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JEE Advanced 2016 Paper 1
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© examsnet.com
Question : 38
Total: 54
The least value of
α
∊
R
for which
4
α
x
2
+
1
x
≥
1
for all
x
>
0
is -
1
64
1
32
1
27
1
25
Validate
Solution:
f
(
x
)
=
4
α
x
2
+
1
x
;
x
>
0
f
′
(
x
)
=
8
α
x
−
1
x
2
=
8
α
x
3
−
1
x
2
f
(
x
)
attains its minimum at
x
=
(
1
8
α
)
1
3
f
(
(
1
8
α
)
1
3
)
=
1
⇒
4
α
(
1
8
α
)
2
3
+
(
8
α
)
1
3
=
1
⇒
3
α
1
3
=
1
⇒
α
=
1
27
© examsnet.com
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