If `f(x)={{:((sqrt(1-cos2x))/(sqrt(2)x)",",xne0),(k",",x=0):}` then which value of k will make function f continuous at x=0 ?

Options

A. 1
B. `-1`
C. 0
D. no value

Answer

To determine the value of \( k \) that will make the function \( f(x) \) continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).

Given:
\[ f(x) = \begin{cases} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2} x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0.

We start by calculating the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2} x} \]

### Step 2: Use the trigonometric identity.

Using the trigonometric identity \( 1 - \cos 2x = 2 \sin^2 x \), we can rewrite the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 x}}{\sqrt{2} x} \]

### Step 3: Simplify the expression.

This simplifies to:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sqrt{2} |\sin x|}{\sqrt{2} x} = \lim_{x \to 0} \frac{|\sin x|}{x} \]

### Step 4: Evaluate the limit.

As \( x \) approaches 0, we know that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Thus,
\[ \lim_{x \to 0} f(x) = 1 \]

### Step 5: Set the limit equal to \( f(0) \).

For the function to be continuous at \( x = 0 \), we need:
\[ \lim_{x \to 0} f(x) = f(0) \]
This means:
\[ 1 = k \]

### Conclusion:

Therefore, the value of \( k \) that makes the function \( f \) continuous at \( x = 0 \) is:
\[ \boxed{1} \]