Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The condition that f(x) = ax3 + bx2 + cx + d has no extreme value is
Q. The condition that
f
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
d
has no extreme value is
2276
220
Application of Derivatives
Report Error
A
b
2
>
3
a
c
21%
B
b
2
=
4
a
c
41%
C
b
3
=
3
a
c
17%
D
b
2
<
3
a
c
21%
Solution:
f
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
d
.
f
′
(
x
)
=
3
a
x
2
+
2
b
x
+
c
For extreme values,
f
′
(
x
)
=
0
∴
3
a
x
2
+
2
b
x
+
c
=
0
It has no roots if
4
b
2
−
12
a
c
<
0
i.e.,
b
2
<
3
a
c
.