Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If x3-2 x2-9 x+18=0 and A=|1 2 3 4 x 6 7 8 9| then the maximum value of A is
Q. If
x
3
−
2
x
2
−
9
x
+
18
=
0
and
A
=
∣
∣
1
4
7
2
x
8
3
6
9
∣
∣
then the maximum value of
A
is
5160
132
KCET
KCET 2021
Determinants
Report Error
A
96
16%
B
36
21%
C
24
29%
D
120
33%
Solution:
(
x
−
2
)
(
x
2
−
9
)
=
0
x
=
2
,
3
,
−
3
f
(
x
)
=
∣
A
∣
=
−
12
x
+
60
Max value at
x
=
−
3
∴
∣
A
∣
=
96