The domain of the function

Question:

The domain of the function

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

  1. $\left[0, \frac{1}{4}\right]$

  2. $[-2,0] \cup\left[\frac{1}{4}, \frac{1}{2}\right]$

  3. $\left[\frac{1}{4}, \frac{1}{2}\right] \cup\{0\}$

  4. $\left[0, \frac{1}{2}\right]$


Correct Option: , 3

Solution:

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

$-1 \leq \frac{x-1}{x+1} \leq 1 \Rightarrow 0 \leq x<\infty$........(1)

$-1 \leq \frac{3 x^{2}+x-1}{(x-1)^{2}} \leq 1 \Rightarrow x \in\left[\frac{-1}{4}, \frac{1}{2}\right] \cup\{0\}$......(2)

(1) & (2)

$\Rightarrow$ Domain $=\left[\frac{1}{4}, \frac{1}{2}\right] \cup\{0\}$

Leave a comment