Let f be a differentiable function from R to R

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

  1. 0

  2. $\frac{1}{2}$

  3. 2

  4. 1


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 $x \rightarrow y$

$\left|f^{\prime}(\mathrm{y})\right| \leq 0 \Rightarrow f^{\prime}(\mathrm{y})=0 \Rightarrow f(\mathrm{y})=\mathrm{c} \Rightarrow f(\mathrm{x})=1$

$\int_{0}^{1} 1 . \mathrm{dx}=1$

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