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Q.
The solution set of the equation $4\{x\} = x +[x],$ where $\{x\}$ and $[x]$ denote the fractional and integral parts of a real number ‘x’ respectively, is
Linear Inequalities
Solution:
Let $x = [x] +\{x\}$ , the equation becomes
$4 \{x\} = [x] + \{x\} + [x] $
$\Rightarrow 3\{x\} = 2[x]$
$ \Rightarrow \{x\} = \frac{2}{3}[x]$ ...(1)
$\because \:0 \leq \{x\} < 1 $
$\Rightarrow 0 \leq \frac{2}{3} [x] < 1$
$ \Rightarrow 0 \leq [x] < \frac{3}{2} $ and [x] in integer
$\therefore $ [x] = 0 or 1, from (1) {x} = 0 or $\frac{2}{3}$
$\therefore \: x = 0 + 0$ or $1 + \frac{2}{3}$
$ \Rightarrow \: x = 0$ or $\frac{5}{3}$
The solution set is $x \in \left\{ 0 , \frac{5}{3} \right\}$