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given aba bbcb cac bbc c0 0 applying c3 c3 c1 c2
Question:
Given:
|
a
b
a
α
+
b
b
c
b
α
+
c
a
c
+
b
b
c
+
c
0
|
=
0
Applying
C
3
→
C
3
−
(
α
C
1
+
C
2
)
If the determinant
|
a
b
0
b
c
0
a
α
+
b
b
c
+
c
−
(
a
c
2
+
2
b
c
+
c
)
|
=
0
⇒
−
(
a
α
2
+
2
b
α
+
c
)
(
a
c
−
b
2
)
=
0
⇒
a
α
2
+
2
b
α
+
c
=
0
or
b
2
=
a
c
∴
α
is root of
a
x
2
+
2
b
x
+
c
or
a
,
b
,
c
are in GP.
WBJEE
Updated On:
Aug 9, 2024
(A) a, b, c are in AP
(B) a, b, c are in GP
(C) a, b, c are in HP
(D) (x -α ) is a factor of ax
2
+ 2bx + c
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The Correct Option is
A,
D
Solution and Explanation
Explanation:
|
a
b
a
c
+
b
b
c
b
c
+
c
a
c
+
b
b
c
+
c
0
|
is equal to zero, then
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