What are we learning ?

Competency in Focus: Limit and Differentiability of a function

This problem from I.S.I. B.Stat. Entrance 2017 , it is based on simple manipulations,  differentiation from first principles and squeeze theorem.

First look at the knowledge graph.

Next understand the problem

Let f : R → R be a continuous function such that for any two real
numbers x and y, \( |f(x)-f(y)| \le {|x-y|}^{201} \).
Source of the problem
I.S.I. B.Stat. Entrance 2017, UGA Problem 5
Key Competency
Differentiation from first principles and squeeze (or sandwichtheorem 
Difficulty Level
Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

Start with hints

Do you really need a hint? Try it first!
Given that \( |f(x)-f(y)| \le {|x-y|}^{201} \) for any two real number x and y . So we can replace x by y+h for any real number y and h .Then we have , \( |f(y+h)-f(y)| \le {|(y+h)-y|}^{201}  \implies |f(y+h)-f(y)| \le {|h|}^{201} \) . Now,  what can we use ?

See that we can write  \(\frac{ |f(y+h)-f(y)|}{h} \le {|h|}^{200} \implies -{h}^{200} \le \frac {|f(y+h)-f(y)|}{h} \le h^{200}\) .  Now taking limit h tends to 0 on both sides gives what ?

See that \(\lim_{h \to 0} -{h}^{200}=\lim_{h \to 0} {h}^{200}=0 \) and differentiation from first principles say’s \(\lim_{h \to 0} \frac{f(y+h)-f(y)}{h} =f'(y) \) .
Now we know the the squeeze theorem, stated as follows   Let I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have
and also suppose that
Then So, what we have ?
f'(y)=0 for all y belongs to real . Thus integrating both sides we have f(y)=c ( constant ) for all y belongs to real . Therefore , f(101)=f(202)=f(200).

Connected Program at Cheenta

Amc 8 Master class

Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.

Similar Problems