For 0<x<1: f(x)=−lnx+x−1, f′(x)=−x1+1=xx−1<0 (decreasing).
For x>1: f(x)=lnx−x+1, f′(x)=x1−1=x1−x<0 (decreasing).
At x=1: LHD =11−1=0, RHD =11−1=0. So f is differentiable at x=1.
(I) TRUE: f is differentiable for all x>0.
(II) FALSE: f is decreasing in (0,1).
(III) TRUE: f is decreasing in (1,∞).
Only (I) and (III) are TRUE.