If the function `f(x)={(kx^2, "if x"le2),(3,"if x"gt2):}` is continuous at x=2, then find the value of k.

Options

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

Answer

To find the value of \( k \) such that 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 will follow these steps:

### Step 1: Understand the condition for continuity
A function is continuous at a point \( x = a \) if:

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

In our case, \( a = 2 \).

### Step 2: Calculate \( f(2) \)
Since \( x = 2 \) falls under the first case of the piecewise function, we have:

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

### Step 3: Calculate the left-hand limit as \( x \) approaches 2
The left-hand limit, \( \lim_{x \to 2^-} f(x) \), is given by the first case of the function:

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

### Step 4: Calculate the right-hand limit as \( x \) approaches 2
The right-hand limit, \( \lim_{x \to 2^+} f(x) \), is given by the second case of the function:

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

### Step 5: Set the left-hand limit equal to the right-hand limit
For the function to be continuous at \( x = 2 \), we set the left-hand limit equal to the right-hand limit:

\[ 4k = 3 \]

### Step 6: Solve for \( k \)
Now, we can solve for \( k \):

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

### Final Answer
Thus, the value of \( k \) is

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