Q. The value of $a$ in order that $f(x) = \sin{x - \cos{x - ax + b}}$ decreases for all real values of $x$ is given by
A) $a \geq \sqrt{2}$
B) $a < \sqrt{2}$
C) $a \geq 1$
D) $a < 1$
Explanation
(a)
Given, $f(x) = \sin{x - \cos{x - ax + b}}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \cos{x + \sin{x - a}}$
For decreasing function, $f^{'}(x) < 0$
$\Rightarrow \ \ \ \cos{x + \sin{x - a < 0}}$
$\Rightarrow \ \ \ \ a > \cos{x + \sin x}$
Since, $- \sqrt{2} \leq \cos{x + \sin x} \leq \sqrt{2}$
$\therefore\ \ \ a \geq \sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.