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Test Index
JEE Main 31-Jan-2024 Shift 2 Solved Paper
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Section:
Mathematics
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© examsnet.com
Question : 22
Total: 90
Let
a
,
b
,
c
be the length of three sides of a triangle satisfying the condition
(
a
2
+
b
2
)
x
2
−
2
b
(
a
+
c
)
.
x
+
(
b
2
+
c
2
)
=
0
. If the set of all possible values of
x
is the interval
(
α
,
β
)
, then
12
(
α
2
+
β
2
)
is equal to____
[31-Jan-2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
(
a
2
+
b
2
)
x
2
−
2
b
(
a
+
c
)
x
+
b
2
+
c
2
=
0
⇒
a
2
x
2
−
2
a
b
x
+
b
2
+
b
2
x
2
−
2
b
c
x
+
c
2
=
0
⇒
(
a
x
−
b
)
2
+
(
b
x
−
c
)
2
=
0
⇒
a
x
−
b
=
0
,
b
x
−
c
=
0
⇒
a
+
b
>
c
b
+
c
>
a
c
+
a
>
b
a
+
ax
>
bx
ax
+
bx
>
a
ax
2
+
a
>
ax
a
+
ax
>
ax
2
ax
+
ax
2
>
a
x
2
−
x
+
1
>
0
x
2
−
x
−
1
<
0
x
2
+
x
−
1
>
0
always true
1
−
√
5
2
<
x
<
1
+
√
5
2
x
<
−
1
−
√
5
2
,
or
x
>
−
1
+
√
5
2
⇒
√
5
−
1
2
<
x
<
√
5
+
1
2
⇒
α
=
√
5
−
1
2
,
β
=
√
5
+
1
2
12
(
α
2
+
β
2
)
=
12
(
(
√
5
−
1
)
2
+
(
√
5
+
1
)
2
4
)
=
36
© examsnet.com
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