Let `f(x)={x/2-1,0lt=xlt=1 1/2,1lt=xlt=2}g(x)=(2x+1)(x-k)+3,0<=x<=oo` then `g(f(x))` is continuous at x=1 if k equal to:

Answer

To find the value of \( k \) such that \( g(f(x)) \) is continuous at \( x = 1 \), we need to analyze the functions \( f(x) \) and \( g(x) \) given in the problem.

### Step 1: Define the functions

The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} \frac{x}{2} - 1 & \text{if } 0 \leq x \leq 1 \\ \frac{1}{2} & \text{if } 1 < x < 2 \end{cases} \]

The function \( g(x) \) is defined as:
\[ g(x) = (2x + 1)(x - k) + 3 \]

### Step 2: Find \( f(1) \)

We first evaluate \( f(1) \):
\[ f(1) = \frac{1}{2} - 1 = -\frac{1}{2} \]

### Step 3: Find \( g(f(1)) \)

Next, we substitute \( f(1) \) into \( g(x) \):
\[ g(f(1)) = g\left(-\frac{1}{2}\right) = (2(-\frac{1}{2}) + 1)(-\frac{1}{2} - k) + 3 \]
Calculating the first part:
\[ 2(-\frac{1}{2}) + 1 = -1 + 1 = 0 \]
Thus,
\[ g(f(1)) = 0 \cdot (-\frac{1}{2} - k) + 3 = 3 \]

### Step 4: Find the left-hand limit as \( x \to 1^- \)

Now, we calculate the left-hand limit of \( g(f(x)) \) as \( x \) approaches 1 from the left:
\[ \lim_{x \to 1^-} f(x) = f(1) = -\frac{1}{2} \]
Then,
\[ \lim_{x \to 1^-} g(f(x)) = g\left(-\frac{1}{2}\right) = 3 \]

### Step 5: Find the right-hand limit as \( x \to 1^+ \)

Next, we calculate the right-hand limit as \( x \) approaches 1 from the right:
\[ \lim_{x \to 1^+} f(x) = \frac{1}{2} \]
Now substitute into \( g(x) \):
\[ g\left(\frac{1}{2}\right) = (2(\frac{1}{2}) + 1)(\frac{1}{2} - k) + 3 \]
Calculating the first part:
\[ 2(\frac{1}{2}) + 1 = 1 + 1 = 2 \]
Thus,
\[ g\left(\frac{1}{2}\right) = 2\left(\frac{1}{2} - k\right) + 3 = 1 - 2k + 3 = 4 - 2k \]

### Step 6: Set the left-hand limit equal to the right-hand limit

For continuity at \( x = 1 \), we set the left-hand limit equal to the right-hand limit:
\[ 3 = 4 - 2k \]

### Step 7: Solve for \( k \)

Rearranging the equation:
\[ 2k = 4 - 3 \implies 2k = 1 \implies k = \frac{1}{2} \]

### Final Answer

The value of \( k \) such that \( g(f(x)) \) is continuous at \( x = 1 \) is:
\[ \boxed{\frac{1}{2}} \]