The correct option is B a = 1, b = 1
f(x)=⎧⎪⎨⎪⎩a|x2−x−2|2+x−x2,x<2x−[x]x−2,x=2⎧⎪⎨⎪⎩−a(x2−x−2)2+x−x2,x<2x−[x]x2,x>2f(x)=⎧⎪
⎪⎨⎪
⎪⎩a,x<2b,x=2x−[x]x−2,x>2
Check at x = 2
LHL=limx→2−f(x)=limh→0f(2−h)=limh→0a=aRHL=limx→2+f(x)=limh→0f(2−h)=limh→02+h−[2+h]2+h−2=limh→02+h−22+h−2=1