Question

Show that for a $\ge 1,f\left(x\right)=\sqrt{3}sinx-cosx-2ax+b$ is decreasing in R.

Answer 1

We have a $\ge 1,f\left(x\right)=\sqrt{3}sinx-cosx-2ax+b{f}^{\prime }\left(x\right)=\sqrt{3}cosx-(-sinx)-2a=\sqrt{3}cosx+sinx-2a=2[\frac{\sqrt{3}}{2}.cosx+\frac{1}{2}sinx]-2a=2[cos\frac{\pi }{6}.cosx+sin\frac{\pi }{6}.sinx]-2a=2(cos\frac{\pi }{6}-x)-2a=2[(cos\frac{\pi }{6}-x)-a]$
We know that, $cosxϵ[-1,1]$
and $a\ge 1$
So, $2[cos(\frac{\pi }{6}-x)-a]\le 0\therefore f^{\prime }\left(x\right)\le 0$
Hence, f(x) is a decreasing function in R.