Let `f(x) = sin^(-1)|sin x| + cos^(-1)( cos x) `. Which of the following statement(s) is / are TRUE ?

Options

A. `f(f(3)) = pi`
B. f (x) is periodic with fundamental period `2 pi`
C. f (x) is neither even nor odd
D. Range of f(x) is `[0,2pi]`

Answer

To solve the problem, we need to analyze the function defined as:

\[ f(x) = \sin^{-1} |\sin x| + \cos^{-1} (\cos x) \]

We will evaluate this function step by step and check the validity of the statements provided.

### Step 1: Understanding the Components of the Function

1. **Understanding \( \sin^{-1} |\sin x| \)**:
- The function \( \sin^{-1} y \) gives the angle whose sine is \( y \).
- Since we have \( |\sin x| \), this means that the output will always be non-negative.
- The range of \( \sin^{-1} |\sin x| \) is from \( 0 \) to \( \frac{\pi}{2} \) for \( x \) in the intervals where \( \sin x \) is non-negative (i.e., \( 0 \leq x \leq \pi \)) and from \( \frac{\pi}{2} \) to \( \pi \) for \( x \) in the intervals where \( \sin x \) is negative (i.e., \( \pi < x < 2\pi \)).

2. **Understanding \( \cos^{-1} (\cos x) \)**:
- The function \( \cos^{-1} y \) gives the angle whose cosine is \( y \).
- The range of \( \cos^{-1} (\cos x) \) is \( [0, \pi] \) for \( x \) in the intervals \( [0, \pi] \) and \( [2\pi, 3\pi] \), and it equals \( 2\pi - x \) for \( x \) in the intervals \( [\pi, 2\pi] \).

### Step 2: Evaluating \( f(x) \) in Different Intervals

1. **For \( x \in [0, \frac{\pi}{2}] \)**:
- Here, \( |\sin x| = \sin x \) and \( \cos x \) is positive.
- Thus, \( f(x) = \sin^{-1}(\sin x) + \cos^{-1}(\cos x) = x + x = 2x \).

2. **For \( x \in [\frac{\pi}{2}, \frac{3\pi}{2}] \)**:
- Here, \( |\sin x| = -\sin x \) (since \( \sin x \) is negative) and \( \cos x \) is negative.
- Thus, \( f(x) = \sin^{-1}(-\sin x) + \cos^{-1}(-\cos x) = \pi - x + \pi - x = 2\pi - 2x \).

3. **For \( x \in [\frac{3\pi}{2}, 2\pi] \)**:
- Here, \( |\sin x| = -\sin x \) and \( \cos x \) is positive.
- Thus, \( f(x) = \sin^{-1}(-\sin x) + \cos^{-1}(\cos x) = \pi - x + x = \pi \).

### Step 3: Analyzing the Periodicity and Other Properties

1. **Periodicity**:
- The function \( f(x) \) is periodic with a fundamental period of \( 2\pi \) since the behavior of \( f(x) \) repeats every \( 2\pi \).

2. **Even/Odd Nature**:
- To check if \( f(x) \) is even or odd, we can evaluate \( f(-x) \) and compare it with \( f(x) \) and \(-f(x)\).
- It can be shown that \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), indicating that \( f(x) \) is neither even nor odd.

3. **Range of \( f(x) \)**:
- The minimum value occurs at \( x = 0 \) (where \( f(0) = 0 \)) and the maximum occurs at \( x = 2\pi \) (where \( f(2\pi) = 2\pi \)).
- Thus, the range of \( f(x) \) is \( [0, 2\pi] \).

### Conclusion

Based on the analysis above, we can conclude the following statements:

- \( f(f(3)) = \pi \) is **True**.
- \( f(x) \) is periodic with a fundamental period of \( 2\pi \) is **True**.
- \( f(x) \) is neither even nor odd is **True**.
- The range of \( f(x) \) is \( [0, 2\pi] \) is **True**.