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JEE Advanced 2018 Paper 2
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© examsnet.com
Question : 51
Total: 54
Let
E
1
=
{
X
∊
ℝ
:
x
≠
1
a
n
d
x
x
−
1
>
0
}
and
E
2
=
{
X
∊
R
1
:
s
i
n
−
1
(
log
e
(
x
x
−
1
)
)
is a real number
}
.(Here the inverse trigonometric function function
s
i
n
−
1
assume values in
[
−
π
2
,
π
2
]
)
Let
f
:
E
1
→
ℝ
be the function defined by
f
(
x
)
=
log
e
(
x
x
−
1
)
and
g
:
E
2
→
ℝ
be the function defined by
g
(
x
)
=
s
i
n
−
1
(
log
(
x
x
−
1
)
)
List - I
List - II
P. The range of f is
1.
(
−
∞
,
1
1
−
e
]
∪
[
e
e
−
1
,
∞
)
Q. The range of g contains
2.
(
0
,
1
)
R. The domain of f contains
3.
[
−
1
2
,
1
2
]
S. The domain of g is
4.
(
−
∞
,
0
]
∪
(
0
,
∞
)
5.
(
−
∞
,
e
e
−
1
]
6.
(
−
∞
,
0
)
∪
(
1
2
,
e
e
−
1
]
The correct options is:
P
Q
R
S
A)
4
2
1
1
b)
3
3
6
5
c)
4
2
1
6
D)
4
3
6
5
A
B
C
D
Validate
Solution:
E
1
:
x
x
−
1
>
0
⇒
E
1
:
x
∊
(
−
∞
,
0
)
∪
(
1
,
∞
)
E
2
:
−
1
≤
ln
(
x
x
+
1
)
≤
1
1
e
≤
x
x
−
1
≤
e
Now
x
x
−
1
=
−
1
e
≥
0
⇒
(
e
−
1
)
x
+
1
e
(
x
+
1
)
≥
0
⇒
x
∊
(
−
∞
,
1
1
−
e
]
∪
(
1
,
∞
)
also
x
x
−
1
−
e
≤
0
(
e
−
1
)
x
−
e
x
−
1
≥
0
⇒
x
∊
(
−
∞
,
1
)
∪
[
e
e
−
1
,
∞
]
So
E
2
:
(
−
∞
,
1
1
−
e
)
∪
[
e
e
−
1
,
∞
]
as Range of
x
x
−
1
is
R
+
−
{
1
}
⇒ Range of f is
R
−
{
0
}
or
(
−
∞
,
0
)
∪
(
0
,
∞
)
Range of g is
[
−
π
2
,
π
2
]
\
{
0
}
or
[
−
π
2
,
0
)
∪
(
0
,
π
2
)
Now P→4,Q→2,R→1,S→1
Hence a is correct
© examsnet.com
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