`int_(-1)^2[([x])/(1+x^2)]dx ,w h e r e[dot]` denotes the greater integer function, is equal to `-2` (b) `-1` `z e ro` (d) none of these

Options

A. `-2`
B. `-1`
C. zero
D. none of these

Answer

B

Explanation

`[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`.