In function `f(x)=cos^(-1)x+cos^(-1)(x/2+(sqrt(3-3x^2))/2)` , then Range of `f(x)` is `[pi/3,(10pi)/3]]` Range of `f(x)` is `[pi/3,5pi]` `f(x)` is one-one for `x in [-1,1/2]` `f(x)` is one-one for `x in [1/2,1]`

Options

A. Range of `f (x) is [(pi)/(3), (10 pi)/(3)]`
B. Rang `f (x) is [(pi)/(3), (5 pi)/(3)]`
C. f (x) is one-one for ` x in [-1, (1)/(2)]`
D. f (x) is one-one for `x in [(1)/(2), 1]`

Answer

To solve the problem, we need to analyze the function \( f(x) = \cos^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2} \right) \) and determine its range and whether it is one-one in the specified intervals.

### Step 1: Rewrite the function
Let \( x = \cos \theta \), where \( \theta \) is in the interval \( [0, \pi] \). Then we can rewrite the function as:
\[ f(x) = \cos^{-1}(\cos \theta) + \cos^{-1}\left(\frac{\cos \theta}{2} + \frac{\sqrt{3 - 3\cos^2 \theta}}{2}\right) \]
This simplifies to:
\[ f(x) = \theta + \cos^{-1}\left(\frac{\cos \theta}{2} + \frac{\sqrt{3 - 3\cos^2 \theta}}{2}\right) \]

### Step 2: Simplify the second term
The term \( \frac{\cos \theta}{2} + \frac{\sqrt{3 - 3\cos^2 \theta}}{2} \) can be recognized as:
\[ \frac{\cos \theta + \sqrt{3 - 3\cos^2 \theta}}{2} \]
Using the identity \( \sqrt{3 - 3\cos^2 \theta} = \sqrt{3(1 - \cos^2 \theta)} = \sqrt{3\sin^2 \theta} = \sqrt{3} \sin \theta \), we can rewrite it as:
\[ \frac{\cos \theta + \sqrt{3} \sin \theta}{2} \]

### Step 3: Analyze the range of \( f(x) \)
For \( \theta \) in the interval \( [0, \pi] \):
- When \( \theta = 0 \):
\[ f(1) = 0 + \cos^{-1}\left(\frac{1 + 0}{2}\right) = \cos^{-1}(0.5) = \frac{\pi}{3} \]
- When \( \theta = \frac{\pi}{3} \):
\[ f\left(\frac{1}{2}\right) = \frac{\pi}{3} + \cos^{-1}(1) = \frac{\pi}{3} + 0 = \frac{\pi}{3} \]
- When \( \theta = \pi \):
\[ f(-1) = \pi + \cos^{-1}(-1) = \pi + \pi = 2\pi \]

### Step 4: Determine the range
From the analysis, we find that:
- The minimum value of \( f(x) \) is \( \frac{\pi}{3} \).
- The maximum value of \( f(x) \) is \( 2\pi \).

Thus, the range of \( f(x) \) is:
\[ \left[\frac{\pi}{3}, 2\pi\right] \]

### Step 5: Check if \( f(x) \) is one-one
To check if \( f(x) \) is one-one, we need to analyze the intervals:
1. For \( x \in [-1, \frac{1}{2}] \):
- The function is increasing since both components \( \cos^{-1} x \) and \( \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2}\right) \) are decreasing functions, making \( f(x) \) one-one in this interval.

2. For \( x \in [\frac{1}{2}, 1] \):
- The function is also increasing, thus it remains one-one in this interval.

### Conclusion
The correct options based on our analysis are:
- The range of \( f(x) \) is \( \left[\frac{\pi}{3}, 2\pi\right] \).
- \( f(x) \) is one-one for \( x \in [-1, \frac{1}{2}] \) and \( x \in [\frac{1}{2}, 1] \).