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JEE Main 24 June 2022 Shift 2 Solved Paper
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© examsnet.com
Question : 5
Total: 90
Let
f
(
x
)
=
{
sin
(
x
−
[
x
]
)
x
−
[
x
]
x
∈
(
−
2
,
−
1
)
max
{
2
x
,
3
[
|
x
|
]
}
|
x
|
<
1
1
othewise
.
where [t] denotes greatest integer
≤
t
. If
m
is the number of points where
f
is not continuous and
n
is the number of points where
f
is not differentiable, then the ordered pair
(
m
,
n
)
is :
[24-Jun-2022-Shift-2]
(
3
,
3
)
(
2
,
4
)
(
2
,
3
)
(
3
,
4
)
Validate
Solution:
👈: Video Solution
f
(
x
)
=
{
sin
(
x
−
[
x
]
)
x
[
x
]
x
∈
(
−
2
,
−
1
)
max
{
2
x
,
3
[
|
x
|
]
}
|
x
|
<
1
1
otherwise
.
f
(
x
)
=
{
sin
(
x
+
2
)
x
+
2
x
∈
(
−
2
,
−
1
)
0
x
∈
(
−
1
,
0
]
2
x
x
∈
(
0
,
1
)
1
otherwise
.
It clearly shows that
f
(
x
)
is discontinuous
At
x
=
−
1
,
1
also non differentiable
and at
x
=
0
,
L
.
H
.
D
=
lim
h
→
0
f
(
0
−
h
)
−
f
(
0
)
−
h
=
0
R. H.D
=
lim
h
→
0
f
(
0
+
h
)
−
f
(
0
)
h
=
2
∴
f
(
x
)
is not differentiable at
x
=
0
∴
m
=
2
,
n
=
3
© examsnet.com
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