f(x)=⎧⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎩−2|x2−1|+1x∈(−2,−1)−|x2−1|+1x∈[−1,0)sin(π3)−1x∈[0,1)|x2−1|+1√2−2x∈[1,2)
limx→−1−f(x)=1 and limx→−1+f(x)=1
Hence continuous at x=−1
limx→0−f(x)=0 and limx→0+f(x)=√32−1
⇒ discontinuous
limx→1−f(x)=√32−1 and limx→1+f(x)=1√2−2
⇒ discontinuous
Hence, 2 points of discontinuity.