The function f(x) = [|x|] - |[x]| where [x] d | vedclass

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

Question

(D) Let $f(x) = [|x|] - |[x]|$.Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$,so $|[x]| = [x]$. Thus,$f(x) = [x] - [x] = 0$.Case $2$: If $x <

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) Let $f(x) = [|x|] - |[x]|$.Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$,so $|[x]| = [x]$. Thus,$f(x) = [x] - [x] = 0$.Case $2$: If $x If $f = 0$,then $x = -n$ (an integer). Then $f(x) = [|-n|] - |[-n]| = [n] - |-n| = n - n = 0$.If $0 $|x| = n + f$. $[|x|] = [n + f] = n$. So,$f(x) = n - (n + 1) = -1$.Thus,$f(x) = 0$ for $x \geq 0$ and $x \in \mathbb{Z}$,and $f(x) = -1$ for $x \in \mathbb{R} \setminus \mathbb{Z}$ where $x Since the function is constant on intervals but jumps at integers for $x The range is $\{0, -1\}$,which is a finite set. Therefore,all given statements are correct.

The function $f(x) = [|x|] - |[x]|$ where $[x]$ denotes the greatest integer function:

Similar Questions

Let $N$ be the set of all natural numbers and $f: N \rightarrow N$ be such that $1990 < f(1990) < 2100$ and satisfies the equation $x-f(x)=19[\frac{x}{19}]-90[\frac{f(x)}{90}]$,where $[y]$ denotes the greatest integer less than or equal to $y$. Then the number of possible values of $f(1990)$ is

Let $f$ and $g$ be increasing and decreasing functions,respectively,from $[0, \infty)$ to $[0, \infty)$. Let $h(x) = f(g(x))$. If $h(0) = 0$,then $h(x) - h(1)$ is:

Let $f: X \rightarrow Y$ be a function and $A, B$ be non-void subsets of $Y$. Which of the following is true?

If $f(x) = x \left( \frac{1}{x-1} + \frac{1}{x} + \frac{1}{x+1} \right)$ for $x > 1$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module