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If derivative of `y=|x-1|^(sinx)` at `x=-(pi)/2` is `(1+api)^(b)` then value of `|1/a+4b|` is |
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Answer» Correct Answer - 6 `y=|x-1|^(sinx)` `y=(1-x)^(sin)` at `x=(pi)/2` `logy=sinx.log(1-x)` `1/y (dy)/(dx)=cosxlog(1-x)-(sinx)/(1-x)` `(dy)/(dx)-|x-1|^(sinx)[cosx.log(1-x)-(sinx)/(1-x)]` `=(1+(pi)/2)^(-1)[0+1/(1+(pi)/2)]=(1+(pi)/2)^(-2)` `implies a=1/2,b=-2` So `1/a+4b=2-8=-6` |
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