Question

The function $f(x)=\sqrt{3}\sin 2x-\cos 2x+4$ 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)

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