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`int_(-1)^(2)[([x])/(1+x^(2))]dx`, where [.] denotes the greatest integer function, is equal toA. `-2`B. `-1`C. zeroD. none of these |
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Answer» Correct Answer - B `[x]=0,AAxepsilon[0,1)` For `x epsilon[1,2),[x]=1` `:.([x])/(1+x^(2))=1/(1+x^(2))lt 1AAxepsilon[1,2)` or `[([x])/(1+x^(2))]=0` For `x epsilon[-1,0),[x]=-1` or `([x])/((1+x)^(2))=-1/(1+x^(2))` Clearly, `2ge1 +x^(2)gt1AAxepsilon[-1,0)` or `1/2le 1(1+x^(2))lt 1` or `-1/2ge - 1/(1+x^(2))gt-1` or `[([x])/(1+x^(2))]=-1AAxepsilon[-1,0)` Thus, the given integral `=-int_(-1)^(0)dx=-1`. |
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