JEE Main 20202 Sep, Morning shiftMathematicsHardTrigonometry118× since 2002The domain of the function f(x) = sin−1(∣x∣+5x2+1){\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)sin−1(x2+1∣x∣+5) is (– ∞\infty∞, -a]∪\cup∪[a, ∞\infty∞). Then a is equal to :AOption A. 17−12{{\sqrt {17} - 1} \over 2}217−1BOption B. 1+172{{1 + \sqrt {17} } \over 2}21+17COption C. 172+1{{\sqrt {17} } \over 2} + 1217+1DOption D. 172{{\sqrt {17} } \over 2}217CheckSkip · show solution
