Given a function
`f(x){{:(-1,if,xle0),(ax+b,if,0ltxlt1),(1,if,xge1):}`
where a, b are constant. The function is continuous everywhere.
What is the value of a?

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find the value of \( a \) in the given piecewise function, we need to ensure that the function is continuous at the points where the definition of the function changes, which are at \( x = 0 \) and \( x = 1 \).

The function is defined as follows:
\[ f(x) = \begin{cases} -1 & \text{if } x \leq 0 \\ ax + b & \text{if } 0 < x < 1 \\ 1 & \text{if } x \geq 1 \end{cases} \]

### Step 1: Check continuity at \( x = 0 \)

To ensure continuity at \( x = 0 \), we need the left-hand limit to equal the right-hand limit at this point.

- **Left-hand limit as \( x \) approaches 0**:
\[ \lim_{x \to 0^-} f(x) = -1 \]

- **Right-hand limit as \( x \) approaches 0**:
\[ \lim_{x \to 0^+} f(x) = a(0) + b = b \]

Setting these equal for continuity:
\[ -1 = b \quad \text{(1)} \]

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

Next, we check continuity at \( x = 1 \).

- **Left-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^-} f(x) = a(1) + b = a + b \]

- **Right-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^+} f(x) = 1 \]

Setting these equal for continuity:
\[ a + b = 1 \quad \text{(2)} \]

### Step 3: Substitute \( b \) from equation (1) into equation (2)

From equation (1), we have \( b = -1 \). Substitute this into equation (2):
\[ a + (-1) = 1 \]
\[ a - 1 = 1 \]
\[ a = 2 \]

### Conclusion

The value of \( a \) is \( 2 \).