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  4. The solution of x-1=(x-[x])(x- x ) (where [x] and x are the integral and fractional part of x is

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Q. The solution of $x-1=(x-[x])(x-\{x\})$ (where $[x]$ and $\{ x \}$ are the integral and fractional part of $x$ is

Linear Inequalities

Solution:

$\because(x-1)=(x-[x])(x-\{x\})$
$\Rightarrow x=1+\{x\}[x] $
$\Rightarrow [x]+\{x\}=1+\{x\}[x\}$
$\Rightarrow (\{x\}-1)([x]-1)=0 ;\{x\}-1 \neq 0 $
$\therefore [x]-1=0$
$\Rightarrow [x]=1 $
$\Rightarrow x \in[1,2)$