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JEE Main 8 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 5
Total: 90
The number of critical points of the function
f
(
x
)
=
(
x
−
2
)
2
∕
3
(
2
x
+
1
)
is :
[8 Apr 2024 Shift 1]
2
0
1
3
Validate
Solution:
👈: Video Solution
f
(
x
)
=
(
x
−
2
)
2
∕
3
(
2
x
+
1
)
f
(
x
)
=
2
3
(
x
−
2
)
−
1
∕
3
(
2
x
+
1
)
+
(
x
−
2
)
2
∕
3
f
(
x
)
=
2
×
(
2
x
+
1
)
+
(
x
−
2
)
3
(
x
−
2
)
1
∕
3
3
x
−
1
(
x
−
2
)
1
∕
3
=
0
Critical points
x
=
1
3
and
x
=
2
© examsnet.com
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