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Tardigrade
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Mathematics
Let α(a) and β(a) be the roots of the equation (√[3]1+a-1) x2+(√1+a-1) x+(√[6]1+a-1)=0 where a>-1. Then lim a arrow o+ α(a) and lim a arrow o+ β(a) are
Q. Let
α
(
a
)
and
β
(
a
)
be the roots of the equation
(
3
1
+
a
−
1
)
x
2
+
(
1
+
a
−
1
)
x
+
(
6
1
+
a
−
1
)
=
0
where
a
>
−
1
. Then
lim
a
→
o
+
α
(
a
)
and
lim
a
→
o
+
β
(
a
)
are
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A
−
2
5
and 1
B
−
2
1
and -1
C
−
2
7
and 2
D
−
2
9
and 3
Solution:
Let
1
+
a
=
y
⇒
(
y
1/3
−
1
)
x
2
+
(
y
1/2
−
1
)
x
+
y
1/6
−
1
=
0
⇒
(
y
−
1
y
1/3
−
1
)
x
2
+
(
y
−
1
y
1/2
−
1
)
x
+
y
−
1
y
1/6
−
1
=
0
Now taking
y
→
1
lim
on both the sides
⇒
3
1
x
2
+
2
1
x
+
6
1
=
0
⇒
2
x
2
+
3
x
+
1
=
0
x
=
−
1
,
−
2
1