1sir can you illustrate how
If f is a real-valued differentiable function satisfying |f(x) - f(y)| ≤ (x-y)2,x,y belong to R and f(0)=0,then f(1) equals
a.1
b.2
c.0
d.-1
-
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3 Answers
3410 \le \left|\frac{f(x+h)-f(x)}{h} \right| \le h
By Sandwich Theorem, f'(x) = 0 i.e. f(x) is a constant
62in the original equation, substitute x=x+h and y=h
you will get the relation given by Prophet sir. :)
