If `f(x) = {{:((sqrt(4+ax)-sqrt(4-ax))/x,-1 le x lt 0),((3x+2)/(x-8), 0 le x le 1):}` continuous in `[-1,1]`, then the value of a is

Options

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

Answer

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

\[ f(x) = \begin{cases} \frac{\sqrt{4 + ax} - \sqrt{4 - ax}}{x} & \text{for } -1 \leq x < 0 \\ \frac{3x + 2}{x - 8} & \text{for } 0 \leq x \leq 1 \end{cases} \]

is continuous on the interval \([-1, 1]\), we need to ensure that the limits from the left and right at \( x = 0 \) are equal.

### Step 1: Find the limit as \( x \) approaches \( 0 \) from the left

For \( x \to 0^- \):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{4 + ax} - \sqrt{4 - ax}}{x} \]

To evaluate this limit, we will rationalize the numerator:

\[ \lim_{x \to 0^-} \frac{\sqrt{4 + ax} - \sqrt{4 - ax}}{x} \cdot \frac{\sqrt{4 + ax} + \sqrt{4 - ax}}{\sqrt{4 + ax} + \sqrt{4 - ax}} = \lim_{x \to 0^-} \frac{(4 + ax) - (4 - ax)}{x(\sqrt{4 + ax} + \sqrt{4 - ax})} \]

This simplifies to:

\[ \lim_{x \to 0^-} \frac{2ax}{x(\sqrt{4 + ax} + \sqrt{4 - ax})} = \lim_{x \to 0^-} \frac{2a}{\sqrt{4 + ax} + \sqrt{4 - ax}} \]

As \( x \to 0 \), \( \sqrt{4 + ax} \to 2 \) and \( \sqrt{4 - ax} \to 2 \):

\[ \lim_{x \to 0^-} f(x) = \frac{2a}{2 + 2} = \frac{2a}{4} = \frac{a}{2} \]

### Step 2: Find the limit as \( x \) approaches \( 0 \) from the right

For \( x \to 0^+ \):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{3x + 2}{x - 8} \]

Substituting \( x = 0 \):

\[ \lim_{x \to 0^+} f(x) = \frac{3(0) + 2}{0 - 8} = \frac{2}{-8} = -\frac{1}{4} \]

### Step 3: Set the limits equal for continuity

For \( f(x) \) to be continuous at \( x = 0 \):

\[ \frac{a}{2} = -\frac{1}{4} \]

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

Multiplying both sides by 2:

\[ a = -\frac{1}{2} \]

### Conclusion

The value of \( a \) is

\[ \boxed{-\frac{1}{2}} \]