Question

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

• A. {0}
• B. $\{0,\frac{5}{3}\}$
• C. $[0,\infty )$
• D. none of these

Answer 1

Answer: (B)

Let $x=[x]+\{x\}$ , the equation becomes
$4\{x\}=[x]+\{x\}+[x]$
$\Rightarrow 3\{x\}=2[x]$
$\Rightarrow \{x\}=\frac{2}{3}[x]$ ...(1)
$\because 0\le \{x\}<1$
$\Rightarrow 0\le \frac{2}{3}[x]<1$
$\Rightarrow 0\le [x]<\frac{3}{2}$ 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 $x\in \left\{0,\frac{5}{3}\right\}$