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NIMCET 2024 Paper
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Section:
Mathematics
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© examsnet.com
Question : 17
Total: 120
If for non-zero
x
,
c
f
(
x
)
+
d
f
(
‌
1
x
)
=
|
log
|
x
|
|
+
3
, where
c
â‰
d
, then
e
∫
1
f
(
x
)
‌
dx
=
:
‌
(
c
−
d
)
(
2
e
−
1
)
c
2
−
d
2
‌
(
c
−
d
)
(
3
e
−
2
)
c
2
−
d
2
‌
(
c
−
d
)
(
3
e
+
2
)
c
2
−
d
2
‌
(
c
−
d
)
(
2
e
+
1
)
c
2
−
d
2
Validate
Solution:
Put
x
=
‌
1
x
in the given equation,
‌
c
f
(
x
)
+
d
f
(
‌
1
x
)
=
|
log
|
x
|
|
+
3
.
.
.
.
.
‌
c
f
(
‌
1
x
)
+
d
f
(
x
)
=
|
log
|
‌
1
x
|
|
+
3
We know that
log
(
‌
1
x
)
=
−
log
‌
x
‌
Hence
‌
c
f
(
‌
1
x
)
+
d
f
(
x
)
=
|
log
|
x
|
|
+
3
.
.
.
.
(
2
)
Eliminating
f
(
‌
1
x
)
from the two equations, we get
‌
f
(
x
)
=
‌
(
c
−
d
)
c
2
−
d
2
(
log
|
x
|
+
3
)
‌
⇒
e
∫
1
f
(
x
)
‌
dx
=
‌
c
−
d
c
2
−
d
2
‌
e
(
log
∫
1
x
+
3
)
‌
dx
Integrating and putting the values of the limits,
‌
‌
c
−
d
c
2
−
d
2
[
x
‌
log
‌
x
−
x
+
3
x
]
1
e
‌
=
‌
c
−
d
c
2
−
d
2
[
3
e
−
2
]
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