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Byju's Answer
Standard XII
Mathematics
Existence of Limit
Let limx→ 0...
Question
Let
lim
x
→
0
sin
x
+
a
e
x
+
b
e
−
x
+
c
ln
(
1
+
x
)
x
2
=
1
The roots of the quadratic equation
a
x
2
+
b
x
+
c
=
0
are
A
both positive
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B
both negative
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C
one positive and one negative
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D
complex conjugate
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Solution
The correct option is
B
both positive
L
=
lim
x
→
0
sin
x
+
a
e
x
+
b
e
−
x
+
c
l
(
1
+
x
)
x
3
=
lim
x
→
0
(
x
−
x
3
31
)
+
a
(
1
+
x
1
!
+
x
2
2
!
+
x
3
3
!
)
x
3
+
b
(
1
−
x
1
!
+
x
2
2
!
−
x
3
3
!
)
+
c
(
x
−
x
2
2
+
x
3
3
)
x
3
=
lim
x
→
0
(
a
+
b
)
+
(
1
+
a
−
b
+
c
)
x
+
(
a
2
+
b
2
−
c
2
)
x
2
x
3
+
(
−
1
3
!
+
a
3
!
−
b
3
!
+
c
3
)
x
3
x
3
or
a
+
b
=
0
,
1
+
a
−
b
+
c
=
0
,
a
2
+
b
2
−
c
2
=
0
and
L
=
−
1
3
!
+
a
3
!
−
a
3
!
+
c
3
Solving the first three equations, we get
c
=
0
,
a
=
−
1
/
2
,
b
=
1
/
2
Then,
L
=
−
1
/
3.
The given equation
a
x
2
+
b
x
+
c
=
0
reduces to
x
2
−
x
=
0
or
x
=
0
,
1.
Suggest Corrections
0
Similar questions
Q.
L
=
lim
x
→
0
sin
x
+
a
e
x
+
b
e
−
x
+
c
ln
(
1
+
x
)
x
3
=
Equation
a
x
2
+
b
x
+
c
=
0
has
Q.
Let
lim
x
→
0
sin
x
+
a
e
x
+
b
e
−
x
+
c
ln
(
1
+
x
)
x
2
=
1
The ordered triplet
(
a
,
b
,
c
)
is
Q.
L
=
lim
x
→
0
sin
x
+
a
e
x
+
b
e
−
x
+
c
ln
(
1
+
x
)
x
3
=
The value of
L
is
Q.
If
L
=
lim
x
→
0
sin
x
+
a
e
x
+
b
e
−
x
+
c
ln
(
1
+
x
)
x
3
exists, then the solution set of the inequality
|
|
x
−
c
|
−
2
a
|
<
4
b
is
Q.
If
α
and
β
are the roots of the quadratic equation
a
x
2
+
b
x
+
c
=
0
,
then
lim
x
→
α
1
−
cos
(
a
x
2
+
b
x
+
c
)
(
x
−
α
)
2
is :
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