Let `g(x)=1+x-[x] and f(x)={-1,x < 0, 0, x=0 1, x > 0.` Then for all `x,f(g(x))` is equal to (where [.] represents the greatest integer function). (a) `x` (b) `1` (c) `f(x)` (d) `g(x)`

Answer

`f(g(x))=1` for all `x`

Explanation

We have `g(x)=1+x-[x]=1+{x},` where `{x}` represent fractional part function
and `f(x)={(-1",",x lt 0),(0 ",",x =0),(1",",x gt 0):}`
`implies f(g(x))={(-1",",1+{x} lt 0),(0 ",",1+{x} =0),(1",",1+{x} gt 0):}`
`implies f(g(x))=1,1+{x} gt 0 " " ( :' 0 le {x} lt 1)`
`implies f(g(x))=1 AA x in R`