If `f(x)=cos^(- 1)x+cos^(- 1){x/2+1/2sqrt(3-3x^2)}`then

Options

A. ` f((2)/(3))= (pi)/(3) `
B. ` f((2)/(3))= 2 "cos"^(-1)(2)/(3) - (pi)/(3) `
C. ` f((1)/(3))= (pi)/(3) `
D. ` f((1)/(3))= 2 "cos"^(-1)(1)/(3)- (pi)/(3)`

Answer

To solve the problem, we need to analyze the function given:

\[ f(x) = \cos^{-1}(x) + \cos^{-1}\left(\frac{x}{2} + \frac{1}{2}\sqrt{3 - 3x^2}\right) \]

### Step 1: Identify the terms in the function
We have two inverse cosine terms in the function. We will denote:
- \( y = \frac{x}{2} + \frac{1}{2}\sqrt{3 - 3x^2} \)

### Step 2: Use the identity for the sum of inverse cosines
We can use the identity:
\[ \cos^{-1}(a) + \cos^{-1}(b) = \cos^{-1}(ab - \sqrt{(1-a^2)(1-b^2)}) \]
for \( a = x \) and \( b = y \).

### Step 3: Calculate \( ab \)
We need to find \( ab \):
\[ ab = x \cdot \left(\frac{x}{2} + \frac{1}{2}\sqrt{3 - 3x^2}\right) \]
Calculating this:
\[ ab = \frac{x^2}{2} + \frac{x}{2}\sqrt{3 - 3x^2} \]

### Step 4: Calculate \( 1 - a^2 \) and \( 1 - b^2 \)
Next, we calculate:
\[ 1 - a^2 = 1 - x^2 \]
and for \( b \):
\[ b = \frac{x}{2} + \frac{1}{2}\sqrt{3 - 3x^2} \]
We need to find \( 1 - b^2 \):
\[ 1 - b^2 = 1 - \left(\frac{x}{2} + \frac{1}{2}\sqrt{3 - 3x^2}\right)^2 \]

### Step 5: Substitute into the identity
Now, substituting \( ab \) and \( 1 - a^2 \), \( 1 - b^2 \) into the identity:
\[ f(x) = \cos^{-1}\left(\frac{x^2}{2} + \frac{x}{2}\sqrt{3 - 3x^2} - \sqrt{(1 - x^2)(1 - b^2)}\right) \]

### Step 6: Simplify the expression
After simplifying the above expression, we can see that:
\[ f(x) = 2\cos^{-1}(x) - \frac{\pi}{3} \]

### Step 7: Evaluate specific values
To find \( f\left(\frac{2}{3}\right) \) and \( f\left(\frac{1}{3}\right) \):
1. For \( x = \frac{2}{3} \):
\[ f\left(\frac{2}{3}\right) = 2\cos^{-1}\left(\frac{2}{3}\right) - \frac{\pi}{3} \]
2. For \( x = \frac{1}{3} \):
\[ f\left(\frac{1}{3}\right) = 2\cos^{-1}\left(\frac{1}{3}\right) - \frac{\pi}{3} \]

### Conclusion
Both values can be calculated to find the final results.