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Byju's Answer
Standard XIII
Mathematics
Higher Order Equations
The least pos...
Question
The least positive value of
a
for which
4
x
−
a
⋅
2
x
−
a
+
3
≤
0
is satisfied by atleast one real value of
x
is
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Solution
Let
f
(
x
)
=
4
x
−
a
⋅
2
x
−
a
+
3
and
x
1
,
x
2
be the roots of
f
(
x
)
=
0
Let
t
=
2
x
,
t
∈
(
0
,
∞
)
∀
x
∈
R
⇒
g
(
t
)
=
t
2
−
a
t
−
a
+
3
Let
t
1
,
t
2
be the roots of
g
(
t
)
=
0
For
f
(
x
)
≤
0
to have atleast one real
x
g
(
t
)
≤
0
should have one real
t
,
where
t
∈
(
0
,
∞
)
Required cases:
Case
I
:
Both roots are positive
So,
(
1
)
D
≥
0
⇒
a
2
+
4
a
−
12
≥
0
⇒
(
a
+
6
)
(
a
−
2
)
≥
0
⇒
a
∈
(
−
∞
,
−
6
]
∪
[
2
,
∞
)
(
2
)
−
B
2
A
>
0
⇒
a
2
>
0
⇒
a
>
0
(
3
)
g
(
0
)
>
0
⇒
−
a
+
3
>
0
⇒
a
<
3
∴
a
∈
[
2
,
3
)
.
.
.
(
i
)
Case
I
I
:
One root is positive other root is negative
0
lies in between the roots
So,
g
(
0
)
<
0
⇒
a
>
3
.
.
.
(
i
i
)
Case
I
I
I
:
One root is
0
and other root is greater than
0
g
(
0
)
=
0
⇒
a
=
3
−
B
2
A
>
0
⇒
a
2
>
0
⇒
a
>
0
∴
a
=
3
.
.
.
(
i
i
i
)
From
(
i
)
,
(
i
i
)
and
(
i
i
i
)
∴
a
∈
[
2
,
∞
)
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0
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