Let `f(x)={{:(,|x|-3,x lt 1),(,|x-2|+a,x ge 1):}`
`g(x)={{:(,2-|x|,x lt 1),(,Sgn(x)-b,x ge 1):}` If h(x)=f(x) is discontinuous at exactly one point, then which of the following are correct?

Options

A. `a=-3, b=0`
B. `a=-3,b=-1`
C. `a=2,b=1`
D. `a=0,b=1`

Answer

To determine the conditions under which the function \( f(x) \) is discontinuous at exactly one point, we will analyze the function piecewise and check the continuity at the point where the definition of \( f(x) \) changes, which is at \( x = 1 \).

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is given as:
\[ f(x) = \begin{cases} |x| - 3 & \text{if } x < 1 \\ |x - 2| + a & \text{if } x \geq 1 \end{cases} \]

### Step 2: Evaluate the left-hand limit (LHL) at \( x = 1 \)
For \( x < 1 \):
\[ f(x) = |x| - 3 \]
At \( x = 1 \), the left-hand limit is:
\[ \lim_{x \to 1^-} f(x) = |1| - 3 = 1 - 3 = -2 \]

### Step 3: Evaluate the right-hand limit (RHL) at \( x = 1 \)
For \( x \geq 1 \):
\[ f(x) = |x - 2| + a \]
At \( x = 1 \), the right-hand limit is:
\[ \lim_{x \to 1^+} f(x) = |1 - 2| + a = 1 + a \]

### Step 4: Set the limits equal for continuity
For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit must equal the right-hand limit:
\[ -2 = 1 + a \]
Solving for \( a \):
\[ a = -2 - 1 = -3 \]

### Step 5: Determine conditions for discontinuity
Since the problem states that \( f(x) \) is discontinuous at exactly one point, \( a \) must not equal \(-3\). Thus, \( a \) can be any value except \(-3\).

### Conclusion
The function \( f(x) \) will be discontinuous at exactly one point (which is \( x = 1 \)) if \( a \neq -3 \). The value of \( b \) in \( g(x) \) does not affect the continuity of \( f(x) \), so it can take any value.

### Final Answer
- \( a \) must not equal \(-3\).
- \( b \) can be any value.