Q. The values of $a$ in order that $f(x) = \sqrt{3}\sin x - \cos x - 2ax + b$ decreases for all real values of $x$, is given by
A) $a < 1$
B) $a \geq 1$
C) $a \leq \sqrt{2}$
D) $a < \sqrt{2}$
Explanation
(b)
Given, $f^{'}(x) < 0,\ \forall\ x \in R$
$\Rightarrow \ \ \ \sqrt{3}\cos x + \sin x - 2a < 0,\ \forall\ x \in R$
$\Rightarrow \ \ \frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x < a,\ \forall\ x \in R$
$\Rightarrow \sin{\left( x + \frac{\pi}{3} \right) < a,\ \forall\ x\ \in R}$
$\Rightarrow \ \ \ a \geq 1\ \left\lbrack \because\ \sin\left( x + \frac{\pi}{3} \right) \leq 1 \right\rbrack\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.