JEE Main 202224 Jun, Morning shiftMathematicsMediumTrigonometry118× since 2002The domain of the function f(x)=cos−1(x2−5x+6x2−9)loge(x2−3x+2)f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}f(x)=loge(x2−3x+2)cos−1(x2−9x2−5x+6) is :AOption A. (−∞,1)∪(2,∞)( - \infty ,1) \cup (2,\infty )(−∞,1)∪(2,∞)BOption B. (2,∞)(2,\infty )(2,∞)COption C. [−12,1)∪(2,∞)\left[ { - {1 \over 2},1} \right) \cup (2,\infty )[−21,1)∪(2,∞)DOption D. [−12,1)∪(2,∞)−{3,3+52,3−52}\left[ { - {1 \over 2},1} \right) \cup (2,\infty ) - \left\{ 3,{{{3 + \sqrt 5 } \over 2},{{3 - \sqrt 5 } \over 2}} \right\}[−21,1)∪(2,∞)−{3,23+5,23−5}CheckSkip · show solution
