let `f(x)={(ax+1, if x le 1),(3, if x=1),(bx^2+1,if x > 1))` if `f(x)` is continuous at `x=1` then value of `a-b` is (A) `0` (B) `1` (C) `2` (D) `4`

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 1 \). The function is defined as follows:

\[ f(x) = \begin{cases} ax + 1 & \text{if } x < 1 \\ 3 & \text{if } x = 1 \\ bx^2 + 1 & \text{if } x > 1 \end{cases} \]

### Step 1: Find the left-hand limit as \( x \) approaches 1
The left-hand limit of \( f(x) \) as \( x \) approaches 1 is given by the expression for \( f(x) \) when \( x < 1 \):

\[ \lim_{x \to 1^-} f(x) = a(1) + 1 = a + 1 \]

### Step 2: Find the right-hand limit as \( x \) approaches 1
The right-hand limit of \( f(x) \) as \( x \) approaches 1 is given by the expression for \( f(x) \) when \( x > 1 \):

\[ \lim_{x \to 1^+} f(x) = b(1^2) + 1 = b + 1 \]

### Step 3: Set the left-hand limit equal to the right-hand limit
For the function to be continuous at \( x = 1 \), the left-hand limit must equal the right-hand limit, and both must equal \( f(1) \):

\[ a + 1 = 3 \quad \text{(since \( f(1) = 3 \))} \]

### Step 4: Solve for \( a \)
From the equation \( a + 1 = 3 \):

\[ a = 3 - 1 = 2 \]

### Step 5: Set the right-hand limit equal to \( f(1) \)
Now we set the right-hand limit equal to \( f(1) \):

\[ b + 1 = 3 \]

### Step 6: Solve for \( b \)
From the equation \( b + 1 = 3 \):

\[ b = 3 - 1 = 2 \]

### Step 7: Calculate \( a - b \)
Now that we have both \( a \) and \( b \):

\[ a - b = 2 - 2 = 0 \]

### Conclusion
Thus, the value of \( a - b \) is:

\[ \boxed{0} \]