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Standard XII
Mathematics
Functions
Question
f
′
(
x
)
=
f
(
x
)
,
f
(
0
)
=
1
, then
∫
d
x
f
(
x
)
+
f
(
−
x
)
l
o
g
(
e
x
+
e
−
x
)
+
C
t
a
n
−
1
(
e
x
)
+
C
None
l
o
g
(
e
2
x
+
1
)
+
C
A
t
a
n
−
1
(
e
x
)
+
C
B
l
o
g
(
e
2
x
+
1
)
+
C
C
None
D
l
o
g
(
e
x
+
e
−
x
)
+
C
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Solution
Verified by Toppr
Given that
f
′
(
x
)
=
f
(
x
)
∫
f
′
(
x
)
f
(
x
)
d
x
=
∫
d
x
⇒
ln
f
(
x
)
=
x
+
C
when
x
=
0
,
f
(
0
)
=
1
Therefore,
C
=
0
f
(
x
)
=
e
x
now,
∫
d
x
f
(
x
)
+
f
(
−
x
)
=
∫
d
x
e
x
+
e
−
x
=
∫
e
x
d
x
1
+
e
2
x
=
tan
−
1
e
x
+
c
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Similar Questions
Q1
f
′
(
x
)
=
f
(
x
)
,
f
(
0
)
=
1
, then
∫
d
x
f
(
x
)
+
f
(
−
x
)
View Solution
Q2
(A) :
∫
e
x
(
log
x
+
x
−
2
)
d
x
=
e
x
(
log
x
−
1
x
)
+
c
(R):
∫
e
x
[
f
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x
)
+
f
′
(
x
)
]
d
x
=
e
x
f
(
x
)
+
c
View Solution
Q3
If
∫
d
x
x
2
(
x
n
+
1
)
(
n
−
1
)
n
=
−
[
f
(
x
)
]
1
n
+
C
then f(x) is
View Solution
Q4
Let
f
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x
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=
s
e
c
−
1
x
+
t
a
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−
1
x
. Then f(x) is real for