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Standard XII
Mathematics
Question
lf
∫
∞
0
e
−
x
2
d
x
=
√
π
2
, then
∫
∞
0
e
−
a
x
2
d
x
,
a
>
0
is
√
π
2
2
√
π
a
√
π
2
a
1
2
√
π
a
A
√
π
2
a
B
2
√
π
a
C
√
π
2
D
1
2
√
π
a
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Solution
Verified by Toppr
∫
∞
0
e
−
a
x
2
d
x
=
∫
∞
0
e
−
(
√
a
x
)
2
d
x
Let,
√
a
.
x
=
1
√
a
.
d
x
=
d
t
d
x
=
d
t
√
a
I
=
∫
∞
0
e
−
t
2
.
d
t
√
a
=
1
√
a
∫
∞
0
e
−
t
2
d
t
=
1
√
a
×
√
π
2
=
1
2
√
π
a
Hence the correct answer is
1
2
√
π
a
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3
Similar Questions
Q1
lf
∫
∞
0
e
−
x
2
d
x
=
√
π
2
, then
∫
∞
0
e
−
a
x
2
d
x
,
a
>
0
is
View Solution
Q2
If
∫
∞
0
e
−
x
2
d
x
=
√
π
2
, then
∫
∞
0
e
−
a
x
2
d
x
where
a
>
0
is?
View Solution
Q3
Let
f
(
x
)
=
A
sin
(
π
x
2
)
+
B
,
f
′
(
1
2
)
=
√
2
and
1
∫
0
f
(
x
)
d
x
=
2
A
π
, then
A
and
B
are
View Solution
Q4
Let
f
(
x
)
=
1
−
tan
x
4
x
−
π
,
x
≠
π
4
,
x
∈
[
0
,
π
4
]
. lf
f
(
x
)
is continuous in
[
0
,
π
2
]
then
f
(
π
4
)
is:
View Solution
Q5
If
z
is a complex number satisfying
|
z
|
=
1
, then the range of
arg
(
1
1
−
z
)
is
View Solution