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Byju's Answer
Standard XII
Physics
Vector Addition
If 3x2 + 4k...
Question
If
3
x
2
+
4
k
x
+
1
>
0
for all real values of
x
, then
k
lies in the interval.
A
(
−
√
3
2
,
√
3
2
)
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B
(
−
1
4
,
1
4
)
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C
(
−
√
3
,
−
√
3
)
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D
(
−
1
2
,
1
2
)
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Solution
The correct option is
A
(
−
√
3
2
,
√
3
2
)
Given
3
x
2
+
4
k
x
+
1
>
0
for all real values of
x
.
This is possible only when,
D
=
b
2
−
4
a
c
<
0
(
4
k
)
2
−
4
(
3
)
<
0
16
k
2
−
12
<
0
4
k
2
−
3
<
0
(
2
k
−
√
3
)
(
2
k
+
√
3
)
<
0
−
√
3
2
<
k
<
√
3
2
Suggest Corrections
0
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