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Byju's Answer
Standard XII
Mathematics
Global Maxima
The minimum v...
Question
The minimum value of
√
e
x
2
−
1
is
Open in App
Solution
f
(
x
)
=
√
(
e
x
2
−
1
)
t
h
e
s
q
u
a
r
e
r
o
o
t
t
e
r
m
s
h
o
u
l
d
a
l
w
a
y
s
b
e
p
o
s
i
t
i
v
e
o
r
z
e
r
o
.
i
t
c
a
n
n
o
t
b
e
n
e
g
a
t
i
v
e
.
s
o
e
x
2
≥
1
a
n
d
w
e
w
a
n
t
t
h
e
m
i
n
i
m
u
m
v
a
l
u
e
o
f
t
h
i
s
f
(
x
)
a
n
d
t
h
a
t
c
a
n
b
e
z
e
r
o
o
n
l
y
a
t
x
=
0.
Suggest Corrections
0
Similar questions
Q.
Minimum value of
√
e
x
2
−
1
is
Q.
Let
∫
e
x
2
⋅
e
x
(
2
x
2
+
x
+
1
)
d
x
=
e
x
2
f
(
x
)
+
c
, where
c
is constant of integration. If the minimum value of
f
(
x
)
is
m
, then the value of
[
−
1
m
]
is
(where [.] represents grestest integer function)
Q.
Let
∫
e
x
2
⋅
e
x
(
2
x
2
+
x
+
1
)
d
x
=
e
x
2
f
(
x
)
+
c
, where
c
is constant of integration. If the minimum value of
f
(
x
)
is
m
, then the value of
[
−
1
m
]
is
(where [.] represents grestest integer function)
Q.
The value of
∫
e
x
(
2
−
x
2
)
(
1
−
x
)
√
1
−
x
2
d
x
is equal to
Q.
The value of
(
Δ
2
E
)
x
2
at the interval
h
=
1
is
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