Question

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

Answer 1

The correct option is C

$f\left(x\right)=1,\forall x\in R$

Explanation for correct option(s)

Option C: $f\left(x\right)=1,\forall x\in R$

Consider the given equation as,

$\left|f\left(x\right)-f\left(y\right)\right|\le \left|{\left(x-y\right)}^2\right|$

Rewrite the above Equation by dividing by $\left(x-y\right)$

$\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \left|\left(x-y\right)\right|......\left(1\right)$

Consider,

$$\begin{gathered} x-y=h \\ x=y+h \end{gathered}$$

Substitute the above value in the Equation (1)

$$\begin{gathered} \underset{h\to 0}{\lim }\left|\frac{f\left(y+h\right)-f\left(y\right)}{h}\right|\le \underset{h\to 0}{\lim }h \\ \left|f\text{'}\left(y\right)\right|\le 0 \\ f\text{'}\left(y\right)=0 \\ f\left(y\right)=k \\ f\left(0\right)=1=k \\ f\left(x\right)=1 \end{gathered}$$

Hence, the correct answer is Option C.