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Let : $\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $|\mathrm{f}(\mathrm{x})-\mathrm{f}(4)| \leq 2|\mathrm{x}-\mathrm{y}|^{\frac{3}{2}} \forall \mathrm{x}, \mathrm{y} \in \mathbb{R}$ If $f(0)=1$, then $\int_0^1 f^2(x) d x=$
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$1$
$\because|f(x)-f(y)| \leq 2|x-y|^{3 / 2}$
$\Rightarrow\left|\frac{f(x)-f(y)}{x-y}\right| \leq 2|x-y|^{\frac{1}{2}}$
as $\mathrm{x} \rightarrow \mathrm{y}$, we get :
$\Rightarrow\left|\mathrm{f}^{\prime}(\mathrm{y})\right| \leq 0$
$\Rightarrow\left|\mathrm{f}^{\prime}(\mathrm{y})\right|=0 \quad\{\because$ absolute value can not be negative $\}$
$\begin{aligned} & \Rightarrow f^{\prime}(y)=0 \\ & \Rightarrow f(y)=c\end{aligned}$
$\because f(0)=1 \Rightarrow 1=c$
$\therefore \mathrm{f}(\mathrm{y})=1$
Now, $\int_0^1 \mathrm{f}^2(\mathrm{x}) \mathrm{dx}=\int_0^1 1 \mathrm{dx}=1-0=1$
$\Rightarrow\left|\frac{f(x)-f(y)}{x-y}\right| \leq 2|x-y|^{\frac{1}{2}}$
as $\mathrm{x} \rightarrow \mathrm{y}$, we get :
$\Rightarrow\left|\mathrm{f}^{\prime}(\mathrm{y})\right| \leq 0$
$\Rightarrow\left|\mathrm{f}^{\prime}(\mathrm{y})\right|=0 \quad\{\because$ absolute value can not be negative $\}$
$\begin{aligned} & \Rightarrow f^{\prime}(y)=0 \\ & \Rightarrow f(y)=c\end{aligned}$
$\because f(0)=1 \Rightarrow 1=c$
$\therefore \mathrm{f}(\mathrm{y})=1$
Now, $\int_0^1 \mathrm{f}^2(\mathrm{x}) \mathrm{dx}=\int_0^1 1 \mathrm{dx}=1-0=1$
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