Let `f(x)={{:(x+1,","x le 1),(3-ax^2,","x gt 1):}` The value of a so that f is continuous is

Options

A. `1//2`
B. 1
C. 2
D. 3

Answer

To determine the value of \( a \) such that the function \( f(x) \) is continuous, we need to ensure that the left-hand limit and the right-hand limit at \( x = 1 \) are equal to the function value at \( x = 1 \).

The function is defined as:
\[ f(x) = \begin{cases} x + 1 & \text{if } x \leq 1 \\ 3 - ax^2 & \text{if } x > 1 \end{cases} \]

### Step 1: Calculate \( f(1) \)
For \( x = 1 \):
\[ f(1) = 1 + 1 = 2 \]

### Step 2: Calculate the left-hand limit as \( x \) approaches 1
The left-hand limit, denoted as \( f(1^-) \), is calculated using the first piece of the function:
\[ f(1^-) = \lim_{x \to 1^-} f(x) = 1 + 1 = 2 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 1
The right-hand limit, denoted as \( f(1^+) \), is calculated using the second piece of the function:
\[ f(1^+) = \lim_{x \to 1^+} f(x) = 3 - a(1^2) = 3 - a \]

### Step 4: Set the left-hand limit equal to the right-hand limit
For the function to be continuous at \( x = 1 \), we need:
\[ f(1^-) = f(1^+) \]
This gives us the equation:
\[ 2 = 3 - a \]

### Step 5: Solve for \( a \)
Rearranging the equation:
\[ a = 3 - 2 = 1 \]

### Conclusion
Thus, the value of \( a \) that makes the function continuous is:
\[ \boxed{1} \]