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Byju's Answer
Standard XII
Mathematics
Continuity of a Function
Let f : R →...
Question
Let
f
:
R
→
R
be a continuous odd function, which vanishes exactly at one point and
f
(
1
)
=
1
2
. Suppose that
F
(
x
)
=
∫
x
−
1
f
(
t
)
d
t
for all
x
∈
[
−
1
,
2
]
and
G
(
x
)
=
∫
x
−
1
t
|
f
(
f
(
t
)
)
|
d
t
for all
x
∈
[
−
1
,
2
]
.
If
lim
x
→
1
F
(
x
)
G
(
x
)
=
1
14
, then the value of
f
(
1
2
)
Open in App
Solution
lim
x
→
1
F
(
x
)
G
(
x
)
=
1
14
lim
x
→
1
∫
x
−
1
f
(
t
)
d
t
∫
x
−
1
t
|
f
(
f
(
t
)
)
|
d
t
=
1
14
Using L-Hospital rule,
lim
x
→
1
f
(
x
)
x
f
(
f
(
x
)
)
=
1
14
⇒
f
(
1
)
f
(
f
(
1
)
)
=
1
14
⇒
1
2
f
(
1
2
)
=
1
14
∴
f
(
1
2
)
=
7
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Similar questions
Q.
Let
f
:
R
→
R
be a continuous odd function, which vanishes exactly at one point and
f
(
1
)
=
1
2
.
Suppose that
F
(
x
)
=
x
∫
−
1
f
(
t
)
d
t
for all
x
∈
[
−
1
,
2
]
and
G
(
x
)
=
x
∫
−
1
t
|
f
(
f
(
t
)
)
|
d
t
for all
x
∈
[
−
1
,
2
]
. If
lim
x
→
1
F
(
x
)
G
(
x
)
=
1
14
, then the value of
f
(
1
2
)
is
Q.
Suppose
f
(
x
)
and
g
(
x
)
are two continuous functions defined for
0
≤
x
≤
1
.
Given,
f
(
x
)
=
∫
1
0
e
x
+
1
.
f
(
t
)
d
t
and
g
(
x
)
=
∫
1
0
e
x
+
t
.
g
(
t
)
d
t
+
x
The value of
f
(
1
)
equals
Q.
Let
f
:
R
→
R
be such that
f
(
2
x
−
1
)
=
f
(
x
)
for all
x
ϵ
R
. If
f
is continuous at
x
=
1
and
f
(
1
)
=
1
, then
Q.
Let
f
:
[
−
1
,
2
]
→
[
0
,
∞
)
be a continuous functlon such that
f
(
x
)
=
f
(
1
−
x
)
for all
x
∈
[
−
1
,
2
]
. Let
R
1
=
∫
2
−
1
x
f
(
x
)
d
x
, and
R
2
be the area of the region bounded by
y
=
f
(
x
)
,
x
=
−
1
,
x
=
2
, and the
x
-axis. Then
Q.
Suppose
f
and
g
are differentiable functions on
(
0
,
∞
)
such that
f
'
(
x
)
=
−
g
(
x
)
x
and
g
'
(
x
)
=
−
f
(
x
)
x
, for all
x
>
0
. Further,
f
(
1
)
=
3
and
g
(
1
)
=
−
1
.
If
f
(
x
)
−
g
(
x
)
=
B
x
m
,
for all
x
>
0
and some constant
B
, then the value of
m
equals
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