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 ${}^{\prime }{x}^{\prime }$ respectively, is

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

Answer 1

Answer: (B)

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