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Byju's Answer
Standard XII
Mathematics
Local Maxima
The function ...
Question
The function
f
(
x
)
=
(
4
sin
2
x
−
1
)
n
(
x
2
−
x
+
1
)
,
n
∈
N
,has a local minimum at
x
=
π
6
, then
A
n
can be any even number
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B
n
can be an odd number
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C
n
can be odd prime number
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D
n
can be any natural number
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Solution
The correct option is
A
n
can be any even number
f
(
π
6
)
=
0
f
(
x
)
=
(
4
sin
2
x
−
1
)
n
(
(
x
−
1
)
2
+
3
4
)
(
x
−
1
)
2
+
3
4
>
0
,
∀
x
∈
R
f
(
π
6
+
)
=
lim
x
→
π
/
6
+
(
4
sin
2
x
−
1
)
n
(
x
2
−
x
+
1
)
lim
x
→
π
/
6
+
4
sin
2
x
=
1
+
⇒
lim
x
→
π
/
6
+
f
(
x
)
=
0
+
f
(
π
6
−
)
=
lim
x
→
π
/
6
−
(
4
sin
2
x
−
1
)
n
(
x
2
−
x
+
1
)
lim
x
→
π
/
6
−
4
sin
2
x
=
1
−
⇒
lim
x
→
π
/
6
−
f
(
x
)
=
0
−
But
x
=
π
6
is a minima
⇒
f
(
π
6
−
)
must be greater than zero. This can only happen for
n
=
2
k
Hence,
n
can be any even number.
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