If `f(x)={{:(ax^(2)+1",",xgt1),(x+a",",xle1):}` is derivable at x=1 , then the value of a is

Options

A. 0
B. 1
C. `(1)/(2)`
D. 2

Answer

To determine the value of \( a \) such that the function

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

is derivable at \( x = 1 \), we need to ensure that the function is both continuous and differentiable at that point.

### Step 1: Check Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit (as \( x \) approaches 1 from the left) must equal the right-hand limit (as \( x \) approaches 1 from the right), and both must equal \( f(1) \).

1. **Calculate \( f(1) \)**:
\[ f(1) = 1 + a \]

2. **Calculate the right-hand limit**:
\[ \lim_{x \to 1^+} f(x) = a(1^2) + 1 = a + 1 \]

3. **Calculate the left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = 1 + a \]

Setting the left-hand limit equal to the right-hand limit for continuity:
\[ a + 1 = 1 + a \]
This condition is satisfied for any \( a \).

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

For \( f(x) \) to be differentiable at \( x = 1 \), the left-hand derivative must equal the right-hand derivative.

1. **Calculate the right-hand derivative**:
\[ f'(x) = \frac{d}{dx}(ax^2 + 1) = 2ax \quad \text{for } x > 1 \]
Thus,
\[ \lim_{x \to 1^+} f'(x) = 2a(1) = 2a \]

2. **Calculate the left-hand derivative**:
\[ f'(x) = \frac{d}{dx}(x + a) = 1 \quad \text{for } x \leq 1 \]
Thus,
\[ \lim_{x \to 1^-} f'(x) = 1 \]

Setting the left-hand derivative equal to the right-hand derivative for differentiability:
\[ 2a = 1 \]

### Step 3: Solve for \( a \)

From the equation \( 2a = 1 \):
\[ a = \frac{1}{2} \]

### Conclusion

The value of \( a \) such that the function \( f(x) \) is derivable at \( x = 1 \) is:
\[ \boxed{\frac{1}{2}} \]