First slide

Introduction to linear inequalities

Question

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

Moderate

A

$x \in R$
B

$x \in R \sim [1, 2)$
C

$x \in [1, 2)$
D

$x \in R \sim [1, 2]$

Solution

$∵$            $(x - 1) = (x - [x]) (x - {x})$
$\Rightarrow$             $x = 1 + {x} [x]$

$\Rightarrow$          $[x] + {x} = 1 + {x} [x]$

$\Rightarrow$          $({x} - 1) ([x] - 1) = 0$

         ${x} - 1 \neq 0, ∴ [x] - 1 = 0$

               $[x] = 1 \Rightarrow x \in [1, 2)$

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