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JEE Main 2024
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Easy

Question

If the domain of the function f(x)=cos1x2x+1sin1(2x12)f(x) = {{{{\cos }^{ - 1}}\sqrt {{x^2} - x + 1} } \over {\sqrt {{{\sin }^{ - 1}}\left( {{{2x - 1} \over 2}} \right)} }} is the interval (α\alpha, β\beta], then α\alpha + β\beta is equal to :

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Solution

Ox2x+11O \le {x^2} - x + 1 \le 1 x2x0 \Rightarrow {x^2} - x \le 0 x[0,1] \Rightarrow x \in [0,1] Also, 0<sin1(2x12)π20 < {\sin ^{ - 1}}\left( {{{2x - 1} \over 2}} \right) \le {\pi \over 2} 0<2x121 \Rightarrow 0 < {{2x - 1} \over 2} \le 1 0<2x12 \Rightarrow 0 < 2x - 1 \le 2 1<2x31 < 2x \le 3 12<x32{1 \over 2} < x \le {3 \over 2} Taking intersection x(12,1]x \in \left( {{1 \over 2},1} \right] α=12,β=1 \Rightarrow \alpha = {1 \over 2},\beta = 1 α+β=32 \Rightarrow \alpha + \beta = {3 \over 2}

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