Question

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

• A. $f(x)$ can take any value in $R$
• B. $f(x)<\, 0, \forall x \in R$
• C. $f(x)=0, \forall x \in R$
• D. $f( x ) >\,0, \forall x\, \in\, R$

Answer 1

Answer: (D)

$\left|\frac{f(x)-f(y)}{(x-y)}\right| \leq|(x-y)|$
$x-y=h\,\,\,$ let $\,\,\,\Rightarrow x=y+h$
$\displaystyle\lim _{x \rightarrow 0}\left|\frac{f(y+h)-f(y)}{h}\right| \leq 0$
$\Rightarrow \left|f^{\prime}(y)\right| \leq 0 \Rightarrow f^{\prime}(y)=0$
$\Rightarrow f( y )= k$ (constant)
and $f(0)=1$ given
So, $f(y)=1 \Rightarrow f(x)=1$