Question

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

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