If f: R rightarrow R is defined by f(x)=x-[x] | vedclass

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

Question

(C) Given the function $f(x) = x - [x] - \frac{1}{2}$.We know that the fractional part of $x$ is defined as $\{x\} = x - [x]$.Thus, $f(x) = \{x\} - \f

Vedclass · Accepted Answer

$\phi$, the empty set

Vedclass · Answer

(C) Given the function $f(x) = x - [x] - \frac{1}{2}$.We know that the fractional part of $x$ is defined as $\{x\} = x - [x]$.Thus, $f(x) = \{x\} - \frac{1}{2}$.We are given $f(x) = \frac{1}{2}$, so $\{x\} - \frac{1}{2} = \frac{1}{2}$.This implies $\{x\} = 1$.However, by definition, the fractional part $\{x\}$ always satisfies $0 \le \{x\} Since $\{x\}$ can never be equal to $1$, there is no value of $x$ that satisfies the equation.Therefore, the set $\{x \in R: f(x) = \frac{1}{2}\}$ is the empty set, denoted by $\phi$.

If $f: R \rightarrow R$ is defined by $f(x)=x-[x]-\frac{1}{2}$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then $\{x \in R: f(x)=\frac{1}{2}\}$ is equal to :

Similar Questions

If $f(x)=|x-1|+|x-2|+|x-3|$ for $2 < x < 3$,then $f$ is

$f(x) = \log \left( \left( \frac{2x^2 - 3}{x} \right) + \sqrt{\frac{4x^4 - 11x^2 + 9}{|x|}} \right)$ is

$f: N \rightarrow N$,is defined by $f(x)=x^6$ then, . . . . . . .

Let $g: N \rightarrow N$ be defined as
$g(3n+1)=3n+2$
$g(3n+2)=3n+3$
$g(3n+3)=3n+1, \text{ for all } n \geq 0$
Then which of the following statements is true?

Let $f, g: N - \{1\} \rightarrow N$ be functions defined by $f(a) = \alpha$,where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha}$ divides $a$,and $g(a) = a + 1$,for all $a \in N - \{1\}$. Then,the function $f + g$ is.

Vedclass Test Series

Exam Paper Generator

Online Exam Module