Question

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

• A. $[-1,\frac{-1}{2}]\cup [0,\frac{1}{2}]\cup \left\{1\right\}$
• B. $[-1,\frac{-1}{2}]\cup [0,\frac{1}{2}]\cup \left\{2\right\}$
• C. $[-1,\frac{-1}{3}]\cup [0,\frac{1}{3}]\cup \left\{3\right\}$
• D. None of these

Answer 1

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

For f to be defined $2\{x{\}}^2-3\{x\}+1\ge 0$
$\left(\right\{x\}-1)\left(2\right\{x\}-1)\ge 0$
$\Rightarrow \left\{x\right\}\in (-\infty ,\frac{1}{2})\cup [1,\infty )$
But we know, $\left\{x\right\}\in [0,1)$
Thus $\left\{x\right\}\in [0,\frac{1}{2})\to \left(1\right)$
Also given that $x\in [-1,1]$
Hence required domain is $[-1,-\frac{1}{2}]\cup [0,\frac{1}{2}],\cup \left\{1\right\}$
Note $\{-x\}=1-\left\{x\right\}$