Question

The function f(x) = \(\sqrt3\) sin2x - cos2x + 4 is one-one in the interval

• A. \([\frac{-\pi}{6},\frac{\pi}{3}]\)
• B. \((\frac{\pi}{6},\frac{-\pi}{3}]\)
• C. \([\frac{-\pi}{2},\frac{\pi}{2}]\)
• D. \([\frac{-\pi}{6},\frac{-\pi}{3})\)

Answer 1

The Correct Option is A

To determine where the function \( f(x) = \sqrt{3} \sin 2x - \cos 2x + 4 \) is one-to-one, we need to examine its monotonicity.
A function is one-to-one if it is either strictly increasing or strictly decreasing. First, compute the derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx} \left( \sqrt{3} \sin 2x - \cos 2x + 4 \right) \] Using the chain rule: \[ f'(x) = \sqrt{3} \cdot 2 \cos 2x + 2 \sin 2x = 2\sqrt{3} \cos 2x + 2 \sin 2x \] To determine the intervals where the function is one-to-one, we look for intervals where \( f'(x) \) is either always positive or always negative.
This will tell us where the function is strictly increasing or decreasing. We notice that the expression \( 2\sqrt{3} \cos 2x + 2 \sin 2x \) has a periodicity of \( \pi \), meaning it repeats every \( \pi \).
By examining the values of \( f'(x) \) over one period, we find that the function is one-to-one in the interval \( \left[ -\frac{\pi}{6}, \frac{\pi}{3} \right] \), as the derivative does not change sign within this interval. 

Therefore, the function \( f(x) \) is one-to-one in the interval \( \left[ -\frac{\pi}{6}, \frac{\pi}{3} \right] \).

Answer 2

Given the function \(f(x) = \sqrt{3}\sin(2x) - \cos(2x) + 4\).

We can rewrite the function in the form \(R\sin(2x - \alpha)\), where \(R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = \sqrt{4} = 2\).

So, \(f(x) = 2(\frac{\sqrt{3}}{2}\sin(2x) - \frac{1}{2}\cos(2x)) + 4\).

We know that \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\) and \(\sin(\frac{\pi}{6}) = \frac{1}{2}\). Thus, we can write

\(f(x) = 2(\cos(\frac{\pi}{6})\sin(2x) - \sin(\frac{\pi}{6})\cos(2x)) + 4\).

Using the identity \(\sin(A-B) = \sin A \cos B - \cos A \sin B\), we get

\(f(x) = 2\sin(2x - \frac{\pi}{6}) + 4\).

For \(f(x)\) to be one-to-one, the argument of the sine function, \(2x - \frac{\pi}{6}\), should lie in an interval of length at most \(\pi\).

Let's analyze the given options:

1. \([\frac{-\pi}{6},\frac{\pi}{3}]\) In this interval, \(x \in [\frac{-\pi}{6},\frac{\pi}{3}]\). Then, \(2x \in [\frac{-\pi}{3},\frac{2\pi}{3}]\). Thus, \(2x - \frac{\pi}{6} \in [\frac{-\pi}{3} - \frac{\pi}{6}, \frac{2\pi}{3} - \frac{\pi}{6}] = [\frac{-2\pi-\pi}{6}, \frac{4\pi-\pi}{6}] = [\frac{-3\pi}{6}, \frac{3\pi}{6}] = [\frac{-\pi}{2}, \frac{\pi}{2}]\). Since the sine function is one-to-one in the interval \([\frac{-\pi}{2},\frac{\pi}{2}]\), this interval works.
2. \((\frac{\pi}{6},\frac{-\pi}{3}]\) - This interval is invalid as the upper bound is smaller than the lower bound.
3. \([\frac{-\pi}{2},\frac{\pi}{2}]\) In this interval, \(2x - \frac{\pi}{6} \in [-\pi - \frac{\pi}{6}, \pi - \frac{\pi}{6}] = [-\frac{7\pi}{6}, \frac{5\pi}{6}]\). The interval size is \(2\pi\).Not injective
4. \([\frac{-\pi}{6},\frac{-\pi}{3})\) This interval is invalid as the upper bound is smaller than the lower bound.

Therefore, the correct interval is \([\frac{-\pi}{6},\frac{\pi}{3}]\).