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AP EAPCET 20-Aug-2021 Shift 2 Paper
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Section:
Mathematics
© examsiri.com
Question :
7 of 160
Marks:
+1
,
-0
If
|
x
x
2
1
+
x
3
y
y
2
1
+
y
3
z
z
2
1
+
z
3
|
=
0
and
x
,
y
,
z
are all distinct, then
x
y
z
=
-1
1
0
3
Validate
Solution:
|
x
x
2
1
y
y
2
1
z
z
2
1
|
(
A
)
+
|
x
x
2
x
3
y
y
2
y
3
z
z
2
z
3
|
(
B
)
=
0
(given) ...(i)
A
=
[
x
x
2
1
y
y
2
1
z
z
2
1
]
On applying
R
3
→
R
3
−
R
1
,
R
2
→
R
2
−
R
1
, we get
A
=
[
x
x
2
1
y
−
x
y
2
−
x
2
0
z
−
x
z
2
−
x
2
0
]
∴
|
A
|
=
(
y
−
x
)
(
z
2
−
x
2
)
−
(
z
−
x
)
(
y
2
−
x
2
)
=
(
y
−
x
)
(
z
−
x
)
{
z
+
x
−
y
−
x
}
=
(
y
−
x
)
(
z
−
x
)
(
z
−
y
)
B
=
[
x
x
2
x
3
y
y
2
y
3
z
z
2
z
3
]
⇒
|
B
|
=
|
x
x
2
x
3
y
y
2
y
3
z
z
2
z
3
|
⇒
|
B
|
=
x
y
z
|
1
x
x
2
1
y
y
2
1
z
z
2
|
R
3
→
R
3
−
R
1
,
R
2
→
R
2
−
R
1
⇒
|
B
|
=
x
y
z
|
1
x
x
2
0
y
−
x
y
2
−
x
2
0
z
−
x
z
2
−
x
2
|
Then,
|
B
|
=
x
y
z
{
(
y
−
x
)
(
z
2
−
x
2
)
−
(
z
−
x
)
(
y
2
−
x
2
)
}
|
B
|
=
x
y
z
(
y
−
x
)
(
z
−
x
)
(
z
−
y
)
From Eq. (i), we get
|
A
|
+
|
B
|
=
0
⇒
(
y
−
x
)
(
z
−
x
)
(
z
−
y
)
+
x
y
z
(
y
−
x
)
(
z
−
x
)
(
z
−
y
)
=
0
⇒
(
y
−
x
)
(
z
−
x
)
(
z
−
y
)
(
1
+
x
y
z
)
=
0
⇒
1
+
x
y
z
=
0
[
∵
x
,
y
,
z
are all distinct
]
⇒
x
y
z
=
−
1
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