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(D) Given the condition $|f(x) - f(y)| \le 2|x - y|^{\frac{3}{2}}$.Dividing both sides by $|x - y|$ (where $x 
eq y$),we get:$\left| \frac{f(x) - f(y

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(D) Given the condition $|f(x) - f(y)| \le 2|x - y|^{\frac{3}{2}}$.Dividing both sides by $|x - y|$ (where $x 
eq y$),we get:$\left| \frac{f(x) - f(y)}{x - y} ight| \le 2|x - y|^{\frac{1}{2}}$.Taking the limit as $x 	o y$ on both sides:$\lim_{x 	o y} \left| \frac{f(x) - f(y)}{x - y} ight| \le \lim_{x 	o y} 2|x - y|^{\frac{1}{2}}$.This implies $|f'(y)| \le 0$.Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which means $f'(y) = 0$ for all $y \in R$.If the derivative is zero everywhere,the function $f(x)$ must be a constant.Given $f(0) = 1$,the constant function is $f(x) = 1$.Now,calculating the integral:$\int_{0}^{1} f^2(x) dx = \int_{0}^{1} (1)^2 dx = \int_{0}^{1} 1 dx = [x]_{0}^{1} = 1 - 0 = 1$.

Let $f$ be a differentiable function from $R$ to $R$ such that $|f(x) - f(y)| \le 2|x - y|^{\frac{3}{2}}$ for all $x, y \in R$. If $f(0) = 1$,then $\int_{0}^{1} f^2(x) dx$ is equal to

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