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JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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Question : 61
Total: 92
Consider the function
f
:
R
→
R
defined by Then,
f
is
f
(
x
)
=
{
[
2
−
sin
(
1
x
)
]
|
x
|
,
x
≠
0
0
x
=
0
.
[2021, 17 March Shift-II]
monotonic on
(
−
∞
,
0
)
∪
(
0
,
∞
)
not monotonic on
(
−
∞
,
0
)
and
(
0
,
∞
)
monotonic on
(
0
,
∞
)
only
monotonic on
(
−
∞
,
0
)
only
Validate
Solution:
Method (1)
Given,
f
(
x
)
=
{
[
2
−
sin
(
1
x
)
]
|
x
|
,
x
≠
0
0
x
=
0
.
Here,
f
′
(
x
)
is an oscillating function which is non-monotonic in
(
−
∞
,
0
)
∪
(
0
,
∞
)
.
Method (II)
∵
f
(
x
)
=
{
−
(
2
−
sin
1
x
)
x
x
<
0
0
x
=
0
(
2
−
sin
1
x
)
x
x
>
0.
From above we observe that,
f
(
x
)
is continuous and
f
(
1
π
)
=
f
(
2
π
)
=
2
π
So,
f
(
x
)
is non-monotonic in
(
0
,
∞
)
.
Further,
lim
x
→
−
∞
(
f
)
→
∞
and
lim
x
→
∞
f
(
x
)
→
∞
and
f
(
0
)
=
0
Hence,
f
(
x
)
is non-monotonic on
(
−
∞
,
0
)
∪
(
0
,
∞
)
.
© examsnet.com
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