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)`.