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Let \mathrm{f(x)=\left[x^2+1\right],}([x] is the greatest integer less than or equal to x ). Then f is continuous
 

Option: 1

 on [1,3]
 


Option: 2

 for all x in [1,3] except four points
 


Option: 3

for all x in [1,3] except seven points
 


Option: 4

 for all x in [1,3] except eight points


Answers (1)

best_answer

f is discontinuous at x for which \mathrm{x^2+1 \in I}. For \mathrm{x \in(1,3)}  we have \mathrm{x^2+1 \in(2,10)} so there are seven values. f is clearly continuous from right at 1 but discontinuous from left at 3 .

Posted by

Ajit Kumar Dubey

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