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If `log_(2)xge0` then `log_(1//pi){sin^(-1)((2x)/(1+x^(2)))+2tan^(-1)x}` is equalA. `log_(1//pi)(4 tan^(-1)x)`B. 0C. `-1`D. none of these |
Answer» We have `log_(2) xge 0rarr xge 2^(@)=1` for `xge1` we have `sin^(-1)(2x)/(1+x^(2))=pi-2 tan^(-1)x` `therefore log_(1//pi){sin^(-1)(2x)/(1+x^(2))+2 tan^(-1)x}` `=log_(1//pi)(pi-2tan^(-1)x+2 tan ^(-1)x)` `=log_(1//pi) pi=-1` |
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