JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f(x)=\left[x^2-x\right]+|-x+[x]|\), where \(x \in R\) and \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is
- A continuous at \(x =0\), but not continuous at \(x =1\)
- B continuous at \(x =0\) and \(x =1\)
- C not continuous at \(x=0\) and \(x=1\)
- D continuous at \(x =1\), but not continuous at \(x =0\)
Answer & Solution
Correct Answer
(D) continuous at \(x =1\), but not continuous at \(x =0\)
Step-by-step Solution
Detailed explanation
Here \(f(x)=[x(x-1)]+\{x\}\) \(f\left(0^{+}\right)=-1+0=-1\) \(f(0)=0\) \(f \left(1^{+}\right)=0+0=0\) \(f (1)=0\) \(f \left(1^{-}\right)=-1+1=0\) \(\therefore \quad f(x)\) is continuous at \(x=1\), discontinuous at \(x=0\)
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