Let `f (x)= {{:(a-3x,,, -2 le x lt 0),( 4x+-3,,, 0 le x lt 1):},if f (x)` has smallest valueat `x=0,` then range of a, is

Options

A. `(-oo, 3)`
B. `(-oo,3]`
C. `(-3,oo)`
D. `(3,oo)`

Answer

To solve the problem, we need to analyze the function \( f(x) \) given in two pieces and determine the range of \( a \) such that the function has its smallest value at \( x = 0 \).

### Step-by-step Solution:

1. **Define the function**:
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} a - 3x & \text{for } -2 \leq x < 0 \\ 4x - 3 & \text{for } 0 \leq x < 1 \end{cases} \]

2. **Evaluate \( f(x) \) at \( x = 0 \)**:
We need to find the value of \( f(0) \):
\[ f(0) = 4(0) - 3 = -3 \]

3. **Evaluate \( f(x) \) as \( x \) approaches 0 from the left**:
For \( x \) approaching 0 from the left (i.e., \( x \to 0^- \)):
\[ f(0^-) = a - 3(0) = a \]

4. **Determine the condition for the smallest value**:
The problem states that \( f(x) \) has its smallest value at \( x = 0 \). Therefore, we need:
\[ f(0) \leq f(0^-) \]
This translates to:
\[ -3 \leq a \]

5. **Conclusion about the range of \( a \)**:
Since \( a \) can take any value greater than or equal to -3, the range of \( a \) is:
\[ a \in [-3, \infty) \]

### Final Answer:
The range of \( a \) is \( [-3, \infty) \).