The domain of the function, f (x) = ^ - 1 ( 3 x ^ 2 + x - 1 ( x - 1...

The domain of the function

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

For, $\cos^{- 1}\left(\dfrac{x - 1}{x + 1}\right)$

$- 1 \le \dfrac{x - 1}{x + 1} \le 1$

$⇒ - 1 \le 1 - \dfrac{2}{x + 1} \le 1$

$⇒ - 2 \le \dfrac{- 2}{x + 1} \le 0$

$⇒ 0 \le \dfrac{1}{x + 1} \le 1$

$⇒ x + 1 \in [ 1, \infty )$

$⇒ x \in [ 0, \infty ) . . . . \left(i\right)$

For, $\sin^{- 1}\left(\dfrac{3{x}^{2} + x - 1}{( x - 1{)}^{2}}\right)$

$- 1 \le \dfrac{3{x}^{2} + x - 1}{{\left(x - 1\right)}^{2}} \le 1$

$⇒ - {\left(x - 1\right)}^{2} \le 3{x}^{2} + x - 1 \le {\left(x - 1\right)}^{2}, x \ne 1$

Now, $- {\left(x - 1\right)}^{2} \le 3{x}^{2} + x - 1, x \ne 1$

$⇒ 4{x}^{2} - x \ge 0, x \ne 1$

$⇒ x\left(4x - 1\right) \ge 0, x \ne 1$

$⇒ x \in ( - \infty , 0 ] \cup \left[\dfrac{1}{4}, \infty\right) - \left\{1\right\} . . . . \left(ii\right)$

And $3{x}^{2} + x - 1 \le {\left(x - 1\right)}^{2}, x \ne 1$

$⇒ 2{x}^{2} + 3x - 2 \le 0, x \ne 1$

$⇒ \left(x + 2\right)\left(2x - 1\right) \le 0, x \ne 1$

$⇒ x \in \left[- 2, \dfrac{1}{2}\right] . . . . \left(iii\right)$

Domain of the function $\sin^{- 1}\left(\dfrac{3{x}^{2} + x - 1}{( x - 1{)}^{2}}\right)$ from the equations \left(ii\right) \text{ and } $\left(iii\right)$ is

$⇒ x \in \left[- 2, 0\right] \cup \left[\dfrac{1}{4}, \dfrac{1}{2}\right] . . . . \left(iv\right)$

Now the domain of the given function will be the intersection of the equation $\left(i\right) \text{ and } \left(iv\right)$

Hence, domain is $x \in \left[\dfrac{1}{4}, \dfrac{1}{2}\right] \cup { 0 }$