#### The domain of the function is :Option: 1Option: 2Option: 3Option: 4

Domain

$-1\leq \frac{3x^{2}+x-1}{\left ( x-1 \right )^{2}}\leq 1;x\neq1\\$

$\Rightarrow -\left ( x-1 \right )^{2}\leq 3x^{2}+x-1\leq \left ( x-1 \right )^{2}\\$

$\Rightarrow 4x^{2}-x\geq 0\; and\; 2x^{2}+3x-2\leq 0\\$

$\Rightarrow 4x\left ( x-\frac{1}{4} \right )\geq 0\; and\; 2x^{2}+4x-x-2\leq 0\\$

$\Rightarrow x\in\left ( -\infty,0 \right ]\cup \left [\frac{1}{4},\infty \right )\; and\; x\in\left [ -2,\frac{1}{2} \right ]\\$

$So\; x\in\; \left [ -2,0 \right ]\cup \left [ \frac{1}{4},\frac{1}{2} \right ]\\$            ..........(1)

$Also\; -1\leq \frac{x-1}{x+1}\leq 1\Rightarrow \frac{x-1}{x+1}+1\geq 0\; and\; \frac{x-1}{x+1}-1\leq 0\\$

$\Rightarrow \frac{2x}{x+1}\geq 0\; and\; \frac{-2}{x+1}\leq 0\Rightarrow \frac{1}{x+1}\geq 0,x\neq -1\\$

$\Rightarrow x \in\left ( -\infty,-1 \right ]\cup \left [ 0,\infty \right )\; and \; x\in\left ( -1,\infty \right )\\$

$\Rightarrow x\in\left [ 0,\infty \right )$                          .......(2)

From (1) and (2)

$x\in \{0\} \cup \left [ \frac{1}{4},\frac{1}{2} \right ]$