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Test Index
JEE Main Math Class 11 Relations and Functions Part 1 Questions
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© examsnet.com
Question : 1
Total: 100
Let
f
:
ℝ
→
ℝ
be defined as
f
(
x
)
=
x
−
1
and
g
:
ℝ
−
{
1
,
−
1
}
→
ℝ
be defined as
g
(
x
)
=
x
2
x
2
−
1
. Then the function fog is :
[26-Jun-2022-Shift-2]
one-one but not onto
onto but not one-one
both one-one and onto
neither one-one nor onto
Validate
Solution:
👈: Video Solution
f
:
R
→
R
defined as
f
(
x
)
=
x
−
1
and
g
:
R
→
{
1
,
−
1
}
→
R
,
g
(
x
)
=
x
2
x
2
−
1
Now
f
o
g
(
x
)
=
x
2
x
2
−
1
−
1
=
1
x
2
−
1
∴
Domain of
f
o
g
(
x
)
=
R
−
{
−
1
,
1
}
And range of
f
o
g
(
x
)
=
(
−
∞
,
−
1
]
∪
(
0
,
∞
)
Now,
d
d
x
(
fog
(
x
)
)
=
−
1
x
2
−
1
⋅
2
x
=
2
x
1
−
x
2
∴
d
d
x
(
f
∘
g
(
x
)
)
>
0
for
2
x
(
1
−
x
)
(
1
+
x
)
>
0
⇒
x
(
x
−
1
)
(
x
+
1
)
<
0
∴
x
∈
(
−
∞
,
−
1
)
∪
(
0
,
1
)
and
d
d
x
(
f
∘
g
(
x
)
)
<
0
for
x
∈
(
−
1
,
0
)
∪
(
1
,
∞
)
∴
f
o
g
(
x
)
is neither one-one nor onto.
© examsnet.com
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