The correct option is C 2
Given that, log0.53−xx+2<0For log0.53−xx+2<0 to be defined3−xx+2>0⇒x−3x+2<0⇒−2<x<3 ....(1]log0.53−xx+2<0⇒3−xx+2>1⇒3−xx+2−1>0⇒3−x−x−2x+2>0or, 2x−1x+2<0or, −2<x<12 ...(2]From (1] and (2] we getx∈(−2,12)The integral values of x are −1 and 0.