If $$f$$ is a real valued differentiable function satisfying $$\left| {f\left( x \right) - f\left( y \right)} \right| \leqslant {\left( {x - y} \right)^2},\,x,\,y\, \in \,R$$        and $$f\left( 0 \right) = 0,$$   then $$f\left( 1 \right)$$  equals-

Solution :
$$\eqalign{ & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & \left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}} \right| \leqslant \mathop {\lim }\limits_{h \to 0} \left| {\frac{{{{\left( h \right)}^2}}}{h}} \right| \cr & \Rightarrow \left| {f'\left( x \right)} \right| \leqslant 0 \cr & \Rightarrow f'\left( x \right) = 0 \cr & \Rightarrow f\left( x \right) = {\text{constant}} \cr & {\text{As }}f\left( 0 \right) = 0\,\, \Rightarrow f\left( 1 \right) = 0 \cr} $$