Question

The domain of the function $f\left(x\right)={\sin }^{-1}\left(\frac{3x^2+x-1}{(x-1{)}^2}\right)+{\cos }^{-1}\left(\frac{x-1}{x+1}\right)$ is

• A. $[0,\frac{1}{2}]$
• B. $[\frac{1}{4},\frac{1}{2}]\cup \left\{0\right\}$
• C. $[0,\frac{1}{4}]$
• D. $[-2,0]\cup [\frac{1}{4},\frac{1}{2}]$

Answer 1

The correct option is B $[\frac{1}{4},\frac{1}{2}]\cup \left\{0\right\}$

Domain: $\left(\frac{3x^2+x-1}{(x-1{)}^2}\right)\in [-1,1]$ and $\frac{x-1}{x+1}\in [-1,1]$

$\frac{3x^2+x-1}{(x-1{)}^2}\ge -1\Rightarrow \frac{4x^2-x}{(x-1{)}^2}\ge 0\Rightarrow \frac{4x(x-\frac{1}{4})}{(x-1{)}^2}\ge 0$

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

And $\frac{3x^2+x-1}{(x-1{)}^2}\le 1$
$\Rightarrow \frac{2(x+2)(x-\frac{1}{2})}{(x-1{)}^2}\le 0$

$\Rightarrow x\in [-2,\frac{1}{2}]\dots \left(2\right)$

And $\frac{x-1}{x+1}\le 1$
$\Rightarrow \frac{-2}{x+1}\le 0\Rightarrow x\in (-1,\infty )\dots \left(3\right)$

And $\frac{x-1}{x+1}\ge -1$
$\Rightarrow \frac{2x}{x+1}\ge 0$

$\Rightarrow x\in (-\infty ,-1)\cup [0,\infty )\dots \left(4\right)$

Now $\left(1\right)\cap \left(2\right)\cap \left(3\right)\cap \left(4\right)$
$x\in [\frac{1}{4},\frac{1}{2}]\cup \left\{0\right\}$