Let `g(x)=1=x-[x] and f(x)={-1, x < 0 , 0, x=0 and 1, x > 0,` then for all `x, f(g(x))` is equal to (i) x (ii) 1 (iii) f(x) (iv) 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`