If `f(x) = 2 sin^(-1) sqrt(1-x) + sin^(-1)(2 sqrt(x (1-x)))` where `x in (0, 1/2)` , then `f'(x)` has the value equal to (i) `2/(xsqrt(1-x))` (ii) `0` (iii) `-2/(xsqrt(1-x))` (iv) `pi`

Options

A. `(2)/(sqrt(x(-1)))`
B. Zero
C. `-(2)/(sqrt(x(-1)))`
D. `pi`

Explanation

To find the value of \( f'(x) \) for the function

\[ f(x) = 2 \sin^{-1}(\sqrt{1-x}) + \sin^{-1}(2 \sqrt{x(1-x)}) \]

where \( x \in (0, \frac{1}{2}) \), we will follow these steps:

### Step 1: Analyze the Function
The function consists of two parts: \( 2 \sin^{-1}(\sqrt{1-x}) \) and \( \sin^{-1}(2 \sqrt{x(1-x)}) \). We need to differentiate both parts with respect to \( x \).

### Step 2: Differentiate the First Part
Using the chain rule, the derivative of \( \sin^{-1}(u) \) is given by \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \).

Let \( u = \sqrt{1-x} \). Then,

\[ \frac{du}{dx} = \frac{-1}{2\sqrt{1-x}}. \]

Now, differentiating the first part:

\[ \frac{d}{dx}[2 \sin^{-1}(\sqrt{1-x})] = 2 \cdot \frac{1}{\sqrt{1 - (1-x)}} \cdot \frac{-1}{2\sqrt{1-x}} = \frac{-1}{\sqrt{x(1-x)}}. \]

### Step 3: Differentiate the Second Part
Now, consider the second part \( \sin^{-1}(2 \sqrt{x(1-x)}) \).

Let \( v = 2 \sqrt{x(1-x)} \). Then,

\[ \frac{dv}{dx} = 2 \cdot \frac{1}{2\sqrt{x(1-x)}} \cdot (1 - 2x) = \frac{1 - 2x}{\sqrt{x(1-x)}}. \]

Now, differentiating the second part:

\[ \frac{d}{dx}[\sin^{-1}(2\sqrt{x(1-x)})] = \frac{1}{\sqrt{1 - (2\sqrt{x(1-x)})^2}} \cdot \frac{dv}{dx}. \]

Calculating \( (2\sqrt{x(1-x)})^2 = 4x(1-x) \), we have:

\[ 1 - 4x(1-x) = 1 - 4x + 4x^2 = (1-2x)^2. \]

Thus,

\[ \sqrt{1 - (2\sqrt{x(1-x)})^2} = |1-2x|. \]

Since \( x \in (0, \frac{1}{2}) \), \( 1-2x > 0 \), so:

\[ \sqrt{1 - (2\sqrt{x(1-x)})^2} = 1 - 2x. \]

Now, we can write the derivative of the second part:

\[ \frac{d}{dx}[\sin^{-1}(2\sqrt{x(1-x)})] = \frac{1}{1-2x} \cdot \frac{1 - 2x}{\sqrt{x(1-x)}} = \frac{1}{\sqrt{x(1-x)}}. \]

### Step 4: Combine the Derivatives
Now, we combine the derivatives of both parts:

\[ f'(x) = \frac{-1}{\sqrt{x(1-x)}} + \frac{1}{\sqrt{x(1-x)}} = 0. \]

### Conclusion
Thus, the value of \( f'(x) \) is:

\[ f'(x) = 0. \]

### Final Answer
The correct option is (ii) \( 0 \).