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Byju's Answer
Standard XIII
Mathematics
Nature of Roots
The number of...
Question
The number of integral values of
k
, for which
(
x
2
−
x
+
1
)
(
k
x
2
−
3
k
x
−
5
)
<
0
is
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Solution
(
x
2
−
x
+
1
)
(
k
x
2
−
3
k
x
−
5
)
<
0
x
2
−
x
+
1
>
0
as
D
<
0
When
k
=
0
, we get
−
5
(
x
2
−
x
+
1
)
<
0
⇒
x
2
−
x
+
1
>
0
Which is true.
Now,
k
x
2
−
3
k
x
−
5
<
0
Therefore,
k
<
0
⋯
(
1
)
D
<
0
⇒
9
k
2
+
20
k
<
0
⇒
k
(
9
k
+
20
)
<
0
⇒
k
∈
(
−
20
9
,
0
)
⋯
(
2
)
From
(
1
)
and
(
2
)
,we get
⇒
k
∈
(
−
20
9
,
0
)
Hence, the integral values are
k
=
−
2
,
−
1
,
0
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0
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