Consider [x]x≤f(x)≤6−x⇒x→2−lim[x]x=12=2 ⇒x→2−lim6−x=2∴x→2−limf(x)=2 [By Sandwich theorem] Now x→2+lim[x]x=1,x→2+lim6−x=2 Hence by Sandwich theorem x→2+limf(x) does not exists. Therefore f is not continuous at x=2. Thus statement-1 is true but statement- 2 is not true