Let ` f(x)= {:{ (x + a , x ge 1 ) , ( ax^(2) + 1, x lt 1) :}` then f(x) is derivable at x =1 , if

Options

A. a=1
B. a=0
C. a =2
D. ` a = 1/2 `

Answer

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

Given the function:
\[ f(x) = \begin{cases} x + a & \text{if } x \geq 1 \\ ax^2 + 1 & \text{if } x < 1 \end{cases} \]

### Step 1: Find the left-hand derivative at \( x = 1 \)

The left-hand derivative at \( x = 1 \) is given by:
\[ \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} \]
For \( x < 1 \), we use \( f(x) = ax^2 + 1 \). First, we need to find \( f(1) \):
\[ f(1) = 1 + a \]
Now, we can compute the left-hand derivative:
\[ \lim_{x \to 1^-} \frac{(ax^2 + 1) - (1 + a)}{x - 1} \]
\[ = \lim_{x \to 1^-} \frac{ax^2 - a + 1 - 1}{x - 1} \]
\[ = \lim_{x \to 1^-} \frac{a(x^2 - 1)}{x - 1} \]
\[ = \lim_{x \to 1^-} \frac{a(x - 1)(x + 1)}{x - 1} \]
For \( x \neq 1 \), we can simplify:
\[ = \lim_{x \to 1^-} a(x + 1) = a(1 + 1) = 2a \]

### Step 2: Find the right-hand derivative at \( x = 1 \)

The right-hand derivative at \( x = 1 \) is given by:
\[ \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} \]
For \( x \geq 1 \), we use \( f(x) = x + a \):
\[ = \lim_{x \to 1^+} \frac{(x + a) - (1 + a)}{x - 1} \]
\[ = \lim_{x \to 1^+} \frac{x - 1}{x - 1} = 1 \]

### Step 3: Set the left-hand derivative equal to the right-hand derivative

For \( f(x) \) to be derivable at \( x = 1 \), we set the left-hand derivative equal to the right-hand derivative:
\[ 2a = 1 \]

### Step 4: Solve for \( a \)

Now, we solve for \( a \):
\[ a = \frac{1}{2} \]

### Conclusion

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