The domain of the function `f(x)=log_e {sgn(9-x^2)}+sqrt([x]^3-4[x])`(where [] represents the greatest integer function isA. `[-2,1)uu[2.3)`B. `[-4,1)uu[2,3)`C. `94,1)uu[2,3)`D. `[2,1)uu[2,3)`

The domain of the function `f(x)=log_e {sgn(9-x^2)}+sqrt([x]^3-4[x])`(

1.	The domain of the function `f(x)=log_e {sgn(9-x^2)}+sqrt([x]^3-4[x])`(where [] represents the greatest integer function isA. `[-2,1)uu[2.3)`B. `[-4,1)uu[2,3)`C. `94,1)uu[2,3)`D. `[2,1)uu[2,3)`
Answer» Correct Answer - A We have `f(x)=log_(e){sgn(9-x^(2))}+sqrt([x]^(3)-4[x])` We must have, `sgn(9-x^(2))gt0` `rArr" "9-x^(2)gt0` `rArr" "x^(2)-9lt0` `rArr" "(x-3)(x+3)lt0` `rArr" "-3ltxlt3` `"Also "[x]^(3)-4[x]ge0` `rArr" "[x]([x]^(2)-4)ge0` `rArr" "[x]([x]-2)([x]+2)le0` `rArr" "[x]ge2 or [x]` lies between -2 and 0, i.e., `[x]=-2,-1or0` Now `[x]ge2 rArrxge2` `[x]=-2rArr-2lexlt1` `[x]=-1 rArr-1lexlt0` `[x]=0 rArr0lexlt1`. Hence `[x]=-2,-1,0rArr -2 lexlt1`. Hence `D_(f)=[-2,1)cup[2,3)`.