Question

Let $[.]$ and $\{.\}$ represent the greatest integer function and the fractional part function respectively. The number of value(s) of $x$ satisfying the equation $|2x-1|=3\left[x\right]+2\left\{x\right\}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$|2x-1|=3\left[x\right]+2\left\{x\right\}$
Let $2x-1\le 0$ i.e., $x\le \frac{1}{2}$.
The given equation yields
$1-2x=3\left[x\right]+2\left\{x\right\}$
$\Rightarrow 1-2\left[x\right]-2\left\{x\right\}=3\left[x\right]+2\left\{x\right\}\Rightarrow 1-5\left[x\right]=4\left\{x\right\}\Rightarrow \left\{x\right\}=\frac{1-5\left[x\right]}{4}\dots \left(1\right)$
$\Rightarrow 0\le \frac{1-5\left[x\right]}{4}<1\Rightarrow 0\le 1-5\left[x\right]<4\Rightarrow -\frac{3}{5}<\left[x\right]\le \frac{1}{5}$
So, $\left[x\right]=0$ as zero is the only integer lying between $-\frac{3}{5}$ and $\frac{1}{5}.$
From $\left(1\right)$, $\left\{x\right\}=\frac{1}{4}$
$\Rightarrow x=\frac{1}{4}$ which is less than $\frac{1}{2}$.
Hence, $x=\frac{1}{4}$ is a solution.

Now, let $2x-1>0$ i.e., $x>\frac{1}{2}$
$\Rightarrow 2x-1=3\left[x\right]+2\left\{x\right\}$
$\Rightarrow 2\left[x\right]+2\left\{x\right\}-1=3\left[x\right]+2\left\{x\right\}$
$\Rightarrow \left[x\right]=-1$
$\Rightarrow -1\le x<0$ which is not a solution as $x>\frac{1}{2}$

Hence, $x=\frac{1}{4}$ is the only solution.