"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 :"

Question

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

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

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

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

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 :

Solution

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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 :

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