If f: R rightarrow R is defined by f(x)=[2x]- | vedclass

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(D) We know that any real number $x$ can be written as $x = [x] + \{x\}$,where $[x]$ is the greatest integer part and $\{x\}$ is the fractional part,w

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$\{0, 1\}$

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(D) We know that any real number $x$ can be written as $x = [x] + \{x\}$,where $[x]$ is the greatest integer part and $\{x\}$ is the fractional part,with $0 \leq \{x\} Multiplying by $2$,we get $2x = 2[x] + 2\{x\}$.Taking the greatest integer on both sides,$[2x] = [2[x] + 2\{x\}] = 2[x] + [2\{x\}]$.Now,the function is $f(x) = [2x] - 2[x] = (2[x] + [2\{x\}]) - 2[x] = [2\{x\}]$.Since $0 \leq \{x\} Therefore,$[2\{x\}]$ can take values based on the interval of $\{x\}$:If $0 \leq \{x\} If $\frac{1}{2} \leq \{x\} Thus,the range of $f$ is $\{0, 1\}$.

If $f: R \rightarrow R$ is defined by $f(x)=[2x]-2[x]$ for $x \in R$,then the range of $f$ is (Here $[x]$ denotes the greatest integer not exceeding $x$)

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