Let `f(x) ={{:((x-4)/(|x-4|)+a, x lt 4),((x-4)/(|x-40|)+b, x gt 4):}`, Then `f(x)` is continuous at x = 4, when

Options

A. a=b=0
B. a=b=1
C. a=-1, b=1
D. a=1, b=1

Answer

To determine the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} \frac{x-4}{|x-4|} + a & \text{if } x < 4 \\ \frac{x-4}{|x-4|} + b & \text{if } x > 4 \end{cases} \]

is continuous at \( x = 4 \), we need to ensure that the following condition holds:

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

### Step 1: Calculate \( \lim_{x \to 4^-} f(x) \)

For \( x < 4 \), we have:

\[ f(x) = \frac{x-4}{|x-4|} + a = \frac{x-4}{-(x-4)} + a = -1 + a \]

Thus,

\[ \lim_{x \to 4^-} f(x) = -1 + a \]

### Step 2: Calculate \( f(4) \)

At \( x = 4 \), the function is not explicitly defined in the piecewise function. However, we can define it as:

\[ f(4) = a + b \]

### Step 3: Calculate \( \lim_{x \to 4^+} f(x) \)

For \( x > 4 \), we have:

\[ f(x) = \frac{x-4}{|x-4|} + b = \frac{x-4}{x-4} + b = 1 + b \]

Thus,

\[ \lim_{x \to 4^+} f(x) = 1 + b \]

### Step 4: Set the limits equal for continuity

Now, we need to set these three expressions equal to each other:

1. From the left limit:
\[ -1 + a = f(4) = a + b \]

2. From the right limit:
\[ f(4) = a + b = 1 + b \]

### Step 5: Solve the equations

From the first equation:

\[ -1 + a = a + b \implies -1 = b \quad \text{(Equation 1)} \]

From the second equation:

\[ a + b = 1 + b \implies a = 1 \quad \text{(Equation 2)} \]

### Step 6: Substitute back to find \( a \) and \( b \)

From Equation 2, we have \( a = 1 \). Substituting \( a = 1 \) into Equation 1:

\[ -1 = b \implies b = -1 \]

### Conclusion

Thus, the values of \( a \) and \( b \) that make \( f(x) \) continuous at \( x = 4 \) are:

\[ a = 1, \quad b = -1 \]