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

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(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

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$\phi$, the empty set

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(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 :

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