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Test Index
JEE Main Math Class 12 Application of Derivatives Part 2 Questions
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© examsnet.com
Question : 23
Total: 100
Let
λ
*
be the largest value of
λ
for which the function
f
λ
(
x
)
=
4
λ
x
3
−
36
λ
x
2
+
36
x
+
48
is increasing for all
x
∈
R
. Then
f
λ
*
(
1
)
+
f
λ
*
(
−
1
)
is equal to :
[24-Jun-2022-Shift-2]
36
48
64
72
Validate
Solution:
👈: Video Solution
∵
f
λ
(
x
)
=
4
λ
x
3
−
36
λ
x
2
+
36
λ
+
48
∴
f
λ
′
(
x
)
=
12
(
λ
x
2
−
6
λ
x
+
3
)
For
f
λ
(
x
)
increasing :
(
6
λ
)
2
−
12
λ
≤
0
∴
λ
∈
[
0
,
1
3
]
∴
λ
*
=
1
3
Now,
f
λ
*
(
x
)
=
4
3
x
3
−
12
x
2
+
36
x
+
48
∴
f
λ
*
(
1
)
+
f
λ
*
(
−
1
)
=
73
1
2
−
1
1
2
=
72
© examsnet.com
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