If $f: R \rightarrow R$ is defined by $f(x)=x-[x]-\frac{1}{2}$ for $x \in R$, where $[x]$ is the greatest…
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 $\left\{x \in R: f(x)=\frac{1}{2}\right\}$ is equal to :
$Z$, the set of all integers
$N$, the set of all natural numbers
$\phi$, the empty set
R
Solution
$\because \quad f(x)=x-[x]-\frac{1}{2}$
Also $\quad f(x)=\frac{1}{2}$
$\therefore \quad \frac{1}{2}=x-[x]-\frac{1}{2}$
$\Rightarrow \quad x-[x]=1$
$\Rightarrow \quad\{x\}=1$ $[\because x=[x]+\{x\}]$
Which is not possible.
$\therefore\left\{x \in R: f(x)=\frac{1}{2}\right\}$ is an empty set.