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JEE Main Math Class 12 Application of Derivatives Part 3 Questions
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© examsnet.com
Question : 18
Total: 40
Let
g
:
R
→
R
be a non constant twice differentiable such that
g
′
(
1
2
)
=
g
′
(
3
2
)
. If a real valued function
f
is defined as
f
(
x
)
=
1
2
[
g
(
x
)
+
g
(
2
−
x
)
]
, then
[30-Jan-2024 Shift 1]
f
′
(
x
)
=
0
for atleast two
x
in
(
0
,
2
)
f
′
′
(
x
)
=
0
for exactly one
x
in
(
0
,
1
)
f
′
(
x
)
=
0
for no
x
in
(
0
,
1
)
f
′
(
3
2
)
+
f
′
(
1
2
)
=
1
Validate
Solution:
👈: Video Solution
f
′
(
x
)
=
g
′
(
x
)
−
g
′
(
2
−
x
)
2
,
f
′
(
3
2
)
=
g
′
(
3
2
)
−
g
′
(
1
2
)
2
=
0
Also
f
′
(
1
2
)
=
g
′
(
1
2
)
−
g
′
(
3
2
)
2
=
0
,
f
′
(
1
2
)
=
0
⇒
f
′
(
3
2
)
=
f
′
(
1
2
)
=
0
⇒
rootsin
(
1
2
,
1
)
and
(
1
,
3
2
)
⇒
f
′
′
(
x
)
is zero at least twice in
(
1
2
,
3
2
)
© examsnet.com
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