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JEE Main 17 March 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 64
Total: 90
The system of equations
k
x
+
y
+
z
=
1
,
x
+
k
y
+
z
=
k
and
x
+
y
+
z
k
=
k
2
has no solution, if
k
is equal to
[17 Mar 2021 Shift 1]
0
1
−
1
−
2
Validate
Solution:
k
x
+
y
+
z
=
1
x
+
k
y
+
z
=
k
x
+
y
+
k
z
=
k
2
For this set of equations to have no solution,
∆
=
0
|
k
1
1
1
k
1
1
1
k
|
=
k
(
k
2
−
1
)
−
1
(
k
−
1
)
+
(
1
−
k
)
=
k
3
−
k
−
k
+
1
+
1
−
k
=
k
3
−
3
k
+
2
Now,
∆
=
0
⇒
k
3
−
3
k
+
2
=
0
⇒
(
k
−
1
)
(
k
2
+
k
−
2
)
=
0
⇒
(
k
−
1
)
(
k
−
1
)
(
k
+
2
)
=
0
∴
k
=
1
,
−
2
x
+
y
+
z
=
1
If
k
=
1
x
+
y
+
z
=
1
x
+
y
+
z
=
1
}
There are same equations and they will have infinite solutions.
So,
K
=
−
2
© examsnet.com
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