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JEE Advanced 2023 Paper 2 Solutions
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© examsnet.com
Question : 40
Total: 52
Let
f
:
(
0
,
1
)
→
ℝ
be the function defined as
f
(
x
)
=
[
4
x
]
(
x
−
1
4
)
2
(
x
−
1
2
)
, where
[
x
]
denotes the greatest integer less than or equal to
x
. Then which of the following is(are) true?
The function
f
is discontinuous exactly at one point in
(
0
,
1
)
There is exactly one point in
(
0
,
1
)
at which the function
f
is continuous but NOT differentiable
The function
f
is NOT differentiable at more than three points in
(
0
,
1
)
The minimum value of the function
f
is
−
1
512
Validate
Solution:
f
:
(
0
,
1
)
→
ℝ
f
(
x
)
=
[
4
x
]
(
x
−
1
4
)
2
(
x
−
1
2
)
⇒
Critical point
=
1
4
,
1
2
,
3
4
Discontinuity at
x
=
3
4
Continuous and differentiable at
x
=
1
4
Continuous but non-differentiable at
x
=
1
2
LHD
(
.
at
x
=
1
4
)
lim
h
→
0
+
0
−
0
−
h
=
0
LHD
(
at
x
=
1
2
)
lim
h
→
0
+
(
1
4
−
h
)
2
(
−
h
)
−
0
−
h
=
1
16
RHD
(
at
x
=
1
4
)
lim
h
→
0
+
h
2
(
−
1
2
+
h
)
h
=
0
RHD
(
at
x
=
1
2
)
lim
h
→
0
+
2
(
1
4
+
h
)
2
h
−
0
h
=
1
8
Minimum -ve value will exist between
1
4
&
1
2
f
(
x
)
=
(
x
−
1
4
)
2
(
x
−
1
2
)
1
4
≤
x
≤
1
2
f
′
(
x
)
=
(
x
−
1
4
)
(
3
x
−
5
4
)
⇒ minima at
x
=
5
12
f
(
5
12
)
=
1
36
×
−
1
12
=
−
1
432
© examsnet.com
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