Consider the function `f(x)={{:((kx)/(|x|)","ifxlt0,),(3","ifxge0,):}`
which is continuous at x=0, then k=?

Options

A. 3
B. 0
C. -3
D. 2

Answer

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

\[ f(x) = \begin{cases} \frac{kx}{|x|} & \text{if } x < 0 \\ 3 & \text{if } x \geq 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the left-hand limit, the right-hand limit, and the function value at \( x = 0 \) are all equal.

### Step 1: Find the value of the function at \( x = 0 \)

The function value at \( x = 0 \) is given by the definition of \( f(x) \) for \( x \geq 0 \):

\[ f(0) = 3 \]

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

For \( x < 0 \), we have:

\[ f(x) = \frac{kx}{|x|} = \frac{kx}{-x} = -k \quad \text{(since } |x| = -x \text{ when } x < 0\text{)} \]

Now, we find the left-hand limit:

\[ \lim_{x \to 0^-} f(x) = -k \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 0

For \( x \geq 0 \), we have:

\[ f(x) = 3 \]

Now, we find the right-hand limit:

\[ \lim_{x \to 0^+} f(x) = 3 \]

### Step 4: Set the limits equal to each other

For the function to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

This gives us the equation:

\[ -k = 3 \]

### Step 5: Solve for \( k \)

Rearranging the equation:

\[ k = -3 \]

Thus, the value of \( k \) that makes the function continuous at \( x = 0 \) is:

\[ \boxed{-3} \]

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