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UPSEE 2017 Solved Paper
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© examsnet.com
Question : 111
Total: 150
The mean value of the function
f
(
x
)
=
2
e
x
+
1
on the interval
[
0
,
2
]
is
2
−
log
e
(
2
e
2
+
1
)
2
+
log
e
(
2
e
2
+
1
)
2
+
log
e
(
2
e
2
−
1
)
−
2
+
log
0
(
2
e
2
−
1
)
Validate
Solution:
Given,
f
(
x
)
=
2
e
x
+
1
We know, if
f
(
x
)
is continuous function on
[
a
,
b
]
,
then
Mean value of the function,
M
=
1
b
−
a
b
∫
a
f
(
x
)
d
x
∴
Mean value of the given function
M
=
1
2
−
0
2
∫
0
2
e
x
+
1
d
x
=
2
∫
0
e
−
x
1
+
e
−
x
d
x
Let
1
+
e
−
x
=
t
On differentiating
e
−
x
(
−
1
)
d
x
=
d
t
∴
M
=
−
1
+
1
e
2
∫
2
d
t
t
=
∫
1
+
1
e
2
d
t
t
[By changing limit of integration]
=
[
log
|
t
|
]
2
1
+
1
e
2
=
log
2
−
log
|
1
+
1
e
2
|
=
log
2
−
log
(
1
+
1
e
2
)
[Here,
1
+
1
e
2
is always positive. So, we can omit modulus size]
=
log
2
−
log
(
1
+
e
2
e
2
)
=
log
2
−
log
(
1
+
e
2
)
+
log
e
2
=
log
2
−
log
(
1
+
e
2
)
+
2
=
2
+
log
(
2
1
+
e
2
)
© examsnet.com
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