Question

Show that for a ≥ 1, f(x) = ${\displaystyle \sqrt{3}}$ sinx – cosx – 2ax + b ∈ is decreasing in R

Answer 1

Given that: f(x) = ${\displaystyle \sqrt{3}}$ sinx – cosx – 2ax + b, a ≥ 1

Differentiating both sides w.r.t. x, we get

f'(x) = ${\displaystyle \sqrt{3}\cos x+\sin x-2\text{a}}$

For decreasing function, f'(x) < 0

∴ ${\displaystyle \sqrt{3}\cos x+\sin x-2\text{a}<0}$

⇒ ${\displaystyle 2(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x)-2\text{a}<0}$

⇒ ${\displaystyle \frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x-\text{a}<0}$

⇒ ${\displaystyle (\cos \frac{\pi }{6}\cos x+\sin \frac{\pi }{6}\sin x)-\text{a}<0}$

⇒ ${\displaystyle \cos (x-\frac{\pi }{6})-\text{a}<0}$

Since cos x ∈ [– 1, 1] and a ≥ 1

∴ f'(x) < 0

Hence, the given function is decreasing in R.