Let f be a differentiable function from


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 :

  1. (1) 1

  2. (2) 2

  3. (3) $\frac{1}{2}$

  4. (4) 0

Correct Option: 1


$\because f: R \rightarrow R$

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

$\int_{0}^{1} f^{2}(x) d x=\int_{0}^{1} 1 d x=[x]_{0}^{1}=1$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now