If `f(x) = sqrt(x+2sqrt(x+1)) + sqrt(x-2sqrt(x-1)), x ge 1`.
Statement-1 : Then, `f(x) = 2`, when `x in [1, 2]`
and
Statement-2 : `sqrt((f(x))^(2)) = {{:(f(x)",", f(x) ge 0),(-f(x)",", f(x) lt 0):}`

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement -1 is False, Statement -2 is False
D. Statement -1 is False, Statement -2 is True

Answer

To solve the problem, we need to analyze the function \( f(x) = \sqrt{x + 2\sqrt{x+1}} + \sqrt{x - 2\sqrt{x-1}} \) for \( x \geq 1 \) and evaluate the two statements provided.

### Step 1: Evaluate \( f(1) \)
Let's first calculate \( f(1) \):
\[ f(1) = \sqrt{1 + 2\sqrt{1+1}} + \sqrt{1 - 2\sqrt{1-1}} \]
Calculating inside the square roots:
\[ = \sqrt{1 + 2\sqrt{2}} + \sqrt{1 - 0} \]
\[ = \sqrt{1 + 2\sqrt{2}} + 1 \]

### Step 2: Evaluate \( f(2) \)
Now, let's calculate \( f(2) \):
\[ f(2) = \sqrt{2 + 2\sqrt{2+1}} + \sqrt{2 - 2\sqrt{2-1}} \]
Calculating inside the square roots:
\[ = \sqrt{2 + 2\sqrt{3}} + \sqrt{2 - 2} \]
\[ = \sqrt{2 + 2\sqrt{3}} + 0 \]
\[ = \sqrt{2 + 2\sqrt{3}} \]

### Step 3: Check if \( f(x) = 2 \) for \( x \in [1, 2] \)
We need to check if \( f(x) = 2 \) for \( x \in [1, 2] \).

From our calculations:
- \( f(1) = \sqrt{1 + 2\sqrt{2}} + 1 \)
- \( f(2) = \sqrt{2 + 2\sqrt{3}} \)

We need to check if these values equal 2.

1. For \( f(1) \):
\[ \sqrt{1 + 2\sqrt{2}} + 1 \neq 2 \quad (\text{since } \sqrt{1 + 2\sqrt{2}} > 1) \]

2. For \( f(2) \):
\[ \sqrt{2 + 2\sqrt{3}} \neq 2 \quad (\text{since } \sqrt{2 + 2\sqrt{3}} > 2) \]

Thus, \( f(x) \neq 2 \) for \( x \in [1, 2] \). Therefore, **Statement 1 is false**.

### Step 4: Analyze Statement 2
The second statement claims:
\[ \sqrt{(f(x))^2} = \begin{cases} f(x) & \text{if } f(x) \geq 0 \\ -f(x) & \text{if } f(x) < 0 \end{cases} \]

Since \( f(x) \) is defined for \( x \geq 1 \) and is always non-negative (as it is a sum of square roots), we can simplify:
\[ \sqrt{(f(x))^2} = |f(x)| = f(x) \quad \text{(since } f(x) \geq 0\text{)} \]

Thus, **Statement 2 is true**.

### Conclusion
- Statement 1 is false.
- Statement 2 is true.