Let f be a differentiable function from R to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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

Similar Questions

If $f(x) = x(\sqrt{x} - \sqrt{x + 1}),$ then

If the function $g(x) = \begin{cases} ae^x, & x \le 0 \\ b\cos x + x, & x > 0 \end{cases}$ is differentiable,then the value of $a^2 + b^2$ is

Let $f: R \rightarrow R$ be a function defined by $f(x) = \max \{x, x^2\}$. Let $S$ denote the set of all points in $R$ where $f$ is not differentiable. Then $S$ is:

If $f: R \rightarrow R$ is defined as $f(x)=|x+1|+|x-1|$, then $f(x)$ is

If $f(x) = \begin{cases} 1, & x < 0 \\ 1 + \sin x, & 0 \le x < \frac{\pi}{2} \end{cases}$,then $f'(0) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module