The value of `f(0)` so that the function `f(x) = (2x - sin^(-1)x)/(2x + tan^(-1)x)` is continuous at each point on its domain is:

Options

A. 2
B. `1//3`
C. `2//3`
D. `-1//3`

Answer

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

\[ f(x) = \frac{2x - \sin^{-1}(x)}{2x + \tan^{-1}(x)} \]

is continuous at each point in its domain, we need to ensure that \( f(0) \) is equal to the limit of \( f(x) \) as \( x \) approaches 0.

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

We start by calculating:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{2x - \sin^{-1}(x)}{2x + \tan^{-1}(x)} \]

### Step 2: Substitute \( x = 0 \) directly

Substituting \( x = 0 \) into the function gives:

\[ f(0) = \frac{2(0) - \sin^{-1}(0)}{2(0) + \tan^{-1}(0)} = \frac{0 - 0}{0 + 0} = \frac{0}{0} \]

This is an indeterminate form, so we need to apply L'Hôpital's rule or simplify the expression.

### Step 3: Simplify the expression

We can factor out \( x \) from both the numerator and denominator:

\[ f(x) = \frac{x \left(2 - \frac{\sin^{-1}(x)}{x}\right)}{x \left(2 + \frac{\tan^{-1}(x)}{x}\right)} \]

This allows us to cancel \( x \) (for \( x \neq 0 \)):

\[ f(x) = \frac{2 - \frac{\sin^{-1}(x)}{x}}{2 + \frac{\tan^{-1}(x)}{x}} \]

### Step 4: Evaluate the limits of the fractions

Now we need to evaluate the limits of \( \frac{\sin^{-1}(x)}{x} \) and \( \frac{\tan^{-1}(x)}{x} \) as \( x \) approaches 0:

\[ \lim_{x \to 0} \frac{\sin^{-1}(x)}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\tan^{-1}(x)}{x} = 1 \]

### Step 5: Substitute the limits back into the expression

Substituting these limits back into our simplified expression gives:

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

### Step 6: Set \( f(0) \) equal to the limit

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

\[ f(0) = \lim_{x \to 0} f(x) = \frac{1}{3} \]

Thus, the value of \( f(0) \) should be:

\[ f(0) = \frac{1}{3} \]

### Final Answer

The value of \( f(0) \) so that the function is continuous at each point on its domain is:

\[ \boxed{\frac{1}{3}} \]