If `f(x)={((log(1+2ax)-log(1-bx))/(x)",", x ne 0),(k",", x =0):}` is continuous at x = 0, then value of k is

Options

A. b + a
B. b - 2a
C. 2a - b
D. 2a + b

Explanation

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

\[ f(x) = \begin{cases} \frac{\log(1 + 2ax) - \log(1 - bx)}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that

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

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

We start by calculating the limit:

\[ \lim_{x \to 0} \frac{\log(1 + 2ax) - \log(1 - bx)}{x}. \]

Using the properties of logarithms, we can combine the logs:

\[ \lim_{x \to 0} \frac{\log\left(\frac{1 + 2ax}{1 - bx}\right)}{x}. \]

### Step 2: Apply L'Hôpital's Rule

Since substituting \( x = 0 \) gives us the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator:

\[ \lim_{x \to 0} \frac{\frac{d}{dx} \left(\log(1 + 2ax) - \log(1 - bx)\right)}{\frac{d}{dx}(x)}. \]

### Step 3: Differentiate the numerator and denominator

The derivative of the numerator is:

\[ \frac{d}{dx} \left(\log(1 + 2ax)\right) = \frac{2a}{1 + 2ax}, \]
and
\[ \frac{d}{dx} \left(\log(1 - bx)\right) = -\frac{b}{1 - bx}. \]

Thus, we have:

\[ \frac{d}{dx} \left(\log(1 + 2ax) - \log(1 - bx)\right) = \frac{2a}{1 + 2ax} + \frac{b}{1 - bx}. \]

The derivative of the denominator \( x \) is simply \( 1 \).

### Step 4: Substitute back into the limit

Now we substitute back into the limit:

\[ \lim_{x \to 0} \left(\frac{2a}{1 + 2ax} + \frac{b}{1 - bx}\right). \]

As \( x \to 0 \):

\[ \frac{2a}{1 + 2a \cdot 0} + \frac{b}{1 - b \cdot 0} = 2a + b. \]

### Step 5: Set the limit equal to \( k \)

For continuity at \( x = 0 \), we set:

\[ k = \lim_{x \to 0} f(x) = 2a + b. \]

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

\[ \boxed{2a + b}. \]