Question

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

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is B $1$

$|f\left(x\right)-f\left(y\right)|\le 2|x-y{|}^{3/2}$
Divide both sides by $|x-y|$
$\Rightarrow \left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le 2|x-y{|}^{1/2}$

Taking $\underset{x\to y}{lim}$ both sides, we get
$\underset{x\to y}{lim}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le 2\underset{x\to y}{lim}|x-y{|}^{1/2}$

$\Rightarrow |f^{\prime }\left(x\right)|\le 0$
$|f^{\prime }\left(x\right)|$ is always non-negative.
$\Rightarrow f^{\prime }\left(x\right)=0$
$\therefore f\left(x\right)$ is a constant function.
As given, $f\left(0\right)=1$
$\therefore f\left(x\right)=1$

Now, $\underset{0}{\overset{1}{\int }}{f}^2\left(x\right)dx=\underset{0}{\overset{1}{\int }}(1{)}^{2}dx$ $=1$