For what value of k, the function
`f(x) ={:{(kx^2", " x le 2 ),(" "5", " xgt2):},`
is continuous at x=2.

Answer

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

\[ f(x) = \begin{cases} kx^2 & \text{if } x \leq 2 \\ 5 & \text{if } x > 2 \end{cases} \]

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

### Step 1: Find the left-hand limit as \( x \) approaches 2.
The left-hand limit is given by:

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

Substituting \( x = 2 \):

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

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

\[ \lim_{x \to 2^+} f(x) = 5 \]

### Step 3: Set the left-hand limit equal to the right-hand limit.
For the function to be continuous at \( x = 2 \), we need:

\[ 4k = 5 \]

### Step 4: Solve for \( k \).
To find \( k \), we solve the equation:

\[ k = \frac{5}{4} \]

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

\[ \boxed{\frac{5}{4}} \]

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