If `f (x)= [{:(e ^(x-1),,, 0 le x le 1),( x+1-{x},', 1 lt x lt 3 ):}and g (x) =x ^(2) -ax +b` such that `f (x)g (x)` is continous `[0,3]` then the ordered pair (a,b) is (where {.} denotes fractional part function):

Options

A. `(2,3)`
B. `(1,2)`
C. `(3,2)`
D. `(2,2)`

Answer

To solve the problem, we need to ensure that the product \( f(x)g(x) \) is continuous on the interval \([0, 3]\). Given the piecewise function \( f(x) \) and the quadratic function \( g(x) \), we will analyze the continuity at the points where \( f(x) \) changes its definition, specifically at \( x = 1 \) and \( x = 2 \).

### Step 1: Define the functions
The function \( f(x) \) is defined as:
- \( f(x) = e^{(x-1)} \) for \( 0 \leq x \leq 1 \)
- \( f(x) = x + 1 - \{x\} \) for \( 1 < x < 3 \)

The function \( g(x) \) is given by:
\[ g(x) = x^2 - ax + b \]

### Step 2: Evaluate \( f(x) \) at the points of interest
We need to evaluate \( f(x) \) at \( x = 1 \) and \( x = 2 \):
- At \( x = 1 \):
\[ f(1) = e^{(1-1)} = e^0 = 1 \]

- At \( x = 2 \):
\[ f(2) = 2 + 1 - \{2\} = 2 + 1 - 0 = 3 \]

### Step 3: Set up continuity conditions for \( g(x) \)
For \( f(x)g(x) \) to be continuous at \( x = 1 \):
\[ f(1)g(1) = 1 \cdot g(1) = g(1) \text{ must be continuous} \]
Thus, we set:
\[ g(1) = 1 - a + b = 0 \quad \text{(1)} \]

For \( x = 2 \):
\[ f(2)g(2) = 3 \cdot g(2) \text{ must be continuous} \]
Thus, we set:
\[ g(2) = 4 - 2a + b = 0 \quad \text{(2)} \]

### Step 4: Solve the system of equations
From equation (1):
\[ -b + a = 1 \quad \text{(1)} \]

From equation (2):
\[ -2a + b = -4 \quad \text{(2)} \]

Now we can solve this system of equations. From (1), we can express \( b \) in terms of \( a \):
\[ b = a - 1 \]

Substituting \( b \) into equation (2):
\[ -2a + (a - 1) = -4 \]
\[ -2a + a - 1 = -4 \]
\[ -a - 1 = -4 \]
\[ -a = -3 \quad \Rightarrow \quad a = 3 \]

Now substituting \( a = 3 \) back into equation (1) to find \( b \):
\[ b = 3 - 1 = 2 \]

### Final Ordered Pair
Thus, the ordered pair \( (a, b) \) is:
\[ (a, b) = (3, 2) \]