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Byju's Answer
Standard XII
Mathematics
Definition of Functions
Let ϕ x, h ...
Question
Let
ϕ
(
x
)
,
h
(
x
)
and
g
(
x
)
be differentiable functions
ϕ
′
(
x
)
=
1
+
1
x
,
ϕ
(
1
)
=
1
;
h
(
x
)
=
e
x
2
and
f
(
x
)
=
h
(
x
)
.
ϕ
(
x
)
∀
x
≥
1
. Let
g
(
x
)
be inverse of
f
(
x
)
.
On the basis of above information answer the following questions:
f
(
2
)
equal to
A
e
4
(
2
+
ln
2
)
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B
e
4
(
4
+
ln
2
)
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C
e
4
(
2
+
ln
4
)
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D
e
4
(
4
+
ln
4
)
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Solution
The correct option is
A
e
4
(
2
+
ln
2
)
ϕ
′
(
x
)
=
1
+
1
x
Integrating we get,
ϕ
(
x
)
=
x
+
ln
x
+
c
Also
ϕ
(
1
)
=
1
⇒
c
=
0
⇒
ϕ
(
x
)
=
x
+
ln
x
⇒
f
(
x
)
=
e
x
2
(
x
+
ln
x
)
∴
f
(
2
)
=
e
4
(
2
+
ln
2
)
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0
Similar questions
Q.
Let f(x),g(x) be two continuesely differentiable functions satisfying the relationships f'(x) = g(x) and f"(x) = - f(x). Let
h
(
x
)
=
[
f
(
x
)
]
2
+
[
g
(
x
)
]
2
. If h(0) = 5, then h(10) =
Q.
For
x
∈
(
0
,
3
2
)
,
let
f
(
x
)
=
√
x
,
g
(
x
)
=
tan
x
and
h
(
x
)
=
1
−
x
2
1
+
x
2
.
If
ϕ
(
x
)
=
(
(
h
o
f
)
o
g
)
(
x
)
,
then
ϕ
(
π
3
)
is equal to
Q.
Let
f
(
x
)
=
[
x
[
x
]
]
,
g
(
x
)
=
[
x
[
1
x
]
]
and
h
(
x
)
=
[
[
x
]
x
]
,
, then
lim
x
→
2
−
f
(
x
)
+
lim
x
→
1
2
+
g
(
x
)
+
lim
x
→
2
+
h
(
x
)
=
Q.
Let
f
(
x
)
=
x
2
+
1
x
2
and
g
(
x
)
=
x
−
1
x
,
x
ϵ
R
−
{
−
1
,
0
,
1
}
. If
h
(
x
)
=
f
(
x
)
g
(
x
)
. Then the local minimum value of
h
(
x
)
is:
Q.
Let
f
(
x
)
=
x
2
−
1
x
2
and
g
(
x
)
=
x
−
1
x
,
x
∈
R
−
−
1
,
0
,
1
. If
h
(
x
)
=
f
(
x
)
g
(
x
)
then the local minimum value of
h
(
x
)
is:
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