Question

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

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

Answer 1

Answer: (A)

$a>1$

Given, $f\left(x\right)=\sqrt{3}\sin x-\cos x-2ax+b$

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

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

$f^{\prime }\left(x\right)=2(\sin (\frac{\pi }{3}+x)-a)$

if $a>1$, then for all real values of $x$ $,f^{\prime }\left(x\right)<0$.

So, $a>1$