To determine the domain of the function \(f(x) = \cos^{-1} \left\{ \frac{4x + 2[x]}{3} \right\}\), we need to examine the range of values that the inside expression, \(\frac{4x + 2[x]}{3}\), can take.
- Since \(\cos^{-1}(x)\) is defined only for \(x \in [-1, 1]\), we need to find when \(\frac{4x + 2[x]}{3} \in [-1, 1]\).
- Firstly, let's analyze the expression \(4x + 2[x]\):
- \([x]\) is the greatest integer function, meaning it gives the greatest integer less than or equal to \(x\).
- Consider \(x = n + \theta\) where \(n\) is an integer and \(0 \leq \theta < 1\). Thus \([x] = n\) and \(x = n + \theta\).
- The expression becomes \(4(n + \theta) + 2n = 6n + 4\theta\).
- Now substitute and simplify:
- \(\frac{6n + 4\theta}{3} = 2n + \frac{4\theta}{3}\).
- We need to find when this expression is within the interval \([-1, 1]\). Consider:
- \(-1 \leq 2n + \frac{4\theta}{3} \leq 1\).
- Solving for the left side: \(-1 \leq 2n + \frac{4\theta}{3} \Rightarrow \frac{4\theta}{3} \geq -2n - 1 \Rightarrow \theta \geq \frac{3(-2n-1)}{4}\).
- Solving for the right side: \(2n + \frac{4\theta}{3} \leq 1 \Rightarrow \frac{4\theta}{3} \leq 1 - 2n \Rightarrow \theta \leq \frac{3(1 - 2n)}{4}\).
- Combining both parts:
- \(\frac{3(-2n-1)}{4} \leq \theta \leq \frac{3(1-2n)}{4}\).
- \(0 \leq \theta < 1\) implies only some specific values of \(n\) will satisfy this condition.
- By testing values of \(n\), when \(\theta = 0\), \(n = -1\) yields:
- \(x \in \left[-\frac{1}{4}, \frac{3}{4}\right]\).
The correct domain for the given function is therefore \([-\frac{1}{4}, \frac{3}{4}]\), which matches the option [-1/4, 3/4].