Question

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

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

Answer 1

The correct option is D $f\left(x\right)>0,\forall x\in R$

$|f\left(x\right)-f\left(y\right)|\le |(x-y{)}^2|,\forall x,y\in R$
$\Rightarrow \left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le |x-y|$
$\Rightarrow \underset{x\to y}{lim}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le 0$
$\Rightarrow |f^{\prime }\left(y\right)|\le 0\Rightarrow f^{\prime }\left(y\right)=0$
$\Rightarrow f\left(y\right)=C$
Since $f\left(0\right)=1\Rightarrow f\left(y\right)=1\forall y\in R$