Given,
f(x)=[x2−x]+∣−x+[x]∣
⇒f(x)=[x2−x]+∣x−[x]∣,as∣−A∣=∣A∣
⇒f(x)=[x2−x]+∣x∣,as[x]+x=x
⇒f(x)=[x2−x]+x(∵x≥0)
Now at x=0, f(0)=0
And f(0+)=−1,asx2−x<0forx→0+
So, function is discontinuous at x=0
Now at x=1, f(1)=0,
f(1+)=0+0=0 and f(1−)=−1+1=0
Hence, the function is continuous at x=1.