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If $f: R \rightarrow R$ is defined by $f(x)=[2 x]-2[x]$ for $x \in R$, then the range of $f$ is (Here $[x]$ denotes the greatest integer not exceeding $x$ )
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Verified Answer
The correct answer is:
$\{0,1\}$
$$
\text { Since, } \begin{aligned}
x & =[x]+\{x\} \\
\Rightarrow \quad 2 x & =2[x]+2\{x\} \\
{[2 x] } & =2[x]+(2[x]) \\
{[2 x] } & = \begin{cases}2[x]+0, & 0 < \{x\} < \frac{1}{2} \\
2[x]+1, & \frac{1}{2} \leq\{x\} < 1\end{cases} \\
\therefore \quad[2 x]-2[x] & = \begin{cases}0, & 0 \leq\{x\} < \frac{1}{2} \\
1, & \frac{1}{2} \leq\{x\} < 1\end{cases}
\end{aligned}
$$
Since,
$$
\therefore \quad[2 x]-2[x]= \begin{cases}0, & 0 \leq\{x\} < \frac{1}{2} \\ 1, & \frac{1}{2} \leq\{x\} < 1\end{cases}
$$
Hence, Range of $f$ is $\{0,1\}$
\text { Since, } \begin{aligned}
x & =[x]+\{x\} \\
\Rightarrow \quad 2 x & =2[x]+2\{x\} \\
{[2 x] } & =2[x]+(2[x]) \\
{[2 x] } & = \begin{cases}2[x]+0, & 0 < \{x\} < \frac{1}{2} \\
2[x]+1, & \frac{1}{2} \leq\{x\} < 1\end{cases} \\
\therefore \quad[2 x]-2[x] & = \begin{cases}0, & 0 \leq\{x\} < \frac{1}{2} \\
1, & \frac{1}{2} \leq\{x\} < 1\end{cases}
\end{aligned}
$$
Since,
$$
\therefore \quad[2 x]-2[x]= \begin{cases}0, & 0 \leq\{x\} < \frac{1}{2} \\ 1, & \frac{1}{2} \leq\{x\} < 1\end{cases}
$$
Hence, Range of $f$ is $\{0,1\}$
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