Show that for `alt=1,f(x)=sqrt(3)` `si nx-cosx-2a x+b` is decreasing on `Rdot`

Explanation

We have `age1, f(x) =sqrt(3)sinx-cosx-2ax+b`
`therefore f^(')(x) = sqrt(3)cosx-(-sinx)-2a`
`=2[sqrt(3)/2.cosx+1/2.sinx]-2a`
`=2(cospi/6-x)-2a` `[therefore cos(A-B) = cosA.cosB+sinA.sinB]`
`=2[(cospi/6-x)-a]`
We know that, `cosx in [-1,1]`
and `age1`
So, `2[cos(pi/6-x)-a]le0`
`therefore f^(')(x)le0`
Hence, f(x) is a decreasing function in R.