Question

Domain $f\left(x\right)=\sqrt{2{\left\{x\right\}}^2-3\left\{x\right\}+1}$ where $\{.\}$ denotes the fractional part, in $\left[-1,1\right]$ is

• A. $\left[-1,1\right]\sim \left(\frac{1}{2},1\right)$
• B. $\left[-1,-\frac{1}{2}\right]\cup \left[0,\frac{1}{2}\right]\cup \left\{1\right\}$
• C. $\left[-1,\frac{1}{2}\right]$
• D. $\left[-\frac{1}{2},1\right]$

Answer 1

The correct option is B $\left[-1,-\frac{1}{2}\right]\cup \left[0,\frac{1}{2}\right]\cup \left\{1\right\}$

$f\left(x\right)=\sqrt{2{\left\{x\right\}}^2-3\left\{x\right\}+1}$

$=\sqrt{2{\left\{x\right\}}^2-2\left\{x\right\}+\left\{x\right\}+1}$

$=\sqrt{(2{\left\{x\right\}}^2-1)\left(\left\{x\right\}\right)-1}$

$\because$ the part under square root can't be negative, so

for domain $\to \left\{x\right\}\notin (1/2,1)$

$\left(B\right)$