For what value of 'k' the function
`{{:(kx^2",",x le 2),(3",",x>2):}` is continuous at x=2

Answer

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

\[ f(x) = \begin{cases} kx^2 & \text{if } x \leq 2 \\ 3 & \text{if } x > 2 \end{cases} \]

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

### Step 1: Find \( f(2) \)
Since \( x = 2 \) falls under the case \( x \leq 2 \), we use the first part of the function:

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

### Step 2: Find the left-hand limit as \( x \) approaches 2
The left-hand limit is given by:

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

### Step 3: Find the right-hand limit as \( x \) approaches 2
The right-hand limit is given by:

\[ \lim_{x \to 2^+} f(x) = 3 \]

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

\[ 4k = 3 \]

### Step 5: Solve for \( k \)
To find \( k \), we can rearrange the equation:

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

### Conclusion
Thus, the value of \( k \) for which the function is continuous at \( x = 2 \) is

\[ \boxed{\frac{3}{4}} \]