Question

The value of $a$ in order that $f\left(x\right)=\sqrt{3}\sin x-\cos x-2ax+b$ decreases for all real values of $x$, is given by

• A. $a<1$
• B. $a\ge 1$
• C. $a\ge \sqrt{2}$
• D. $a<\sqrt{2}$

Answer 1

Answer: (B)

$a\ge 1$

If $f\left(x\right)$ is decreasing, Then $f^{\prime }\left(x\right)<0Forx\in R$

$f^{\prime }\left(x\right)=\sqrt{3}\sin x+\cos x-2a\le 0$

$\Rightarrow \sqrt{3}sinx+\cos x\le 2a$

Maximum value of $\sqrt{3}sinx+cosx$ is $\sqrt{3+1}=2$

$\therefore 2a\ge 2$

$a\ge 1$