If `f(x)={|x|-3 x < 1|x-2|+a x >= 1 & g(x)={2-|x| x < 2 sgn(x)-b x >= 2. if h(x)=f(x)+g(x)` is discontinuous at exactly one point, then -

Options

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

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the question, and then find the conditions under which the function \( h(x) = f(x) + g(x) \) is discontinuous at exactly one point.

### Step 1: Define the functions \( f(x) \) and \( g(x) \)

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

The function \( g(x) \) is defined as:
\[ g(x) = \begin{cases} 2 - |x| & \text{if } x < 2 \\ \text{sgn}(x) - b & \text{if } x \geq 2 \end{cases} \]

### Step 2: Analyze \( f(x) \)

1. For \( x < 1 \):
\[ f(x) = -x - 3 \]
2. For \( x \geq 1 \):
- If \( x < 2 \):
\[ f(x) = |x - 2| + a = -x + 2 + a \]
- If \( x \geq 2 \):
\[ f(x) = x - 2 + a \]

### Step 3: Analyze \( g(x) \)

1. For \( x < 2 \):
\[ g(x) = 2 - |x| \]
- If \( x < 0 \):
\[ g(x) = 2 + x \]
- If \( 0 \leq x < 2 \):
\[ g(x) = 2 - x \]
2. For \( x \geq 2 \):
\[ g(x) = \text{sgn}(x) - b = 1 - b \]

### Step 4: Define \( h(x) \)

Now, we can define \( h(x) = f(x) + g(x) \):

1. For \( x < 0 \):
\[ h(x) = (-x - 3) + (2 + x) = -1 \]
2. For \( 0 \leq x < 1 \):
\[ h(x) = (-x - 3) + (2 - x) = -1 \]
3. For \( 1 \leq x < 2 \):
\[ h(x) = (-x + 2 + a) + (2 - x) = -2x + 4 + a \]
4. For \( x \geq 2 \):
\[ h(x) = (x - 2 + a) + (1 - b) = x - 1 + a - b \]

### Step 5: Check continuity at points \( x = 1 \) and \( x = 2 \)

**At \( x = 1 \)**:
- Left-hand limit:
\[ h(1^-) = -1 \]
- Right-hand limit:
\[ h(1^+) = -2(1) + 4 + a = 2 + a \]
For continuity at \( x = 1 \):
\[ -1 = 2 + a \implies a = -3 \]

**At \( x = 2 \)**:
- Left-hand limit:
\[ h(2^-) = -2(2) + 4 + a = -4 + 4 - 3 = -3 \]
- Right-hand limit:
\[ h(2^+) = 2 - 1 + a - b = 1 + a - b \]
For continuity at \( x = 2 \):
\[ -3 = 1 + a - b \implies a - b = -4 \]

### Step 6: Conditions for discontinuity

1. If \( h(x) \) is discontinuous at \( x = 2 \), then:
\[ a = -3 \quad \text{and} \quad b \neq 1 \]

2. If \( h(x) \) is discontinuous at \( x = 1 \), then:
\[ a \neq -3 \quad \text{and} \quad b = 1 \]

### Conclusion

The values of \( a \) and \( b \) that satisfy the conditions for \( h(x) \) to be discontinuous at exactly one point are:
- \( a = -3 \) and \( b \neq 1 \) (discontinuous at \( x = 2 \))
- \( a \neq -3 \) and \( b = 1 \) (discontinuous at \( x = 1 \))