Consider a function f defined by `f(x)=sin^(-1) sin ((x+sinx)/2) AA x in [0,pi]`which satisfies `f(x)+f(2pi-x)=pi AA x in [pi, 2pi]` and `f(x)=f( 4pi-x)`for all `x in [2pi,4pi]` then If `alpha` is the length of the largest interval on which f(x) is increasing, then `alpha` =

Options

A. ` (pi)/(2)`
B. `pi`
C. `2pi`
D. `4pi`

Answer

To solve the problem, we will analyze the function \( f(x) = \sin^{-1}(\sin((x + \sin x)/2)) \) for \( x \in [0, \pi] \) and the conditions given for \( f(x) \).

### Step 1: Simplifying the Function
The function is defined as:
\[ f(x) = \sin^{-1}(\sin((x + \sin x)/2)) \]
For \( x \in [0, \pi] \), since \( \sin^{-1}(\sin y) = y \) when \( y \) is in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\), we need to check if \((x + \sin x)/2\) falls within this range.

### Step 2: Analyzing the Range of \((x + \sin x)/2\)
For \( x \in [0, \pi] \):
- At \( x = 0 \), \( (0 + \sin 0)/2 = 0 \).
- At \( x = \pi \), \( (\pi + \sin \pi)/2 = \pi/2 \).

Since \( (x + \sin x)/2 \) is continuous and increases from \( 0 \) to \( \pi/2 \) as \( x \) goes from \( 0 \) to \( \pi \), we can conclude:
\[ f(x) = \frac{x + \sin x}{2} \quad \text{for } x \in [0, \pi]. \]

### Step 3: Finding the Derivative
Next, we differentiate \( f(x) \):
\[ f'(x) = \frac{1}{2} + \frac{1}{2} \cos x. \]
This is derived from:
\[ f'(x) = \frac{d}{dx} \left(\frac{x + \sin x}{2}\right) = \frac{1}{2} + \frac{1}{2} \cos x. \]

### Step 4: Setting the Derivative Greater Than Zero
To find where \( f(x) \) is increasing, we need:
\[ f'(x) \geq 0. \]
This gives us:
\[ \frac{1}{2} + \frac{1}{2} \cos x \geq 0 \implies 1 + \cos x \geq 0 \implies \cos x \geq -1. \]
Since \( \cos x \) is always greater than or equal to \(-1\) for all \( x \), \( f'(x) \) is non-negative for \( x \in [0, \pi] \).

### Step 5: Finding the Largest Interval
Since \( f'(x) \) is non-negative throughout the interval \( [0, \pi] \), \( f(x) \) is increasing on the entire interval. Thus, the length of the largest interval on which \( f(x) \) is increasing is:
\[ \alpha = \pi - 0 = \pi. \]

### Final Answer
The length of the largest interval on which \( f(x) \) is increasing is:
\[ \alpha = \pi. \]