Question:
Let $\mathrm{f}$ be a differentiable function from $\mathrm{R}$ to $\mathrm{R}$ such
that $|f(x)-f(y)| \leq 2|x-y|^{\frac{3}{2}}$, for all $x, y \varepsilon R$. If
$f(0)=1$ then $\int_{0}^{1} f^{2}(\mathrm{x}) \mathrm{dx}$ is equal to
Correct Option: , 4
Solution:
$|f(\mathrm{x})-f(\mathrm{y})| \leq 2|\mathrm{x}-\mathrm{y}|^{3 / 2}$
divide both sides by $|x-y|$
$\left|\frac{f(x)-f(y)}{x-y}\right| \leq 2 .|x-y|^{1 / 2}$
apply limit $x \rightarrow y$
$\left|f^{\prime}(\mathrm{y})\right| \leq 0 \Rightarrow f^{\prime}(\mathrm{y})=0 \Rightarrow f(\mathrm{y})=\mathrm{c} \Rightarrow f(\mathrm{x})=1$
$\int_{0}^{1} 1 . \mathrm{dx}=1$