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If `A(x+y)=A(x)A(y) and A(0) ne 0 and B(x)=(A(x))/(1+(A(x))^(2))`, thenA. `int_(-2010)^(2010)B(x)dx=int_(0)^(2011)B(x)dx`B. `int_(-2010)^(2011)B(x)dx-int_(0)^(2010)B(x)dx=int_(0)^(2011)B(x)dx`C. `int_(-2010)^(2011)B(x)dx=0`D. `int_(-2010)^(2010)(2B(-x)-B(x))dx=2int_(0)^(2010)B(x)dx` |
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Answer» Correct Answer - B::D `A(x+y)=A(x)A(y)` `rArr" "A(0+0)=A(0)A(0)` `rArr" "A(0)=1` Put `y=-x,` we get `A(0)=A(x)A(-x)" (i)"` `B(-x)=(A(-x))/(1+(A(-x))^(2))` `=((1)/(A(x)))/(1+(1)/((A(x))^(2)))` `=(A(x))/(1+(A(x))^(2))` = B(x) Thus, B(x) is even. `int_(-2010)^(2011)B(x)dx=int_(-2010)^(2010)B(x)dx+int_(2010)^(2011)B(x)dx` `=2int_(0)^(2010)B(x)dx+int_(2010)^(2011)B(x)dx` `=int_(0)^(2010)B(x)dx+int_(0)^(2011)B(x)dx` |
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