The value of 'k' for which `f(x)= {(kx^2", " x ge 2 ),(12", "x lt 2):}`
continuous at x=2 is :

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

\[ f(x) = \begin{cases} kx^2 & \text{if } x \geq 2 \\ 12 & \text{if } x < 2 \end{cases} \]

is continuous at \( x = 2 \), we need to ensure that the left-hand limit, right-hand limit, and the function value at \( x = 2 \) are all equal.

### Step 1: Find \( f(2) \)
Since \( x = 2 \) falls under the case where \( x \geq 2 \), we can find \( f(2) \) using the first case:

\[ f(2) = k(2^2) = 4k \]

### Step 2: Find the left-hand limit as \( x \) approaches 2
For \( x < 2 \), the function is defined as \( f(x) = 12 \). Therefore, the left-hand limit as \( x \) approaches 2 is:

\[ \lim_{x \to 2^-} f(x) = 12 \]

### Step 3: Find the right-hand limit as \( x \) approaches 2
For \( x \geq 2 \), the function is defined as \( f(x) = kx^2 \). Thus, the right-hand limit as \( x \) approaches 2 is:

\[ \lim_{x \to 2^+} f(x) = k(2^2) = 4k \]

### Step 4: Set the left-hand limit equal to the right-hand limit and the function value
For the function to be continuous at \( x = 2 \), we need:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \]

This gives us the equation:

\[ 12 = 4k \]

### Step 5: Solve for \( k \)
Now, we can solve for \( k \):

\[ 4k = 12 \]

Dividing both sides by 4:

\[ k = \frac{12}{4} = 3 \]

### Conclusion
The value of \( k \) for which \( f(x) \) is continuous at \( x = 2 \) is:

\[ \boxed{3} \]