JEE Main 202120 Jul, Evening shiftMathematicsEasySets, Relations and Functions168× since 2002Let f:R−{α6}→Rf:R - \left\{ {{\alpha \over 6}} \right\} \to Rf:R−{6α}→R be defined by f(x)=5x+36x−αf(x) = {{5x + 3} \over {6x - \alpha }}f(x)=6x−α5x+3. Then the value of α\alphaα for which (fof)(x) = x, for all x∈R−{α6}x \in R - \left\{ {{\alpha \over 6}} \right\}x∈R−{6α}, is :AOption A. No such α\alphaα existsBOption B. 5COption C. 8DOption D. 6CheckSkip · show solution
