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Byju's Answer
Standard XII
Mathematics
Monotonically Increasing Functions
lf fx=0 is ...
Question
lf
f
(
x
)
=
0
is a reciprocal equation of first type and odd degree, then a factor of
f
(
x
)
is:
A
x
−
2
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B
x
−
1
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C
x
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D
x
+
1
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Solution
The correct option is
D
x
+
1
When the reciprocal equation is of an odd degree,
x
=
−
1
is always a solution.
So,
x
+
1
is a factor.
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Q.
lf
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f
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is a reciprocal equation of second type and fifth degree, then a root of
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Q.
The root of the reciprocal equation of first type and of odd degree is:
Q.
lf
f
(
x
)
=
x
2
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≤
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≤
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,
f
(
x
)
=
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x
2
−
3
x
+
(
3
/
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)
, lf
1
≤
x
≤
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f
′′
(
x
)
is
Q.
lf
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=
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, if
0
≤
x
≤
1
,
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x
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