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
Create a free account to view solution
View Solution FreeMore Set, Relation and Function Questions
Let g : → be a differentiable functions with g(0) = 0, g′(0) = 0 and g′(1) ≠ 0. Let and h(x) = e...The set S = {1, 2, 3, ...., 12) is to be partitioned into three sets A, B, C of equal size. Thus, A∪B∪C = S,...A function f from the set of natural numbers to integers defined by f(n) = is...Two points A and B in a plane are related if OA = OB, where O is a fixed point. This relation is...Let R be a relation defined on Z × Z0 (Z is the set of all integers and Z0 is the set of all non-zero integers) by ...