If `f(x) = {((sin3x)/(x), x !=0),(k/2,x=0):}` is continuous at x = 0, then the value of k is

Options

A. 12
B. 9
C. 6
D. 2

Answer

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} \frac{\sin(3x)}{x} & \text{if } x \neq 0 \\ \frac{k}{2} & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).

### Step 1: Find \( f(0) \)

From the definition of the function, we have:

\[ f(0) = \frac{k}{2} \]

### Step 2: Calculate the limit as \( x \) approaches 0

We need to find:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(3x)}{x} \]

To evaluate this limit, we can use the standard limit \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \).

### Step 3: Rewrite the limit

We can manipulate the limit:

\[ \lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 \]

### Step 4: Apply the limit

Now, applying the limit:

\[ \lim_{x \to 0} \frac{\sin(3x)}{3x} = 1 \]

Thus,

\[ \lim_{x \to 0} \frac{\sin(3x)}{x} = 3 \cdot 1 = 3 \]

### 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 gives us:

\[ 3 = \frac{k}{2} \]

### Step 6: Solve for \( k \)

To find \( k \), we multiply both sides by 2:

\[ k = 6 \]

### Final Answer

Thus, the value of \( k \) is \( 6 \).

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