To find: Number of points where
f(x)=|x.sgn(1−x2)| is non- differentiable
For x∈(−1,1)→(1−x2)>0→sgn(1−x2)=1
For x=(−1,1)→(1−x2)=0⟹sgn(1−x2)=0
For x∈(−∞,−1)∪(1,∞)→(1−x2)<0⟹sgn(1−x2)=−1
⟹f(x)=⎧⎨⎩|x|;x∈(−1,1)0;x∈(−1,−1)|−x|=|x|;x∈(−∞,−1)∪(1,∞)
Hence, f(x)=|x| with discontinuity at x=−1,&1
⟹ Total points of non-differentiability =3