JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 12)
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 :
$${3 \over 2}$$
2
$${1 \over 2}$$
1
Explanation
$$O \le {x^2} - x + 1 \le 1$$
$$ \Rightarrow {x^2} - x \le 0$$
$$ \Rightarrow x \in [0,1]$$
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
$$x \in \left( {{1 \over 2},1} \right]$$
$$ \Rightarrow \alpha = {1 \over 2},\beta = 1$$
$$ \Rightarrow \alpha + \beta = {3 \over 2}$$
$$ \Rightarrow {x^2} - x \le 0$$
$$ \Rightarrow x \in [0,1]$$
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
$$x \in \left( {{1 \over 2},1} \right]$$
$$ \Rightarrow \alpha = {1 \over 2},\beta = 1$$
$$ \Rightarrow \alpha + \beta = {3 \over 2}$$
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