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JEE Main Math Class 12 Application of Derivatives Part 2 Questions
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© examsnet.com
Question : 76
Total: 100
The minimum value of
α
for which the equation
4
sin
x
+
1
1
−
sin
x
=
α
has at least one solution in
(
0
,
π
2
)
is
24 Feb 2021 Shift 1
Your Answer:
Validate
Solution:
Let
f
(
x
)
=
4
sin
x
+
1
1
−
sin
x
⇒
f
′
(
x
)
=
0
⇒
sin
x
=
2
∕
3
∴
f
(
x
)
min
=
4
2
∕
3
+
1
1
−
2
∕
3
=
9
f
(
x
)
max
.
→
∞
f
(
x
)
is continuous function
∴
α
min
=
9
© examsnet.com
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