Let `g(x) = 1 + x-[x] and f(x)=-1` if `x <0=0` if `x > 0`, then `f|g(x)]=1 x >0`

Let g(x)=1+x-[x] and f(x)=-1 if x lt 0 =0 if x=0 then f[g(x)]= =1 if x gt 0

Let g(x) =1+x-[x] and f(x) ={{:(-1, if, x lt 0),(0, if, x=0),(1, if, x gt 0):} , then (f(g(2009)))/(g(f(2009)) =

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)

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)

Let g(x)=1+x-[x] and f(x)={:{(-1","x 0):} Then for all x,f(g(x)) is equal to :

Let g(x)=1+x-[x] and f(x)={{:(-1",", x lt 0),(0",",x=0),(1",", x gt 0):} , then for all x , f[g(x)] is equal to

Let g(x)=1+x-[x] and f(x)={{:(-1",", x lt 0),(0",",x=0),(1",", x gt 0):} , then for all x , f[g(x)] is equal to

Let g(x)=f(x)+f(1-x) and f''(x)<0 , when x in (0,1) . Then f(x) is

Let g(x)=f(x)+f(1-x) and f''(x)<0 , when x in (0,1) . Then f(x) is