Let `f(x)={(x-4)/(|x-4|)+a ,x<4a+b ,(x-4)/(|x-4|)+b ,x >4` Then `f(x)` is continous at `x=4` when `a=0,b=0` b. `a=1,b=1` c. `a=-1,b=1` d. `a=-1,b=-1`

Options

A. a=0,b=0
B. a=1,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

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

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

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

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

For \( x < 4 \):

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

Thus,

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

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

At \( x = 4 \):

\[ f(4) = b \]

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

For \( x > 4 \):

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

Thus,

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

### Step 4: Set the limits equal to each other

For continuity at \( x = 4 \):

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

This gives us the equation:

\[ -1 + a = b = 1 + b \]

### Step 5: Solve the equations

From the first equation:

\[ -1 + a = b \quad (1) \]

From the second equation:

\[ b = 1 + b \quad (2) \]

From equation (2), we can see that it simplifies to \( 0 = 1 \), which is not possible. Therefore, we only need to focus on equation (1):

Substituting \( b \) from equation (1) into itself gives:

\[ -1 + a = 1 + (-1 + a) \]

This simplifies to:

\[ -1 + a = 1 - 1 + a \]

This equation is always true, which means we need to find specific values for \( a \) and \( b \).

### Step 6: Substitute values from the options

Let’s check the options provided:

1. **Option a:** \( a = 0, b = 0 \)

\[ -1 + 0 = 0 \quad (False) \]

2. **Option b:** \( a = 1, b = 1 \)

\[ -1 + 1 = 1 \quad (True) \]

3. **Option c:** \( a = -1, b = 1 \)

\[ -1 - 1 = 1 \quad (False) \]

4. **Option d:** \( a = -1, b = -1 \)

\[ -1 - 1 = -1 \quad (False) \]

### Conclusion

The only option that satisfies the continuity condition is:

**Option b:** \( a = 1, b = 1 \)