The function f(x) = begin{cases} sgn([x]) & x | vedclass

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

Question

(B) Let us analyze the function $f(x)$ for different intervals.For $x 
otin I$,$f(x) = sgn([x])$.If $x \in (0, 1)$,$[x] = 0$,so $f(x) = sgn(0) = 0$.I

Vedclass · Accepted Answer

Discontinuous at every integer

Vedclass · Answer

(B) Let us analyze the function $f(x)$ for different intervals.For $x 
otin I$,$f(x) = sgn([x])$.If $x \in (0, 1)$,$[x] = 0$,so $f(x) = sgn(0) = 0$.If $x \in (1, 2)$,$[x] = 1$,so $f(x) = sgn(1) = 1$.If $x \in (-1, 0)$,$[x] = -1$,so $f(x) = sgn(-1) = -1$.For $x \in I$,$f(x) = [sgn(x)]$.If $x = 0$,$f(0) = [sgn(0)] = [0] = 0$.If $x > 0$ and $x \in I$,$f(x) = [sgn(x)] = [1] = 1$.If $x Combining these,the function is discontinuous at every integer $n \in I$ because the left-hand limit and right-hand limit do not match the function value at integers. Specifically,for any integer $n$,$\lim_{x 	o n^-} f(x) = sgn(n-1)$ and $\lim_{x 	o n^+} f(x) = sgn(n)$. These are not equal to $f(n) = [sgn(n)]$. Thus,the function is discontinuous at every integer.

The function $f(x) = \begin{cases} sgn([x]) & x \notin I \\ [sgn(x)] & x \in I \end{cases}$ is (where $sgn()$ denotes the signum function and $[.]$ denotes the greatest integer function):

Similar Questions

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

If $f(x) = \begin{cases} x^2, & \text{when } x \le 1 \\ x + 5, & \text{when } x > 1 \end{cases}$,then

If $f(x) = \begin{cases} e^{1/x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then:

$f(x) = \begin{cases} [x^2] - [-x^2], & x \neq 3 \\ k, & x = 3 \end{cases}$ is continuous at $x = 3$,then $k = $ where $[\cdot]$ is the greatest integer function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module