Let [x] denote the greatest integer less than | vedclass

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

Question

(C) For the function $f(x)$ to be continuous at $x=0$, we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(x) = \frac{\sin [-x^2]}{[-x^2]}$ for $x 
eq

Vedclass · Accepted Answer

$\alpha=\sin (1)$

Vedclass · Answer

(C) For the function $f(x)$ to be continuous at $x=0$, we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(x) = \frac{\sin [-x^2]}{[-x^2]}$ for $x 
eq 0$.As $x 	o 0$, $x^2$ approaches $0$ from the positive side, so $-x^2$ approaches $0$ from the negative side (i.e.,$-x^2 \in (-1, 0)$).Therefore, the greatest integer function $[-x^2]$ will take the value $-1$ as $x 	o 0$.Thus, $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{\sin [-x^2]}{[-x^2]} = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1)$.Since $f(0) = \alpha$, for continuity, we must have $\alpha = \sin(1)$.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then, the value of $\alpha$ for which the function $f(x)=\begin{cases} \frac{\sin [-x^2]}{[-x^2]}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$ is:

Similar Questions

Let $a, b, c$ be three real numbers. If the function $f(x) = \begin{cases} \cos(2x + \pi) & \text{if } x \leq 0 \\ ax^2 + b & \text{if } 0 < x < 1 \\ cx + 4 & \text{if } 1 \leq x \leq 2 \\ 3a + 1 & \text{if } x \geq 2 \end{cases}$ is continuous everywhere, then $b^2 - bc + c^2 =$

If the function $f$ is continuous at $x = 1$, where $f(x) = \frac{1 + \cos(\pi x)}{\pi(1-x)^2}$ for $x \neq 1$, then the value of $f(1)$ is....

The function $f(x)=\sqrt{\frac{3 x^2-5 x-2}{2 x^2-7 x+5}}$ has discontinuous points at $x=$

The function $f(x) = \frac{x+1}{9x+x^3}$ is

If the function $f(x) = \begin{cases} \frac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2}, & x \neq \pi \\ k, & x = \pi \end{cases}$ is continuous at $x = \pi$,then $k$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module