Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The domain of f(x)=( log (x+1)(x-2)/e2 log e x-(2 x+3)), x ∈ R is
Q. The domain of
f
(
x
)
=
e
2
l
o
g
e
x
−
(
2
x
+
3
)
l
o
g
(
x
+
1
)
(
x
−
2
)
,
x
∈
R
is
2718
164
JEE Main
JEE Main 2023
Relations and Functions
Report Error
A
(
2
,
∞
)
−
{
3
}
40%
B
R
−
{
3
}
14%
C
R
−
{
−
1
,
3
}
37%
D
(
−
1
,
∞
)
−
{
3
}
10%
Solution:
x
−
2
>
0
⇒
x
>
2
x
+
1
>
0
⇒
x
>
−
1
x
+
1
=
1
⇒
x
=
0
and
x
>
0
Denominator
x
2
−
2
x
−
3
=
0
(
x
−
3
)
(
x
+
1
)
=
0
x
=
−
1
,
3
So Ans
(
2
,
∞
)
−
{
3
}