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Test Index
JEE Mains 28-July-2022 Shift 2 Solved Paper
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© examsnet.com
Question : 5
Total: 90
The function
f
:
ℝ
→
ℝ
defined by
f
(
x
)
=
lim
n
→
∞
cos
(
2
π
x
)
−
x
2
n
s
i
n
(
x
−
1
)
1
+
x
2
n
+
1
−
x
2
n
is continuous for all
x
in :
[28-Jul-2022-Shift-2]
R
−
{
−
1
}
ℝ
−
{
−
1
,
1
}
R
−
{
1
}
R
−
{
0
}
Validate
Solution:
f
(
x
)
=
lim
n
→
∞
cos
(
2
π
x
)
−
x
2
n
s
i
n
(
x
−
1
)
1
+
x
2
n
+
1
−
x
2
n
For
|
x
|
<
1
,
f
(
x
)
=
cos
2
π
x
, continuous function
|
x
|
>
1
,
f
(
x
)
=
lim
n
→
∞
1
x
2
n
cos
2
π
x
−
s
i
n
(
x
−
1
)
1
x
2
n
+
x
−
1
=
−
s
i
n
(
x
−
1
)
x
−
1
, continuous
For
|
x
|
=
1
,
f
(
x
)
=
{
1
if
x
=
1
−
(
1
+
s
i
n
2
)
if
x
=
−
1.
Now,
lim
x
→
1
+
f
(
x
)
=
−
1
,
lim
x
→
1
−
f
(
x
)
=
1
, so discontinuous at
x
=
1
lim
x
→
1
+
f
(
x
)
=
1
,
lim
x
→
−
1
−
f
(
x
)
=
−
s
i
n
2
2
, so discontinuous at
x
=
−
1
∴
f
(
x
)
is continuous for all
x
∈
R
−
{
−
1
,
1
}
© examsnet.com
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