If derivative of `y=|x-1|^(sinx)` at `x=-(pi)/2` is `(1+api)^(b)` then value of `|1/a+4b|` is

Answer

6

Explanation

`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`