Question

The function $f(x)=\sqrt{3} \sin 2 x-\cos 2 x+4$ is one-one in the interval

• A. $\left[\frac{-\pi}{6}, \frac{\pi}{3}\right]$
• B. $\left[\frac{\pi}{6}, \frac{-\pi}{3}\right]$
• C. $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
• D. $\left[\frac{-\pi}{6}, \frac{-\pi}{3}\right]$

Answer 1

Answer: (A)

$f=\sqrt{3} \sin 2 x-\cos 2 x+4=2\left[\sin \left(2 x-\frac{\pi}{6}\right)\right]+4$
$f$ is one-one
$-\frac{\pi}{2} \leq 2 x-\frac{\pi}{6} \leq \frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$
${\left[-\frac{\pi}{6}, \frac{\pi}{3}\right]}$