Let f be any function defined on R and let it | vedclass

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(D) Given the condition $|f(x) - f(y)| \leq |(x - y)^2|$.Dividing both sides by $|x - y|$ (for $x 
eq y$),we get $\left| \frac{f(x) - f(y)}{x - y}

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$f(x) > 0, \forall x \in R$

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(D) Given the condition $|f(x) - f(y)| \leq |(x - y)^2|$.Dividing both sides by $|x - y|$ (for $x 
eq y$),we get $\left| \frac{f(x) - f(y)}{x - y} ight| \leq |x - y|$.Taking the limit as $x 	o y$,we have $\lim_{x 	o y} \left| \frac{f(x) - f(y)}{x - y} ight| \leq \lim_{x 	o y} |x - y|$.This implies $|f'(y)| \leq 0$.Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which means $f'(y) = 0$ for all $y \in R$.$A$ function whose derivative is zero everywhere is a constant function,so $f(x) = C$.Given $f(0) = 1$,we find $C = 1$.Thus,$f(x) = 1$ for all $x \in R$.Since $1 > 0$,the correct statement is $f(x) > 0, \forall x \in R$.

Let $f$ be any function defined on $R$ and let it satisfy the condition $|f(x) - f(y)| \leq |(x - y)^2|$,for all $(x, y) \in R$. If $f(0) = 1$,then:

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