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 are looking for $x$ such that $f(x) = \frac{1}{2}$.Substituting the expression for $f(x)$,we

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$\phi$,the empty set

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(C) Given the function $f(x) = x - [x] - \frac{1}{2}$.We are looking for $x$ such that $f(x) = \frac{1}{2}$.Substituting the expression for $f(x)$,we get $\frac{1}{2} = x - [x] - \frac{1}{2}$.Adding $\frac{1}{2}$ to both sides,we get $x - [x] = 1$.By definition of the fractional part function,${x} = x - [x]$,where $0 \le \{x\} Thus,the equation becomes ${x} = 1$.Since the fractional part of any real number $x$ must be strictly less than $1$,there is no real number $x$ that satisfies this condition.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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