If `f(x) = (log (1+ax)-log (1-bx))/(x)` for `x ne 0 and f(0) = k ` and f(x) is continuous at x = 0 then k is equal to

Options

A. `a+b`
B. `a-b`
C. a
D. b

Answer

To solve the problem, we need to find the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 0 \). The function is given by:

\[ f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x} \quad \text{for } x \neq 0 \]

and \( f(0) = k \).

### Step 1: Determine the limit of \( f(x) \) as \( x \) approaches 0

To ensure continuity at \( x = 0 \), we need to find:

\[ \lim_{x \to 0} f(x) = f(0) = k \]

### Step 2: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \( x \) approaches 0, we can apply L'Hôpital's Rule. This rule states that if the limit results in an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and denominator.

The numerator is:

\[ \log(1 + ax) - \log(1 - bx) = \log\left(\frac{1 + ax}{1 - bx}\right) \]

Differentiating the numerator:

\[ \frac{d}{dx} \left[\log(1 + ax) - \log(1 - bx)\right] = \frac{a}{1 + ax} + \frac{b}{1 - bx} \]

The denominator is simply:

\[ \frac{d}{dx}(x) = 1 \]

### Step 3: Evaluate the limit using L'Hôpital's Rule

Now we can evaluate the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{a}{1 + ax} + \frac{b}{1 - bx} \]

Substituting \( x = 0 \):

\[ = \frac{a}{1 + 0} + \frac{b}{1 - 0} = a + b \]

### Step 4: Set the limit equal to \( k \)

Since \( f(0) = k \) and we have found that:

\[ \lim_{x \to 0} f(x) = a + b \]

We can equate this to \( k \):

\[ k = a + b \]

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{a + b} \]