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JEE Main 5 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 1
Total: 90
Let
f
:
[
−
1
,
2
]
⟶
R
be given by
f
(
x
)
=
2
x
2
+
x
+
[
x
2
]
−
[
x
]
, where
[
t
]
denotes the greatest integer less than or equal to
t
. The number of points, where
f
is not continuous, is :
[5 Apr 2024 Shift 2]
6
3
4
5
Validate
Solution:
👈: Video Solution
Doubtful points :
−
1
,
0
,
1
,
√
2
,
√
3
,
2
at
x
=
√
2
,
√
3
at
x
=
√
2
,
√
3
f
(
x
)
=
(
2
x
2
+
↓
Cont
.
x
−
[
x
]
)
+
[
x
2
]
↓
Cont
.
=
Discount
at
x
=
−
1
:
RHL
⇒
f
(
x
)
=
(
2
−
1
−
(
−
1
)
)
+
0
=
2
f
(
−
1
)
=
2
−
1
−
(
−
1
)
+
1
=
3
}
Dis.
at
x
=
2
:
LHL
⇒
f
(
x
)
=
8
+
2
−
1
+
3
=
12
f
(
2
)
=
8
+
2
−
2
+
4
=
12
}
Cont.
at
x
=
0
:
LHL
⇒
0
+
0
−
(
−
1
)
+
0
=
1
f
(
0
)
=
0
}
Dis.
at
x
=
1
LHL
⇒
2
+
1
−
0
+
0
=
3
f
(
1
)
=
3
−
1
+
1
=
3
RHL
⇒
2
+
1
−
1
+
1
=
3
}
Cont.
© examsnet.com
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