The function f(x) = sqrt{3} sin 2x - cos 2x + | vedclass

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(A) Given function,$f(x) = \sqrt{3} \sin 2x - \cos 2x + 4$. We can rewrite this as: $f(x) = 2(\sin 2x \cdot \frac{\sqrt{3}}{2} - \cos 2x \cdot \frac{1

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$[-\frac{\pi}{6}, \frac{\pi}{3}]$

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(A) Given function,$f(x) = \sqrt{3} \sin 2x - \cos 2x + 4$. We can rewrite this as: $f(x) = 2(\sin 2x \cdot \frac{\sqrt{3}}{2} - \cos 2x \cdot \frac{1}{2}) + 4$ $f(x) = 2(\sin 2x \cos \frac{\pi}{6} - \cos 2x \sin \frac{\pi}{6}) + 4$ $f(x) = 2 \sin(2x - \frac{\pi}{6}) + 4$. $A$ function $f(x) = \sin(	heta)$ is one-one when the argument $	heta$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Therefore,for $f(x)$ to be one-one: $-\frac{\pi}{2} \leq 2x - \frac{\pi}{6} \leq \frac{\pi}{2}$ Adding $\frac{\pi}{6}$ to all parts: $-\frac{\pi}{2} + \frac{\pi}{6} \leq 2x \leq \frac{\pi}{2} + \frac{\pi}{6}$ $-\frac{2\pi}{6} \leq 2x \leq \frac{4\pi}{6}$ $-\frac{\pi}{3} \leq 2x \leq \frac{2\pi}{3}$ Dividing by $2$: $-\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$. Thus,the function is one-one in the interval $[-\frac{\pi}{6}, \frac{\pi}{3}]$.

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

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