Set, Relation and FunctionHard
Question
If f is a real valued function defined on [0,1] such that f(0) = f(1) = 0 and |f(x1) - f(x2)| < |x1 - x2| for all distinct x1, x2 ∈ [0,1]. Then which of the following is incorrect -
Options
A.|f(x1) - f(x2)| > 
, where f(x1) ≠ f(x2)
, where f(x1) ≠ f(x2)B.|f(x)| < x, ∀ x ∈ (0, 1]
C.|f(x)| < 1 - x, ∀ x ∈ [0, 1)
D.|f(x1) - f(x2)| <
Solution
Let x2 > x7
|f(x1) - f(x2)| ≤ |f(x1)| + |f(x2)|
⇒ |f(x1) - f(x2)| ≤ |f(x1) - f(0)| + |f(x2) - f(1)|
⇒ |f(x1) - f(x2)| < |(x1 - 0)| + |x2 - 1|
⇒ |f(x1) - f(x2)| < x1 + 1 - x2
⇒ |f(x1) - f(x2)| < 1 + x1 - x2 .........(1)
Also |f(x1) - f(x2)| < |x1 - x2|
or |f(x1) - f(x2)| < x2 - x1 .........(1)
adding (1) & (2)
2|f(x1) - f(x2)| < 1 ⇒ |f(x1) - f(x2)| <
So A is incorrect & is correct
Also |f(x) - f(0)| < |x - 0| ⇒ |f(x)| < x
SoB is correct
& |f(x) - f(1)| < |x - 1| ⇒ |f(x)| < 1 - x
So C is correct
|f(x1) - f(x2)| ≤ |f(x1)| + |f(x2)|
⇒ |f(x1) - f(x2)| ≤ |f(x1) - f(0)| + |f(x2) - f(1)|
⇒ |f(x1) - f(x2)| < |(x1 - 0)| + |x2 - 1|
⇒ |f(x1) - f(x2)| < x1 + 1 - x2
⇒ |f(x1) - f(x2)| < 1 + x1 - x2 .........(1)
Also |f(x1) - f(x2)| < |x1 - x2|
or |f(x1) - f(x2)| < x2 - x1 .........(1)
adding (1) & (2)
2|f(x1) - f(x2)| < 1 ⇒ |f(x1) - f(x2)| <
So A is incorrect & is correct
Also |f(x) - f(0)| < |x - 0| ⇒ |f(x)| < x
SoB is correct
& |f(x) - f(1)| < |x - 1| ⇒ |f(x)| < 1 - x
So C is correct
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