The value of `f(0),` so that the function `f(x)=(sqrt(a^2-a x+x^2)-sqrt(a^2+a x+x^2))/(sqrt(a+x)-sqrt(a-x))` becomes continuous for all `x ,` given by `a^(3/2)` (b) `a^(1/2)` (c) `-a^(1/2)` (d) `-a^(3/2)`

Options

A. `asqrt(a)`
B. `-sqrt(a)`
C. `sqrt(a)`
D. `-asqrt(a)`

Answer

To determine the value of \( f(0) \) such that the function

\[ f(x) = \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}} \]

is continuous for all \( x \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Evaluate \( f(0) \)**:
First, we substitute \( x = 0 \) into the function:
\[ f(0) = \frac{\sqrt{a^2 - a(0) + 0^2} - \sqrt{a^2 + a(0) + 0^2}}{\sqrt{a + 0} - \sqrt{a - 0}} = \frac{\sqrt{a^2} - \sqrt{a^2}}{\sqrt{a} - \sqrt{a}} = \frac{0}{0} \]
This results in an indeterminate form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

3. **Differentiate the Numerator**:
Let \( N(x) = \sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2} \).
The derivative \( N'(x) \) is:
\[ N'(x) = \frac{1}{2\sqrt{a^2 - ax + x^2}}(-a + 2x) - \frac{1}{2\sqrt{a^2 + ax + x^2}}(a + 2x) \]

4. **Differentiate the Denominator**:
Let \( D(x) = \sqrt{a + x} - \sqrt{a - x} \).
The derivative \( D'(x) \) is:
\[ D'(x) = \frac{1}{2\sqrt{a + x}}(1) + \frac{1}{2\sqrt{a - x}}(1) \]

5. **Evaluate the Limit**:
Now we need to evaluate:
\[ \lim_{x \to 0} \frac{N'(x)}{D'(x)} \]
Plugging \( x = 0 \) into the derivatives:
- For \( N'(0) \):
\[ N'(0) = \frac{1}{2\sqrt{a^2}}(-a + 0) - \frac{1}{2\sqrt{a^2}}(a + 0) = \frac{-a}{2a} - \frac{a}{2a} = -\frac{1}{2} - \frac{1}{2} = -1 \]
- For \( D'(0) \):
\[ D'(0) = \frac{1}{2\sqrt{a}} + \frac{1}{2\sqrt{a}} = \frac{1}{\sqrt{a}} \]

6. **Final Calculation**:
Therefore,
\[ \lim_{x \to 0} f(x) = \frac{-1}{\frac{1}{\sqrt{a}}} = -\sqrt{a} \]

7. **Set \( f(0) \) for Continuity**:
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ f(0) = -\sqrt{a} \]

### Conclusion:
Thus, the value of \( f(0) \) that makes the function continuous for all \( x \) is:

\[ f(0) = -a^{1/2} \]