If `f(x)=(2x+sin^(-1)x)/(2x-tan^(-1)x)` is continuous for all x in (-1,1) , then value of f (0) is

Options

A. 2
B. 3
C. `-3`
D. `(3)/(2)`

Answer

To find the value of \( f(0) \) for the function

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

given that it is continuous for all \( x \) in the interval \( (-1, 1) \), we will follow these steps:

### Step 1: Understand Continuity at \( x = 0 \)

Since \( f(x) \) is continuous at \( x = 0 \), we know that:

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

### Step 2: Substitute \( f(x) \) into the Limit

We substitute \( f(x) \) into the limit:

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

### Step 3: Evaluate the Limit

To evaluate the limit, we can divide both the numerator and the denominator by \( x \):

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

### Step 4: Use Known Limits

We know from calculus that:

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

Applying these limits, we get:

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

### Conclusion

Thus, the value of \( f(0) \) is:

\[ \boxed{3} \]