"Let f be any function defined on R and let it satisfy the condition : $|f(x) - f(y)|\, \le \,|{(x - y)^2}|,\forall (x,y) \in R$If f(0) = 1, then :"

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"$|f(x) - f(y)|\, \le \,|{(x - y)^2}|$

$\Rightarrow \left| {{{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|x - y|$

$$ \Rightarrow \left| {\mathop {\lim }\limits_{x \to y} {{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|\mathop {\lim }\limits_{x \to y} (x - y)|$$

$\Rightarrow |f'(x)|\, \le 0$

$\Rightarrow f'(x) = 0$

$\Rightarrow f(x)$ is constant function.

$\because$ $f(0)$ = 1 then f(x) = 1"

Let f be any function defined on R and let it satisfy the condition : $|f(x) - f(y)|\, \le \,|{(x - y)^2}|,\forall (x,y) \in R$

If f(0) = 1, then :

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

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More Application of Derivatives questions

Drill 25 more like these. Every day. Free.

Let f be any function defined on R and let it satisfy the condition : $|f(x) - f(y)|\, \le \,|{(x - y)^2}|,\forall (x,y) \in R$

If f(0) = 1, then :