Let f : R to R be a function such that |f(x)| | vedclass

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

Question

(B) Given $|f(x)| \leq x^2$ for all $x \in R$.At $x = 0$,$|f(0)| \leq 0^2 = 0$,which implies $f(0) = 0$.To check differentiability at $x = 0$,we evalu

Vedclass · Accepted Answer

continuous as well as differentiable

Vedclass · Answer

(B) Given $|f(x)| \leq x^2$ for all $x \in R$.At $x = 0$,$|f(0)| \leq 0^2 = 0$,which implies $f(0) = 0$.To check differentiability at $x = 0$,we evaluate the limit of the derivative:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h - 0} = \lim_{h 	o 0} \frac{f(h)}{h}$.From the given inequality,$|f(h)| \leq h^2$,so $|\frac{f(h)}{h}| \leq |h|$.By the Sandwich Theorem,as $h 	o 0$,$|h| 	o 0$,therefore $\lim_{h 	o 0} \frac{f(h)}{h} = 0$.Thus,$f'(0) = 0$,which means $f$ is differentiable at $x = 0$.Since differentiability implies continuity,$f$ is also continuous at $x = 0$.Therefore,$f$ is continuous as well as differentiable at $x = 0$.

Let $f : R \to R$ be a function such that $|f(x)| \leq x^2$,for all $x \in R$. Then,at $x = 0$,$f$ is

Similar Questions

Let $S = \{(\lambda, \mu) \in R \times R : f(t) = (\|\lambda\|e^{\|t\|} - \mu) \sin(2\|t\|), t \in R\}$ be a differentiable function. Then $S$ is a subset of?

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
Then:

Let $f(x) = \begin{cases} (x - 1) \sin \frac{1}{x - 1}, & x \neq 1 \\ 0, & x = 1 \end{cases}$. Then which one of the following is true?

If the function $f(x) = \begin{cases} -x, & x < 1 \\ a + \cos^{-1}(x + b), & 1 \le x \le 2 \end{cases}$ is differentiable at $x = 1$,then $\frac{a}{b}$ is equal to:

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} \frac{\sin(x^2)}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$. Then,at $x=0$,$f$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module