Let `f(x)=sin x and cg(x)={max {f(t), 0lexlepi},` and `(1-cosx)/2,` for `x >pi` Then , `g(x)` is

Let f(x) = sin x and "c g(x)" = {{:(max {f(t)","0 le x le pi},"for", 0 le x le pi),((1-cos x)/(2)",","for",x gt pi):} Then, g(x) is

Let f(x) = sin x and " g(x)" = {{:(max {f(t)","0 le x le pi},"for", 0 le x le pi),((1-cos x)/(2)",","for",x gt pi):} Then, g(x) is

Let f(x) = sin x and " g(x)" = {{:(max {f(t)","0 le x le pi},"for", 0 le x le pi),((1-cos x)/(2)",","for",x gt pi):} Then, g(x) is

Let f(x)=x^(3)-x^(2)+x+1 and g(x)={("max "f(t) 0letlex 0lexle1),(3-x 1ltxle2):} then

Let f(x)=sin x,g(x)={{max f(t),0 pi Then number of point in (0,oo) where f(x) is not differentiable is

Let f(x)=sin x+cosx and g(x)=x^(2)-1 , then g{f(x)} is invertible if -

Let f (x)= x^3+ x^2+ x+ 1 and g(x)=max[f(t)] , 0<=t<=x , 0<=x<=1 and f(x)=3-x , 1 < x<=2 then function g(x) is

Let f(x)=x^(3)-x^(2)+x+1 and g(x) be a function defined by g(x)={{:(,"Max"{f(t):0le t le x},0 le x le 1),(,3-x,1 le x le 2):} Then, g(x) is

Let f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):} Then, g(x) in [0, 2] is

Let f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):} Then, g(x) in [0, 2] is