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JEE Main Math Class 12 Application of Derivatives Part 2 Questions
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© examsnet.com
Question : 11
Total: 100
If the maximum value of
a
, for which the function
f
a
(
x
)
=
tan
−
1
2
x
−
3
a
x
+
7
is non-decreasing in
(
−
π
6
,
π
6
)
, is
a
, then
f
a
(
π
8
)
is equal to
[26-Jul-2022-Shift-2]
8
−
9
π
4
(
9
+
π
2
)
8
−
4
π
9
(
4
+
π
2
)
8
(
1
+
π
2
9
+
π
2
)
8
−
π
4
Validate
Solution:
f
a
(
x
)
=
tan
−
1
2
x
−
3
a
x
+
7
∵
f
a
(
x
)
is non-decreasing in
(
−
π
6
,
π
6
)
∴
f
a
′
(
x
)
≥
0
⇒
2
1
+
4
x
2
−
3
a
≥
0
⇒
3
a
≤
2
1
+
4
x
2
So,
a
max
=
2
3
(
1
1
+
4
×
π
2
36
)
=
6
9
+
π
2
=
a
∴
f
a
(
π
8
)
=
1
−
3
π
8
⋅
6
9
+
π
2
+
7
=
8
−
9
π
4
(
9
+
π
2
)
© examsnet.com
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