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JEE Main Math Class 12 Integrals Part 3 Questions
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© examsnet.com
Question : 15
Total: 100
The minimum value of the function
f
(
x
)
=
2
∫
0
e
|
x
−
t
|
dt
is is
[25-Jan-2023 Shift 1]
2
(
e
−
1
)
2
e
−
1
2
e
(
e
−
1
)
Validate
Solution:
For
x
≤
0
f
(
x
)
=
2
∫
0
e
t
−
x
dt
=
e
−
x
(
e
2
−
1
)
For
0
<
x
<
2
f
(
x
)
=
x
∫
0
e
x
−
t
dt
+
2
∫
x
e
t
−
x
dt
=
e
x
+
e
2
−
x
−
2
For
x
≥
2
f
(
x
)
=
2
∫
0
e
x
−
t
dt
=
e
x
−
2
(
e
2
−
1
)
For
x
≤
0
,
f
(
x
)
is
↓
and
x
≥
2
,
f
(
x
)
is
↑
∴
Minimum value of
f
(
x
)
lies in
x
∈
(
0
,
2
)
Applying A.M
≥
G.M
minimum value of
f
(
x
)
is
2
(
e
−
1
)
© examsnet.com
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