Q. If $f:R \rightarrow S$, defined by $f(x) = \sin{x - \sqrt{3}\cos{x - 1}}$, is onto, then the intervel of $s$ is
A) [0, 3]
B) [-1, 1]
C) [0, 1]
D) [-1, 3]
Explanation
(d)
$\sin{x - \sqrt{3}\cos{x = 2\sin\left( x - \frac{\pi}{3} \right)}}$
Since, $\ \ \ - 2 \leq 2\sin{\left( x - \frac{\pi}{3} \right) \leq 2}$
$\Rightarrow \ \ \ \ - 1 \leq 1 + 2\sin{\left( x - \frac{\pi}{3} \right) \leq 3}$
$\therefore$ Range of $S = \lbrack - 1,\ 3\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.