If the domain of the function [f(x) = ^ - 1 x^2 - x + 1 ^ - 1( 2x - 1 2 ) ] is the interva…

Question

If the domain of the function $$f(x) = {{{{\cos }^{ - 1}}\sqrt {{x^2} - x + 1} } \over {\sqrt {{{\sin }^{ - 1}}\left( {{{2x - 1} \over 2}} \right)} }}$$ is the interval ($$\alpha$$, $$\beta$$], then $$\alpha$$ + $$\beta$$ is equal to :

Accepted Answer

$$O \le {x^2} - x + 1 \le 1$$

$$ \Rightarrow {x^2} - x \le 0$$

$$ \Rightarrow x \in [0,1]$$

Also, $$0 < {\sin ^{ - 1}}\left( {{{2x - 1} \over 2}} \right) \le {\pi  \over 2}$$

$$ \Rightarrow 0 < {{2x - 1} \over 2} \le 1$$

$$ \Rightarrow 0 < 2x - 1 \le 2$$

$$1 < 2x \le 3$$

$${1 \over 2} < x \le {3 \over 2}$$

Taking intersection

$$x \in \left( {{1 \over 2},1} \right]$$

$$ \Rightarrow \alpha  = {1 \over 2},\beta  = 1$$

$$ \Rightarrow \alpha  + \beta  = {3 \over 2}$$

Solutions · 1