Question:
Let $f$ be a differentiable function from $\mathbf{R}$ to $\mathbf{R}$ such that $|f(x)-f(y)| \leq 2|x-y|^{3 / 2}$, for all $x, y, \in \mathbf{R}$. If $f(0)=1$ then
$\int_{0}^{1} f^{2}(x) d x$ is equal to :
Correct Option: 1
Solution:
$\because f: R \rightarrow R$
and $|f(x)-f(y)| \leq 2 \cdot|x-y|^{3 / 2}$
$\Rightarrow \quad\left|\frac{f(x) \quad f(y)}{x \quad y}\right| \leq \sqrt{x-y}$
$\Rightarrow \quad \lim _{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right| \leq \lim _{x \rightarrow y} 2 \sqrt{x-y}$
$\Rightarrow \quad\left|f^{\prime}(x)\right|=0$
$\therefore f(x)$ is a constant function.
Given $f(0)=1 \quad \Rightarrow \quad f(x)=1$
Hence, the integral
$\int_{0}^{1} f^{2}(x) d x=\int_{0}^{1} 1 d x=[x]_{0}^{1}=1$