Question

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

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$0$

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

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

Taking limit as $y\to x$, we get

$\underset{y\to x}{lim}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \underset{y\to x}{lim}|x-y|$

$\Rightarrow \left|\underset{y\to x}{lim}\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \left|\underset{y\to x}{lim}(x-y)\right|$

$\Rightarrow \left|f^{\prime }\left(x\right)\right|\le 0\Rightarrow \left|f^{\prime }\left(x\right)\right|=0[\because \left|f^{\prime }\left(x\right)\right|\ge 0]$

$\therefore f^{\prime }\left(x\right)=0\Rightarrow f\left(x\right)=c$

$\therefore h\left(x\right)=\int f\left(x\right)dx=\int cdx=cx+d$

where $d$ is constant of integraion
Therefore $h\left(x\right)$ is a linear function of $x$
which is constant for all $x.$

$f\left(0\right)=0$ ........given

$\therefore f\left(1\right)=0$