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Byju's Answer
Standard XII
Mathematics
Monotonically Increasing Functions
Let f:R→ R ...
Question
Let
f
:
R
→
R
is a function satisfying
f
(
2
−
x
)
=
f
(
2
+
x
)
and
f
(
20
−
x
)
=
f
(
x
)
,
∀
x
∈
R
. For this function
f
answer the following question.
If
f
(
2
)
≠
f
(
6
)
, then
A
Fundamental period of
f
(
x
)
is 16
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B
Fundamental period of
f
(
x
)
may be 12
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C
Period of
f
(
x
)
can't be 12
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D
Fundamental period of
f
(
x
)
is 8
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Solution
The correct option is
C
Fundamental period of
f
(
x
)
is 16
f
(
2
−
x
)
=
f
(
2
+
x
)
........
(
i
)
and
f
(
20
−
x
)
=
f
(
x
)
......
(
i
i
)
,
∀
x
ϵ
R
Replacing
x
by
2
−
x
in equation
(
i
)
f
(
x
)
=
f
(
4
−
x
)
........
(
i
i
i
)
By comparing
(
i
i
)
&
(
i
i
i
)
f
(
20
−
x
)
=
f
(
4
−
x
)
r
e
p
l
a
c
e
b
y
4
−
x
⇒
f
(
x
)
=
f
(
x
+
16
)
Hence fundamental period of
f
(
x
)
is 16.
Therefore,
(
A
)
is the correct ans
Suggest Corrections
0
Similar questions
Q.
Let
f
:
R
→
R
is a function satisfying
f
(
2
−
x
)
=
f
(
2
+
x
)
and
f
(
20
−
x
)
=
f
(
x
)
,
∀
x
∈
R
. For this function
f
answer the following question.
Graph of
y
=
f
(
x
)
is
Q.
Let
f
:
R
→
R
is a function satisfying
f
(
2
−
x
)
=
f
(
2
+
x
)
and
f
(
20
−
x
)
=
f
(
x
)
,
∀
x
∈
R
. For this function
f
answer the following question.
If
f
(
0
)
=
5
, then minimum possible number of values of
x
satisfying
f
(
x
)
=
5
, for
x
∈
[
0
,
170
]
, is
Q.
Let
f
:
R
→
R
is a function satisfying
f
(
2
−
x
)
=
f
(
2
+
x
)
and
f
(
20
−
x
)
=
f
(
x
)
∀
x
∈
R
If
f
(
0
)
=
5
then the minimum possible no. of values of
x
satisfying
f
(
x
)
=
5
for
x
=
[
0.
,
70
]
, is
Q.
I
f
f
:
R
→
R
be a function satisfying the functional Rule
f
(
x
+
f
(
y
)
)
=
f
(
x
)
+
x
+
f
(
x
−
y
)
;
∀
x
,
y
∈
R
then
Column I
Column II
(
P
)
f
(
0
)
(
A
)
1
(
Q
)
|
f
(
1
)
+
f
(
2
)
|
(
B
)
3
(
R
)
|
f
(
2
)
+
f
(
−
3
)
|
(
C
)
0
(
S
)
|
f
(
1
)
+
f
(
−
3
)
|
(
D
)
2
Q.
Let
f
be a non-zero real valued continuous function satisfying
f
(
x
+
y
)
=
f
(
x
)
.
f
(
y
)
for all
x
,
y
ϵ
R
. If
f
(
2
)
=
9
then
f
(
6
)
=
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