxima or minima
⇒
100
−
t
2
=
0
⇒
t
=
±
10
Now, again differentiating Eq. (ii), w.r.t.
x
p
′
′
(
t
)
=
1000
×
[
(
100
+
t
2
)
2
(
−
2
t
)
−
(
100
−
t
2
)
2
(
100
+
t
2
)
2
t
(
100
+
t
2
)
4
]
=
1000
t
[
(
100
+
t
2
)
(
−
2
)
−
(
100
−
t
2
)
(
4
)
]
(
100
+
t
2
)
3
=
−
1000
t
[
600
−
2
t
2
]
(
100
+
t
2
)
3
At
t
=
10
p
′
′
(
t
)
<
0
∴ The maximum value is
p
(
10
)
=
1000
+
10000
100
+
100
=
1000
+
10000
200
=
1000
+
50
= 1050
© examsnet.com
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