Question

If $f\left(x\right)=\sqrt{3}\sin x-\cos x-2ax+b$ decreases for all real values of $x$ then

• A. $a<1$
• B. $a>1$
• C. $a<\frac{1}{2}$
• D. $a>\frac{1}{2}$

Answer 1

Answer: (B)

$a>1$

$f\left(x\right)=\sqrt{3}\sin x-\cos x-2a+b{f}^1\left(x\right)=\sqrt{3}\cos x+\sin x-2a$

$f\left(x\right)$ decreases for all values of x

$\therefore f^1\left(x\right)<0\sqrt{3}\cos x+\sin x-2a<0\Rightarrow 2a>\sqrt{3}+\sin x2a>\sqrt{(\sqrt{3}{)}^2+1^2}\Rightarrow 2a>\sqrt{3+1}\Rightarrow 2a>\sqrt{4}\Rightarrow 2a>2\Rightarrow a>1or2a<-2\Rightarrow a<-1\therefore aϵ(-\infty ,-1)°(1,\infty )$