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Question
Mathematics
The maximum value of the function f(x)=2 x3-15 x2+36 x-48 on the set A= x|x2+20 ≤ 9 x| is
Q. The maximum value of the function
f
(
x
)
=
2
x
3
−
15
x
2
+
36
x
−
48
on the set
A
=
{
x
∣
∣
x
2
+
20
≤
9
x
∣
∣
}
is
1498
180
JEE Advanced
JEE Advanced 2009
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Answer:
7
Solution:
f
′
(
x
)
=
6
(
x
−
2
)
(
x
−
3
)
so
f
(
x
)
is increasing
in
(
3
,
∞
)
Also,
A
=
{
4
≤
x
≤
5
}
∴
f
m
a
x
=
f
(
5
)
=
7