Let b be a nonzero real number , Suppose `f : R to R ` is differentiable function such that f(0) = 1 . If the derivative `f'` of `f` satisfies the equation
`f'(x) = (f(x))/(b^(2)+x^(2))`
for all ` x in R ` , then which of the following statements is/are True ?

Options

A. If ` b gt 0 ` then f is an increasing function
B. If `b lt 0 ` then f is a decreasing function
C. `f(x) f(-x) = 1 ` for all `x in RR `
D. `f(x) - f(-x) = 0 ` for all `x in RR`

Answer

To solve the problem, we start with the given differential equation:

\[ f'(x) = \frac{f(x)}{b^2 + x^2} \]

We can rearrange this equation as follows:

\[ \frac{f'(x)}{f(x)} = \frac{1}{b^2 + x^2} \]

Next, we can integrate both sides. The left side requires the use of the natural logarithm, while the right side can be integrated using the arctangent function:

\[ \int \frac{f'(x)}{f(x)} \, dx = \int \frac{1}{b^2 + x^2} \, dx \]

The left side integrates to:

\[ \ln |f(x)| + C_1 \]

The right side integrates to:

\[ \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right) + C_2 \]

Combining these results, we have:

\[ \ln |f(x)| = \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right) + C \]

where \( C = C_2 - C_1 \).

Exponentiating both sides gives us:

\[ f(x) = e^{C} \cdot e^{\frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)} \]

Let \( e^{C} = k \), where \( k \) is a constant. Thus, we can write:

\[ f(x) = k \cdot e^{\frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)} \]

Now, we know that \( f(0) = 1 \):

\[ f(0) = k \cdot e^{\frac{1}{b} \tan^{-1}(0)} = k \cdot e^0 = k \]

Thus, \( k = 1 \) and we have:

\[ f(x) = e^{\frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)} \]

Next, we need to analyze the behavior of \( f(x) \) to determine if it is increasing or decreasing. We find the derivative \( f'(x) \):

Using the chain rule:

\[ f'(x) = e^{\frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)} \cdot \frac{1}{b} \cdot \frac{1}{1 + \left(\frac{x}{b}\right)^2} \cdot \frac{1}{b} \]

This simplifies to:

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

Since \( e^{\frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)} > 0 \) for all \( x \) and \( b > 0 \), and \( 1 + \frac{x^2}{b^2} > 0 \), we conclude that:

\[ f'(x) > 0 \quad \text{for all } x \]

This indicates that \( f(x) \) is an increasing function.

### Summary of Statements
1. \( f(x) > 0 \) for all \( x \) (True)
2. \( f(x) \) is an increasing function (True)
3. \( f(x) \) is a decreasing function (False)
4. \( f(-x) = f(x) \) (True)