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Test Index
JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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Question : 67
Total: 92
Let
f
be any function defined on
R
and let it satisfy the condition
|
f
(
x
)
−
f
(
y
)
|
≤
|
(
x
−
y
)
2
|
,
∀
(
x
,
y
)
∈
R
If
f
(
0
)
=
1
, then
[2021,26 Feb. Shift-1]
f
(
x
)
can take any value in
R
f
(
x
)
<
0
,
∀
x
∈
R
f
(
x
)
=
0
,
∀
x
∈
R
f
(
x
)
>
0
,
∀
x
∈
R
Validate
Solution:
Given,
|
f
(
x
)
−
f
(
y
)
|
≤
x
−
y
|
2
⇒
|
f
(
x
)
−
f
(
y
)
|
|
x
−
y
|
≤
x
−
y
|
Now, taking the limit,
lim
x
→
y
|
f
(
x
)
−
f
(
y
)
x
−
y
|
≤
lim
x
→
y
|
x
−
y
|
⇒
f
′
(
y
)
|
≤
0
[ using the definition of
f
′
(
y
)
]
⇒
f
′
(
y
)
=
0
[since, modulus value can never be less than 0 ]
On integrating it, we get
f
(
y
)
=
c
(constant)
Given,
f
(
0
)
=
1
gives
c
=
1
∴
f
(
y
)
=
1
∀
y
∈
R
From given options,
f
(
x
)
>
0
∀
x
∈
R
is satisfied only.
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