JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsEasyQuestionIf the domain of the function f(x)=cos−1x2−x+1sin−1(2x−12)f(x) = {{{{\cos }^{ - 1}}\sqrt {{x^2} - x + 1} } \over {\sqrt {{{\sin }^{ - 1}}\left( {{{2x - 1} \over 2}} \right)} }}f(x)=sin−1(22x−1)cos−1x2−x+1 is the interval (α\alphaα, β\betaβ], then α\alphaα + β\betaβ is equal to :OptionsA32{3 \over 2}23B2C12{1 \over 2}21D1Check AnswerHide SolutionSolutionO≤x2−x+1≤1O \le {x^2} - x + 1 \le 1O≤x2−x+1≤1 ⇒x2−x≤0 \Rightarrow {x^2} - x \le 0⇒x2−x≤0 ⇒x∈[0,1] \Rightarrow x \in [0,1]⇒x∈[0,1] Also, 0<sin−1(2x−12)≤π20 < {\sin ^{ - 1}}\left( {{{2x - 1} \over 2}} \right) \le {\pi \over 2}0<sin−1(22x−1)≤2π ⇒0<2x−12≤1 \Rightarrow 0 < {{2x - 1} \over 2} \le 1⇒0<22x−1≤1 ⇒0<2x−1≤2 \Rightarrow 0 < 2x - 1 \le 2⇒0<2x−1≤2 1<2x≤31 < 2x \le 31<2x≤3 12<x≤32{1 \over 2} < x \le {3 \over 2}21<x≤23 Taking intersection x∈(12,1]x \in \left( {{1 \over 2},1} \right]x∈(21,1] ⇒α=12,β=1 \Rightarrow \alpha = {1 \over 2},\beta = 1⇒α=21,β=1 ⇒α+β=32 \Rightarrow \alpha + \beta = {3 \over 2}⇒α+β=23