Statement-1 : `lim_(x to 0) (sqrt(1 - cos 2x))/(x)` at (x = 0) does not exist.
Statement -2 :Right hand limit `!=` Left hand limit
i) Statement - 1 is True, Statement-2 is True, Statement-2 is a correct explanation for statement-1
ii)Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for statement-1
iii)Statement-1 is True, Statement-2 is False
iv)Statement -1 is False, Statement-2 is True

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 True, Statement-2 is False
D. Statement -1 is False, Statement-2 is True

Answer

To solve the limit problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{\sqrt{1 - \cos(2x)}}{x} \]

### Step 1: Rewrite the expression
We can use the trigonometric identity \(1 - \cos(2x) = 2\sin^2(x)\). Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sqrt{2\sin^2(x)}}{x} \]

### Step 2: Simplify the expression
This simplifies to:

\[ \lim_{x \to 0} \frac{\sqrt{2} |\sin(x)|}{x} \]

### Step 3: Evaluate the right-hand limit
Now, we evaluate the right-hand limit as \(x\) approaches 0 from the positive side:

\[ \lim_{x \to 0^+} \frac{\sqrt{2} \sin(x)}{x} \]

Using the fact that \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\), we find:

\[ \lim_{x \to 0^+} \frac{\sqrt{2} \sin(x)}{x} = \sqrt{2} \cdot 1 = \sqrt{2} \]

### Step 4: Evaluate the left-hand limit
Next, we evaluate the left-hand limit as \(x\) approaches 0 from the negative side:

\[ \lim_{x \to 0^-} \frac{\sqrt{2} \sin(x)}{x} \]

Since \(\sin(x)\) is an odd function, \(\sin(-x) = -\sin(x)\), we have:

\[ \lim_{x \to 0^-} \frac{\sqrt{2} \sin(x)}{x} = \lim_{x \to 0^-} \frac{\sqrt{2} (-\sin(-x))}{-x} = \sqrt{2} \cdot 1 = \sqrt{2} \]

### Step 5: Compare the limits
Since both the right-hand limit and left-hand limit are equal:

\[ \lim_{x \to 0^+} \frac{\sqrt{2} \sin(x)}{x} = \sqrt{2} \quad \text{and} \quad \lim_{x \to 0^-} \frac{\sqrt{2} \sin(x)}{x} = \sqrt{2} \]

This indicates that the limit exists and is equal to \(\sqrt{2}\).

### Conclusion
Thus, the limit exists, and Statement-1 is false. Since the right-hand limit equals the left-hand limit, Statement-2 is also false. Therefore, the correct answer is:

**iv) Statement -1 is False, Statement-2 is True.**