For what value of k is the function
`f(x)={{:(2x+(1)/(4),xlt0),(k,x=0),((x+(1)/(2))^(2),xgt0):}` continuous?

Options

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

Answer

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

\[ f(x) = \begin{cases} 2x + \frac{1}{4} & \text{if } x < 0 \\ k & \text{if } x = 0 \\ \left(x + \frac{1}{2}\right)^2 & \text{if } x > 0 \end{cases} \]

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

### Step 1: Calculate the Left-Hand Limit (LHL)

The left-hand limit as \( x \) approaches 0 from the left (i.e., \( x \to 0^- \)) is given by:

\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left(2x + \frac{1}{4}\right) \]

Substituting \( x = 0 \):

\[ \text{LHL} = 2(0) + \frac{1}{4} = \frac{1}{4} \]

### Step 2: Calculate the Right-Hand Limit (RHL)

The right-hand limit as \( x \) approaches 0 from the right (i.e., \( x \to 0^+ \)) is given by:

\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(x + \frac{1}{2}\right)^2 \]

Substituting \( x = 0 \):

\[ \text{RHL} = \left(0 + \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 3: Set Limits Equal to Each Other

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

\[ \text{LHL} = \text{RHL} = f(0) \]

From the calculations, we have:

\[ \frac{1}{4} = k \]

### Conclusion

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

\[ k = \frac{1}{4} \]