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Byju's Answer
Standard XII
Mathematics
One - One function
Let fx = |x...
Question
Let
f
(
x
)
=
|
x
−
2
|
, where
x
is a real number. Which one of the following is true?
A
f
is periodic
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B
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
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C
f
is an odd function
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D
f
is not one-one function
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E
f
is an even function
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Solution
The correct option is
D
f
is not one-one function
From the graph of the function it can be observed clearly that
f
is not one-one function.
So option
D
is correct.
Suggest Corrections
0
Similar questions
Q.
Assertion :
f
is even,
g
is odd then
f
g
(
g
≠
0
)
is an odd function. Reason: If
f
(
−
x
)
=
−
f
(
x
)
for every
x
of its domain then
f
(
x
)
is called odd function and if
f
(
−
x
)
=
f
(
x
)
for every
x
of its domain, then
f
(
x
)
is called even function.
Q.
If
f
(
x
)
is an odd function then-
(i)
f
(
−
x
)
+
f
(
x
)
2
is an even function
(ii)
[
∣
f
(
x
)
∣
+
1
]
is even where [.] denotes greatest integer function.
(iii)
f
(
x
)
−
f
(
−
x
)
2
is neither even nor odd
(iv)
f
(
x
)
+
f
(
−
x
)
is neither even nor odd
Which of these statements are correct
Q.
Let
f
:
(
−
π
2
,
π
2
)
→
R
be given by f(x) =
[
l
o
g
(
s
e
c
x
+
t
a
n
x
)
]
3
. Then
Q.
Let
f
be a differentiable function on
R
and satisfies
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
∀
x
,
y
ϵ
R
. If
f
′
(
0
)
=
2
, then the value of
f
(
4
)
is
Q.
Assertion :The function
f
(
x
)
=
∫
x
0
√
1
+
t
2
d
t
is an odd function and
g
(
x
)
=
f
′
(
x
)
is an even function. Reason: For a differentiable function
f
(
x
)
if
f
′
′
(
x
)
is an even function, then
f
(
x
)
is an odd function.
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