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Let,f(x)=\left[x^2-x\right]+|-x+[x]| where x\epsilon \mathbb{R}  and [t] denotes the greatest integer less than or equal to t. Then,f is

Option: 1

not continuous at x = 0 and x = 1
 


Option: 2

continuous at x = 0 and x = 1
 


Option: 3

continuous at x = 1, but not continuous at x = 0
 


Option: 4

continuous at x = 0, but not continuous at x = 1


Answers (1)

best_answer

Here $f(x)=[x(x-1)]+\{x\}$ \begin{array}{ll} \mathrm{f}\left(0^{+}\right)=-1+0=-1 & \mathrm{f}\left(1^{+}\right)=0+0=0 \\ \mathrm{f}(0)=0 & \mathrm{f}(1)=0 \\ & \mathrm{f}\left(1^{-}\right)=-1+1=0 \end{array} \therefore \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1, discontinuous at \mathrm{x}=0

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HARSH KANKARIA

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